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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.
3
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
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
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