A258938 - OEIS

%I #16 Nov 12 2015 06:02:11

%S 2,3,4,6,7,8,9,11,14,18,19,21,23,25,30,32,33,42,44,47,57,59,63,65,71,

%T 76,77,82,83,85,86,93,97,107,108,111,116,117,118,121,127,131,135,137,

%U 139,143,151,153,154,162,164,167,175,177,183,185,188,189,194,195

%N Index of the Lucas numbers in A258740.

%C This sequence is probably finite with 228 terms. The probable last term of the sequence is a(228) = 6903, and the corresponding Lucas number A000032(6903) = 4382083468497... 10084254004 contains 1443 decimal digits.

%H Michel Lagneau, <a href="/A258938/b258938.txt">Table of n, a(n) for n = 1..228</a>

%e 7 is in the sequence because L(7)= A000032(7) = 29 = A258740(5) where 2+9=11 > 7.

%p with(combinat,fibonacci):with(numtheory):

%p a:=n->2*fibonacci(n-1)+fibonacci(n):

%p    for m from 1 to 400 do:

%p       x:=a(m):y:=convert(x,base,10):n1:=nops(y):

%p       s:=sum('y[i]', 'i'=1..n1):

%p        if s>m

%p         then

%p         printf(`%d, `,m):

%p         else

%p        fi:

%p      od:

%o (PARI) lista(nn) = {for(n=1, nn, if (sumdigits(l = fibonacci(n+1) + fibonacci(n-1)) > n, print1(n, ", "))); } \\ _Altug Alkan_, Nov 08 2015

%Y Cf. A000032, A258740.

%K nonn,base

%O 1,1

%A _Michel Lagneau_, Nov 07 2015