login
a(n) = largest m such that m! divides n-th Fibonacci number; a(n) = A055881(A000045(n)).
9

%I #9 Dec 17 2013 09:18:09

%S 1,1,2,1,1,2,1,1,2,1,1,4,1,1,2,1,1,2,1,1,2,1,1,4,1,1,2,1,1,2,1,1,2,1,

%T 1,4,1,1,2,1,1,2,1,1,2,1,1,4,1,1,2,1,1,2,1,1,2,1,1,6,1,1,2,1,1,2,1,1,

%U 2,1,1,4,1,1,2,1,1,2,1,1,2,1,1,4,1,1,2,1,1,2,1,1,2,1,1,4,1,1,2,1,1,2,1,1,2,1,1,4,1,1,2,1,1,2,1,1,2,1,1,7

%N a(n) = largest m such that m! divides n-th Fibonacci number; a(n) = A055881(A000045(n)).

%C The lengths of palindromic prefixes begin as:

%C 1, 2, 5, 8, 11, 23, 35, 47, 59, 119, 239, 359, 479, 959, 1439, 1919, ...

%C +1 results: 2, 3, 6, 9, 12, 24, 36, 48, 60, 120, 240, 360, 480, 960, 1440, 1920, ...

%H Antti Karttunen, <a href="/A233285/b233285.txt">Table of n, a(n) for n = 1..12600</a>

%Y Differs from A233284 for the first time at n=120, where a(120)=7, while A233284(120)=12.

%Y Cf. A055881, A000045, A001175-A001177, A233283, A233281.

%K nonn

%O 1,3

%A _Antti Karttunen_, Dec 06 2013