HofstadterConway $10,000 sequence
You are encouraged to solve this task according to the task description, using any language you may know.
The definition of the sequence is colloquially described as:
 Starting with the list [1,1],
 Take the last number in the list so far: 1, I'll call it x.
 Count forward x places from the beginning of the list to find the first number to add (1)
 Count backward x places from the end of the list to find the second number to add (1)
 Add the two indexed numbers from the list and the result becomes the next number in the list (1+1)
 This would then produce [1,1,2] where 2 is the third element of the sequence.
Note that indexing for the description above starts from alternately the left and right ends of the list and starts from an index of one.
A less wordy description of the sequence is:
a(1)=a(2)=1 a(n)=a(a(n1))+a(na(n1))
The sequence begins:
1, 1, 2, 2, 3, 4, 4, 4, 5, ...
Interesting features of the sequence are that:
 a(n)/n tends to 0.5 as n grows towards infinity.
 a(n)/n where n is a power of 2 is 0.5
 For n>4 the maximal value of a(n)/n between successive powers of 2 decreases.
The sequence is so named because John Conway offered a prize of $10,000 to the first person who could
find the first position, p in the sequence where
│a(n)/n│ < 0.55 for all n > p
It was later found that Hofstadter had also done prior work on the sequence.
The 'prize' was won quite quickly by Dr. Colin L. Mallows who proved the properties of the sequence and allowed him to find the value of n (which is much smaller than the 3,173,375,556 quoted in the NYT article).
 Task
 Create a routine to generate members of the HofstadterConway $10,000 sequence.
 Use it to show the maxima of a(n)/n between successive powers of two up to 2**20
 As a stretch goal: compute the value of n that would have won the prize and confirm it is true for n up to 2**20
 Also see
 Conways Challenge Sequence, Mallows' own account.
 Mathworld Article.
11l
V last = 1 << 20
V a_list = [0] * (last+1)
a_list[0] = 50'000
a_list[1] = 1
a_list[2] = 1
V v = a_list[2]
V k1 = 2
V lg2 = 1
V amax = 0.0
L(n) 3..last
v = a_list[v] + a_list[n  v]
a_list[n] = v
amax = max(amax, Float(v) / n)
I (k1 [&] n) == 0
print(‘Maximum between 2^#. and 2^#. was #.6’.format(lg2, lg2 + 1, amax))
amax = 0
lg2++
k1 = n
 Output:
Maximum between 2^1 and 2^2 was 0.666667 Maximum between 2^2 and 2^3 was 0.666667 Maximum between 2^3 and 2^4 was 0.636364 Maximum between 2^4 and 2^5 was 0.608696 ... Maximum between 2^17 and 2^18 was 0.536020 Maximum between 2^18 and 2^19 was 0.534645 Maximum between 2^19 and 2^20 was 0.533779
360 Assembly
For maximum compatibility, this program uses only the basic instruction set (S/360)
with 2 ASSIST macros (XDECO,XPRNT).
The program addresses the problem for l=2**12 (4K). For l=2**20 (1M) you must
allocate dynamic storage instead using static storage.
* HofstadterConway $10,000 sequence 07/05/2016
HOFSTADT START
B 72(R15) skip savearea
DC 17F'0' savearea
STM R14,R12,12(R13) save registers
ST R13,4(R15) link backward SA
ST R15,8(R13) link forward SA
LR R13,R15 establish addressability
USING HOFSTADT,R13 set base register
LA R4,2 pow2=2
LA R8,4 p2=2**pow2
MVC A+0,=F'1' a(1)=1
MVC A+4,=F'1' a(2)=1
LA R6,3 n=3
LOOPN C R6,UPRDIM do n=3 to uprdim
BH ELOOPN
LR R1,R6 n
SLA R1,2
L R5,A8(R1) a(n1)
LR R1,R5
SLA R1,2
L R2,A4(R1) a(a(n1))
LR R1,R6 n
SR R1,R5 na(n1)
SLA R1,2
L R3,A4(R1) a(na(n1)
AR R2,R3 a(a(n1))+a(na(n1))
LR R1,R6 n
SLA R1,2
ST R2,A4(R1) a(n)=a(a(n1))+a(na(n1))
LR R1,R6 n
SLA R1,2
L R2,A4(R1) a(n)
MH R2,=H'10000' fixed point 4dec
SRDA R2,32
DR R2,R6 /n
LR R7,R3 r=a(n)/n
C R7,=F'5500' if r>=0.55
BL EIF1
LR R9,R6 mallows=n
EIF1 C R7,PEAK if r>peak
BNH EIF2
ST R7,PEAK peak=r
ST R6,PEAKPOS peakpos=n
EIF2 CR R6,R8 if n=p2
BNE EIF3
LR R1,R4 pow2
BCTR R1,0 pow21
XDECO R1,XDEC edit pow21
MVC PG1+18(2),XDEC+10
XDECO R4,XDEC edit pow2
MVC PG1+27(2),XDEC+10
L R1,PEAK peak
XDECO R1,XDEC edit peak
MVC PG1+35(4),XDEC+8
L R1,PEAKPOS peakpos
XDECO R1,XDEC edit peakpos
MVC PG1+45(5),XDEC+7
XPRNT PG1,80 print buffer
LA R4,1(R4) pow2=pow2+1
SLA R8,1 p2=2**pow2
MVC PEAK,=F'5000' peak=0.5
EIF3 LA R6,1(R6) n=n+1
B LOOPN
ELOOPN L R1,L l
XDECO R1,XDEC edit l
MVC PG2+6(2),XDEC+10
XDECO R9,XDEC edit mallows
MVC PG2+29(5),XDEC+7
XPRNT PG2,80 print buffer
RETURN L R13,4(0,R13) restore savearea pointer
LM R14,R12,12(R13) restore registers
XR R15,R15 return code = 0
BR R14 return to caller
LTORG
L DC F'12'
UPRDIM DC F'4096' 2^L
PEAK DC F'5000' 0.5 fixed point 4dec
PEAKPOS DC F'0'
XDEC DS CL12
PG1 DC CL80'maximum between 2^xx and 2^xx is 0.xxxx at n=xxxxx'
PG2 DC CL80'for l=xx : mallows number is xxxxx'
A DS 4096F array a(uprdim)
REGEQU
END HOFSTADT
 Output:
maximum between 2^ 1 and 2^ 2 is 0.6666 at n= 3 maximum between 2^ 2 and 2^ 3 is 0.6666 at n= 6 maximum between 2^ 3 and 2^ 4 is 0.6363 at n= 11 maximum between 2^ 4 and 2^ 5 is 0.6086 at n= 23 maximum between 2^ 5 and 2^ 6 is 0.5909 at n= 44 maximum between 2^ 6 and 2^ 7 is 0.5760 at n= 92 maximum between 2^ 7 and 2^ 8 is 0.5674 at n= 178 maximum between 2^ 8 and 2^ 9 is 0.5594 at n= 370 maximum between 2^ 9 and 2^10 is 0.5549 at n= 719 maximum between 2^10 and 2^11 is 0.5501 at n= 1487 maximum between 2^11 and 2^12 is 0.5474 at n= 2897 for l=12 : mallows number is 1489
Ada
 Ada95 version
 Allocation of arrays on the heap
with Ada.Text_IO; use Ada.Text_IO;
with Unchecked_Deallocation;
procedure Conway is
package Real_io is new Float_IO (Float);
Maxrange : constant := 2 ** 20;
type Sequence is array (Positive range 1 .. Maxrange) of Positive;
type Sequence_Ptr is access all Sequence;
procedure Free is new Unchecked_Deallocation (Sequence, Sequence_Ptr);
S : Sequence_Ptr := new Sequence;
type Ratio_Array is array (Positive range 1 .. Maxrange) of Float;
type Ratio_Ptr is access all Ratio_Array;
procedure Free is new Unchecked_Deallocation (Ratio_Array, Ratio_Ptr);
Ratio : Ratio_Ptr := new Ratio_Array;
Mallows : Positive;
M : Natural := 0;
begin
S (1) := 1;
S (2) := 1;
for K in 3 .. Maxrange loop
S (K) := S (S (K  1)) + S (K  S (K  1));
end loop;
for k in 1 .. Maxrange loop
Ratio (k) := Float (S (k)) / Float (k);
end loop;
for N in 1 .. 19 loop
declare
Max : Float := 0.0;
Where : Positive;
begin
for K in 2 ** N .. 2 ** (N + 1) loop
if Max < Ratio (K) then
Max := Ratio (K);
Where := K;
end if;
end loop;
if (M = 0 and Max < 0.55) then
M := N  1;
end if;
Put
("Maximun of a(n)/n between 2^" &
Integer'Image (N) &
" and 2^" &
Integer'Image (N + 1) &
" was ");
Real_io.Put (Max, Fore => 1, Aft => 8, Exp => 0);
Put_Line (" at" & Integer'Image (Where));
end;
end loop;
 Calculate Mallows number
for I in reverse 2 ** M .. 2 ** (M + 1) loop
if (Ratio (I) > 0.55) then
Mallows := I;
exit;
end if;
end loop;
Put_Line ("Mallows number" & Integer'Image (Mallows));
Free (S);
Free (Ratio);
end Conway;
Sample output:
Maximun of a(n)/n between 2^ 1 and 2^ 2 was 0.66666669 at 3 Maximun of a(n)/n between 2^ 2 and 2^ 3 was 0.66666669 at 6 Maximun of a(n)/n between 2^ 3 and 2^ 4 was 0.63636363 at 11 Maximun of a(n)/n between 2^ 4 and 2^ 5 was 0.60869563 at 23 Maximun of a(n)/n between 2^ 5 and 2^ 6 was 0.59090906 at 44 Maximun of a(n)/n between 2^ 6 and 2^ 7 was 0.57608694 at 92 Maximun of a(n)/n between 2^ 7 and 2^ 8 was 0.56741571 at 178 Maximun of a(n)/n between 2^ 8 and 2^ 9 was 0.55945945 at 370 Maximun of a(n)/n between 2^ 9 and 2^ 10 was 0.55493742 at 719 Maximun of a(n)/n between 2^ 10 and 2^ 11 was 0.55010086 at 1487 Maximun of a(n)/n between 2^ 11 and 2^ 12 was 0.54746288 at 2897 Maximun of a(n)/n between 2^ 12 and 2^ 13 was 0.54414475 at 5969 Maximun of a(n)/n between 2^ 13 and 2^ 14 was 0.54244268 at 11651 Maximun of a(n)/n between 2^ 14 and 2^ 15 was 0.54007107 at 22223 Maximun of a(n)/n between 2^ 15 and 2^ 16 was 0.53878403 at 45083 Maximun of a(n)/n between 2^ 16 and 2^ 17 was 0.53704363 at 89516 Maximun of a(n)/n between 2^ 17 and 2^ 18 was 0.53602004 at 181385 Maximun of a(n)/n between 2^ 18 and 2^ 19 was 0.53464544 at 353683 Maximun of a(n)/n between 2^ 19 and 2^ 20 was 0.53377920 at 722589 Mallows number 1489
ALGOL 68
PROC do sqnc = (INT max)INT:
BEGIN
[max]INT a list;
INT k1 := 2,
lg2 := 1,
v := a list[1] := a list[2] := 1; # Concurrent declaration and assignment in declarations are allowed #
INT nmax;
LONG REAL amax := 0.0;
INT mallows number;
FOR n FROM 3 TO max DO
v := a list[n] := a list[v] + a list[nv];
( amax < v/n  amax := v/n; nmax := n ); # When given a Boolean as the 1st expression, (  ) is the short form of IF...THEN...FI #
IF v/n >= 0.55 THEN # This is the equivalent full form of the above construct #
mallows number := n
FI;
IF ABS(BIN k1 AND BIN n) = 0 THEN
# 'BIN' converts an INT type to a BITS type; In this context, 'ABS' reverses that operation #
printf(($"Maximum between 2^"g(0)" and 2^"g(0)" is about "g(10,8)" at "g(0)l$, lg2,lg2+1, amax, nmax));
amax := 0;
lg2 PLUSAB 1 # 'PLUSAB' (plusandbecomes) has the short form +:= #
FI;
k1 := n
OD;
mallows number # the result of the last expression evaluated is returned as the result of the PROC #
END;
INT mallows number = do sqnc(2**20); # This definition of 'mallows number' does not clash with the variable
of the same name inside PROC do sqnc  they are in different scopes#
printf(($"You too might have won $1000 with an answer of n = "g(0)$, mallows number))
Output:
Maximum between 2^1 and 2^2 is about 0.66666667 at 3 Maximum between 2^2 and 2^3 is about 0.66666667 at 6 Maximum between 2^3 and 2^4 is about 0.63636364 at 11 Maximum between 2^4 and 2^5 is about 0.60869565 at 23 Maximum between 2^5 and 2^6 is about 0.59090909 at 44 Maximum between 2^6 and 2^7 is about 0.57608696 at 92 Maximum between 2^7 and 2^8 is about 0.56741573 at 178 Maximum between 2^8 and 2^9 is about 0.55945946 at 370 Maximum between 2^9 and 2^10 is about 0.55493741 at 719 Maximum between 2^10 and 2^11 is about 0.55010087 at 1487 Maximum between 2^11 and 2^12 is about 0.54746289 at 2897 Maximum between 2^12 and 2^13 is about 0.54414475 at 5969 Maximum between 2^13 and 2^14 is about 0.54244271 at 11651 Maximum between 2^14 and 2^15 is about 0.54007110 at 22223 Maximum between 2^15 and 2^16 is about 0.53878402 at 45083 Maximum between 2^16 and 2^17 is about 0.53704366 at 89516 Maximum between 2^17 and 2^18 is about 0.53602007 at 181385 Maximum between 2^18 and 2^19 is about 0.53464543 at 353683 Maximum between 2^19 and 2^20 is about 0.53377923 at 722589 You too might have won $1000 with an answer of n = 1489
AppleScript
Like the other solutions here, this takes the linguistically confusing expression "the first position, p in the sequence where │a(n)/n│ < 0.55 for all n > p" to mean the last position where the result's not < 0.55.
on HC10000(ceiling)
script o
property lst : {1, 1}
property maxima : {}
end script
set p to missing value
set max to 0
set maxPos to 0
set power to 0
set x to end of o's lst
repeat with n from 3 to ceiling
set x to (item x of o's lst) + (item x of o's lst)
set end of o's lst to x
set ann to x / n
if (ann is not less than 0.55) then set p to n
if (ann > max) then
set max to ann
set maxPos to n
end if
if (ann is 0.5) then
set power to power + 1
set end of o's maxima to {<powers>:{power, power + 1}, n:maxPos, max:max}
set max to 0
end if
end repeat
return {p:p, maxima:o's maxima}
end HC10000
HC10000(2 ^ 20)
 Output:
{p:1489, maxima:{{<powers>:{1, 2}, n:3, max:0.666666666667}, {<powers>:{2, 3}, n:6, max:0.666666666667}, {<powers>:{3, 4}, n:11, max:0.636363636364}, {<powers>:{4, 5}, n:23, max:0.608695652174}, {<powers>:{5, 6}, n:44, max:0.590909090909}, {<powers>:{6, 7}, n:92, max:0.576086956522}, {<powers>:{7, 8}, n:178, max:0.567415730337}, {<powers>:{8, 9}, n:370, max:0.559459459459}, {<powers>:{9, 10}, n:719, max:0.554937413074}, {<powers>:{10, 11}, n:1487, max:0.550100874243}, {<powers>:{11, 12}, n:2897, max:0.547462892648}, {<powers>:{12, 13}, n:5969, max:0.544144747864}, {<powers>:{13, 14}, n:11651, max:0.54244270878}, {<powers>:{14, 15}, n:22223, max:0.540071097512}, {<powers>:{15, 16}, n:45083, max:0.538784020584}, {<powers>:{16, 17}, n:89516, max:0.537043657}, {<powers>:{17, 18}, n:181385, max:0.536020067812}, {<powers>:{18, 19}, n:353683, max:0.534645431078}, {<powers>:{19, 20}, n:722589, max:0.533779229963}}}
AutoHotkey
Progress, b2 w150 zh0 fs9, CreateLists ...
CreateLists(2 ** (Max:=20))
Progress,, Find Maxima ...
Loop, % Max  1
msg .= "Maximum between 2^" A_Index " and 2^" A_Index + 1
. " is " GetMax(2 ** A_Index, 2 ** (A_Index + 1), n)
. " for n = " n "`n"
Progress,, Find Mallows Number ...
Loop, % 2 ** Max
If (n_%A_Index% > 0.55)
MallowsNumber := A_Index
msg .= "Mallows Number = " MallowsNumber
Progress, Off
MsgBox, %msg%
;
GetMax(a, b, ByRef Item) { ; return max value of a(n)/n between a and b
;
Loop {
IfGreater, a, %b%, Break
If (Maximum < n_%a%)
Maximum := n_%a%, Item := a
a++
}
Return, Maximum
}
;
CreateLists(Lenght) { ; HofstadterConway sequences (using lookups)
;
; create the sequence a_%A_Index% [ a(n) ]
; and the sequence n_%A_Index% [ a(n)/n ]
;
global
a_1 := a_2 := n_1 := 1, n_2 := 1 / 2
Loop, %Lenght% {
IfLess, A_Index, 3, Continue
n1 := A_Index  1
an1 := a_%n1%
nan1 := A_Index  an1
a_%A_Index% := a_%an1% + a_%nan1%
n_%A_Index% := a_%A_Index% / A_Index
}
}
Message box shows:
Maximum between 2^1 and 2^2 is 0.666667 for n = 3 Maximum between 2^2 and 2^3 is 0.666667 for n = 6 Maximum between 2^3 and 2^4 is 0.636364 for n = 11 Maximum between 2^4 and 2^5 is 0.608696 for n = 23 Maximum between 2^5 and 2^6 is 0.590909 for n = 44 Maximum between 2^6 and 2^7 is 0.576087 for n = 92 Maximum between 2^7 and 2^8 is 0.567416 for n = 178 Maximum between 2^8 and 2^9 is 0.559459 for n = 370 Maximum between 2^9 and 2^10 is 0.554937 for n = 719 Maximum between 2^10 and 2^11 is 0.550101 for n = 1487 Maximum between 2^11 and 2^12 is 0.547463 for n = 2897 Maximum between 2^12 and 2^13 is 0.544145 for n = 5969 Maximum between 2^13 and 2^14 is 0.542443 for n = 11651 Maximum between 2^14 and 2^15 is 0.540071 for n = 22223 Maximum between 2^15 and 2^16 is 0.538784 for n = 45083 Maximum between 2^16 and 2^17 is 0.537044 for n = 89516 Maximum between 2^17 and 2^18 is 0.536020 for n = 181385 Maximum between 2^18 and 2^19 is 0.534645 for n = 353683 Maximum between 2^19 and 2^20 is 0.533779 for n = 722589 Mallows Number = 1489
AWK
Iterative approach:
#!/usr/bin/awk f
BEGIN {
NN = 20;
iterativeHCsequence(2^NN+1,Q);
for (K=1; K<NN; K++) {
m = 0;
for (n=2^K+1; n<=2^(K+1); n++) {
v = Q[n]/n;
if (m < v) {nn=n; m = v};
}
printf "Maximum a(n)/n between 2^%i and 2^%i is %f at n=%i\n",K,K+1,m,nn;
}
print "number of Q(n)<Q(n+1) for n<=100000 : " NN;
}
function iterativeHCsequence(N,Q) {
Q[1] = 1;
Q[2] = 1;
for (n=3; n<=N; n++) {
Q[n] = Q[Q[n1]]+Q[nQ[n1]];
}
}
Recursive variant:
#!/usr/bin/awk f
BEGIN {
Q[1] = 1;
Q[2] = 1;
S[1] = 1;
S[2] = 1;
NN = 20;
for (K=1; K<NN; K++) {
m = 0;
for (n=2^K+1; n<=2^(K+1); n++) {
v = HCsequence(n,Q,S)/n;
if (m < v) {nn=n; m = v};
}
printf "Maximum between 2^%i and 2^%i is %f at n=%i\n",K,K+1,m,nn;
}
}
function HCsequence(n,Q,S) {
## recursive definition
if (S[n]==0) {
k = n1;
if (S[k]==0) {
HCsequence(k,Q,S);
}
k = Q[n1];
if (S[k]==0) {
HCsequence(k,Q,S);
}
k = nQ[n1];
if (S[k]==0) {
HCsequence(k,Q,S);
}
}
Q[n] = Q[Q[n1]]+Q[nQ[n1]];
S[n] = 1;
return (Q[n]);
}
Output:
Maximum between 2^1 and 2^2 is 0.666667 at n=3 Maximum between 2^2 and 2^3 is 0.666667 at n=6 Maximum between 2^3 and 2^4 is 0.636364 at n=11 Maximum between 2^4 and 2^5 is 0.608696 at n=23 Maximum between 2^5 and 2^6 is 0.590909 at n=44 Maximum between 2^6 and 2^7 is 0.576087 at n=92 Maximum between 2^7 and 2^8 is 0.567416 at n=178 Maximum between 2^8 and 2^9 is 0.559459 at n=370 Maximum between 2^9 and 2^10 is 0.554937 at n=719 Maximum between 2^10 and 2^11 is 0.550101 at n=1487 Maximum between 2^11 and 2^12 is 0.547463 at n=2897 Maximum between 2^12 and 2^13 is 0.544145 at n=5969 Maximum between 2^13 and 2^14 is 0.542443 at n=11651 Maximum between 2^14 and 2^15 is 0.540071 at n=22223 Maximum between 2^15 and 2^16 is 0.538784 at n=45083 Maximum between 2^16 and 2^17 is 0.537044 at n=89516 Maximum between 2^17 and 2^18 is 0.536020 at n=181385 Maximum between 2^18 and 2^19 is 0.534645 at n=353683 Maximum between 2^19 and 2^20 is 0.533779 at n=722589
BASIC
BASIC256
arraybase 1
pow2 = 2
p2 = 2 ^ pow2
peak = .5
dim a(2 ^ 20)
a[1] = 1
a[2] = 1
for n = 3 to 2 ^ 20
a[n] = a[a[n1]] + a[na[n1]]
r = a[n] / n
if r >= .55 then mallows = n
if r > peak then peak = r : peakpos = n
if n = p2 then
print "Maximum between 2 ^ "; rjust((pow2  1),2); " and 2 ^ "; rjust(pow2,2); " is "; ljust(peak,13,"0"); " at n = "; peakpos
pow2 += 1
p2 = 2 ^ pow2
peak = .5
end if
next n
print
print "Mallows number is "; mallows
 Output:
Same as FreeBASIC entry.
BBC BASIC
HIMEM=LOMEM+1E7 : REM Reserve enough memory for a 4 MB array, plus other code
DIM a%(2^20)
a%(1)=1
a%(2)=1
pow2%=2
p2%=2^pow2%
peak=0.5
peakpos%=0
FOR n%=3 TO 2^20
a%(n%)=a%(a%(n%1))+a%(n%a%(n%1))
r=a%(n%)/n%
IF r>=0.55 THEN Mallows%=n%
IF r>peak THEN peak=r:peakpos%=n%
IF n%=p2% THEN
PRINT "Maximum between 2^";pow2%1;" and 2^";pow2%;" is ";peak;" at n=";peakpos%
pow2%+=1
p2%=2^pow2%
peak=0.5
ENDIF
NEXT n%
PRINT "Mallows number is ";Mallows%
Results
Maximum between 2^1 and 2^2 is 0.666666667 at n=3 Maximum between 2^2 and 2^3 is 0.666666667 at n=6 Maximum between 2^3 and 2^4 is 0.636363637 at n=11 Maximum between 2^4 and 2^5 is 0.608695652 at n=23 Maximum between 2^5 and 2^6 is 0.590909091 at n=44 Maximum between 2^6 and 2^7 is 0.576086957 at n=92 Maximum between 2^7 and 2^8 is 0.56741573 at n=178 Maximum between 2^8 and 2^9 is 0.55945946 at n=370 Maximum between 2^9 and 2^10 is 0.554937413 at n=719 Maximum between 2^10 and 2^11 is 0.550100874 at n=1487 Maximum between 2^11 and 2^12 is 0.547462893 at n=2897 Maximum between 2^12 and 2^13 is 0.544144748 at n=5969 Maximum between 2^13 and 2^14 is 0.542442709 at n=11651 Maximum between 2^14 and 2^15 is 0.540071098 at n=22223 Maximum between 2^15 and 2^16 is 0.538784021 at n=45083 Maximum between 2^16 and 2^17 is 0.537043657 at n=89516 Maximum between 2^17 and 2^18 is 0.536020068 at n=181385 Maximum between 2^18 and 2^19 is 0.534645431 at n=353683 Maximum between 2^19 and 2^20 is 0.53377923 at n=722589 Mallows number is 1489
FreeBASIC
' version 13072018
' compile with: fbc s console
Dim As UInteger a(), pow2 = 2, p2 = 2 ^ pow2, peakpos, n, mallows
Dim As Double peak = 0.5, r
ReDim a(2 ^ 20)
a(1) = 1
a(2) = 1
For n = 3 To 2 ^ 20
a(n) = a(a(n 1)) + a(n  a(n 1))
r = a(n) / n
If r >= 0.55 Then mallows = n
If r > peak Then peak = r : peakpos = n
If n = p2 Then
Print Using "Maximum between 2 ^ ## and 2 ^ ## is"; pow2 1; pow2;
Print Using " #.##### at n = "; peak;
Print peakpos
pow2 += 1
p2 = 2 ^ pow2
peak = 0.5
End If
Next
Print
Print "Mallows number is "; mallows
' empty keyboard buffer
While Inkey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End
 Output:
Maximum between 2 ^ 1 and 2 ^ 2 is 0.66667 at n = 3 Maximum between 2 ^ 2 and 2 ^ 3 is 0.66667 at n = 6 Maximum between 2 ^ 3 and 2 ^ 4 is 0.63636 at n = 11 Maximum between 2 ^ 4 and 2 ^ 5 is 0.60870 at n = 23 Maximum between 2 ^ 5 and 2 ^ 6 is 0.59091 at n = 44 Maximum between 2 ^ 6 and 2 ^ 7 is 0.57609 at n = 92 Maximum between 2 ^ 7 and 2 ^ 8 is 0.56742 at n = 178 Maximum between 2 ^ 8 and 2 ^ 9 is 0.55946 at n = 370 Maximum between 2 ^ 9 and 2 ^ 10 is 0.55494 at n = 719 Maximum between 2 ^ 10 and 2 ^ 11 is 0.55010 at n = 1487 Maximum between 2 ^ 11 and 2 ^ 12 is 0.54746 at n = 2897 Maximum between 2 ^ 12 and 2 ^ 13 is 0.54414 at n = 5969 Maximum between 2 ^ 13 and 2 ^ 14 is 0.54244 at n = 11651 Maximum between 2 ^ 14 and 2 ^ 15 is 0.54007 at n = 22223 Maximum between 2 ^ 15 and 2 ^ 16 is 0.53878 at n = 45083 Maximum between 2 ^ 16 and 2 ^ 17 is 0.53704 at n = 89516 Maximum between 2 ^ 17 and 2 ^ 18 is 0.53602 at n = 181385 Maximum between 2 ^ 18 and 2 ^ 19 is 0.53465 at n = 353683 Maximum between 2 ^ 19 and 2 ^ 20 is 0.53378 at n = 722589 Mallows number is 1489
PureBasic
If OpenConsole()
Define.i upperlim, i=1, k1=2, n=3, v=1
Define.d Maximum
Print("Enter limit (ENTER gives 2^20=1048576): "): upperlim=Val(Input())
If upperlim<=0: upperlim=1048576: EndIf
Dim tal(upperlim)
If ArraySize(tal())=1
PrintN("Could not allocate needed memory!"): Input(): End
EndIf
tal(1)=1: tal(2)=1
While n<=upperlim
v=tal(v)+tal(nv)
tal(n)=v
If Maximum<(v/n): Maximum=v/n: EndIf
If Not n&k1
PrintN("Maximum between 2^"+Str(i)+" and 2^"+Str(i+1)+" was "+StrD(Maximum,6))
Maximum=0.0
i+1
EndIf
k1=n
n+1
Wend
Print(#CRLF$+"Press ENTER to exit."): Input()
CloseConsole()
EndIf
Run BASIC
input "Enter upper limit between 1 and 20 (ENTER 20 gives 2^20):"); uprLim
if uprLim < 1 or uprLim > 20 then uprLim = 20
dim a(2^uprLim)
a(1) = 1
a(2) = 1
pow2 = 2
p2 = 2^pow2
p = 0.5
pPos = 0
for n = 3 TO 2^uprLim
a(n) = a(a(n1)) + a(na(n1))
r = a(n)/n
if r >= 0.55 THEN Mallows = n
if r > p THEN
p = r
pPos = n
end if
if n = p2 THEN
print "Maximum between";chr$(9);" 2^";pow21;" and 2^";pow2;chr$(9);" is ";p;chr$(9);" at n = ";pPos
pow2 = pow2 + 1
p2 = 2^pow2
p = 0.5
end IF
next n
print "Mallows number is ";Mallows
Enter upper limit between 1 and 20 (ENTER 20 gives 2^20): ?20 Maximum between 2^1 and 2^2 is 0.666666698 at n = 3 Maximum between 2^2 and 2^3 is 0.666666698 at n = 6 Maximum between 2^3 and 2^4 is 0.636363601 at n = 11 Maximum between 2^4 and 2^5 is 0.608695602 at n = 23 Maximum between 2^5 and 2^6 is 0.590909051 at n = 44 Maximum between 2^6 and 2^7 is 0.57608695 at n = 92 Maximum between 2^7 and 2^8 is 0.567415714 at n = 178 Maximum between 2^8 and 2^9 is 0.559459447 at n = 370 Maximum between 2^9 and 2^10 is 0.55493741 at n = 719 Maximum between 2^10 and 2^11 is 0.550100851 at n = 1487 Maximum between 2^11 and 2^12 is 0.547462892 at n = 2897 Maximum between 2^12 and 2^13 is 0.544144725 at n = 5969 Maximum between 2^13 and 2^14 is 0.542442655 at n = 11651 Maximum between 2^14 and 2^15 is 0.540071058 at n = 22223 Maximum between 2^15 and 2^16 is 0.538784027 at n = 45083 Maximum between 2^16 and 2^17 is 0.537043619 at n = 89516 Maximum between 2^17 and 2^18 is 0.53602004 at n = 181385 Maximum between 2^18 and 2^19 is 0.534645414 at n = 353683 Maximum between 2^19 and 2^20 is 0.533779191 at n = 722589 Mallows number is 1489
True BASIC
LET pow2 = 2
LET p2 = 2^pow2
LET peak = 0.5
DIM a(0)
MAT REDIM a(2^20)
LET a(1) = 1
LET a(2) = 1
FOR n = 3 TO 2^20
LET a(n) = a(a(n1))+a(na(n1))
LET r = a(n)/n
IF r >= 0.55 THEN LET mallows = n
IF r > peak THEN
LET peak = r
LET peakpos = n
END IF
IF n = p2 THEN
PRINT USING "Maximum between 2 ^ ## and 2 ^ ## is": pow21, pow2;
PRINT USING " #.##### at n = ": peak;
PRINT peakpos
LET pow2 = pow2+1
LET p2 = 2^pow2
LET peak = 0.5
END IF
NEXT n
PRINT
PRINT "Mallows number is "; mallows
END
 Output:
Same as FreeBASIC entry.
Yabasic
pow2 = 2
p2 = 2 ^ pow2
peak = .5
dim a(2 ^ 20)
a(1) = 1
a(2) = 1
for n = 3 to 2 ^ 20
a(n) = a(a(n  1)) + a(n  a(n  1))
r = a(n) / n
if r >= .55 mallows = n
if r > peak then peak = r : peakpos = n : fi
if n = p2 then
print "Maximum between 2 ^ ", pow2  1 using("##"), " and 2 ^ ", pow2 using("##"), " is ", peak using("#.#####"), " at n = ", peakpos
pow2 = pow2 + 1
p2 = 2 ^ pow2
peak = .5
end if
next n
print "\nMallows number is ", mallows
 Output:
Same as FreeBASIC entry.
ZX Spectrum Basic
Nine first results.
10 DIM a(2000)
20 LET a(1)=1: LET a(2)=1
30 LET pow2=2: LET p2=2^pow2
40 LET peak=0.5: LET peakpos=0
50 FOR n=3 TO 2000
60 LET a(n)=a(a(n1))+a(na(n1))
70 LET r=a(n)/n
80 IF r>0.55 THEN LET Mallows=n
90 IF r>peak THEN LET peak=r: LET peakpos=n
100 IF n=p2 THEN PRINT "Maximum (2^";pow21;", 2^";pow2;") is ";peak;" at n=";peakpos: LET pow2=pow2+1: LET p2=2^pow2: LET peak=0.5
110 NEXT n
120 PRINT "Mallows number is ";Mallows
Bracmat
( ( a
=
. !arg:(12)&1
 (as..find)$!arg:(?.?arg)&!arg
 (as..insert)
$ ( !arg
. a$(a$(!arg+1))+a$(!arg+1*a$(!arg+1)):?arg
)
& !arg
)
& new$hash:?as
& 0:?n:?maxan/n
& 1:?pow
& whl
' ( 1+!n:?n
& !pow:~>20
& ( 2^!pow:~!n
 out
$ ( str
$ ( "Between 2^"
!pow+1
" and 2^"
!pow
" the maximum value of a(n)/n is reached for n = "
!maxn
" with the value "
!maxan/n
)
)
& 0:?maxan/n
& 1+!pow:?pow
)
& a$!n*!n^1:?an/n
& ( !an/n:>!maxan/n:?maxan/n
& !n:?maxn

)
& ( !an/n:~<11/20:?Man/n&!n:?Mallows

)
)
& out
$ ( str
$ ( "Mallows number is "
!Mallows
", where a("
!Mallows
")/"
!Mallows
" == "
!Man/n
", which is greater than 0.55 by "
!Man/n+11/20
)
)
)
Output:
Between 2^0 and 2^1 the maximum value of a(n)/n is reached for n = 1 with the value 1 Between 2^1 and 2^2 the maximum value of a(n)/n is reached for n = 3 with the value 2/3 Between 2^2 and 2^3 the maximum value of a(n)/n is reached for n = 6 with the value 2/3 Between 2^3 and 2^4 the maximum value of a(n)/n is reached for n = 11 with the value 7/11 Between 2^4 and 2^5 the maximum value of a(n)/n is reached for n = 23 with the value 14/23 Between 2^5 and 2^6 the maximum value of a(n)/n is reached for n = 44 with the value 13/22 Between 2^6 and 2^7 the maximum value of a(n)/n is reached for n = 92 with the value 53/92 Between 2^7 and 2^8 the maximum value of a(n)/n is reached for n = 178 with the value 101/178 Between 2^8 and 2^9 the maximum value of a(n)/n is reached for n = 370 with the value 207/370 Between 2^9 and 2^10 the maximum value of a(n)/n is reached for n = 719 with the value 399/719 Between 2^10 and 2^11 the maximum value of a(n)/n is reached for n = 1487 with the value 818/1487 Between 2^11 and 2^12 the maximum value of a(n)/n is reached for n = 2897 with the value 1586/2897 Between 2^12 and 2^13 the maximum value of a(n)/n is reached for n = 5969 with the value 3248/5969 Between 2^13 and 2^14 the maximum value of a(n)/n is reached for n = 11651 with the value 6320/11651 Between 2^14 and 2^15 the maximum value of a(n)/n is reached for n = 22223 with the value 12002/22223 Between 2^15 and 2^16 the maximum value of a(n)/n is reached for n = 45083 with the value 24290/45083 Between 2^16 and 2^17 the maximum value of a(n)/n is reached for n = 89516 with the value 24037/44758 Between 2^17 and 2^18 the maximum value of a(n)/n is reached for n = 181385 with the value 97226/181385 Between 2^18 and 2^19 the maximum value of a(n)/n is reached for n = 353683 with the value 189095/353683 Between 2^19 and 2^20 the maximum value of a(n)/n is reached for n = 722589 with the value 385703/722589 Mallows number is 1489, where a(1489)/1489 == 819/1489, which is greater than 0.55 by 1/29780
C
#include <stdio.h>
#include <stdlib.h>
int a_list[1<<20 + 1];
int doSqnc( int m)
{
int max_df = 0;
int p2_max = 2;
int v, n;
int k1 = 2;
int lg2 = 1;
double amax = 0;
a_list[0] = 50000;
a_list[1] = a_list[2] = 1;
v = a_list[2];
for (n=3; n <= m; n++) {
v = a_list[n] = a_list[v] + a_list[nv];
if ( amax < v*1.0/n) amax = v*1.0/n;
if ( 0 == (k1&n)) {
printf("Maximum between 2^%d and 2^%d was %f\n", lg2,lg2+1, amax);
amax = 0;
lg2++;
}
k1 = n;
}
return 1;
}
Results
Maximum between 2^1 and 2^2 was 0.666667 Maximum between 2^2 and 2^3 was 0.666667 Maximum between 2^3 and 2^4 was 0.636364 Maximum between 2^4 and 2^5 was 0.608696 .... Maximum between 2^18 and 2^19 was 0.534645 Maximum between 2^19 and 2^20 was 0.533779
C#
using System;
using System.Linq;
namespace HofstadterConway
{
class Program
{
static int[] GenHofstadterConway(int max)
{
int[] result = new int[max];
result[0]=result[1]=1;
for (int ix = 2; ix < max; ix++)
result[ix] = result[result[ix  1]  1] + result[ix  result[ix  1]];
return result;
}
static void Main(string[] args)
{
double[] adiv = new double[1 << 20];
{
int[] a = GenHofstadterConway(1 << 20);
for (int i = 0; i < 1 << 20; i++)
adiv[i] = a[i] / (double)(i + 1);
}
for (int p = 2; p <= 20; p++)
{
var max = Enumerable.Range(
(1 << (p  1))  1,
(1 << p)  (1 << (p  1))
)
.Select(ix => new { I = ix + 1, A = adiv[ix] })
.OrderByDescending(x => x.A)
.First();
Console.WriteLine("Maximum from 2^{0} to 2^{1} is {2} at {3}",
p  1, p, max.A, max.I);
}
Console.WriteLine("The winning number is {0}.",
Enumerable.Range(0, 1 << 20)
.Last(i => (adiv[i] > 0.55)) + 1
);
}
}
}
Output:
Maximum from 2^1 to 2^2 is 0.666666666666667 at 3 Maximum from 2^2 to 2^3 is 0.666666666666667 at 6 Maximum from 2^3 to 2^4 is 0.636363636363636 at 11 Maximum from 2^4 to 2^5 is 0.608695652173913 at 23 Maximum from 2^5 to 2^6 is 0.590909090909091 at 44 Maximum from 2^6 to 2^7 is 0.576086956521739 at 92 Maximum from 2^7 to 2^8 is 0.567415730337079 at 178 Maximum from 2^8 to 2^9 is 0.559459459459459 at 370 Maximum from 2^9 to 2^10 is 0.554937413073713 at 719 Maximum from 2^10 to 2^11 is 0.550100874243443 at 1487 Maximum from 2^11 to 2^12 is 0.547462892647566 at 2897 Maximum from 2^12 to 2^13 is 0.544144747863964 at 5969 Maximum from 2^13 to 2^14 is 0.542442708780362 at 11651 Maximum from 2^14 to 2^15 is 0.540071097511587 at 22223 Maximum from 2^15 to 2^16 is 0.538784020584256 at 45083 Maximum from 2^16 to 2^17 is 0.537043656999866 at 89516 Maximum from 2^17 to 2^18 is 0.536020067811561 at 181385 Maximum from 2^18 to 2^19 is 0.534645431078112 at 353683 Maximum from 2^19 to 2^20 is 0.533779229963368 at 722589 The winning number is 1489.
C++
#include <deque>
#include <iostream>
int hcseq(int n)
{
static std::deque<int> seq(2, 1);
while (seq.size() < n)
{
int x = seq.back();
seq.push_back(seq[x1] + seq[seq.size()x]);
}
return seq[n1];
}
int main()
{
int pow2 = 1;
for (int i = 0; i < 20; ++i)
{
int pow2next = 2*pow2;
double max = 0;
for (int n = pow2; n < pow2next; ++n)
{
double anon = hcseq(n)/double(n);
if (anon > max)
max = anon;
}
std::cout << "maximum of a(n)/n between 2^" << i
<< " (" << pow2 << ") and 2^" << i+1
<< " (" << pow2next << ") is " << max << "\n";
pow2 = pow2next;
}
}
Output:
maximum of a(n)/n between 2^0 (1) and 2^1 (2) is 1 maximum of a(n)/n between 2^1 (2) and 2^2 (4) is 0.666667 maximum of a(n)/n between 2^2 (4) and 2^3 (8) is 0.666667 maximum of a(n)/n between 2^3 (8) and 2^4 (16) is 0.636364 maximum of a(n)/n between 2^4 (16) and 2^5 (32) is 0.608696 maximum of a(n)/n between 2^5 (32) and 2^6 (64) is 0.590909 maximum of a(n)/n between 2^6 (64) and 2^7 (128) is 0.576087 maximum of a(n)/n between 2^7 (128) and 2^8 (256) is 0.567416 maximum of a(n)/n between 2^8 (256) and 2^9 (512) is 0.559459 maximum of a(n)/n between 2^9 (512) and 2^10 (1024) is 0.554937 maximum of a(n)/n between 2^10 (1024) and 2^11 (2048) is 0.550101 maximum of a(n)/n between 2^11 (2048) and 2^12 (4096) is 0.547463 maximum of a(n)/n between 2^12 (4096) and 2^13 (8192) is 0.544145 maximum of a(n)/n between 2^13 (8192) and 2^14 (16384) is 0.542443 maximum of a(n)/n between 2^14 (16384) and 2^15 (32768) is 0.540071 maximum of a(n)/n between 2^15 (32768) and 2^16 (65536) is 0.538784 maximum of a(n)/n between 2^16 (65536) and 2^17 (131072) is 0.537044 maximum of a(n)/n between 2^17 (131072) and 2^18 (262144) is 0.53602 maximum of a(n)/n between 2^18 (262144) and 2^19 (524288) is 0.534645 maximum of a(n)/n between 2^19 (524288) and 2^20 (1048576) is 0.533779
Clojure
(ns rosettacode.hofstaderconway
(:use [clojure.math.numerictower :only [expt]]))
;; A literal transcription of the definition, with memoize doing the heavy lifting
(def conway
(memoize
(fn [x]
(if (< x 3)
1
(+ (> x dec conway conway)
(>> x dec conway ( x) conway))))))
(let [N (drop 1 (range))
pow2 (map #(expt 2 %) N)
; Split the natural numbers into groups at each power of two
groups (partitionby (fn [x] (filter #(> % x) pow2)) N)
maxima (>> (map #(map conway %) groups)
(map #(map / %2 %1) groups) ; Each conway number divided by its index
(map #(apply max %)))
m20 (take 20 maxima)]
(println
(take 4 maxima) "\n"
(apply >= m20) "\n"
(map double m20))) ; output the decimal forms
Common Lisp
(defparameter *hofcon*
(makearray '(2) :initialcontents '(1 1) :adjustable t
:elementtype 'integer :fillpointer 2))
(defparameter *hofconratios*
(makearray '(2) :initialcontents '(1.0 0.5) :adjustable t
:elementtype 'singlefloat :fillpointer 2))
(defun hofcon (n)
(let ((l (length *hofcon*)))
(if (<= n l) (aref *hofcon* (1 n))
(extendhofconsequence l n))))
(defun extendhofconsequence (l n)
(loop for i from l below n do
(let* ((x (aref *hofcon* (1 i)))
(hc (+ (aref *hofcon* (1 x))
(aref *hofcon* ( i x)))))
(vectorpushextend hc *hofcon*)
(vectorpushextend (/ hc (+ i 1.0)) *hofconratios*)))
(aref *hofcon* (1 n)))
(defun maxinarrayrange (arr id1 id2)
(let ((m 0) (id 0))
(loop for i from (1 id1) to (1 id2) do
(let ((n (aref arr i)))
(if (> n m) (setq m n id i))))
(values m (1+ id))))
(defun maxima (po2)
(hofcon (expt 2 po2))
(loop for i from 1 below po2 do
(let ((id1 (expt 2 i)) (id2 (expt 2 (1+ i))))
(multiplevaluebind (m id)
(maxinarrayrange *hofconratios* id1 id2)
(format t "Local maximum in [~A .. ~A]: ~A at n = ~A~%" id1 id2 m id)))))
(defun mallows (po2)
(let ((n (expt 2 po2)))
(hofcon n)
(do ((i (1 n) (1 i)))
((> (aref *hofconratios* i) 0.55) (+ i 1)))))
Sample session:
ROSETTA> (maxima 20) Local maximum in [2 .. 4]: 0.6666667 at n = 3 Local maximum in [4 .. 8]: 0.6666667 at n = 6 Local maximum in [8 .. 16]: 0.6363636 at n = 11 Local maximum in [16 .. 32]: 0.6086956 at n = 23 Local maximum in [32 .. 64]: 0.59090906 at n = 44 Local maximum in [64 .. 128]: 0.57608694 at n = 92 Local maximum in [128 .. 256]: 0.5674157 at n = 178 Local maximum in [256 .. 512]: 0.55945945 at n = 370 Local maximum in [512 .. 1024]: 0.5549374 at n = 719 Local maximum in [1024 .. 2048]: 0.55010086 at n = 1487 Local maximum in [2048 .. 4096]: 0.5474629 at n = 2897 Local maximum in [4096 .. 8192]: 0.54414475 at n = 5969 Local maximum in [8192 .. 16384]: 0.5424427 at n = 11651 Local maximum in [16384 .. 32768]: 0.54007107 at n = 22223 Local maximum in [32768 .. 65536]: 0.538784 at n = 45083 Local maximum in [65536 .. 131072]: 0.53704363 at n = 89516 Local maximum in [131072 .. 262144]: 0.53602004 at n = 181385 Local maximum in [262144 .. 524288]: 0.53464544 at n = 353683 Local maximum in [524288 .. 1048576]: 0.5337792 at n = 722589 NIL ROSETTA> (mallows 20) 1489
D
import std.stdio, std.algorithm;
void hofstadterConwaySequence(in int m) {
auto alist = new int[m + 1];
alist[0 .. 2] = 1;
auto v = alist[2];
int k1 = 2, lg2 = 1;
double amax = 0.0;
foreach (n; 2 .. m + 1) {
v = alist[n] = alist[v] + alist[n  v];
amax = max(amax, v * 1.0 / n);
if ((k1 & n) == 0) {
writefln("Max in [2^%d, 2^%d]: %f", lg2, lg2 + 1, amax);
amax = 0;
lg2++;
}
k1 = n;
}
}
void main() {
hofstadterConwaySequence(2 ^^ 20);
}
Output:
Max in [2^1, 2^2]: 0.666667 Max in [2^2, 2^3]: 0.666667 Max in [2^3, 2^4]: 0.636364 Max in [2^4, 2^5]: 0.608696 Max in [2^5, 2^6]: 0.590909 Max in [2^6, 2^7]: 0.576087 Max in [2^7, 2^8]: 0.567416 Max in [2^8, 2^9]: 0.559459 Max in [2^9, 2^10]: 0.554937 Max in [2^10, 2^11]: 0.550101 Max in [2^11, 2^12]: 0.547463 Max in [2^12, 2^13]: 0.544145 Max in [2^13, 2^14]: 0.542443 Max in [2^14, 2^15]: 0.540071 Max in [2^15, 2^16]: 0.538784 Max in [2^16, 2^17]: 0.537044 Max in [2^17, 2^18]: 0.536020 Max in [2^18, 2^19]: 0.534645 Max in [2^19, 2^20]: 0.533779
EasyLang
numfmt 4 0
a[] = [ 1 1 ]
x = 1
n = 2
mallow = 0
for p = 1 to 19
max = 0
nextPot = n * 2
while n < nextPot
n = len a[] + 1
x = a[x] + a[n  x]
a[] &= x
f = x / n
max = higher max f
if f >= 0.55
mallow = n
.
.
print "max between 2^" & p & " and 2^" & p + 1 & " was " & max
.
print "winning number " & mallow
EchoLisp
(decimals 4)
(cachesize 2000000)
(define (a n)
(+ (a (a (1 n))) (a ( n (a (1 n))))))
(remember 'a #(0 1 1)) ;; memoize
;; prints max a(n)/n in [2**i 2**i+1] intervals
;; return Mallows number checked up to 2**20
(define (task (maxv) (start 1) (end 2) (v) (mrange 0))
(for ((i (inrange 1 21)))
(set! maxv 0)
(for ((n (inrange start end)))
(set! v (// (a n) n))
#:when (> v maxv)
(set! maxv v))
(when (and (zero? mrange) (< maxv 0.55)) (set! mrange end))
(printf "[%d .. %d] → max a(n)/n: %d " start end maxv)
(set! start end)
(set! end (* start 2)))
;; mallows
(for ((n (inrange mrange 2 1)))
#:break (> (// (a n) n) 0.55) => n )
)
 Output:
(task) [1 .. 2] → max a(n)/n: 1 [2 .. 4] → max a(n)/n: 0.6667 [4 .. 8] → max a(n)/n: 0.6667 [8 .. 16] → max a(n)/n: 0.6364 [16 .. 32] → max a(n)/n: 0.6087 [32 .. 64] → max a(n)/n: 0.5909 [64 .. 128] → max a(n)/n: 0.5761 [128 .. 256] → max a(n)/n: 0.5674 [256 .. 512] → max a(n)/n: 0.5595 [512 .. 1024] → max a(n)/n: 0.5549 [1024 .. 2048] → max a(n)/n: 0.5501 [2048 .. 4096] → max a(n)/n: 0.5475 [4096 .. 8192] → max a(n)/n: 0.5441 [8192 .. 16384] → max a(n)/n: 0.5424 [16384 .. 32768] → max a(n)/n: 0.5401 [32768 .. 65536] → max a(n)/n: 0.5388 [65536 .. 131072] → max a(n)/n: 0.537 [131072 .. 262144] → max a(n)/n: 0.536 [262144 .. 524288] → max a(n)/n: 0.5346 [524288 .. 1048576] → max a(n)/n: 0.5338 → 1489 ;; Mallows number
Eiffel
class
APPLICATION
create
make
feature
make
Tests the feature sequence.
local
j, n, exp: INTEGER
max: REAL_64
do
exp := 15
n := (2 ^ exp).floor
sequence (n)
across
1 .. (exp  1) as c
loop
max := 0
from
j := (2 ^ c.item).floor
until
j > 2 ^ (c.item + 1)
loop
if members [j] / j > max then
max := members [j] / j
end
j := j + 1
end
io.put_string ("Between 2^" + c.item.out + "and 2^" + (c.item + 1).out + " the max is: " + max.out)
io.new_line
end
end
feature {NONE}
members: LINKED_LIST [INTEGER]
 Members of the Hofstadter Conway $10000 sequence.
sequence (n: INTEGER)
 Hofstadter Conway $10000 sequence up to 'n' in 'members'.
require
n_positive: n > 0
local
last: INTEGER
do
create members.make
members.extend (1)
members.extend (1)
across
3 .. n as c
loop
last := members.last
members.extend (members [last] + members [c.item  last])
end
end
end
 Output:
As the run time is quite slow, the test output is shown only up to 2^15.
Between 2^1 and 2^2 the max is: 0.66666666666666663 at 3. Between 2^2 and 2^3 the max is: 0.66666666666666663 at 6. Between 2^3 and 2^4 the max is: 0.63636363636363635 at 11. Between 2^4 and 2^5 the max is: 0.60869565217391308 at 23. Between 2^5 and 2^6 the max is: 0.59090909090909094 at 44. Between 2^6 and 2^7 the max is: 0.57608695652173914 at 92. Between 2^7 and 2^8 the max is: 0.56741573033707871 at 178. Between 2^8 and 2^9 the max is: 0.55945945945945941 at 370. Between 2^9 and 2^10 the max is: 0.55493741307371347 at 719. Between 2^10 and 2^11 the max is: 0.55010087424344323 at 1487. Between 2^11 and 2^12 the max is: 0.54746289264756642 at 2897. Between 2^12 and 2^13 the max is: 0.54414474786396383 at 5969. Between 2^13 and 2^14 the max is: 0.54244270878036216 at 11651. Between 2^14 and 2^15 the max is: 0.54007109754458711 at 22223.
Erlang
module( hofstadter_conway ).
export( [sequence/1, sequence_div_n/1, task/0] ).
record( power_of_2, {div_n=0, max=4, min=2, n=0} ).
sequence( 1 ) > [1];
sequence( 2 ) > [1, 1];
sequence( Up_to ) when Up_to >= 3 >
From_3 = lists:seq( 3, Up_to ),
Dict = lists:foldl( fun sequence_dict/2, dict:from_list([{1, 1}, {2, 1}]), From_3 ),
[1, 1  [dict:fetch(X, Dict)  X < From_3]].
sequence_div_n( N ) >
Sequence = sequence( N ),
[{X, Y / X}  {X, Y} < lists:zip(lists:seq(1, N), Sequence)].
task() >
[_First  Rest] = sequence_div_n( erlang:round(math:pow(2, 20)) ),
{_Power, Powers} = lists:foldl( fun max_between_power_of_2/2, {#power_of_2{}, []}, Rest ),
[io:fwrite( "Maximum between ~p and ~p is ~p for n=~p~n", [X#power_of_2.min, X#power_of_2.max, X#power_of_2.div_n, X#power_of_2.n])  X < Powers].
max_between_power_of_2( {N, _Div_n}, {#power_of_2{max=N}=P, Acc} ) >
{#power_of_2{max=N * 2, min=N}, [P  Acc]};
max_between_power_of_2( {N, Larger_div_n}, {#power_of_2{div_n=Div_n}=P, Acc} ) when Larger_div_n > Div_n >
{P#power_of_2{n=N, div_n=Larger_div_n}, Acc};
max_between_power_of_2( _, Both ) > Both.
sequence_dict( Key, Dict ) >
Last_number = dict:fetch( Key  1, Dict ),
At_begining = dict:fetch( Last_number, Dict ),
At_end = dict:fetch( Key  Last_number, Dict ),
dict:store( Key, At_begining + At_end, Dict ).
 Output:
17> hofstadter_conway:task(). Maximum between 524288 and 1048576 is 0.5337792299633678 for n=722589 Maximum between 262144 and 524288 is 0.5346454310781124 for n=353683 Maximum between 131072 and 262144 is 0.5360200678115611 for n=181385 Maximum between 65536 and 131072 is 0.5370436569998659 for n=89516 Maximum between 32768 and 65536 is 0.5387840205842557 for n=45083 Maximum between 16384 and 32768 is 0.5400710975115871 for n=22223 Maximum between 8192 and 16384 is 0.5424427087803622 for n=11651 Maximum between 4096 and 8192 is 0.5441447478639638 for n=5969 Maximum between 2048 and 4096 is 0.5474628926475664 for n=2897 Maximum between 1024 and 2048 is 0.5501008742434432 for n=1487 Maximum between 512 and 1024 is 0.5549374130737135 for n=719 Maximum between 256 and 512 is 0.5594594594594594 for n=370 Maximum between 128 and 256 is 0.5674157303370787 for n=178 Maximum between 64 and 128 is 0.5760869565217391 for n=92 Maximum between 32 and 64 is 0.5909090909090909 for n=44 Maximum between 16 and 32 is 0.6086956521739131 for n=23 Maximum between 8 and 16 is 0.6363636363636364 for n=11 Maximum between 4 and 8 is 0.6666666666666666 for n=6 Maximum between 2 and 4 is 0.6666666666666666 for n=3
Euler Math Toolbox
>function hofstadter (n) ...
$v=ones(1,n);
$ loop 2 to n1
$ k=v{#};
$ v{#+1}=v{k}+v{#k+1};
$ end
$ return v
$endfunction
>v=hofstadter(2^20);
>k=1:256; plot2d(v[k]/k):
>function hsmaxima (v,k) ...
$ w=zeros(1,k);
$ for j=1 to k
$ i=2^(j1):2^j;
$ w[j]=max(v[i]/i);
$ end;
$ return w;
$endfunction
>w=hsmaxima(v,20)
[ 1 0.666666666667 0.666666666667 0.636363636364 0.608695652174
0.590909090909 0.576086956522 0.567415730337 0.559459459459
0.554937413074 0.550100874243 0.547462892648 0.544144747864
0.54244270878 0.540071097512 0.538784020584 0.537043657
0.536020067812 0.534645431078 0.533779229963 ]
>v1=flipx(cummax(flipx(v/(1:cols(v)))));
>max(nonzeros(v1>0.55))
1489
F#
let a = ResizeArray[0; 1; 1]
while a.Count <= (1 <<< 20) do
a.[a.[a.Count  1]] + a.[a.Count  a.[a.Count  1]] > a.Add
for p = 1 to 19 do
Seq.max [for i in 1 <<< p .. 1 <<< p+1 > float a.[i] / float i]
> printf "Maximum in %6d..%7d is %g\n" (1 <<< p) (1 <<< p+1)
let mallows, _ = a
> List.ofSeq
> List.mapi (fun i n > i, n)
> List.rev
> List.find (fun (i, n) > float(n) / float(i) > 0.55)
printfn "Mallows number is %d" mallows
Outputs:
Maximum in 2.. 4 is 0.666667 Maximum in 4.. 8 is 0.666667 Maximum in 8.. 16 is 0.636364 Maximum in 16.. 32 is 0.608696 Maximum in 32.. 64 is 0.590909 Maximum in 64.. 128 is 0.576087 Maximum in 128.. 256 is 0.567416 Maximum in 256.. 512 is 0.559459 Maximum in 512.. 1024 is 0.554937 Maximum in 1024.. 2048 is 0.550101 Maximum in 2048.. 4096 is 0.547463 Maximum in 4096.. 8192 is 0.544145 Maximum in 8192.. 16384 is 0.542443 Maximum in 16384.. 32768 is 0.540071 Maximum in 32768.. 65536 is 0.538784 Maximum in 65536.. 131072 is 0.537044 Maximum in 131072.. 262144 is 0.53602 Maximum in 262144.. 524288 is 0.534645 Maximum in 524288..1048576 is 0.533779 Mallows number is 1489
Factor
USING: combinators formatting io kernel locals math math.ranges
prettyprint sequences splitting ;
MEMO:: a ( n  a(n) ) ! memoize the recurrence relation
n {
{ 1 [ 1 ] }
{ 2 [ 1 ] }
[ 1  a a n n 1  a  a + ]
} case ;
20 2^ [1,b] [ [ a ] [ 1 + / ] bi* ] mapindex
[
{ 1/2 } split harvest restslice
[ supremum ] map 1 19 [a,b]
[ dup 1 + [ 2^ ] bi@ "%f max in (%d, %d)\n" printf ]
2each
]
[ "Mallow's number: " write [ 0.55 >= ] findlast drop 1 + . ]
bi
 Output:
0.666667 max in (2, 4) 0.666667 max in (4, 8) 0.636364 max in (8, 16) 0.608696 max in (16, 32) 0.590909 max in (32, 64) 0.576087 max in (64, 128) 0.567416 max in (128, 256) 0.559459 max in (256, 512) 0.554937 max in (512, 1024) 0.550101 max in (1024, 2048) 0.547463 max in (2048, 4096) 0.544145 max in (4096, 8192) 0.542443 max in (8192, 16384) 0.540071 max in (16384, 32768) 0.538784 max in (32768, 65536) 0.537044 max in (65536, 131072) 0.536020 max in (131072, 262144) 0.534645 max in (262144, 524288) 0.533779 max in (524288, 1048576) Mallow's number: 1489
Fortran
program conway
implicit none
integer :: a(2**20) ! The sequence a(n)
real :: b(2**20) ! The sequence a(n)/n
real :: v ! Max value in the range [2*i, 2**(i+1)]
integer :: nl(1) ! The location of v in the array b(n)
integer :: i, N, first, second, last, m
! Populate a(n) and b(n)
a(1:2) = [1, 1]
b(1:2) = [1.0e0, 0.5e0]
N = 2
do i=1,2**20
last = a(N)
first = a(last)
second = a(Nlast+1)
N = N+1
a(N:N) = first + second
b(N:N) = a(N:N)/real(N)
end do
! Calculate the max values in the logarithmic ranges
m = 0
do i=1,19
v = maxval(b(2**i:2**(i+1)))
nl = maxloc(b(2**i:2**(i+1)))
write(*,'(2(a,i0),a,f8.6,a,i0)') &
'Max. between 2**', i, &
' and 2**', (i+1), &
' is ', v, &
' at n = ', 2**i+nl(1)1
if (m == 0 .and. v < 0.55e0) then
m = i1
end if
end do
! Calculate Mallows number
do i=2**(m+1), 2**m,1
if (b(i) > 0.55e0) then
exit
end if
end do
write(*,'(a,i0)') 'Mallows number = ',i
end program conway
Output:
Max. between 2**1 and 2**2 is 0.666667 at n = 3 Max. between 2**2 and 2**3 is 0.666667 at n = 6 Max. between 2**3 and 2**4 is 0.636364 at n = 11 Max. between 2**4 and 2**5 is 0.608696 at n = 23 Max. between 2**5 and 2**6 is 0.590909 at n = 44 Max. between 2**6 and 2**7 is 0.576087 at n = 92 Max. between 2**7 and 2**8 is 0.567416 at n = 178 Max. between 2**8 and 2**9 is 0.559459 at n = 370 Max. between 2**9 and 2**10 is 0.554937 at n = 719 Max. between 2**10 and 2**11 is 0.550101 at n = 1487 Max. between 2**11 and 2**12 is 0.547463 at n = 2897 Max. between 2**12 and 2**13 is 0.544145 at n = 5969 Max. between 2**13 and 2**14 is 0.542443 at n = 11651 Max. between 2**14 and 2**15 is 0.540071 at n = 22223 Max. between 2**15 and 2**16 is 0.538784 at n = 45083 Max. between 2**16 and 2**17 is 0.537044 at n = 89516 Max. between 2**17 and 2**18 is 0.536020 at n = 181385 Max. between 2**18 and 2**19 is 0.534645 at n = 353683 Max. between 2**19 and 2**20 is 0.533779 at n = 722589 Mallows number = 1489
Fōrmulæ
Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text. Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for storage and transfer purposes more than visualization and edition.
Programs in Fōrmulæ are created/edited online in its website.
In this page you can see and run the program(s) related to this task and their results. You can also change either the programs or the parameters they are called with, for experimentation, but remember that these programs were created with the main purpose of showing a clear solution of the task, and they generally lack any kind of validation.
Solution
This is the function according to the definition. It is very inefficient:
If a sequence is desired, it is much better to store the already calculated terms:
Test case 1. Show the maxima of between successive powers of two up to 2^{20}
Test case 2. compute the value of n that would have won the prize and confirm it is true for n up to 2^{20}
FutureBasic
window 1
// Set width of tab
def tab 9
dim as long Mallows, n, pow2, p2, pPos, uprLim
dim as double p
print
// Adjust array elements depending on size of sequence
_maxArrayElements = 1200000
input "Enter upper limit between 1 and 20 (Enter 20 gives 2^20): "; uprLim
dim as double r
dim as long a( _maxArrayElements )
if uprLim < 1 or uprLim > 20 then uprLim = 20
a(1) = 1
a(2) = 1
pow2 = 2
p2 = 2 ^ pow2
p = 0.5
pPos = 0
print
for n = 3 to 2 ^ uprLim
a(n) = a( a( n1 ) ) + a( na( n1 ) )
r = a(n) / n
if r >= 0.55 then Mallows = n
if r > p
p = r
pPos = n
end if
if n == p2
print "Maximum of a(n)/n between", " 2^"; pow21; " and 2^"; pow2," is "; p;, " at n = "; pPos
pow2 = pow2 + 1
p2 = 2 ^ pow2
p = 0.5
end if
next
print
print "Dr. Mallow's winning number is:"; Mallows
HandleEvents
Output:
Enter upper limit between 1 and 20 (Enter 20 gives 2^20): 20 Maximum of a(n)/n between 2^ 1 and 2^ 2 is 0.6666666667 at n = 3 Maximum of a(n)/n between 2^ 2 and 2^ 3 is 0.6666666667 at n = 6 Maximum of a(n)/n between 2^ 3 and 2^ 4 is 0.6363636364 at n = 11 Maximum of a(n)/n between 2^ 4 and 2^ 5 is 0.6086956522 at n = 23 Maximum of a(n)/n between 2^ 5 and 2^ 6 is 0.5909090909 at n = 44 Maximum of a(n)/n between 2^ 6 and 2^ 7 is 0.5760869565 at n = 92 Maximum of a(n)/n between 2^ 7 and 2^ 8 is 0.5674157303 at n = 178 Maximum of a(n)/n between 2^ 8 and 2^ 9 is 0.5594594595 at n = 370 Maximum of a(n)/n between 2^ 9 and 2^ 10 is 0.5549374131 at n = 719 Maximum of a(n)/n between 2^ 10 and 2^ 11 is 0.5501008742 at n = 1487 Maximum of a(n)/n between 2^ 11 and 2^ 12 is 0.5474628926 at n = 2897 Maximum of a(n)/n between 2^ 12 and 2^ 13 is 0.5441447479 at n = 5969 Maximum of a(n)/n between 2^ 13 and 2^ 14 is 0.5424427088 at n = 11651 Maximum of a(n)/n between 2^ 14 and 2^ 15 is 0.5400710975 at n = 22223 Maximum of a(n)/n between 2^ 15 and 2^ 16 is 0.5387840206 at n = 45083 Maximum of a(n)/n between 2^ 16 and 2^ 17 is 0.537043657 at n = 89516 Maximum of a(n)/n between 2^ 17 and 2^ 18 is 0.5360200678 at n = 181385 Maximum of a(n)/n between 2^ 18 and 2^ 19 is 0.5346454311 at n = 353683 Maximum of a(n)/n between 2^ 19 and 2^ 20 is 0.53377923 at n = 722589 Dr. Mallow's winning number is: 1489
Go
package main
import (
"fmt"
)
func main() {
a := []int{0, 1, 1} // ignore 0 element. work 1 based.
x := 1 // last number in list
n := 2 // index of last number in list = len(a)1
mallow := 0
for p := 1; p < 20; p++ {
max := 0.
for nextPot := n*2; n < nextPot; {
n = len(a) // advance n
x = a[x]+a[nx]
a = append(a, x)
f := float64(x)/float64(n)
if f > max {
max = f
}
if f >= .55 {
mallow = n
}
}
fmt.Printf("max between 2^%d and 2^%d was %f\n", p, p+1, max)
}
fmt.Println("winning number", mallow)
}
Output:
max between 2^1 and 2^2 was 0.666667 max between 2^2 and 2^3 was 0.666667 max between 2^3 and 2^4 was 0.636364 max between 2^4 and 2^5 was 0.608696 max between 2^5 and 2^6 was 0.590909 max between 2^6 and 2^7 was 0.576087 max between 2^7 and 2^8 was 0.567416 max between 2^8 and 2^9 was 0.559459 max between 2^9 and 2^10 was 0.554937 max between 2^10 and 2^11 was 0.550101 max between 2^11 and 2^12 was 0.547463 max between 2^12 and 2^13 was 0.544145 max between 2^13 and 2^14 was 0.542443 max between 2^14 and 2^15 was 0.540071 max between 2^15 and 2^16 was 0.538784 max between 2^16 and 2^17 was 0.537044 max between 2^17 and 2^18 was 0.536020 max between 2^18 and 2^19 was 0.534645 max between 2^19 and 2^20 was 0.533779 winning number 1489
Haskell
import Data.List
import Data.Ord
import Data.Array
import Text.Printf
hc :: Int > Array Int Int
hc n = arr
where arr = listArray (1, n) $ 1 : 1 : map (f (arr!)) [3 .. n]
f a i = a (a $ i  1) + a (i  a (i  1))
printMaxima :: (Int, (Int, Double)) > IO ()
printMaxima (n, (pos, m)) =
printf "Max between 2^%2d and 2^%2d is %1.5f at n = %6d\n"
n (n + 1) m pos
main = do
mapM_ printMaxima maxima
printf "Mallows's number is %d\n" mallows
where
hca = hc $ 2^20
hc' n = fromIntegral (hca!n) / fromIntegral n
maxima = zip [0..] $ map max powers
max seq = maximumBy (comparing snd) $ zip seq (map hc' seq)
powers = map (\n > [2^n .. 2^(n + 1)  1]) [0 .. 19]
mallows = last.takeWhile ((< 0.55) . hc') $ [2^20, 2^20  1 .. 1]
Icon and Unicon
Output:
>hc Max between 2^0 and 2^1 is 1.0 at n = 1 Max between 2^1 and 2^2 is 0.6666666667 at n = 3 Max between 2^2 and 2^3 is 0.6666666667 at n = 6 Max between 2^3 and 2^4 is 0.6363636364 at n = 11 Max between 2^4 and 2^5 is 0.6086956522 at n = 23 Max between 2^5 and 2^6 is 0.5909090909 at n = 44 Max between 2^6 and 2^7 is 0.5760869565 at n = 92 Max between 2^7 and 2^8 is 0.5674157303 at n = 178 Max between 2^8 and 2^9 is 0.5594594595 at n = 370 Max between 2^9 and 2^10 is 0.5549374131 at n = 719 Max between 2^10 and 2^11 is 0.5501008742 at n = 1487 Max between 2^11 and 2^12 is 0.5474628926 at n = 2897 Max between 2^12 and 2^13 is 0.5441447479 at n = 5969 Max between 2^13 and 2^14 is 0.5424427088 at n = 11651 Max between 2^14 and 2^15 is 0.5400710975 at n = 22223 Max between 2^15 and 2^16 is 0.5387840206 at n = 45083 Max between 2^16 and 2^17 is 0.537043657 at n = 89516 Max between 2^17 and 2^18 is 0.5360200678 at n = 181385 Max between 2^18 and 2^19 is 0.5346454311 at n = 353683 Max between 2^19 and 2^20 is 0.53377923 at n = 722589 Mallow's number is 1489 >
J
Solution (tacit):
hc10k =: , ] +/@:{~ (,&<: .)@{: NB. Actual sequence a(n)
AnN =: % 1+i.@:# NB. a(n)/n
MxAnN =: >./;.1~ 2 (=<.)@:^. 1+i.@# NB. Maxima of a(n)/n between successive powers of 2
Alternative solution (exponential growth):
The first, naive, formulation of hc10k
grows by a single term every iteration; in this one, the series grows exponentially in the iterations.
hc10kE =: 1 1 , expand@tail
expand =: 2+I.@;
tail =: copies&.>^:(<@>:`(<@,@2:))
copies =: >: .@(#!.1 .)~ 1 j. #;.1 #^:_1 ::1:~ ]~:{.
Example:
] A=:1 1 hc10k @]^:[~ 2^20x
1 1 2 2 3 4 4 4 5 6 7 7 8 8 8 8 9 ...
AnN A
1 0.5 0.666667 0.5 0.6 0.666667 ...
MxAnN@AnN A
1 0.666667 0.666667 0.636364 ...
MxAnN@AnN@hc10kE 20
1 0.666667 0.666667 0.636364 ...
Java
Remove translation and provide native Java implementation.
// Title: HofstadterConway $10,000 sequence
public class HofstadterConwaySequence {
private static int MAX = (int) Math.pow(2, 20) + 1;
private static int[] HCS = new int[MAX];
static {
HCS[1] = 1;
HCS[2] = 1;
for ( int n = 3 ; n < MAX ; n++ ) {
int nm1 = HCS[n  1];
HCS[n] = HCS[nm1] + HCS[n  nm1];
}
}
public static void main(String[] args) {
int mNum = 0;
for ( int m = 1 ; m < 20 ; m++ ) {
int min = (int) Math.pow(2, m);
int max = min * 2;
double maxRatio = 0.0;
int nVal = 0;
for ( int n = min ; n <= max ; n ++ ) {
double ratio = (double) HCS[n] / n;
if ( ratio > maxRatio ) {
maxRatio = Math.max(ratio, maxRatio);
nVal = n;
}
if ( ratio >= 0.55 ) {
mNum = n;
}
}
System.out.printf("Max ratio between 2^%d and 2^%d is %f at n = %,d%n", m, m+1, maxRatio, nVal);
}
System.out.printf("Mallow's number is %d.%n", mNum);
}
}
 Output:
Max ratio between 2^1 and 2^2 is 0.666667 at n = 3 Max ratio between 2^2 and 2^3 is 0.666667 at n = 6 Max ratio between 2^3 and 2^4 is 0.636364 at n = 11 Max ratio between 2^4 and 2^5 is 0.608696 at n = 23 Max ratio between 2^5 and 2^6 is 0.590909 at n = 44 Max ratio between 2^6 and 2^7 is 0.576087 at n = 92 Max ratio between 2^7 and 2^8 is 0.567416 at n = 178 Max ratio between 2^8 and 2^9 is 0.559459 at n = 370 Max ratio between 2^9 and 2^10 is 0.554937 at n = 719 Max ratio between 2^10 and 2^11 is 0.550101 at n = 1,487 Max ratio between 2^11 and 2^12 is 0.547463 at n = 2,897 Max ratio between 2^12 and 2^13 is 0.544145 at n = 5,969 Max ratio between 2^13 and 2^14 is 0.542443 at n = 11,651 Max ratio between 2^14 and 2^15 is 0.540071 at n = 22,223 Max ratio between 2^15 and 2^16 is 0.538784 at n = 45,083 Max ratio between 2^16 and 2^17 is 0.537044 at n = 89,516 Max ratio between 2^17 and 2^18 is 0.536020 at n = 181,385 Max ratio between 2^18 and 2^19 is 0.534645 at n = 353,683 Max ratio between 2^19 and 2^20 is 0.533779 at n = 722,589 Mallow's number is 1489.
JavaScript
var hofst_10k = function(n) {
var memo = [1, 1];
var a = function(n) {
var result = memo[n1];
if (typeof result !== 'number') {
result = a(a(n1))+a(na(n1));
memo[n1] = result;
}
return result;
}
return a;
}();
var maxima_between_twos = function(exp) {
var current_max = 0;
for(var i = Math.pow(2,exp)+1; i < Math.pow(2,exp+1); i += 1) {
current_max = Math.max(current_max, hofst_10k(i)/i);
}
return current_max;
}
for(var i = 1; i <= 20; i += 1) {
console.log("Maxima between 2^"+i+"2^"+(i+1)+" is: "+maxima_between_twos(i)+"\n");
}
Output:
Maxima between 2^12^2 is: 0.6666666666666666 Maxima between 2^22^3 is: 0.6666666666666666 Maxima between 2^32^4 is: 0.6363636363636364 Maxima between 2^42^5 is: 0.6086956521739131 Maxima between 2^52^6 is: 0.5909090909090909 ... Maxima between 2^182^19 is: 0.5346454310781124 Maxima between 2^192^20 is: 0.5337792299633678 Maxima between 2^202^21 is: 0.5326770563524978
jq
def hcSequence($limit):
reduce range(3; $limit) as $n ([0,1,1];
.[$n1] as $p
 .[$n] = .[$p] + .[$n$p]);
def task($limit):
hcSequence($limit) as $a
 " Range Maximum",
" ",
(foreach range(2; $limit) as $n ( {max: $a[1], pow2: 1, p:1 };
($a[$n] / $n) as $r
 .emit = null
 if $r > .max then .max=$r else . end
 if ($n == .pow2 * 2)
then .emit = "2 ^ \(.p1) to 2 ^ \(.p) \(.max)"
 .pow2 *= 2
 .p += 1
 .max = $r
else . end;
select(.emit).emit)),
( (first( range( $limit1; 0; 1) as $n
 if ($a[$n]/$n >= 0.55) then $n else empty end) ) // 0
 "\nMallows' number = \(.)") ;
task( pow(2;20) + 1 )
 Output:
Range Maximum   2 ^ 0 to 2 ^ 1 1 2 ^ 1 to 2 ^ 2 0.6666666666666666 2 ^ 2 to 2 ^ 3 0.6666666666666666 2 ^ 3 to 2 ^ 4 0.6363636363636364 2 ^ 4 to 2 ^ 5 0.6086956521739131 2 ^ 5 to 2 ^ 6 0.5909090909090909 2 ^ 6 to 2 ^ 7 0.5760869565217391 2 ^ 7 to 2 ^ 8 0.5674157303370787 2 ^ 8 to 2 ^ 9 0.5594594594594594 2 ^ 9 to 2 ^ 10 0.5549374130737135 2 ^ 10 to 2 ^ 11 0.5501008742434432 2 ^ 11 to 2 ^ 12 0.5474628926475664 2 ^ 12 to 2 ^ 13 0.5441447478639638 2 ^ 13 to 2 ^ 14 0.5424427087803622 2 ^ 14 to 2 ^ 15 0.5400710975115871 2 ^ 15 to 2 ^ 16 0.5387840205842557 2 ^ 16 to 2 ^ 17 0.5370436569998659 2 ^ 17 to 2 ^ 18 0.5360200678115611 2 ^ 18 to 2 ^ 19 0.5346454310781124 2 ^ 19 to 2 ^ 20 0.5337792299633678 Mallows' number = 1489
Julia
# v0.6
# Task 1
function hofstadterconway(n::Integer)::Vector{Int}
rst = fill(1, n)
for i in 3:n
rst[i] = rst[rst[i  1]] + rst[i  rst[i  1]]
end
return rst
end
function hcfraction(n::Integer)::Vector{Float64}
rst = Array{Float64}(n)
for (i, a) in enumerate(hofstadterconway(n))
rst[i] = abs(a) / i
end
return rst
end
# Task 2
seq = hcfraction(2 ^ 20)
for i in 1:20
a, b = 2 ^ (i  1), 2 ^ i
@printf("max value of a(n)/n in %6i < n < %7i = %5.3f\n", a, b, maximum(seq[a:b]))
end
# Task 3
function lastindex(val::Float64)
r, p = 1, 0
# Find the range of 2 power in which the maximum is < val
seq = hcfraction(2 ^ 15)
while maximum(seq[2^p:2^(p+1)]) > val; p += 1 end
r = 2 ^ (p + 1)
while seq[r] < val; r = 1 end
return r
end
println("You too might have won \$1000 with the mallows number of ", lastindex(0.55))
 Output:
max value of a(n)/n in 1 < n < 2 = 1.000 max value of a(n)/n in 2 < n < 4 = 0.667 max value of a(n)/n in 4 < n < 8 = 0.667 max value of a(n)/n in 8 < n < 16 = 0.636 max value of a(n)/n in 16 < n < 32 = 0.609 max value of a(n)/n in 32 < n < 64 = 0.591 max value of a(n)/n in 64 < n < 128 = 0.576 max value of a(n)/n in 128 < n < 256 = 0.567 max value of a(n)/n in 256 < n < 512 = 0.559 max value of a(n)/n in 512 < n < 1024 = 0.555 max value of a(n)/n in 1024 < n < 2048 = 0.550 max value of a(n)/n in 2048 < n < 4096 = 0.547 max value of a(n)/n in 4096 < n < 8192 = 0.544 max value of a(n)/n in 8192 < n < 16384 = 0.542 max value of a(n)/n in 16384 < n < 32768 = 0.540 max value of a(n)/n in 32768 < n < 65536 = 0.539 max value of a(n)/n in 65536 < n < 131072 = 0.537 max value of a(n)/n in 131072 < n < 262144 = 0.536 max value of a(n)/n in 262144 < n < 524288 = 0.535 max value of a(n)/n in 524288 < n < 1048576 = 0.534 You too might have won $1000 with the mallows number of 1489
Kotlin
// version 1.1.2
fun main(args: Array<String>) {
val limit = (1 shl 20) + 1
val a = IntArray(limit)
a[1] = 1
a[2] = 1
for (n in 3 until limit) {
val p = a[n  1]
a[n] = a[p] + a[n  p]
}
println(" Range Maximum")
println(" ")
var pow2 = 1
var p = 1
var max = a[1].toDouble()
for (n in 2 until limit) {
val r = a[n].toDouble() / n
if (r > max) max = r
if (n == pow2 * 2) {
println("2 ^ ${"%2d".format(p  1)} to 2 ^ ${"%2d".format(p)} ${"%f".format(max)}")
pow2 *= 2
p++
max = r
}
}
var prize = 0
for (n in limit  1 downTo 1) {
if (a[n].toDouble() / n >= 0.55) {
prize = n
break
}
}
println("\nMallows' number = $prize")
}
 Output:
Range Maximum   2 ^ 0 to 2 ^ 1 1.000000 2 ^ 1 to 2 ^ 2 0.666667 2 ^ 2 to 2 ^ 3 0.666667 2 ^ 3 to 2 ^ 4 0.636364 2 ^ 4 to 2 ^ 5 0.608696 2 ^ 5 to 2 ^ 6 0.590909 2 ^ 6 to 2 ^ 7 0.576087 2 ^ 7 to 2 ^ 8 0.567416 2 ^ 8 to 2 ^ 9 0.559459 2 ^ 9 to 2 ^ 10 0.554937 2 ^ 10 to 2 ^ 11 0.550101 2 ^ 11 to 2 ^ 12 0.547463 2 ^ 12 to 2 ^ 13 0.544145 2 ^ 13 to 2 ^ 14 0.542443 2 ^ 14 to 2 ^ 15 0.540071 2 ^ 15 to 2 ^ 16 0.538784 2 ^ 16 to 2 ^ 17 0.537044 2 ^ 17 to 2 ^ 18 0.536020 2 ^ 18 to 2 ^ 19 0.534645 2 ^ 19 to 2 ^ 20 0.533779 Mallows' number = 1489
Lua
I solved it, using a coroutine to generate the Hofstadter numbers, because it's fun to do so. This can be done differently, but not with so much fun, I guess. It's counting from the first number 1, the second 1, the third 2 and so on. This doesn't change anything about the outcome, but I guess it's better like this and consistent with such things like fibonacci numbers.
local fmt, write=string.format,io.write
local hof=coroutine.wrap(function()
local yield=coroutine.yield
local a={1,1}
yield(a[1], 1)
yield(a[2], 2)
local n=a[#a]
repeat
n=a[n]+a[1+#an]
a[#a+1]=n
yield(n, #a)
until false
end)
local mallows, mdiv=0,0
for p=1,20 do
local max, div, num, last, fdiv=0,0,0,0,0
for i=2^(p1),2^p1 do
h,n=hof()
div=h/n
if div>max then
max=div
num=n
end
if div>0.55 then
last=n
fdiv=div
end
end
write(fmt("From 2^%2d to 2^%2d the max is %.4f the %6dth Hofstadter number.\n",
p1, p, max, num))
if max>.55 and p>4 then
mallows, mdiv=last, fdiv
end
end
write("So Mallows number is ", mallows, " with ", fmt("%.4f",mdiv), ", yay, just wire me my $10000 now!\n")
 Output:
From 2^0 to 2^1 the max is 1.0000 the 1th Hofstadter number. From 2^1 to 2^2 the max is 0.6667 the 3th Hofstadter number. From 2^2 to 2^3 the max is 0.6667 the 6th Hofstadter number. From 2^3 to 2^4 the max is 0.6364 the 11th Hofstadter number. From 2^4 to 2^5 the max is 0.6087 the 23th Hofstadter number. From 2^5 to 2^6 the max is 0.5909 the 44th Hofstadter number. From 2^6 to 2^7 the max is 0.5761 the 92th Hofstadter number. From 2^7 to 2^8 the max is 0.5674 the 178th Hofstadter number. From 2^8 to 2^9 the max is 0.5595 the 370th Hofstadter number. From 2^9 to 2^10 the max is 0.5549 the 719th Hofstadter number. From 2^10 to 2^11 the max is 0.5501 the 1487th Hofstadter number. From 2^11 to 2^12 the max is 0.5475 the 2897th Hofstadter number. From 2^12 to 2^13 the max is 0.5441 the 5969th Hofstadter number. From 2^13 to 2^14 the max is 0.5424 the 11651th Hofstadter number. From 2^14 to 2^15 the max is 0.5401 the 22223th Hofstadter number. From 2^15 to 2^16 the max is 0.5388 the 45083th Hofstadter number. From 2^16 to 2^17 the max is 0.5370 the 89516th Hofstadter number. From 2^17 to 2^18 the max is 0.5360 the 181385th Hofstadter number. From 2^18 to 2^19 the max is 0.5346 the 353683th Hofstadter number. From 2^19 to 2^20 the max is 0.5338 the 722589th Hofstadter number. So Mallows number is 1489 with 0.5500, yay, just wire me my $10000 now!
Mathematica / Wolfram Language
a[1] := 1; a[2] := 1;
a[n_] := a[n] = a[a[n1]] + a[na[n1]]
Map[Print["Max value: ",Max[Table[a[n]/n//N,{n,2^#,2^(#+1)}]]," for n between 2^",#," and 2^",(#+1)]& , Range[19]]
n=2^20; While[(a[n]/n//N)<0.55,n]; Print["Mallows number: ",n]
Outputs:
Max value: 0.666667 for n between 2^1 and 2^2 Max value: 0.666667 for n between 2^2 and 2^3 Max value: 0.636364 for n between 2^3 and 2^4 Max value: 0.608696 for n between 2^4 and 2^5 Max value: 0.590909 for n between 2^5 and 2^6 Max value: 0.576087 for n between 2^6 and 2^7 Max value: 0.567416 for n between 2^7 and 2^8 Max value: 0.559459 for n between 2^8 and 2^9 Max value: 0.554937 for n between 2^9 and 2^10 Max value: 0.550101 for n between 2^10 and 2^11 Max value: 0.547463 for n between 2^11 and 2^12 Max value: 0.544145 for n between 2^12 and 2^13 Max value: 0.542443 for n between 2^13 and 2^14 Max value: 0.540071 for n between 2^14 and 2^15 Max value: 0.538784 for n between 2^15 and 2^16 Max value: 0.537044 for n between 2^16 and 2^17 Max value: 0.53602 for n between 2^17 and 2^18 Max value: 0.534645 for n between 2^18 and 2^19 Max value: 0.533779 for n between 2^19 and 2^20 Mallows number: 1489
MATLAB / Octave
function Q = HCsequence(N)
Q = zeros(1,N);
Q(1:2) = 1;
for n = 3:N,
Q(n) = Q(Q(n1))+Q(nQ(n1));
end;
end;
The function can be tested in this way:
NN = 20;
Q = HCsequence(2^NN+1);
V = Q./(1:2^NN);
for k=1:NN,
[m,i] = max(V(2^k:2^(k+1)));
i = i + 2^k  1;
printf('Maximum between 2^%i and 2^%i is %f at n=%i\n',k,k+1,m,i);
end;
Output:
Maximum between 2^1 and 2^2 is 0.666667 at n=3 Maximum between 2^2 and 2^3 is 0.666667 at n=6 Maximum between 2^3 and 2^4 is 0.636364 at n=11 Maximum between 2^4 and 2^5 is 0.608696 at n=23 Maximum between 2^5 and 2^6 is 0.590909 at n=44 Maximum between 2^6 and 2^7 is 0.576087 at n=92 Maximum between 2^7 and 2^8 is 0.567416 at n=178 Maximum between 2^8 and 2^9 is 0.559459 at n=370 Maximum between 2^9 and 2^10 is 0.554937 at n=719 Maximum between 2^10 and 2^11 is 0.550101 at n=1487 Maximum between 2^11 and 2^12 is 0.547463 at n=2897 Maximum between 2^12 and 2^13 is 0.544145 at n=5969 Maximum between 2^13 and 2^14 is 0.542443 at n=11651 Maximum between 2^14 and 2^15 is 0.540071 at n=22223 Maximum between 2^15 and 2^16 is 0.538784 at n=45083 Maximum between 2^16 and 2^17 is 0.537044 at n=89516 Maximum between 2^17 and 2^18 is 0.536020 at n=181385 Maximum between 2^18 and 2^19 is 0.534645 at n=353683 Maximum between 2^19 and 2^20 is 0.533779 at n=722589
Nim
import strutils
const last = 1 shl 20
var aList: array[last + 1, int]
aList[0..2] = [50_000, 1, 1]
var
v = aList[2]
k1 = 2
lg2 = 1
aMax = 0.0
for n in 3..last:
v = aList[v] + aList[nv]
aList[n] = v
aMax = max(aMax, v.float / n.float)
if (k1 and n) == 0:
echo "Maximum between 2^$# and 2^$# was $#".format(lg2, lg2+1, aMax)
aMax = 0
inc lg2
k1 = n
Output:
Maximum between 2^1 and 2^2 was 0.6666666666666666 Maximum between 2^2 and 2^3 was 0.6666666666666666 Maximum between 2^3 and 2^4 was 0.6363636363636364 Maximum between 2^4 and 2^5 was 0.6086956521739131 ... Maximum between 2^18 and 2^19 was 0.5346454310781124 Maximum between 2^19 and 2^20 was 0.5337792299633678
Objeck
bundle Default {
class HofCon {
function : Main(args : String[]) ~ Nil {
DoSqnc(1<<20);
}
function : native : DoSqnc(m : Int) ~ Nil {
a_list := Int>New[m + 1];
max_df := 0;
p2_max := 2;
k1 := 2;
lg2 := 1;
amax := 0.0;
a_list[0] := 1;
a_list[1] := 1;
v := a_list[2];
for(n := 2; n <= m; n+=1;) {
r := a_list[v] + a_list[n  v];
v := r;
a_list[n] := r;
if(amax < v * 1.0 / n) {
amax := v * 1.0 / n;
};
if(0 = (k1 and n)) {
IO.Console>Print("Maximum between 2^")>Print(lg2)
>Print(" and 2^")>Print(lg2 + 1)>Print(" was ")>PrintLine(amax);
amax := 0;
lg2+=1;
};
k1 := n;
};
}
}
}
Maximum between 2^1 and 2^2 was 0.666666667 Maximum between 2^2 and 2^3 was 0.666666667 Maximum between 2^3 and 2^4 was 0.636363636 Maximum between 2^4 and 2^5 was 0.608695652 Maximum between 2^5 and 2^6 was 0.590909091 Maximum between 2^6 and 2^7 was 0.576086957 Maximum between 2^7 and 2^8 was 0.56741573 Maximum between 2^8 and 2^9 was 0.559459459 Maximum between 2^9 and 2^10 was 0.554937413 Maximum between 2^10 and 2^11 was 0.550100874 Maximum between 2^11 and 2^12 was 0.547462893 Maximum between 2^12 and 2^13 was 0.544144748 Maximum between 2^13 and 2^14 was 0.542442709 Maximum between 2^14 and 2^15 was 0.540071098 Maximum between 2^15 and 2^16 was 0.538784021 Maximum between 2^16 and 2^17 was 0.537043657 Maximum between 2^17 and 2^18 was 0.536020068 Maximum between 2^18 and 2^19 was 0.534645431 Maximum between 2^19 and 2^20 was 0.53377923
Oforth
: hofstadter(n)
 l i 
ListBuffer newSize(n) dup add(1) dup add(1) >l
n 2  loop: i [ l at(l last) l at(l size l last  1+ ) + l add ]
l dup freeze ;
: hofTask
 h m i 
2 20 pow >m
hofstadter(m) m seq zipWith(#[ tuck asFloat / swap Pair new ]) >h
19 loop: i [
i . "^2 ==>" .
h extract(2 i pow , 2 i 1+ pow) reduce(#maxKey) println
]
"Mallows number ==>" . h reverse detect(#[ first 0.55 >= ], true) println
;
 Output:
1^2 ==> [0.666666666666667, 3] 2^2 ==> [0.666666666666667, 6] 3^2 ==> [0.636363636363636, 11] 4^2 ==> [0.608695652173913, 23] 5^2 ==> [0.590909090909091, 44] 6^2 ==> [0.576086956521739, 92] 7^2 ==> [0.567415730337079, 178] 8^2 ==> [0.559459459459459, 370] 9^2 ==> [0.554937413073713, 719] 10^2 ==> [0.550100874243443, 1487] 11^2 ==> [0.547462892647566, 2897] 12^2 ==> [0.544144747863964, 5969] 13^2 ==> [0.542442708780362, 11651] 14^2 ==> [0.540071097511587, 22223] 15^2 ==> [0.538784020584256, 45083] 16^2 ==> [0.537043656999866, 89516] 17^2 ==> [0.536020067811561, 181385] 18^2 ==> [0.534645431078112, 353683] 19^2 ==> [0.533779229963368, 722589] Mallows number ==> [0.550033579583613, 1489]
Oz
A direct implementation of the recursive definition with explicit memoization using a mutable map (dictionary):
declare
local
Cache = {Dictionary.new}
Cache.1 := 1
Cache.2 := 1
in
fun {A N}
if {Not {Dictionary.member Cache N}} then
Cache.N := {A {A N1}} + {A N{A N1}}
end
Cache.N
end
end
Float = Int.toFloat
for I in 0..19 do
Range = {List.number {Pow 2 I} {Pow 2 I+1} 1}
RelativeValues = {Map Range
fun {$ N}
{Float {A N}}
/ {Float N}
end}
Maximum = {FoldL RelativeValues Max 0.0}
in
{System.showInfo "Max. between 2^"#I#" and 2^"#I+1#": "#Maximum}
end
Output:
Max. between 2^0 and 2^1: 1.0 Max. between 2^1 and 2^2: 0.66667 Max. between 2^2 and 2^3: 0.66667 Max. between 2^3 and 2^4: 0.63636 Max. between 2^4 and 2^5: 0.6087 Max. between 2^5 and 2^6: 0.59091 Max. between 2^6 and 2^7: 0.57609 Max. between 2^7 and 2^8: 0.56742 Max. between 2^8 and 2^9: 0.55946 Max. between 2^9 and 2^10: 0.55494 Max. between 2^10 and 2^11: 0.5501 Max. between 2^11 and 2^12: 0.54746 Max. between 2^12 and 2^13: 0.54414 Max. between 2^13 and 2^14: 0.54244 Max. between 2^14 and 2^15: 0.54007 Max. between 2^15 and 2^16: 0.53878 Max. between 2^16 and 2^17: 0.53704 Max. between 2^17 and 2^18: 0.53602 Max. between 2^18 and 2^19: 0.53465 Max. between 2^19 and 2^20: 0.53378
PARI/GP
HC(n)=my(a=vectorsmall(n));a[1]=a[2]=1;for(i=3,n,a[i]=a[a[i1]]+a[ia[i1]]);a;
maxima(n)=my(a=HC(1<<n),m);vector(n1,k,m=0;for(i=1<<k+1,1<<(k+1)1,m=max(m,a[i]/i));m);
forstep(i=#a,1,1,if(a[i]/i>=.55,return(i)))
Output:
%1 = [2/3, 2/3, 7/11, 14/23, 13/22, 53/92, 101/178, 207/370, 399/719, 818/1487, 1586/2897, 3248/5969, 6320/11651, 12002/22223, 24290/45083, 24037/44758, 97226/181385, 189095/353683, 385703/722589] %2 = 1489
Pascal
tested with freepascal 3.1.1 64 Bit.
program HofStadterConway;
const
Pot2 = 20;// tested with 30 > 4 GB;
type
tfeld = array[0..1 shl Pot2] of LongWord;
tpFeld = ^tFeld;
tMaxPos = record
mpMax : double;
mpValue,
mpPos : longWord;
end;
tArrMaxPos = array[0..Pot21] of tMaxPos;
var
a : tpFeld;
MaxPos : tArrMaxPos;
procedure Init(a:tpFeld);
var
n,k: LongWord;
begin
a^[1]:= 1;
a^[2]:= 1;
//a[n] := a[a[n1]]+a[na[n1]];
//k := a[n1]
k := a^[2];
For n := 3 to High(a^) do
Begin
k := a^[k]+a^[nk];
a^[n] := k;
end;
end;
function GetMax(a:tpFeld;starts,ends:LongWord):tMaxPos;
var
posMax : LongWord;
r,
max : double;
Begin
posMax:= starts;
max := 0.0;
repeat
r := a^[starts]/ starts;
IF max < r then
Begin
max := r;
posMax := starts;
end;
inc(starts);
until starts >= ends;
with GetMax do
Begin
mpPos:= posMax;
mpValue := a^[posMax];
mpMax:= max;
end;
end;
procedure SearchMax(a:tpFeld);
var
ends,idx : LongWord;
begin
idx := 0;
ends := 2;
while ends <= High(a^) do
Begin
MaxPos[idx]:=GetMax(a,ends shr 1,ends);
ends := 2*ends;
inc(idx);
end;
end;
procedure OutputMax;
var
i : integer;
begin
For i := Low(MaxPos) to High(MaxPOs) do
with MaxPos[i] do
Begin
Write('Max between 2^',i:2,' and 2^',i+1:2);
writeln(mpMax:14:11,' at ',mpPos:9,' value :',mpValue:10);
end;
writeln;
end;
function SearchLastPos(a:tpFeld;limit: double):LongInt;
var
i,l : LongInt;
Begin
Limit := limit;
IF (Limit>1.0 ) OR (Limit < 0.5) then
Begin
SearchLastPos := 1;
EXIT;
end;
i := 0;
while (i<=High(MaxPos)) AND (MaxPos[i].mpMax > Limit) do
inc(i);
dec(i);
l := MaxPos[i].mpPos;
i := 1 shl (i+1);
while (l< i) AND (a^[i]/i < limit) do
dec(i);
SearchLastPos := i;
end;
var
p : Pointer;
l : double;
Begin
//using getmem because FPCs new is limited to 2^311 Byte for the test 2^30 )
getmem(p,SizeOf(tfeld));
a := p;
Init(a);
SearchMax(a);
outputMax;
l:= 0.55;
writeln('Mallows number with limit ',l:10:8,' at ',SearchLastPos(a,l));
freemem(p);
end.
 output
Max between 2^ 0 and 2^ 1 1.00000000000 at 1 value : 1 Max between 2^ 1 and 2^ 2 0.66666666667 at 3 value : 2 Max between 2^ 2 and 2^ 3 0.66666666667 at 6 value : 4 Max between 2^ 3 and 2^ 4 0.63636363636 at 11 value : 7 Max between 2^ 4 and 2^ 5 0.60869565217 at 23 value : 14 ...... Max between 2^16 and 2^17 0.53704365700 at 89516 value : 48074 Max between 2^17 and 2^18 0.53602006781 at 181385 value : 97226 Max between 2^18 and 2^19 0.53464543108 at 353683 value : 189095 Max between 2^19 and 2^20 0.53377922996 at 722589 value : 385703 Mallows number with limit 0.55000000 at 1489
Perl
#!/usr/bin/perl
use warnings ;
use strict ;
my $limit = 2 ** 20 ;
my @numbers = ( 0 , 1 , 1 ) ;
my $mallows ;
my $max_i ;
foreach my $i ( 3..$limit ) {
push ( @numbers , $numbers[ $numbers[ $i  1 ]] + $numbers[ $i  $numbers[ $i  1 ] ] ) ;
}
for ( my $rangelimit = 1 ; $rangelimit < 20 ; $rangelimit++ ) {
my $max = 0 ;
for ( my $i = 2 ** $rangelimit ; $i < ( 2 ** ( $rangelimit + 1 ) ) ; $i++ ) {
my $rat = $numbers[ $i ] / $i ;
$mallows = $i if $rat >= 0.55 ;
if ( $rat > $max ) {
$max = $rat ;
$max_i = $i ;
}
}
my $upperlimit = $rangelimit + 1 ;
print "Between 2 ^ $rangelimit and 2 ^ $upperlimit the maximum value is $max at $max_i !\n" ;
}
print "The prize would have been won at $mallows !\n"
Output:
Between 2 ^ 1 and 2 ^ 2 the maximum value is 0.666666666666667 at 3 ! Between 2 ^ 2 and 2 ^ 3 the maximum value is 0.666666666666667 at 6 ! Between 2 ^ 3 and 2 ^ 4 the maximum value is 0.636363636363636 at 11 ! Between 2 ^ 4 and 2 ^ 5 the maximum value is 0.608695652173913 at 23 ! Between 2 ^ 5 and 2 ^ 6 the maximum value is 0.590909090909091 at 44 ! Between 2 ^ 6 and 2 ^ 7 the maximum value is 0.576086956521739 at 92 ! Between 2 ^ 7 and 2 ^ 8 the maximum value is 0.567415730337079 at 178 ! Between 2 ^ 8 and 2 ^ 9 the maximum value is 0.559459459459459 at 370 ! Between 2 ^ 9 and 2 ^ 10 the maximum value is 0.554937413073713 at 719 ! Between 2 ^ 10 and 2 ^ 11 the maximum value is 0.550100874243443 at 1487 ! Between 2 ^ 11 and 2 ^ 12 the maximum value is 0.547462892647566 at 2897 ! Between 2 ^ 12 and 2 ^ 13 the maximum value is 0.544144747863964 at 5969 ! Between 2 ^ 13 and 2 ^ 14 the maximum value is 0.542442708780362 at 11651 ! Between 2 ^ 14 and 2 ^ 15 the maximum value is 0.540071097511587 at 22223 ! Between 2 ^ 15 and 2 ^ 16 the maximum value is 0.538784020584256 at 45083 ! Between 2 ^ 16 and 2 ^ 17 the maximum value is 0.537043656999866 at 89516 ! Between 2 ^ 17 and 2 ^ 18 the maximum value is 0.536020067811561 at 181385 ! Between 2 ^ 18 and 2 ^ 19 the maximum value is 0.534645431078112 at 353683 ! Between 2 ^ 19 and 2 ^ 20 the maximum value is 0.533779229963368 at 722589 ! The prize would have been won at 1489 !
Phix
with javascript_semantics sequence a = {1,1} function q(integer n) for l=length(a)+1 to n do a &= a[a[l1]]+a[la[l1]] end for return a[n] end function integer mallows = 1, max_n for p=0 to 19 do atom max_an = 0.5 integer l = power(2,p), h=l*2 for n=l to h do atom an = q(n)/n if an>=max_an then max_an = an max_n = n end if if an>0.55 then mallows = n end if end for printf(1,"Maximum in range %6d to %7d occurs at %6d: %f\n",{l,h,max_n,max_an}) end for printf(1,"Mallows number is %d\n",{mallows})
In this particular case the for loop of q() only ever iterates 0 or 1 times.
 Output:
Maximum in range 1 to 2 occurs at 1: 1.000000 Maximum in range 2 to 4 occurs at 3: 0.666667 Maximum in range 4 to 8 occurs at 6: 0.666667 Maximum in range 8 to 16 occurs at 11: 0.636364 Maximum in range 16 to 32 occurs at 23: 0.608696 Maximum in range 32 to 64 occurs at 44: 0.590909 Maximum in range 64 to 128 occurs at 92: 0.576087 Maximum in range 128 to 256 occurs at 178: 0.567416 Maximum in range 256 to 512 occurs at 370: 0.559459 Maximum in range 512 to 1024 occurs at 719: 0.554937 Maximum in range 1024 to 2048 occurs at 1487: 0.550101 Maximum in range 2048 to 4096 occurs at 2897: 0.547463 Maximum in range 4096 to 8192 occurs at 5969: 0.544145 Maximum in range 8192 to 16384 occurs at 11651: 0.542443 Maximum in range 16384 to 32768 occurs at 22223: 0.540071 Maximum in range 32768 to 65536 occurs at 45083: 0.538784 Maximum in range 65536 to 131072 occurs at 89516: 0.537044 Maximum in range 131072 to 262144 occurs at 181385: 0.536020 Maximum in range 262144 to 524288 occurs at 353683: 0.534645 Maximum in range 524288 to 1048576 occurs at 722589: 0.533779 Mallows number is 1489
Picat
go =>
foreach(N in 0..19)
[Val,Ix] = argmax({a(I)/I : I in 2**N..2**(N+1)}),
printf("Max from 2^%2d..2^%2d is %0.8f at %d\n",N,N+1,Val,Ix+2**N1)
end,
println(mallows_number=mallows_number()),
nl.
% The sequence definition
table
a(1) = 1.
a(2) = 1.
a(N) = a(a(N1))+a(Na(N1)).
% argmax: find the (first) index for the max value(s) of L.
argmax(L) = [Max,MaxIxFirst] =>
Max = max(L),
MaxIxFirst = {I : I in 1..L.length, L[I] == Max}.first.
% Calculate the Mallows number separately.
mallows_number() = Mallow =>
Mallow = _,
foreach(M in 1..19)
Min = 2**M,
Max = Min*2,
MaxRatio = 0,
NVal = 0,
foreach(N in Min..Max)
Ratio = a(N)/N,
if Ratio > MaxRatio then
MaxRatio := Ratio,
NVal := N
end,
if Ratio > 0.55 then
Mallow := N
end
end
end.
 Output:
Max from 2^ 0..2^1 is 1.00000000 at 1 Max from 2^ 1..2^2 is 0.66666667 at 3 Max from 2^ 2..2^3 is 0.66666667 at 6 Max from 2^ 3..2^4 is 0.63636364 at 11 Max from 2^ 4..2^5 is 0.60869565 at 23 Max from 2^ 5..2^6 is 0.59090909 at 44 Max from 2^ 6..2^7 is 0.57608696 at 92 Max from 2^ 7..2^8 is 0.56741573 at 178 Max from 2^ 8..2^9 is 0.55945946 at 370 Max from 2^ 9..2^10 is 0.55493741 at 719 Max from 2^10..2^11 is 0.55010087 at 1487 Max from 2^11..2^12 is 0.54746289 at 2897 Max from 2^12..2^13 is 0.54414475 at 5969 Max from 2^13..2^14 is 0.54244271 at 11651 Max from 2^14..2^15 is 0.54007110 at 22223 Max from 2^15..2^16 is 0.53878402 at 45083 Max from 2^16..2^17 is 0.53704366 at 89516 Max from 2^17..2^18 is 0.53602007 at 181385 Max from 2^18..2^19 is 0.53464543 at 353683 Max from 2^19..2^20 is 0.53377923 at 722589 mallows_number = 1489
PicoLisp
(de hofcon (N)
(cache '(NIL) N
(if (>= 2 N)
1
(+
(hofcon (hofcon (dec N)))
(hofcon ( N (hofcon (dec N)))) ) ) ) )
(scl 20)
(de sequence (M)
(let (Lim 4 Max 0 4k$ 0)
(for (N 3 (>= M N) (inc N))
(let V (*/ (hofcon N) 1.0 N)
(setq Max (max Max V))
(when (>= V 0.55)
(setq 4k$ N) )
(when (= N Lim)
(prinl
"Maximum between " (/ Lim 2)
" and " Lim
" was " (format Max `*Scl) )
(inc 'Lim Lim)
(zero Max) ) ) )
(prinl
"Win with " 4k$
" (the task requests 'n > p' now)" ) ) )
(sequence (** 2 20))
Output:
Maximum between 2 and 4 was 0.66666666666666666667 Maximum between 4 and 8 was 0.66666666666666666667 Maximum between 8 and 16 was 0.63636363636363636364 Maximum between 16 and 32 was 0.60869565217391304348 Maximum between 32 and 64 was 0.59090909090909090909 Maximum between 64 and 128 was 0.57608695652173913043 Maximum between 128 and 256 was 0.56741573033707865169 Maximum between 256 and 512 was 0.55945945945945945946 Maximum between 512 and 1024 was 0.55493741307371349096 Maximum between 1024 and 2048 was 0.55010087424344317418 Maximum between 2048 and 4096 was 0.54746289264756644805 Maximum between 4096 and 8192 was 0.54414474786396381303 Maximum between 8192 and 16384 was 0.54244270878036220067 Maximum between 16384 and 32768 was 0.54007109751158709445 Maximum between 32768 and 65536 was 0.53878402058425570614 Maximum between 65536 and 131072 was 0.53704365699986594575 Maximum between 131072 and 262144 was 0.53602006781156104419 Maximum between 262144 and 524288 was 0.53464543107811232092 Maximum between 524288 and 1048576 was 0.53377922996336783427 Win with 1489 (the task requests 'n > p' now)
PL/I
/* First part: */
declare L (10000) fixed static initial ((1000) 0);
L(1), L(2) = 1;
do i = 3 to 10000;
k = L(i);
L(i) = L(ik) + L(1+k);
end;
Python
from __future__ import division
def maxandmallows(nmaxpower2):
nmax = 2**nmaxpower2
mx = (0.5, 2)
mxpow2 = []
mallows = None
# HofstadterConway sequence starts at hc[1],
# hc[0] is not part of the series.
hc = [None, 1, 1]
for n in range(2, nmax + 1):
ratio = hc[n] / n
if ratio > mx[0]:
mx = (ratio, n)
if ratio >= 0.55:
mallows = n
if ratio == 0.5:
print("In the region %7i < n <= %7i: max a(n)/n = %6.4f at n = %i" %
(n//2, n, mx[0], mx[1]))
mxpow2.append(mx[0])
mx = (ratio, n)
hc.append(hc[hc[n]] + hc[hc[n]])
return hc, mallows if mxpow2 and mxpow2[1] < 0.55 and n > 4 else None
if __name__ == '__main__':
hc, mallows = maxandmallows(20)
if mallows:
print("\nYou too might have won $1000 with the mallows number of %i" % mallows)
Sample output
In the region 1 < n <= 2: max a(n)/n = 0.5000 at n = 2 In the region 2 < n <= 4: max a(n)/n = 0.6667 at n = 3 In the region 4 < n <= 8: max a(n)/n = 0.6667 at n = 6 In the region 8 < n <= 16: max a(n)/n = 0.6364 at n = 11 In the region 16 < n <= 32: max a(n)/n = 0.6087 at n = 23 In the region 32 < n <= 64: max a(n)/n = 0.5909 at n = 44 In the region 64 < n <= 128: max a(n)/n = 0.5761 at n = 92 In the region 128 < n <= 256: max a(n)/n = 0.5674 at n = 178 In the region 256 < n <= 512: max a(n)/n = 0.5595 at n = 370 In the region 512 < n <= 1024: max a(n)/n = 0.5549 at n = 719 In the region 1024 < n <= 2048: max a(n)/n = 0.5501 at n = 1487 In the region 2048 < n <= 4096: max a(n)/n = 0.5475 at n = 2897 In the region 4096 < n <= 8192: max a(n)/n = 0.5441 at n = 5969 In the region 8192 < n <= 16384: max a(n)/n = 0.5424 at n = 11651 In the region 16384 < n <= 32768: max a(n)/n = 0.5401 at n = 22223 In the region 32768 < n <= 65536: max a(n)/n = 0.5388 at n = 45083 In the region 65536 < n <= 131072: max a(n)/n = 0.5370 at n = 89516 In the region 131072 < n <= 262144: max a(n)/n = 0.5360 at n = 181385 In the region 262144 < n <= 524288: max a(n)/n = 0.5346 at n = 353683 In the region 524288 < n <= 1048576: max a(n)/n = 0.5338 at n = 722589 You too might have won $1000 with the mallows number of 1489
If you don't create enough terms in the sequence, no mallows number is returned.
Quackery
[ $ "bigrat.qky" loadfile ] now!
[ ' [ 1 1 ]
swap 2  times
[ dup 1 peek 2dup 1  peek
dip [ dip dup negate peek ]
+ join ] ] is hofcon ( n > [ )
[ behead do rot
witheach
[ do 2over 2over v v0<
if 2swap
2drop ] ] is maximum ( [ > n n )
20 dup bit 1  hofcon
[] swap witheach
[ i^ 1+ join nested join ]
swap 1  bit times
[ i^ 1+ dup 2dup
say "Maximum in range " echo
say " to " 1 << 1  echo
step split swap maximum
say " at n = "
dup echo
say " is " 10 point$ echo$ cr ]
drop
 Output:
Maximum in range 1 to 1 at n = 1 is 1 Maximum in range 2 to 3 at n = 3 is 0.6666666667 Maximum in range 4 to 7 at n = 6 is 0.6666666667 Maximum in range 8 to 15 at n = 11 is 0.6363636364 Maximum in range 16 to 31 at n = 23 is 0.6086956522 Maximum in range 32 to 64 at n = 44 is 0.5909090909 Maximum in range 64 to 127 at n = 92 is 0.5760869565 Maximum in range 128 to 255 at n = 178 is 0.5674157303 Maximum in range 256 to 511 at n = 370 is 0.5594594595 Maximum in range 512 to 1023 at n = 719 is 0.5549374131 Maximum in range 1024 to 2047 at n = 1487 is 0.5501008742 Maximum in range 2048 to 4095 at n = 2897 is 0.5474628926 Maximum in range 4096 to 8191 at n = 5969 is 0.5441447479 Maximum in range 8192 to 16383 at n = 11651 is 0.5424427088 Maximum in range 16384 to 32767 at n = 22223 is 0.5400710975 Maximum in range 32768 to 65535 at n = 45083 is 0.5387840206 Maximum in range 65536 to 131071 at n = 89516 is 0.537043657 Maximum in range 131072 to 262143 at n = 181385 is 0.5360200678 Maximum in range 262144 to 524287 at n = 353683 is 0.5346454311 Maximum in range 524288 to 1048575 at n = 722589 is 0.53377923
R
A memoizing function to compute individual elements of the sequence could be written like this:
f = local(
{a = c(1, 1)
function(n)
{if (is.na(a[n]))
a[n] << f(f(n  1)) + f(n  f(n  1))
a[n]}})
But a more straightforward way to get the local maxima and the Mallows point is to begin by generating as much of the sequence as we need.
hofcon = c(1, 1, rep(NA, 2^20  2))
for (n in 3 : (2^20))
{hofcon[n] =
hofcon[hofcon[n  1]] +
hofcon[n  hofcon[n  1]]}
We can now quickly finish the task with vectorized operations.
ratios = hofcon / seq_along(hofcon)
message("Maxima:")
print(sapply(1 : 20, function(pwr)
max(ratios[2^(pwr  1) : 2^pwr])))
message("Prizewinning point:")
print(max(which(ratios >= .55)))
 Output:
Maxima: [1] 1.0000000 0.6666667 0.6666667 0.6363636 0.6086957 [6] 0.5909091 0.5760870 0.5674157 0.5594595 0.5549374 [11] 0.5501009 0.5474629 0.5441447 0.5424427 0.5400711 [16] 0.5387840 0.5370437 0.5360201 0.5346454 0.5337792 Prizewinning point: [1] 1489
Racket
The macro define/memoize1 creates an 1argument procedure and handles all the details about memorization. We use it to define (conway n) as a transcription of the definition.
#lang racket/base
(definesyntaxrule (define/memoize1 (proc x) body ...)
(define proc
(let ([cache (makehash)]
[direct (lambda (x) body ...)])
(lambda (x)
(hashref! cache x (lambda () (direct x)))))))
(define/memoize1 (conway n)
(if (< n 3)
1
(+ (conway (conway (sub1 n)))
(conway ( n (conway (sub1 n)))))))
The macro for/max1 is like for, but the result is the maximum of the values produced by the body. The result also includes the position of the maximum in the sequence. We use this to find the maximum in each powerof2sector.
(definesyntaxrule (for/max1 ([i sequence]) body ...)
(for/fold ([max inf.0] [argmax #f]) ([i sequence])
(define val (begin body ...))
(if (< max val)
(values val i)
(values max argmax))))
(for ([i (inrange 0 20)])
(define lowb (expt 2 i))
(define upb (expt 2 (add1 i)))
(definevalues (max argmax) (for/max1 ([k (inrange lowb upb)])
(/ (conway k) k)))
(printf "Max. between 2^~a and 2^~a is ~a at ~a ~n" i (add1 i) (real>decimalstring max 5) argmax))
The macro for/prev is like for/and, it stops when it finds the first #f, but the result is previous value produced by the body. We use this to find the first powerof2sector that has no ratio avobe .55. The previous result is the Mallows number.
(definesyntaxrule (for/prev (sequences ...) body ...)
(for/fold ([prev #f]) (sequences ...)
(define val (let () body ...))
#:break (not val)
val))
(define mallows (for/prev ([i (innaturals)])
(define lowb (expt 2 i))
(define upb (expt 2 (add1 i)))
(for/last ([k (inrange lowb upb)]
#:when (>= (/ (conway k) k) .55))
k)))
(printf "Mallows number: ~a~n" mallows)
Sample Output:
Max. between 2^0 and 2^1 is 1.00000 at 1 Max. between 2^1 and 2^2 is 0.66667 at 3 Max. between 2^2 and 2^3 is 0.66667 at 6 Max. between 2^3 and 2^4 is 0.63636 at 11 Max. between 2^4 and 2^5 is 0.60870 at 23 Max. between 2^5 and 2^6 is 0.59091 at 44 Max. between 2^6 and 2^7 is 0.57609 at 92 Max. between 2^7 and 2^8 is 0.56742 at 178 Max. between 2^8 and 2^9 is 0.55946 at 370 Max. between 2^9 and 2^10 is 0.55494 at 719 Max. between 2^10 and 2^11 is 0.55010 at 1487 Max. between 2^11 and 2^12 is 0.54746 at 2897 Max. between 2^12 and 2^13 is 0.54414 at 5969 Max. between 2^13 and 2^14 is 0.54244 at 11651 Max. between 2^14 and 2^15 is 0.54007 at 22223 Max. between 2^15 and 2^16 is 0.53878 at 45083 Max. between 2^16 and 2^17 is 0.53704 at 89516 Max. between 2^17 and 2^18 is 0.53602 at 181385 Max. between 2^18 and 2^19 is 0.53465 at 353683 Max. between 2^19 and 2^20 is 0.53378 at 722589 Mallows number: 1489
Raku
(formerly Perl 6)
Note that @a is a lazy array, and the Z variants are "zipwith" operators.
my $n = 3;
my @a = (0,1,1, > $p { @a[$p] + @a[$n++  $p] } ... *);
@a[2**20]; # precalculate sequence
my $last55;
for 1..19 > $power {
my @range := 2**$power .. 2**($power+1)1;
my @ratios = (@a[@range] Z/ @range) Z=> @range;
my $max = @ratios.max;
($last55 = .value if .key >= .55 for @ratios) if $max.key >= .55;
say $power.fmt('%2d'), @range.min.fmt("%10d"), '..', @range.max.fmt("%10d"),
$max.key, ' at ', $max.value;
}
say "Mallows' number would appear to be ", $last55;
 Output:
1 2..3 0.666666666666667 at 3 2 4..7 0.666666666666667 at 6 3 8..15 0.636363636363636 at 11 4 16..31 0.608695652173913 at 23 5 32..63 0.590909090909091 at 44 6 64..127 0.576086956521739 at 92 7 128..255 0.567415730337079 at 178 8 256..511 0.559459459459459 at 370 9 512..1023 0.554937413073713 at 719 10 1024..2047 0.550100874243443 at 1487 11 2048..4095 0.547462892647566 at 2897 12 4096..8191 0.544144747863964 at 5969 13 8192..16383 0.542442708780362 at 11651 14 16384..32767 0.540071097511587 at 22223 15 32768..65535 0.538784020584256 at 45083 16 65536..131071 0.537043656999866 at 89516 17 131072..262143 0.536020067811561 at 181385 18 262144..524287 0.534645431078112 at 353683 19 524288..1048575 0.533779229963368 at 722589 Mallows' number would appear to be 1489
REXX
/*REXX program solves the Hofstadter─Conway sequence $10,000 prize (puzzle). */
@pref= 'Maximum of a(n) ÷ n between ' /*a prologue for the text of message. */
H.=.; H.1=1; H.2=1; !.=0; @.=0 /*initialize some REXX variables. */
win=0
do k=0 to 20; p.k=2**k; maxp=p.k /*build an array of the powers of two. */
end /*k*/
r=1 /*R: is the range of the power of two.*/
do n=1 for maxp; if n> p.r then r=r+1 /*for golf coders, same as: r=r+(n>p.r)*/
_=H(n)/n; if _>=.55 then win=n /*get next seq number; if ≥.55, a win? */
if _<=@.r then iterate /*less than previous? Then keep looking*/
@.r=_; !.r=n /*@.r and !.r are like ginkgo biloba.*/
end /*n*/ /* ··· or in other words, memoization.*/
do j=1 for 20; range= '2**'right(j1, 2) "───► 2**"right( j, 2)
say @pref range '(inclusive) is ' left(@.j, 9) " at n="right(!.j, 7)
end /*j*/
say
say 'The winning number is: ' win /*and the money shot is ··· */
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
H: procedure expose H.; parse arg z
if H.z==. then do; m=z1; $=H.m; _=z$; H.z=H.$+H._; end
return H.z
output
Maximum of a(n) ÷ n between 2** 0 ───► 2** 1 (inclusive) is 1 at n= 1 Maximum of a(n) ÷ n between 2** 1 ───► 2** 2 (inclusive) is 0.6666666 at n= 3 Maximum of a(n) ÷ n between 2** 2 ───► 2** 3 (inclusive) is 0.6666666 at n= 6 Maximum of a(n) ÷ n between 2** 3 ───► 2** 4 (inclusive) is 0.6363636 at n= 11 Maximum of a(n) ÷ n between 2** 4 ───► 2** 5 (inclusive) is 0.6086956 at n= 23 Maximum of a(n) ÷ n between 2** 5 ───► 2** 6 (inclusive) is 0.5909090 at n= 44 Maximum of a(n) ÷ n between 2** 6 ───► 2** 7 (inclusive) is 0.5760869 at n= 92 Maximum of a(n) ÷ n between 2** 7 ───► 2** 8 (inclusive) is 0.5674157 at n= 178 Maximum of a(n) ÷ n between 2** 8 ───► 2** 9 (inclusive) is 0.5594594 at n= 370 Maximum of a(n) ÷ n between 2** 9 ───► 2**10 (inclusive) is 0.5549374 at n= 719 Maximum of a(n) ÷ n between 2**10 ───► 2**11 (inclusive) is 0.5501008 at n= 1487 Maximum of a(n) ÷ n between 2**11 ───► 2**12 (inclusive) is 0.5474628 at n= 2897 Maximum of a(n) ÷ n between 2**12 ───► 2**13 (inclusive) is 0.5441447 at n= 5969 Maximum of a(n) ÷ n between 2**13 ───► 2**14 (inclusive) is 0.5424427 at n= 11651 Maximum of a(n) ÷ n between 2**14 ───► 2**15 (inclusive) is 0.5400710 at n= 22223 Maximum of a(n) ÷ n between 2**15 ───► 2**16 (inclusive) is 0.5387840 at n= 45083 Maximum of a(n) ÷ n between 2**16 ───► 2**17 (inclusive) is 0.5370436 at n= 89516 Maximum of a(n) ÷ n between 2**17 ───► 2**18 (inclusive) is 0.5360200 at n= 181385 Maximum of a(n) ÷ n between 2**18 ───► 2**19 (inclusive) is 0.5346454 at n= 353683 Maximum of a(n) ÷ n between 2**19 ───► 2**20 (inclusive) is 0.5337792 at n= 722589 The winning number is: 1489
Ring
decimals(9)
size = 15
a = list(pow(2,size))
a[1]=1
a[2]=1
power=2
p2=pow(2,power)
peak=0.5
peakpos=0
for n=3 to pow(2,size)
a[n]=a[a[n1]]+a[na[n1]]
r=a[n]/n
if r>=0.55 mallows=n ok
if r>peak peak=r peakpos=n ok
if n=p2
see "maximum between 2^" + (power  1) + " and 2^" + power + " is " + peak + " at n=" + peakpos + nl
power += 1
p2=pow(2,power)
peak=0.5 ok
next
see "mallows number is : " + mallows + nl
Output:
maximum between 2^1 and 2^2 is 0.666666667 at n=3 maximum between 2^2 and 2^3 is 0.666666667 at n=6 maximum between 2^3 and 2^4 is 0.636363636 at n=11 maximum between 2^4 and 2^5 is 0.608695652 at n=23 maximum between 2^5 and 2^6 is 0.590909091 at n=44 maximum between 2^6 and 2^7 is 0.576086957 at n=92 maximum between 2^7 and 2^8 is 0.567415730 at n=178 maximum between 2^8 and 2^9 is 0.559459459 at n=370 maximum between 2^9 and 2^10 is 0.554937413 at n=719 maximum between 2^10 and 2^11 is 0.550100874 at n=1487 maximum between 2^11 and 2^12 is 0.547462893 at n=2897 maximum between 2^12 and 2^13 is 0.544144748 at n=5969 maximum between 2^13 and 2^14 is 0.542442709 at n=11651 maximum between 2^14 and 2^15 is 0.540071098 at n=22223 mallows number is : 1489
Ruby
class HofstadterConway10000
def initialize
@sequence = [nil, 1, 1]
end
def [](n)
raise ArgumentError, "n must be >= 1" if n < 1
a = @sequence
a.length.upto(n) {i a[i] = a[a[i1]] + a[ia[i1]] }
a[n]
end
end
hc = HofstadterConway10000.new
mallows = nil
(1...20).each do i
j = i + 1
max_n, max_v = 1, 1
(2**i .. 2**j).each do n
v = hc[n].to_f / n
max_n, max_v = n, v if v > max_v
# Mallows number
mallows = n if v >= 0.55
end
puts "maximum between 2^%2d and 2^%2d occurs at%7d: %.8f" % [i, j, max_n, max_v]
end
puts "the mallows number is #{mallows}"
 Output:
maximum between 2^ 1 and 2^ 2 occurs at 3: 0.66666667 maximum between 2^ 2 and 2^ 3 occurs at 6: 0.66666667 maximum between 2^ 3 and 2^ 4 occurs at 11: 0.63636364 maximum between 2^ 4 and 2^ 5 occurs at 23: 0.60869565 maximum between 2^ 5 and 2^ 6 occurs at 44: 0.59090909 maximum between 2^ 6 and 2^ 7 occurs at 92: 0.57608696 maximum between 2^ 7 and 2^ 8 occurs at 178: 0.56741573 maximum between 2^ 8 and 2^ 9 occurs at 370: 0.55945946 maximum between 2^ 9 and 2^10 occurs at 719: 0.55493741 maximum between 2^10 and 2^11 occurs at 1487: 0.55010087 maximum between 2^11 and 2^12 occurs at 2897: 0.54746289 maximum between 2^12 and 2^13 occurs at 5969: 0.54414475 maximum between 2^13 and 2^14 occurs at 11651: 0.54244271 maximum between 2^14 and 2^15 occurs at 22223: 0.54007110 maximum between 2^15 and 2^16 occurs at 45083: 0.53878402 maximum between 2^16 and 2^17 occurs at 89516: 0.53704366 maximum between 2^17 and 2^18 occurs at 181385: 0.53602007 maximum between 2^18 and 2^19 occurs at 353683: 0.53464543 maximum between 2^19 and 2^20 occurs at 722589: 0.53377923 the mallows number is 1489
Rust
struct HofstadterConway {
current: usize,
sequence: Vec<usize>,
}
impl HofstadterConway {
/// Define a constructor for the struct.
fn new() > HofstadterConway {
HofstadterConway {
current: 0,
sequence: vec![1, 1],
}
}
}
impl Default for HofstadterConway {
fn default() > Self {
Self::new()
}
}
/// Implement the hofstadter q iteration sequence.
impl Iterator for HofstadterConway {
type Item = usize;
/// This gets called to fetch the next item of the iterator.
fn next(&mut self) > Option<usize> {
let max_index = self.sequence.len()  1;
let last_value = self.sequence[max_index];
if self.current > max_index {
let new_x = self.sequence[last_value  1] + self.sequence[max_index  last_value + 1];
self.sequence.push(new_x);
}
self.current += 1;
Some(self.sequence[self.current  1])
}
}
#[allow(clippy::cast_precision_loss)]
fn main() {
let mut hof = HofstadterConway::new();
let mut winning_num = 0_usize;
for p in 0..20 {
let max_hof = (2_usize.pow(p)..2_usize.pow(p + 1))
.map(n (n, hof.next().unwrap() as f64 / n as f64))
.fold(f64::NAN, a, (n, b) {
if b >= 0.55 {
winning_num = n;
}
a.max(b)
});
println!("2^{:>2}2^{:>2}, {:>.8}", p, p + 1, max_hof);
}
println!("Winning number: {}", winning_num);
}
 Output:
2^ 02^ 1, 1.00000000 2^ 12^ 2, 0.66666667 2^ 22^ 3, 0.66666667 2^ 32^ 4, 0.63636364 2^ 42^ 5, 0.60869565 2^ 52^ 6, 0.59090909 2^ 62^ 7, 0.57608696 2^ 72^ 8, 0.56741573 2^ 82^ 9, 0.55945946 2^ 92^10, 0.55493741 2^102^11, 0.55010087 2^112^12, 0.54746289 2^122^13, 0.54414475 2^132^14, 0.54244271 2^142^15, 0.54007110 2^152^16, 0.53878402 2^162^17, 0.53704366 2^172^18, 0.53602007 2^182^19, 0.53464543 2^192^20, 0.53377923 Winning number: 1489
Scala
object HofstadterConway {
def pow2(n: Int): Int = (Iterator.fill(n)(2)).product
def makeHCSequence(max: Int): Seq[Int] =
(0 to max  1).foldLeft (Vector[Int]()) { (v, idx) =>
if (idx <= 1) v :+ 1 else v :+ (v(v(idx  1)  1) + v(idx  v(idx  1)))
}
val max = pow2(20)
val maxSeq = makeHCSequence(max)
def hcRatio(n: Int, seq: Seq[Int]): Double = seq(n  1).toDouble / n
def maximumHCRatioBetween(a: Int, b: Int): (Int, Double) =
Iterator.range(a, b + 1) map (n => (n, hcRatio(n, maxSeq))) maxBy (_._2)
lazy val mallowsNumber: Int =
((max to 1 by 1) takeWhile (hcRatio(_, maxSeq) < 0.55) last)  1
def main(args: Array[String]): Unit = {
for (n < 1 to 19) {
val (value, ratio) = maximumHCRatioBetween(pow2(n), pow2(n+1))
val message = "Maximum of a(n)/n between 2^%s and 2^%s was %s at %s"
println(message.format(n, n+1, ratio, value))
}
println("Mallow's number = %s".format(mallowsNumber))
}
}
Output
Maximum of a(n)/n between 2^1 and 2^2 was 0.6666666666666666 at 3 Maximum of a(n)/n between 2^2 and 2^3 was 0.6666666666666666 at 6 Maximum of a(n)/n between 2^3 and 2^4 was 0.6363636363636364 at 11 Maximum of a(n)/n between 2^4 and 2^5 was 0.6086956521739131 at 23 Maximum of a(n)/n between 2^5 and 2^6 was 0.5909090909090909 at 44 Maximum of a(n)/n between 2^6 and 2^7 was 0.5760869565217391 at 92 Maximum of a(n)/n between 2^7 and 2^8 was 0.5674157303370787 at 178 Maximum of a(n)/n between 2^8 and 2^9 was 0.5594594594594594 at 370 Maximum of a(n)/n between 2^9 and 2^10 was 0.5549374130737135 at 719 Maximum of a(n)/n between 2^10 and 2^11 was 0.5501008742434432 at 1487 Maximum of a(n)/n between 2^11 and 2^12 was 0.5474628926475664 at 2897 Maximum of a(n)/n between 2^12 and 2^13 was 0.5441447478639638 at 5969 Maximum of a(n)/n between 2^13 and 2^14 was 0.5424427087803622 at 11651 Maximum of a(n)/n between 2^14 and 2^15 was 0.5400710975115871 at 22223 Maximum of a(n)/n between 2^15 and 2^16 was 0.5387840205842557 at 45083 Maximum of a(n)/n between 2^16 and 2^17 was 0.5370436569998659 at 89516 Maximum of a(n)/n between 2^17 and 2^18 was 0.5360200678115611 at 181385 Maximum of a(n)/n between 2^18 and 2^19 was 0.5346454310781124 at 353683 Maximum of a(n)/n between 2^19 and 2^20 was 0.5337792299633678 at 722589 Mallow's number = 1489
Scheme
(import (scheme base)
(scheme write)
(only (srfi 1) iota))
;; maximum size of sequence to consider, as a power of 2
(define *maxpower* 20)
(define *size* (expt 2 *maxpower*))
;; Task 1: Generate members of the sequence
(define *seq* (makevector (+ 1 *size*))) ; add 1, to use 1indexing into sequence
(vectorset! *seq* 1 1)
(vectorset! *seq* 2 1)
(foreach
(lambda (n)
(let ((x (vectorref *seq* ( n 1))))
(vectorset! *seq* n (+ (vectorref *seq* x)
(vectorref *seq* ( n x))))))
(iota ( *size* 2) 3))
;; Task 2: Show maxima of a(n)/n between successive powers of two
(foreach
(lambda (power)
(let ((startidx (+ (expt 2 ( power 1)) 1))
(endidx (expt 2 power)))
(do ((i startidx (+ 1 i))
(maximum 0 (max maximum (/ (vectorref *seq* i)
i))))
((> i endidx)
(display
(stringappend
"Maximum between 2^" (number>string ( power 1))
" and 2^" (number>string power)
" = " (number>string (inexact maximum))
"\n"))))))
(iota ( *maxpower* 1) 2))
;; Task 3: Find value of p where a(n)/n < 0.55 for all n > p (in our sequence)
(do ((idx *size* ( idx 1)))
((or (zero? idx) ; safety net
(> (/ (vectorref *seq* idx) idx)
0.55))
(display (stringappend "\np=" (number>string idx) "\n"))))
 Output:
Maximum between 2^1 and 2^2 = 0.6666666666666666 Maximum between 2^2 and 2^3 = 0.6666666666666666 Maximum between 2^3 and 2^4 = 0.6363636363636364 Maximum between 2^4 and 2^5 = 0.6086956521739131 Maximum between 2^5 and 2^6 = 0.5909090909090909 Maximum between 2^6 and 2^7 = 0.5760869565217391 Maximum between 2^7 and 2^8 = 0.5674157303370787 Maximum between 2^8 and 2^9 = 0.5594594594594594 Maximum between 2^9 and 2^10 = 0.5549374130737135 Maximum between 2^10 and 2^11 = 0.5501008742434432 Maximum between 2^11 and 2^12 = 0.5474628926475664 Maximum between 2^12 and 2^13 = 0.5441447478639638 Maximum between 2^13 and 2^14 = 0.5424427087803622 Maximum between 2^14 and 2^15 = 0.5400710975115871 Maximum between 2^15 and 2^16 = 0.5387840205842557 Maximum between 2^16 and 2^17 = 0.5370436569998659 Maximum between 2^17 and 2^18 = 0.5360200678115611 Maximum between 2^18 and 2^19 = 0.5346454310781124 Maximum between 2^19 and 2^20 = 0.5337792299633678 p=1489
Sidef
class HofstadterConway10000 {
has sequence = [nil, 1, 1]
method term(n {.is_pos}) {
var a = sequence
{i a[i] = a[a[i1]]+a[ia[i1]] } << a.len..n
a[n]
}
}
var hc = HofstadterConway10000()
var mallows = nil
for i in (1..19) {
var j = i+1
var (max_n, max_v) = (1, 1)
for n in (1<<i .. 1<<j) {
var v = (hc.term(n) / n)
(max_n, max_v) = (n, v) if (v > max_v)
mallows = n if (v >= 0.55)
}
say ("maximum between 2^%2d and 2^%2d occurs at%7d: %.8f" % (i, j, max_n, max_v))
}
say "the mallows number is #{mallows}"
 Output:
maximum between 2^ 1 and 2^ 2 occurs at 3: 0.66666667 maximum between 2^ 2 and 2^ 3 occurs at 6: 0.66666667 maximum between 2^ 3 and 2^ 4 occurs at 11: 0.63636364 maximum between 2^ 4 and 2^ 5 occurs at 23: 0.60869565 maximum between 2^ 5 and 2^ 6 occurs at 44: 0.59090909 maximum between 2^ 6 and 2^ 7 occurs at 92: 0.57608696 maximum between 2^ 7 and 2^ 8 occurs at 178: 0.56741573 maximum between 2^ 8 and 2^ 9 occurs at 370: 0.55945946 maximum between 2^ 9 and 2^10 occurs at 719: 0.55493741 maximum between 2^10 and 2^11 occurs at 1487: 0.55010087 maximum between 2^11 and 2^12 occurs at 2897: 0.54746289 maximum between 2^12 and 2^13 occurs at 5969: 0.54414475 maximum between 2^13 and 2^14 occurs at 11651: 0.54244271 maximum between 2^14 and 2^15 occurs at 22223: 0.54007110 maximum between 2^15 and 2^16 occurs at 45083: 0.53878402 maximum between 2^16 and 2^17 occurs at 89516: 0.53704366 maximum between 2^17 and 2^18 occurs at 181385: 0.53602007 maximum between 2^18 and 2^19 occurs at 353683: 0.53464543 maximum between 2^19 and 2^20 occurs at 722589: 0.53377923 the mallows number is 1489
Swift
func doSqnc(m:Int) {
var aList = [Int](count: m + 1, repeatedValue: 0)
var k1 = 2
var lg2 = 1
var amax:Double = 0
aList[0] = 1
aList[1] = 1
var v = aList[2]
for n in 2...m {
let add = aList[v] + aList[n  v]
aList[n] = add
v = aList[n]
if amax < Double(v) * 1.0 / Double(n) {
amax = Double(v) * 1.0 / Double(n)
}
if (k1 & n == 0) {
println("Maximum between 2^\(lg2) and 2^\(lg2 + 1) was \(amax)")
amax = 0
lg2++
}
k1 = n
}
}
doSqnc(1 << 20)
 Output:
Maximum between 2^1 and 2^2 was 0.666666666666667 Maximum between 2^2 and 2^3 was 0.666666666666667 Maximum between 2^3 and 2^4 was 0.636363636363636 Maximum between 2^4 and 2^5 was 0.608695652173913 Maximum between 2^5 and 2^6 was 0.590909090909091 Maximum between 2^6 and 2^7 was 0.576086956521739 Maximum between 2^7 and 2^8 was 0.567415730337079 Maximum between 2^8 and 2^9 was 0.559459459459459 Maximum between 2^9 and 2^10 was 0.554937413073713 Maximum between 2^10 and 2^11 was 0.550100874243443 Maximum between 2^11 and 2^12 was 0.547462892647566 Maximum between 2^12 and 2^13 was 0.544144747863964 Maximum between 2^13 and 2^14 was 0.542442708780362 Maximum between 2^14 and 2^15 was 0.540071097511587 Maximum between 2^15 and 2^16 was 0.538784020584256 Maximum between 2^16 and 2^17 was 0.537043656999866 Maximum between 2^17 and 2^18 was 0.536020067811561 Maximum between 2^18 and 2^19 was 0.534645431078112 Maximum between 2^19 and 2^20 was 0.533779229963368
Tcl
The routine to return the n^{th} member of the sequence.
package require Tcl 8.5
set hofcon10k {1 1}
proc hofcon10k n {
global hofcon10k
if {$n < 1} {error "n must be at least 1"}
if {$n <= [llength $hofcon10k]} {
return [lindex $hofcon10k [expr {$n1}]]
}
while {$n > [llength $hofcon10k]} {
set i [lindex $hofcon10k end]
set a [lindex $hofcon10k [expr {$i1}]]
# Don't use endbased indexing here; faster to compute manually
set b [lindex $hofcon10k [expr {[llength $hofcon10k]$i}]]
lappend hofcon10k [set c [expr {$a + $b}]]
}
return $c
}
The code to explore the sequence, looking for maxima in the ratio.
for {set p 1} {$p<20} {incr p} {
set end [expr {2**($p+1)}]
set maxI 0; set maxV 0
for {set i [expr {2**$p}]} {$i<=$end} {incr i} {
set v [expr {[hofcon10k $i] / double($i)}]
if {$v > $maxV} {set maxV $v; set maxI $i}
}
puts "max in 2**$p..2**[expr {$p+1}] at $maxI : $maxV"
}
Output:
max in 2**1..2**2 at 3 : 0.6666666666666666 max in 2**2..2**3 at 6 : 0.6666666666666666 max in 2**3..2**4 at 11 : 0.6363636363636364 max in 2**4..2**5 at 23 : 0.6086956521739131 max in 2**5..2**6 at 44 : 0.5909090909090909 max in 2**6..2**7 at 92 : 0.5760869565217391 max in 2**7..2**8 at 178 : 0.5674157303370787 max in 2**8..2**9 at 370 : 0.5594594594594594 max in 2**9..2**10 at 719 : 0.5549374130737135 max in 2**10..2**11 at 1487 : 0.5501008742434432 max in 2**11..2**12 at 2897 : 0.5474628926475664 max in 2**12..2**13 at 5969 : 0.5441447478639638 max in 2**13..2**14 at 11651 : 0.5424427087803622 max in 2**14..2**15 at 22223 : 0.5400710975115871 max in 2**15..2**16 at 45083 : 0.5387840205842557 max in 2**16..2**17 at 89516 : 0.5370436569998659 max in 2**17..2**18 at 181385 : 0.5360200678115611 max in 2**18..2**19 at 353683 : 0.5346454310781124 max in 2**19..2**20 at 722589 : 0.5337792299633678
VBA
Function q rewritten to sub.
Public q() As Long
Sub make_q()
ReDim q(2 ^ 20)
q(1) = 1
q(2) = 1
Dim l As Long
For l = 3 To 2 ^ 20
q(l) = q(q(l  1)) + q(l  q(l  1))
Next l
End Sub
Public Sub hcsequence()
Dim mallows As Long: mallows = 1
Dim max_n As Long, n As Long
Dim l As Long, h As Long
make_q
For p = 0 To 19
max_an = 0.5
l = 2 ^ p: h = l * 2
For n = l To h
an = q(n) / n
If an >= max_an Then
max_an = an
max_n = n
End If
If an > 0.55 Then
mallows = n
End If
Next n
Debug.Print "Maximum in range"; Format(l, "@@@@@@@"); " to"; h; String$(7  Len(CStr(h)), " ");
Debug.Print "occurs at"; Format(max_n, "@@@@@@@"); ": "; Format(max_an, "0.000000")
Next p
Debug.Print "Mallows number is"; mallows
End Sub
 Output:
Maximum in range 1 to 2 occurs at 1: 1,000000 Maximum in range 2 to 4 occurs at 3: 0,666667 Maximum in range 4 to 8 occurs at 6: 0,666667 Maximum in range 8 to 16 occurs at 11: 0,636364 Maximum in range 16 to 32 occurs at 23: 0,608696 Maximum in range 32 to 64 occurs at 44: 0,590909 Maximum in range 64 to 128 occurs at 92: 0,576087 Maximum in range 128 to 256 occurs at 178: 0,567416 Maximum in range 256 to 512 occurs at 370: 0,559459 Maximum in range 512 to 1024 occurs at 719: 0,554937 Maximum in range 1024 to 2048 occurs at 1487: 0,550101 Maximum in range 2048 to 4096 occurs at 2897: 0,547463 Maximum in range 4096 to 8192 occurs at 5969: 0,544145 Maximum in range 8192 to 16384 occurs at 11651: 0,542443 Maximum in range 16384 to 32768 occurs at 22223: 0,540071 Maximum in range 32768 to 65536 occurs at 45083: 0,538784 Maximum in range 65536 to 131072 occurs at 89516: 0,537044 Maximum in range 131072 to 262144 occurs at 181385: 0,536020 Maximum in range 262144 to 524288 occurs at 353683: 0,534645 Maximum in range 524288 to 1048576 occurs at 722589: 0,533779 Mallows number is 1489
V (Vlang)
fn main() {
mut a := [0, 1, 1] // ignore 0 element. work 1 based.
mut x := 1 // last number in list
mut n := 2 // index of last number in list = a.len1
mut mallow := 0
for p in 1..20 {
mut max := 0.0
for next_pot := n*2; n < next_pot; {
n = a.len // advance n
x = a[x]+a[nx]
a << x
f := f64(x)/f64(n)
if f > max {
max = f
}
if f >= .55 {
mallow = n
}
}
println("max between 2^$p and 2^${p+1} was ${max:.6}")
}
println("winning number $mallow")
}
 Output:
max between 2^1 and 2^2 was 0.666667 max between 2^2 and 2^3 was 0.666667 max between 2^3 and 2^4 was 0.636364 max between 2^4 and 2^5 was 0.608696 max between 2^5 and 2^6 was 0.590909 max between 2^6 and 2^7 was 0.576087 max between 2^7 and 2^8 was 0.567416 max between 2^8 and 2^9 was 0.559459 max between 2^9 and 2^10 was 0.554937 max between 2^10 and 2^11 was 0.550101 max between 2^11 and 2^12 was 0.547463 max between 2^12 and 2^13 was 0.544145 max between 2^13 and 2^14 was 0.542443 max between 2^14 and 2^15 was 0.540071 max between 2^15 and 2^16 was 0.538784 max between 2^16 and 2^17 was 0.537044 max between 2^17 and 2^18 was 0.536020 max between 2^18 and 2^19 was 0.534645 max between 2^19 and 2^20 was 0.533779 winning number 1489
Wren
import "./fmt" for Fmt
var limit = 1<<20 + 1
var a = List.filled(limit, 0)
a[1] = 1
a[2] = 1
for (n in 3...limit) {
var p = a[n1]
a[n] = a[p] + a[np]
}
System.print(" Range Maximum")
System.print(" ")
var pow2 = 1
var p = 1
var max = a[1]
for (n in 2...limit) {
var r = a[n] / n
if (r > max) max = r
if (n == pow2 * 2) {
Fmt.print("2 ^ $2d to 2 ^ $2d $f", p  1, p, max)
pow2 = pow2 * 2
p = p + 1
max = r
}
}
var prize = 0
for (n in limit1..1) {
if (a[n]/n >= 0.55) {
prize = n
break
}
}
System.print("\nMallows' number = %(prize)")
 Output:
Range Maximum   2 ^ 0 to 2 ^ 1 1.000000 2 ^ 1 to 2 ^ 2 0.666667 2 ^ 2 to 2 ^ 3 0.666667 2 ^ 3 to 2 ^ 4 0.636364 2 ^ 4 to 2 ^ 5 0.608696 2 ^ 5 to 2 ^ 6 0.590909 2 ^ 6 to 2 ^ 7 0.576087 2 ^ 7 to 2 ^ 8 0.567416 2 ^ 8 to 2 ^ 9 0.559459 2 ^ 9 to 2 ^ 10 0.554937 2 ^ 10 to 2 ^ 11 0.550101 2 ^ 11 to 2 ^ 12 0.547463 2 ^ 12 to 2 ^ 13 0.544145 2 ^ 13 to 2 ^ 14 0.542443 2 ^ 14 to 2 ^ 15 0.540071 2 ^ 15 to 2 ^ 16 0.538784 2 ^ 16 to 2 ^ 17 0.537044 2 ^ 17 to 2 ^ 18 0.536020 2 ^ 18 to 2 ^ 19 0.534645 2 ^ 19 to 2 ^ 20 0.533779 Mallows' number = 1489
X86 Assembly
Using FASM syntax.
; HofstadterConway $10,000 sequence
call a.memorization
call Mallows_Number
; ECX is the $1000 #
int3
a.memorization:
; skip [a] to make it one based
mov [a+1*4],1
mov [a+2*4],1
mov ecx,3
@@:
mov eax,ecx
mov edx,[a+(ecx1)*4] ; a[n1]
sub eax,edx ; na[n1]
mov eax,[a+eax*4] ; a[na[n1]]
add eax,[a+edx*4] ;+a[a[n1]]
mov [a+ecx*4],eax
inc ecx
cmp ecx,1 shl 20
jnz @B
retn
_0.55 equ ((55 shl 32)/100) ; Floor[55 * 2^N / 100], for N=32
Mallows_Number: ; $5D1
mov ecx,1 shl 20
@@: dec ecx
mov edx,[a+ecx*4]
xor eax,eax
div ecx
cmp eax,_0.55 + 1
jc @B
retn
a rd 1 shl 20
XPL0
int A(1 + 1<<20), N, Power2, WinningN;
real Max, Member;
[A(1):= 1; A(2):= 1;
N:= 3; Power2:= 2; Max:= 0.;
Text(0, " Range Maximum^m^j");
Format(1, 6);
repeat A(N):= A(A(N1)) + A(NA(N1));
Member:= float(A(N)) / float(N);
if Member >= Max then Max:= Member;
if Member >= 0.55 then WinningN:= N;
if N & 1<<Power2 then
[Text(0, "2^^"); IntOut(0, Power21);
Text(0, " to 2^^"); IntOut(0, Power2);
ChOut(0, 9\tab\);
RlOut(0, Max);
CrLf(0);
Power2:= Power2+1;
Max:= 0.;
];
N:= N+1;
until N > 1<<20;
IntOut(0, WinningN);
Text(0, " is the winning position.^m^j");
]
 Output:
Range Maximum 2^1 to 2^2 0.666667 2^2 to 2^3 0.666667 2^3 to 2^4 0.636364 2^4 to 2^5 0.608696 2^5 to 2^6 0.590909 2^6 to 2^7 0.576087 2^7 to 2^8 0.567416 2^8 to 2^9 0.559459 2^9 to 2^10 0.554937 2^10 to 2^11 0.550101 2^11 to 2^12 0.547463 2^12 to 2^13 0.544145 2^13 to 2^14 0.542443 2^14 to 2^15 0.540071 2^15 to 2^16 0.538784 2^16 to 2^17 0.537044 2^17 to 2^18 0.536020 2^18 to 2^19 0.534645 2^19 to 2^20 0.533779 1489 is the winning position.
zkl
fcn hofstadterConwaySequence(m){
a:=List.createLong(m + 1,0);
a[0]=a[1]=1;
v,p2,lg2, amax, mallow := a[2],4,1, 0.0, Void;
foreach n in ([2 .. m]){
v=a[n]=a[v] + a[n  v];
f:=1.0*v/n;
if(f>=0.55) mallow=n;
amax=amax.max(f);
if(n==p2){
println("Max in [2^%d, 2^%d]: %f".fmt(lg2, lg2+1, amax));
amax,p2 = 0.0, (n+1).nextPowerOf2;
lg2+=1;
}
}
if(mallow) println("Winning number = ",mallow);
}
hofstadterConwaySequence((2).pow(20));
 Output:
Max in [2^1, 2^2]: 0.666667 Max in [2^2, 2^3]: 0.666667 Max in [2^3, 2^4]: 0.636364 Max in [2^4, 2^5]: 0.608696 Max in [2^5, 2^6]: 0.590909 Max in [2^6, 2^7]: 0.576087 Max in [2^7, 2^8]: 0.567416 Max in [2^8, 2^9]: 0.559459 Max in [2^9, 2^10]: 0.554937 Max in [2^10, 2^11]: 0.550101 Max in [2^11, 2^12]: 0.547463 Max in [2^12, 2^13]: 0.544145 Max in [2^13, 2^14]: 0.542443 Max in [2^14, 2^15]: 0.540071 Max in [2^15, 2^16]: 0.538784 Max in [2^16, 2^17]: 0.537044 Max in [2^17, 2^18]: 0.536020 Max in [2^18, 2^19]: 0.534645 Max in [2^19, 2^20]: 0.533779 Winning number = 1489
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