A105802 Smallest m such that the m-th Fibonacci number has exactly n divisors that are also Fibonacci numbers.
1, 3, 6, 15, 12, 45, 24, 36, 48, 405, 60, 315, 192, 144, 120, 945, 180, 1575, 240, 576, 3072, 295245, 360, 1296, 12288, 900, 960, 25515, 720, 14175, 840, 9216, 196608, 5184, 1260, 17325, 786432, 36864, 1680, 31185, 2880, 127575, 15360, 3600, 99225
Offset: 1
Keywords
Examples
n=6: a(6) = 45, A076985(6) = A000045(45) = 1134903170,
A076984(45) = #{1,2,5,34,109441,1134903170} = #{fib(1),fib(2),fib(5),fib(9),fib(21),fib(45)} = 6.
Links
- Eric Weisstein's World of Mathematics, Fibonacci Number
Crossrefs
Cf. A068499.
Programs
-
Mathematica
t=Table[s=DivisorSigma[0, n]; If[OddQ[n], s, s-1], {n, 1000000}]; lst={}; n=1; While[pos=Flatten[Position[t, n, 1, 1]]; Length[pos]>0, AppendTo[lst, pos[[1]]]; n++ ]; lst (Noe)
Formula
Conjecture: a(2k+1) = 3*2^(Prime[k-1]-1) for k>3. It appears that a(2k+1) = 3*2^k for k = {1,2,3,4,6,10,12,16,18,...} = A068499[n] Numbers n such that n! reduced modulo (n+1) is not zero. - Alexander Adamchuk, Sep 15 2006
Extensions
More terms from T. D. Noe, Jan 18 2006
Comments