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A351435 If n = Product (p_j^k_j) then a(n) = Product ((p_j + 1)^(k_j + 1)).

%I A351435 #17 Dec 15 2023 06:19:48
%S A351435 1,9,16,27,36,144,64,81,64,324,144,432,196,576,576,243,324,576,400,
%T A351435 972,1024,1296,576,1296,216,1764,256,1728,900,5184,1024,729,2304,2916,
%U A351435 2304,1728,1444,3600,3136,2916,1764,9216,1936,3888,2304,5184,2304,3888,512,1944,5184,5292,2916
%N A351435 If n = Product (p_j^k_j) then a(n) = Product ((p_j + 1)^(k_j + 1)).
%H A351435 <a href="/index/Eu#epf">Index entries for sequences computed from exponents in factorization of n</a>.
%H A351435 <a href="/index/Pri#prime_indices">Index entries for sequences computed from indices in prime factorization</a>.
%F A351435 a(n) = A003959(n) * A048250(n).
%F A351435 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
%F A351435 Sum_{n>=1} 1/a(n) = zeta(2)/zeta(3) (A306633). - _Amiram Eldar_, Dec 15 2023
%p A351435 a:= n-> mul((i[1]+1)^(i[2]+1), i=ifactors(n)[2]):
%p A351435 seq(a(n), n=1..53);  # _Alois P. Heinz_, Feb 11 2022
%t A351435 f[p_, e_] := (p + 1)^(e + 1); a[1] = 1; a[n_] := Times @@ (f @@@ FactorInteger[n]); Table[a[n], {n, 1, 53}]
%o A351435 (PARI) a(n) = my(f=factor(n)); for (k=1, #f~, f[k,1]++; f[k,2]++); factorback(f); \\ _Michel Marcus_, Feb 11 2022
%Y A351435 Cf. A003557, A003958, A003959, A048250, A064549, A065478, A306633, A326297, A327564, A351434.
%K A351435 nonn,mult,easy
%O A351435 1,2
%A A351435 _Ilya Gutkovskiy_, Feb 11 2022