A111419 a(n) is the smallest positive integer for which Fibonacci(n + a(n)) == Fibonacci(n) (mod n).
1, 2, 2, 6, 5, 15, 2, 9, 6, 10, 1, 12, 2, 9, 10, 24, 2, 24, 1, 5, 6, 7, 2, 12, 25, 15, 18, 48, 1, 15, 1, 11, 14, 19, 10, 12, 2, 15, 34, 60, 1, 15, 2, 30, 30, 25, 2, 12, 14, 50, 42, 78, 2, 24, 10, 24, 30, 13, 1, 60, 1, 27, 18, 96, 10, 120, 2, 36, 6, 25, 1, 12, 2, 39, 50, 18, 6, 39, 1, 35
Offset: 1
Keywords
Examples
a(3) = 2 because Fibonacci(3+2) - Fibonacci(3) = 5 - 2 == 0 (mod 3) and 2 is the smallest integer for which this is true.
Links
- Michael De Vlieger, Table of n, a(n) for n = 1..10000
Crossrefs
Cf. A002708.
Programs
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Mathematica
Array[Block[{k = 1}, While[Mod[Fibonacci[# + k], #] != Mod[Fibonacci@ #, #], k++]; k] &, 80] (* Michael De Vlieger, Dec 17 2017 *) -
MuPAD
for n from 1 to 100 do an := 0; repeat an := an+1; until (numlib::fibonacci(n+an)-numlib::fibonacci(n)) mod n = 0 end_repeat; print(an); end_for;
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PARI
a(n) = {my(k = 1); while(Mod(fibonacci(n + k), n) != Mod(fibonacci(n), n), k++); k;} \\ Michel Marcus, Dec 18 2017
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