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a(0) = 1; a(n) = sum_{k=1 to d(n)} [a(n-k)] where d(n) is number of positive divisors of n.
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%I #10 Apr 09 2014 10:15:16

%S 1,1,2,3,6,9,20,29,64,113,226,339,791,1130,2486,4746,9492,14238,32883,

%T 47121,110966,205208,396178,601386,1417472,2415036,4830072,9263966,

%U 18924110,28188076,66036296,94224372,221466892,409915636,791643196

%N a(0) = 1; a(n) = sum_{k=1 to d(n)} [a(n-k)] where d(n) is number of positive divisors of n.

%H Harvey P. Dale, <a href="/A055873/b055873.txt">Table of n, a(n) for n = 0..1000</a>

%e Since d(8) = 4, a(8) = a(7) +a(6) +a(5) +a(4) {4 terms}. So a(8) = 29 +20 +9 +6 =64.

%t Module[{lst={1},n=1},Do[AppendTo[lst,Total[Take[lst,-DivisorSigma[0,n]]]];n++,{40}];lst] (* _Harvey P. Dale_, May 01 2012 *)

%K easy,nonn

%O 0,3

%A _Leroy Quet_, Jul 16 2000