A053031 Numbers with 1 zero in Fibonacci numbers mod m.

1, 2, 4, 11, 19, 22, 29, 31, 38, 44, 58, 59, 62, 71, 76, 79, 101, 116, 118, 121, 124, 131, 139, 142, 151, 158, 179, 181, 191, 199, 202, 209, 211, 229, 236, 239, 242, 251, 262, 271, 278, 284, 302, 311, 316, 319, 331, 341, 349, 358, 359, 361, 362, 379, 382, 398

Offset

1,2

Comments

Conjecture: m is on this list iff m is an odd number all of whose factors are on this list or m is 2 or 4 times such an odd number.

A001176(a(n)) = A128924(a(n),1) = 1. - Reinhard Zumkeller, Jan 16 2014

Also numbers n such that A001175(n) = A001177(n). - Daniel Suteu, Aug 08 2018

Links

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000

Brennan Benfield and Michelle Manes, The Fibonacci Sequence is Normal Base 10, arXiv:2202.08986 [math.NT], 2022.

M. Renault, Fibonacci sequence modulo m

Mathematica

With[{s = {1}~Join~Table[Count[Drop[NestWhile[Append[#, Mod[Total@ Take[#, -2], n]] &, {1, 1}, If[Length@ # < 3, True, Take[#, -2] != {1, 1}] &], -2], 0], {n, 2, 400}]}, Position[s, 1][[All, 1]] ] (* Michael De Vlieger, Aug 08 2018 *)

Prog

(Haskell)
a053031 n = a053031_list !! (n-1)
a053031_list = filter ((== 1) . a001176) [1..]
-- Reinhard Zumkeller, Jan 16 2014
(PARI) entryp(p)=my(k=p+[0, -1, 1, 1, -1][p%5+1], f=factor(k)); for(i=1, #f[, 1], for(j=1, f[i, 2], if((Mod([1, 1; 1, 0], p)^(k/f[i, 1]))[1, 2], break); k/=f[i, 1])); k
entry(n)=if(n==1, return(1)); my(f=factor(n), v); v=vector(#f~, i, if(f[i, 1]>1e14, entryp(f[i, 1]^f[i, 2]), entryp(f[i, 1])*f[i, 1]^(f[i, 2]-1))); if(f[1, 1]==2&&f[1, 2]>1, v[1]=3<<max(f[1, 2]-2, 1)); lcm(v)
fibmod(n, m)=((Mod([1, 1; 1, 0], m))^n)[1, 2]
is(n)=fibmod(entry(n)+1, n)==1 \\ Charles R Greathouse IV, Dec 14 2016

Crossrefs

Cf. A001176, A053032.

Cf. A053029 (with 4 zeros), A053030 (with 2 zeros).

Sequence in context: A056394 A056395 A288621 * A018674 A076518 A139785

Adjacent sequences: A053028 A053029 A053030 * A053032 A053033 A053034

Keyword

nonn

Author

Henry Bottomley, Feb 23 2000

Status

approved