A224523 Let {f_n(k)} be the n-th sequence of Fibonacci-like numbers defined by recursion f_n(0) = 0, f_n(1) = 1 and, for k>=2, f_n(k) = f_n(k-1) + f_n(k-2) divided by maximal possible powers of primes >= prime(n). a(n) is length of the smallest period of {f_n(k)}, and a(n)=0, if {f_n(k)} is not eventually periodic.

1, 3, 6, 9, 9, 12, 15, 27, 12, 12, 15, 15, 15, 15, 15, 15, 15, 42, 42, 42, 42, 42, 42, 90, 72, 36, 36, 36, 36, 36, 36, 36, 36, 36, 54, 54, 66, 102, 102, 102, 102, 102, 102, 102, 102, 36, 36, 36, 36, 36, 36, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24, 24

Offset

1,2

Comments

Conjecture. For n>=1, a(n)>0.

If Conjecture is true, then we have an answer in affirmative on question in A078414.

Links

Peter J. C. Moses, Table of n, a(n) for n = 1..4000

Example

Let n=9, prime(9)=23. Sequence {f_9(k)} begins 0,1,1,2,3,5,8,13,21,34,55. Now 34 + 55 = 89 is prime >=23, so the following terms are 89/89 = 1,56,57. Further, since 56 + 57 = 113 is prime >=23, then the following term is 113/113 = 1 and, since 57 + 1 = 58=29*2, then the following term is 58/29 = 2. Now we have period {1,2,3,5,8,13,21,34,55,1,56,57} with length 12. Thus a(9)=12.

Mathematica

seqPosition[{list_, seqtofind_}]:=Position[Partition[list, Length[#], 1], Flatten[{___, #, ___}]]&[seqtofind]; Table[Clear[a]; a[0]:=0; a[1]:=1; a[n_]:=a[n]=#/(Product[Prime[i]^IntegerExponent[#, Prime[i]], {i, z, PrimePi[#]+1}])&[(a[n-1]+a[n-2])]; NestWhile[#+2&, 24, Length[diff=Flatten[seqPosition[{#, Take[#, -2]}]]&[Map[a, Range[0, #]]]]<=1&]; (#[[2]]-#[[1]])&[diff], {z, 1, 50}] (* Peter J. C. Moses, Apr 10 2013 *)

Crossrefs

Sequence in context: A329513 A329213 A007844 * A159785 A057338 A337603

Adjacent sequences: A224520 A224521 A224522 * A224524 A224525 A224526

Keyword

nonn

Author

Vladimir Shevelev, Apr 09 2013

Extensions

For n >= 9 the terms were calculated by Peter J. C. Moses, Apr 09 2013

Status

approved