Pisano period: Difference between revisions

Content added Content deleted
(added RPL)
(Added Easylang)
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168 336 48 364 180 72 264 348 168 400 120 232 132 178 120</pre>
168 336 48 364 180 72 264 348 168 400 120 232 132 178 120</pre>
=={{header|EasyLang}}==
{{trans|Go}}
<syntaxhighlight>
fastfunc isprim num .
if num mod 2 = 0
if num = 2
return 1
.
return 0
.
if num mod 3 = 0
if num = 3
return 1
.
return 0
.
i = 5
while i <= sqrt num
if num mod i = 0
return 0
.
i += 2
if num mod i = 0
return 0
.
i += 4
.
return 1
.
func gcd a b .
if b = 0
return a
.
return gcd b (a mod b)
.
func lcm a b .
return a / gcd a b * b
.
func ipow x p .
prod = 1
while p > 0
if p mod 2 = 1
prod *= x
.
p = p div 2
x *= x
.
return prod
.
proc getprims n . prims[] .
prims[] = [ ]
for i = 2 to n
d = n / i
m = n mod i
if m = 0
prims[] &= i
prims[] &= 0
while m = 0
prims[-1] += 1
n = d
d = n div i
m = n mod i
.
.
.
.
func pisanoPeriod m .
c = 1
for i = 1 to m * m
swap p c
c = (p + c) mod m
if p = 0 and c = 1
return i
.
.
return 1
.
func pisanoPrime p k .
if isprim p = 0 or k = 0
return 0
.
return ipow p (k - 1) * pisanoPeriod p
.
func pisano m .
getprims m p[]
for i = 1 step 2 to len p[] - 1
pps[] &= pisanoPrime p[i] p[i + 1]
.
if len pps[] = 0
return 1
.
if len pps[] = 1
return pps[1]
.
f = pps[1]
for i = 2 to len pps[]
f = lcm f pps[i]
.
return f
.
proc main . .
for p = 2 to 14
pp = pisanoPrime p 2
if pp > 0
print "pisanoPrime(" & p & ": 2) = " & pp
.
.
print ""
for p = 2 to 179
pp = pisanoPrime p 1
if pp > 0
print "pisanoPrime(" & p & ": 1) = " & pp
.
.
print ""
numfmt 0 3
print "pisano(n) for integers 'n' from 1 to 180 are:"
for n = 1 to 180
write pisano (n) & " "
if n mod 15 = 0
print ""
.
.
.
main
</syntaxhighlight>

=={{header|Factor}}==
=={{header|Factor}}==
{{works with|Factor|0.99 2020-01-23}}
{{works with|Factor|0.99 2020-01-23}}