A080651 -id:A080651 - cp's OEIS frontend

A160685 Numbers k >= 1 such that A000005(A000045(k))/A000005(k) is an integer.

1, 3, 5, 6, 7, 8, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 24, 25, 26, 27, 29, 30, 31, 33, 34, 35, 37, 38, 39, 40, 41, 42, 43, 46, 47, 50, 51, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 65, 66, 67, 69, 70, 71, 72, 73, 74, 75, 77, 78, 79, 82, 83, 84, 85, 86, 87, 88, 89, 91, 93, 94

Offset: 1

Ctibor O. Zizka, May 23 2009

A080651 is a subsequence. - R. J. Mathar, May 25 2009

Robert Israel, Table of n, a(n) for n = 1..1073

Cf. A000045, A000005.

A000045 := proc(n) combinat[fibonacci](n) ; end: A000005 := proc(n) numtheory[tau](n) ; end: A063375 := proc(n) A000005(A000045(n)) ; end: for n from 1 to 130 do if A063375(n) mod A000005(n) = 0 then printf("%d,",n) ; fi; od: # R. J. Mathar, May 25 2009

okQ[n_] := With[{tau = DivisorSigma[0, #]&},
  IntegerQ[tau[Fibonacci[n]]/tau[n]]];
Select[Range[100], okQ] (* Jean-François Alcover, May 17 2023 *)

{n: A000005(n) | A063375(n)}. - R. J. Mathar, May 25 2009

2, 12 removed and sequence extended by R. J. Mathar, May 25 2009

A330777 Numbers k such that k and Lucas(k) have the same number of divisors.

Original entry on oeis.org

1, 2, 5, 7, 10, 11, 13, 14, 17, 19, 24, 26, 28, 31, 37, 38, 40, 41, 47, 53, 61, 62, 71, 79, 86, 88, 113, 152, 178, 202, 248, 313, 353, 458, 488, 503, 586, 613, 617, 856, 863, 914, 1082, 1097, 1306, 1318, 1361, 1784

Offset: 1

Chai Wah Wu, Dec 31 2019

All prime terms of A001606 (i.e., terms in A001606 that are not nontrivial powers of 2) are terms of this sequence.
Conjecture: all terms are of the form 2^k*p for k >= 0 and p prime.
It is unknown whether 1816 is a term (the smallest number for which membership in the sequence is unknown); it depends on whether Lucas(1816)/47 is a semiprime or not. The following composite numbers are terms of the sequence: 3106, 3928, 4006, 5414, 5498, 14318, 20578. - Chai Wah Wu, Jan 03 2020

Blair Kelly, Fibonacci and Lucas Factorizations.

Cf. A000032, A001606, A080651.

Select[Range[100],DivisorSigma[0,#]==DivisorSigma[0,LucasL[#]]&] (* Metin Sariyar, Jan 03 2020 *)

for(k=1,320,if(numdiv(k)==numdiv(fibonacci(k+1)+fibonacci(k-1)),print1(k,", "))) \\ Hugo Pfoertner, Jan 03 2020

cp's OEIS Frontend

A160685 Numbers k >= 1 such that A000005(A000045(k))/A000005(k) is an integer.

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