Hofstadter-Conway $10,000 sequence
From Rosetta Code
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(n-1))+a(n-a(n-1))
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)
The task is to:
- Create a routine to generate members of the Hofstadter-Conway $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
References:
- Conways Challenge Sequence, Mallows' own account.
- Mathworld Article.
[edit] 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
[edit] 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[n-v];
( 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$, lg2,lg2+1, amax, nmax));
amax := 0;
lg2 PLUSAB 1 # 'PLUSAB' (plus-and-becomes) has the short form +:= #
FI;
k1 := n
OD;
mallows number # the result of the last expression evaluated is returned as the result of the PROC #
END;
INT mallows number = do sqnc(2**20); # This definition of 'mallows number' does not clash with the variable
of the same name inside PROC do sqnc - they are in different scopes#
printf(($"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
[edit] 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) { ; Hofstadter-Conway 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
[edit] 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[n-1]]+Q[n-Q[n-1]];
}
}
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 = n-1;
if (S[k]==0) {
HCsequence(k,Q,S);
}
k = Q[n-1];
if (S[k]==0) {
HCsequence(k,Q,S);
}
k = n-Q[n-1];
if (S[k]==0) {
HCsequence(k,Q,S);
}
}
Q[n] = Q[Q[n-1]]+Q[n-Q[n-1]];
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
[edit] 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
[edit] 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[n-v];
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
[edit] 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[x-1] + seq[seq.size()-x]);
}
return seq[n-1];
}
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
[edit] 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.
[edit] Clojure
(use 'clojure.contrib.math) ; for expt
;; A literal transcription of the definition, with memoize doing the heavy lifting
(def conway
(memoize
(fn [x]
(if (< x 3)
1
(+ (conway (conway (dec x)))
(conway (- x (conway (dec x)))))))))
(def N (drop 1 (range))) ; natural numbers
;; This is enough to compute them all. The rest is grouping and max-finding
(def all-conways (map conway N))
;; All the powers of two
(def pow2 (map #(expt 2 %) N))
;; Find the lowest power of two > n
(defn lowest-pow-higher [n]
(some #(and (> % n) %) pow2))
;; Split the natural numbers into groups at each power of two
(def groups (partition-by lowest-pow-higher N))
;; The conway numbers of each number in the group
(def C (map #(map conway %) groups))
;; Each conway number divided by its index
(def ratios (map #(map / %1 %2) C groups))
;; The largest value in each group of ratios
(def maxima (map #(apply max %) ratios))
(take 4 maxima) ; no rounding errors: still using ratios
;; yields (1 2/3 2/3 7/11)
(def first-20 (take 20 maxima))
(apply >= first-20) ; yields true - the sequence is decreasing
(map double first-20) ; to get them in decimal form
[edit] Common Lisp
(defparameter *hof-con*
(make-array '(2) :initial-contents '(1 1) :adjustable t
:element-type 'integer :fill-pointer 2))
(defparameter *hof-con-ratios*
(make-array '(2) :initial-contents '(1.0 0.5) :adjustable t
:element-type 'single-float :fill-pointer 2))
(defun hof-con (n)
(let ((l (length *hof-con*)))
(if (<= n l) (aref *hof-con* (1- n))
(extend-hof-con-sequence l n))))
(defun extend-hof-con-sequence (l n)
(loop for i from l below n do
(let* ((x (aref *hof-con* (1- i)))
(hc (+ (aref *hof-con* (1- x))
(aref *hof-con* (- i x)))))
(vector-push-extend hc *hof-con*)
(vector-push-extend (/ hc (+ i 1.0)) *hof-con-ratios*)))
(aref *hof-con* (1- n)))
(defun max-in-array-range (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)
(hof-con (expt 2 po2))
(loop for i from 1 below po2 do
(let ((id1 (expt 2 i)) (id2 (expt 2 (1+ i))))
(multiple-value-bind (m id)
(max-in-array-range *hof-con-ratios* 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)))
(hof-con n)
(do ((i (1- n) (1- i)))
((> (aref *hof-con-ratios* 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
[edit] 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
[edit] Euler Math Toolbox
>function hofstadter (n) ...
$v=ones(1,n);
$ loop 2 to n-1
$ 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^(j-1):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
[edit] 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(N-last+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 = i-1
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
[edit] 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
[edit] 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[n-x]
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
[edit] 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]
[edit] Icon and Unicon
procedure main(args)
m := integer(!args) | 20
nextNum := create put(A := [], 1 | 1 | |A[A[*A]]+A[-A[*A]])[*A]
p2 := 2 ^ (p := 1)
maxv := 0
every n := 1 to (2^m) do {
if maxv <:= (x := @nextNum / real(n)) then maxm := n
if x >= 0.55 then mallows := n # Want *this* n, not next one!
if n = p2 then {
write("Max between 2^",p-1," and 2^",p," is ",maxv," at n = ",maxm)
p2 := 2 ^ (p +:= 1)
maxv := 0
}
}
write("Mallows's number is ",\mallows | "NOT found!")
end
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 ->
[edit] 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@tailExample:
expand =: 2+I.@;
tail =: copies&.>^:(<@>:`(<@,@2:))
copies =: >: |.@(#!.1 |.)~ 1 j. #;.1 #^:_1 ::1:~ ]~:{.
] 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 ...
[edit] Java
with corrections to 0 indexing.public class HofCon
{
public static void main(final String[] args)
{
doSqnc(1<<20);
}
public static void doSqnc(int m)
{
int[] a_list = new int[m + 1];
int max_df = 0;
int p2_max = 2;
int k1 = 2;
int lg2 = 1;
double amax = 0;
a_list[0] = a_list[1] = 1;
int v = a_list[2];
for (int n = 2; n <= m; n++)
{
v = a_list[n] = a_list[v] + a_list[n - v];
if (amax < v * 1.0 / n)
amax = v * 1.0 / n;
if (0 == (k1 & n))
{
System.out.printf("Maximum between 2^%d and 2^%d was %f\n", 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 2^18 and 2^19 was 0.534645 Maximum between 2^19 and 2^20 was 0.533779
[edit] JavaScript
var hofst_10k = function(n) {
var memo = [1, 1];
var a = function(n) {
var result = memo[n-1];
if (typeof result !== 'number') {
result = a(a(n-1))+a(n-a(n-1));
memo[n-1] = 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 2^1-2^2 is: 0.6666666666666666 Maxima between 2^2-2^3 is: 0.6666666666666666 Maxima between 2^3-2^4 is: 0.6363636363636364 Maxima between 2^4-2^5 is: 0.6086956521739131 Maxima between 2^5-2^6 is: 0.5909090909090909 ... Maxima between 2^18-2^19 is: 0.5346454310781124 Maxima between 2^19-2^20 is: 0.5337792299633678 Maxima between 2^20-2^21 is: 0.5326770563524978
[edit] Mathematica
a[1] := 1; a[2] := 1;
a[n_] := a[n] = a[a[n-1]] + a[n-a[n-1]]
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
[edit] MATLAB / Octave
function Q = HCsequence(N)
Q = zeros(1,N);
Q(1:2) = 1;
for n = 3:N,
Q(n) = Q(Q(n-1))+Q(n-Q(n-1));
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
[edit] 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
[edit] 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 N-1}} + {A N-{A N-1}}
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
[edit] PARI/GP
HC(n)=my(a=vectorsmall(n));a[1]=a[2]=1;for(i=3,n,a[i]=a[a[i-1]]+a[i-a[i-1]]);a;
maxima(n)=my(a=HC(1<<n),m);vector(n-1,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
[edit] 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 !
[edit] Perl 6
This is written in relatively low-level primitives (for Perl 6) because rakudo still has some performance issues with long lists. One interesting feature, however, is the use of the max infix operator as an assignment operator, and using it on a Pair so that the maximum is calculated on the key of the Pair, while the value of the Pair carries the index of the maximum through to the end without having to keep a second variable.
my $POW = 20;
my $top = 2**$POW;
my @a = (0,1,1);
@a[$top] = 0; # pre-extend array
my $n = 3;
my $p = 1;
loop ($n = 3; $n <= $top; $n++) {
@a[$n] = $p = @a[$p] + @a[$n - $p];
}
my $last55;
for 1 ..^ $POW -> $power {
my $beg = 2 ** $power;
my $end = $beg * 2 - 1;
my $max;
my $ratio;
loop (my $n = $beg; $n <= $end; $n++) {
my $ratio = @a[$n] / $n;
$last55 = $n if $ratio * 100 >= 55;
$max max= $ratio => $n;
}
say $power.fmt('%2d'), $beg.fmt("%10d"), '..', $end.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
Incidentally, this takes three hours or so currently. We expect this to get much faster.
Here is a more list-oriented version. 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] } ... *);
my $last55;
for 1..19 -> $power {
my @range := 2**$power .. 2**($power+1)-1;
my @ratios = (@a[@range] Z/ @range) Z=> @range;
my $max = [max] @ratios;
($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 ", $last55;
[edit] PicoLisp
(de hofcon (N)
(cache '(NIL) (pack (char (hash N)) N)
(if (>= 2 N)
1
(+
(hofcon (hofcon (dec N)))
(hofcon (- N (hofcon (dec N)))) ) ) ) )
(scl 20)
(de sequence (M)
(let (Lim 4 Max 0 4k$ 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)
[edit] 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(i-k) + L(1+k);
end;
[edit] 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(n-v)
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
[edit] Python
from __future__ import division
def maxandmallows(nmaxpower2=20):
# Note: The first hc number is returned in hc[1];
# hc[0] is not part of the series.
nmax = 2**nmaxpower2
hc, mallows, mx, mxpow2 = [None, 1, 1], None, (0.5, 2), []
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 = %s" %
((n)//2, n, "%6.4f at n = %i" % mx))
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.
[edit] REXX
/*REXX program solves the Hofstadter-Conway $10,000 prize. */
hC.=; i.=0; m.=0; n.=0; w=0; wi=0; few=70; L=
do i=1 for few; L=L hc(i) /*build first 70 numbers in seq. */
end /*i*/
/*I wear a belt and suspenders. */
say 'The first' few "numbers in the Hofstadter-Conway sequence:"
say strip(L) /*show&tell, no trees have to die*/
say /*show a blank line for the eyes.*/
do k=0 to 20 /*build an array, powers of two.*/
p.k=2**k /*Bang-bang!. Er, I mean pow-pow*/
maxp=p.k /*and remember who's da big 'un. */
end /*k*/
r=1 /*R: the range of the power of 2.*/
do n=1 for maxp /*heck, let's get cracking then.*/
if n>p.r then r=r+1 /*for golf players: r=r+(n>p.r) */
_=hc(n)/n; if _>=.55 then w=n /*get next seq #; if ≥.55, a win?*/
if _<=m.r then iterate /*less than? Then keep truckin'.*/
m.r=_; i.r=n /*m.r & i.r is for ginkgo biloba*/
end /*n*/
pref='Maximum of a(n) ÷ n between ' /*prefix text of message.*/
do j=1 for 20; range='2**'right(j-1,2) "───► 2**"right(j,2)
say pref range '(inclusive) is ' left(m.j,8) ' at n='right(i.j,7)
end /*j*/
say
say 'The winning number is: ' w /*and the money shot is ... */
exit /*stick a fork in it, we're done.*/
/*─────────────────────────────────────hC [Hofstadter-Conway] subroutine*/
hC: procedure expose hC.; parse arg n; if n<3 then return 1
if hC.n=='' then hC.n = hC(hC(n-1)) + hC(n-hC(n-1))
return hC.n /*return with the goodie stuff. */
output
The first 70 numbers in the Hofstadter-Conway sequence: 1 1 2 2 3 4 4 4 5 6 7 7 8 8 8 8 9 10 11 12 12 13 14 14 15 15 15 16 16 16 16 16 17 18 19 20 21 21 22 23 24 24 25 26 26 27 27 27 28 29 29 30 30 30 31 31 31 31 32 32 32 32 32 32 33 34 35 36 37 38 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 0.666666 at n= 3 Maximum of a(n) ÷ n between 2** 2 ───► 2** 3 (inclusive) is 0.666666 at n= 6 Maximum of a(n) ÷ n between 2** 3 ───► 2** 4 (inclusive) is 0.636363 at n= 11 Maximum of a(n) ÷ n between 2** 4 ───► 2** 5 (inclusive) is 0.608695 at n= 23 Maximum of a(n) ÷ n between 2** 5 ───► 2** 6 (inclusive) is 0.590909 at n= 44 Maximum of a(n) ÷ n between 2** 6 ───► 2** 7 (inclusive) is 0.576086 at n= 92 Maximum of a(n) ÷ n between 2** 7 ───► 2** 8 (inclusive) is 0.567415 at n= 178 Maximum of a(n) ÷ n between 2** 8 ───► 2** 9 (inclusive) is 0.559459 at n= 370 Maximum of a(n) ÷ n between 2** 9 ───► 2**10 (inclusive) is 0.554937 at n= 719 Maximum of a(n) ÷ n between 2**10 ───► 2**11 (inclusive) is 0.550100 at n= 1487 Maximum of a(n) ÷ n between 2**11 ───► 2**12 (inclusive) is 0.547462 at n= 2897 Maximum of a(n) ÷ n between 2**12 ───► 2**13 (inclusive) is 0.544144 at n= 5969 Maximum of a(n) ÷ n between 2**13 ───► 2**14 (inclusive) is 0.542442 at n= 11651 Maximum of a(n) ÷ n between 2**14 ───► 2**15 (inclusive) is 0.540071 at n= 22223 Maximum of a(n) ÷ n between 2**15 ───► 2**16 (inclusive) is 0.538784 at n= 45083 Maximum of a(n) ÷ n between 2**16 ───► 2**17 (inclusive) is 0.537043 at n= 89516 Maximum of a(n) ÷ n between 2**17 ───► 2**18 (inclusive) is 0.536020 at n= 181385 Maximum of a(n) ÷ n between 2**18 ───► 2**19 (inclusive) is 0.534645 at n= 353683 Maximum of a(n) ÷ n between 2**19 ───► 2**20 (inclusive) is 0.533779 at n= 722589 The winning number is: 1489
[edit] 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("Prize-winning point:")
print(max(which(ratios >= .55)))
[edit] Ruby
class HofstadterConway10000
def initialize
@sequence = [nil, 1, 1]
end
attr_reader :sequence
def [](n)
raise ArgumentError, "n must be >= 1" if n < 1
a = @sequence
a.length.upto(n) {|i| a[i] = a[a[i-1]] + a[i-a[i-1]] }
a[n]
end
end
hc = HofstadterConway10000.new
I run the sequence backwards to make it easier to find the Mallows number.
The first call to the hc.[] method will pass the value 2^20, pre-populating the entire sequence.
p = -1
20.downto(1).each_cons(2) do |j, i|
max_n, max_v = -1, -1
(2**j).downto(2**i).each do |n|
v = hc[n].to_f / n
max_n, max_v = n, v if v > max_v
# Mallows number
p = n if p == -1 and v >= 0.55
end
puts "maximum between 2^#{i} and 2^#{j} occurs at #{max_n}: #{max_v}"
end
puts "the mallows number is #{p}"
output
maximum between 2^19 and 2^20 occurs at 722589: 0.533779229963368 maximum between 2^18 and 2^19 occurs at 353683: 0.534645431078112 maximum between 2^17 and 2^18 occurs at 181385: 0.536020067811561 maximum between 2^16 and 2^17 occurs at 89516: 0.537043656999866 maximum between 2^15 and 2^16 occurs at 45083: 0.538784020584256 maximum between 2^14 and 2^15 occurs at 22223: 0.540071097511587 maximum between 2^13 and 2^14 occurs at 11651: 0.542442708780362 maximum between 2^12 and 2^13 occurs at 5969: 0.544144747863964 maximum between 2^11 and 2^12 occurs at 2897: 0.547462892647566 maximum between 2^10 and 2^11 occurs at 1487: 0.550100874243443 maximum between 2^9 and 2^10 occurs at 719: 0.554937413073713 maximum between 2^8 and 2^9 occurs at 370: 0.559459459459459 maximum between 2^7 and 2^8 occurs at 178: 0.567415730337079 maximum between 2^6 and 2^7 occurs at 92: 0.576086956521739 maximum between 2^5 and 2^6 occurs at 44: 0.590909090909091 maximum between 2^4 and 2^5 occurs at 23: 0.608695652173913 maximum between 2^3 and 2^4 occurs at 11: 0.636363636363636 maximum between 2^2 and 2^3 occurs at 6: 0.666666666666667 maximum between 2^1 and 2^2 occurs at 3: 0.666666666666667 the mallows number is 1489
[edit] 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(n-1)) + a(n-a(n-1))
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^";pow2-1;" 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
[edit] 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
[edit] Tcl
The routine to return the nth 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 $hofcon10k [expr {$n-1}]]
}
while {$n > [llength $hofcon10k]} {
set i [lindex $hofcon10k end]
set a [lindex $hofcon10k [expr {$i-1}]]
# Don't use end-based 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
[edit] X86 Assembly
Using FASM syntax.
; Hofstadter-Conway $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+(ecx-1)*4] ; a[n-1]
sub eax,edx ; n-a[n-1]
mov eax,[a+eax*4] ; a[n-a[n-1]]
add eax,[a+edx*4] ;+a[a[n-1]]
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
