A351435 If n = Product (p_j^k_j) then a(n) = Product ((p_j + 1)^(k_j + 1)).
1, 9, 16, 27, 36, 144, 64, 81, 64, 324, 144, 432, 196, 576, 576, 243, 324, 576, 400, 972, 1024, 1296, 576, 1296, 216, 1764, 256, 1728, 900, 5184, 1024, 729, 2304, 2916, 2304, 1728, 1444, 3600, 3136, 2916, 1764, 9216, 1936, 3888, 2304, 5184, 2304, 3888, 512, 1944, 5184, 5292, 2916
Offset: 1
Links
Crossrefs
Programs
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Maple
a:= n-> mul((i[1]+1)^(i[2]+1), i=ifactors(n)[2]): seq(a(n), n=1..53); # Alois P. Heinz, Feb 11 2022
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Mathematica
f[p_, e_] := (p + 1)^(e + 1); a[1] = 1; a[n_] := Times @@ (f @@@ FactorInteger[n]); Table[a[n], {n, 1, 53}] -
PARI
a(n) = my(f=factor(n)); for (k=1, #f~, f[k,1]++; f[k,2]++); factorback(f); \\ Michel Marcus, Feb 11 2022
Formula
Sum_{k=1..n} a(k) ~ c * n^3, where c = 1/(3 * Product_{p prime} (1 - p/(p^3-1))) = 1 /(3 * A065478) = 0.5787439255... . - Amiram Eldar, Nov 19 2022
Sum_{n>=1} 1/a(n) = zeta(2)/zeta(3) (A306633). - Amiram Eldar, Dec 15 2023