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Let { d_1, d_2, ..., d_k } be the divisors of n. Then a(n) = d_k^1 + d_(k-1)^2 + ... + d_1^k.
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%I #20 Jan 01 2016 08:09:39

%S 1,3,4,9,6,24,8,33,19,44,12,226,14,72,68,161,18,429,20,534,98,152,24,

%T 3858,51,204,136,856,30,6534,32,1089,182,332,210,22965,38,408,236,

%U 12886,42,14262,44,2148,1868,584,48,128338,99,2333,368,3214,54,21810,302

%N Let { d_1, d_2, ..., d_k } be the divisors of n. Then a(n) = d_k^1 + d_(k-1)^2 + ... + d_1^k.

%H Carlos Eduardo Olivieri, <a href="http://oeis.org/wiki/User:Carlos_Eduardo_Olivieri/ideas/polyfromdivisors">A polynomial from divisors of n</a>

%e For n = 4: a(4) = 4^1 + 2^2 + 1^3 = 9.

%e For n = 5: a(5) = 5^1 + 1^2 = 6.

%e For n = 6: a(6) = 6^1 + 3^2 + 2^3 + 1^4 = 24.

%t a[n_] := Sum[Sort[Divisors[n], #1 > #2 &][[i]]^i, {i, DivisorSigma[0, n]}]; Table[a[n], {n, 60}]

%o (PARI) a(n) = my(d = divisors(n)); sum(k=1, #d, d[k]^(#d-k+1)); \\ _Michel Marcus_, Jan 01 2016

%Y Cf. A027750, A180851.

%K nonn,easy

%O 1,2

%A _Carlos Eduardo Olivieri_, Nov 29 2015