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a(n) = Sum_{d|n} d^tau(n/d).
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%I #14 Jun 09 2023 22:38:09

%S 1,3,4,9,6,20,8,33,19,40,12,108,14,68,50,161,18,225,20,278,80,148,24,

%T 850,51,200,136,600,30,1114,32,1089,164,328,110,3387,38,404,218,2450,

%U 42,3214,44,1892,558,580,48,12596,99,1409,350,2958,54,8630,202,6370,428,904,60,33042

%N a(n) = Sum_{d|n} d^tau(n/d).

%C If p is a prime, a(p) = Sum_{d|p} d^tau(p/d) = 1^2 + p^1 = p + 1.

%H Seiichi Manyama, <a href="/A345270/b345270.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = Sum_{d|n} (n/d)^tau(d). - _Wesley Ivan Hurt_, Jun 09 2023

%e a(10) = Sum_{d|10} d^tau(10/d) = 1^4 + 2^2 + 5^2 + 10^1 = 40.

%t Table[Sum[k^DivisorSigma[0, n/k] (1 - Ceiling[n/k] + Floor[n/k]), {k, n}], {n, 60}]

%o (PARI) a(n) = sumdiv(n, d, d^numdiv(n/d)); \\ _Michel Marcus_, Oct 08 2021

%Y Cf. A000005 (tau), A174937, A345271.

%K nonn

%O 1,2

%A _Wesley Ivan Hurt_, Jun 12 2021