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a(1) = 1; a(n+1) = sum{k|n k<=sqrt(n)} a(k) where sum is over the positive divisors k of n with k <= sqrt(n).
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%I #16 Mar 11 2014 01:32:11

%S 1,1,1,1,2,1,2,1,2,2,2,1,3,1,2,2,3,1,3,1,3,2,2,1,4,3,2,2,3,1,5,1,3,2,

%T 2,3,5,1,2,2,5,1,4,1,3,4,2,1,5,3,4,2,3,1,4,3,5,2,2,1,7,1,2,4,4,3,4,1,

%U 3,2,6,1,6,1,2,4,3,3,4,1,6,4,2,1,7,3,2,2,4,1,8,3,3,2,2,3,6,1,4,4,7,1,4,1,4

%N a(1) = 1; a(n+1) = sum{k|n k<=sqrt(n)} a(k) where sum is over the positive divisors k of n with k <= sqrt(n).

%H T. D. Noe, <a href="/A068108/b068108.txt">Table of n, a(n) for n=1..10000</a>

%F a(n) = A038548(n-1), 2<=n<=25. - _Omar E. Pol_, Feb 05 2014

%t a[1]=1; a[n_] := a[n] = Module[{d=Divisors[n-1], s=0, i=1}, While[i<=Length[d] && d[[i]]<=Sqrt[n-1], s=s+a[d[[i]]]; i++ ]; s]; Table[a[n], {n, 150}] (T. D. Noe)

%Y Different from A038548.

%K nonn

%O 1,5

%A _Leroy Quet_, Mar 22 2002

%E More terms from _T. D. Noe_, Nov 03 2004