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
Keywords
Examples
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.
Links
- Peter J. C. Moses, Table of n, a(n) for n = 1..4000
Programs
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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 *)
Extensions
For n >= 9 the terms were calculated by Peter J. C. Moses, Apr 09 2013
Comments