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A321874 If n = Product (p_j^k_j) then a(n) = Product (prime(p_j)^prime(k_j)). 5

%I #16 Jan 20 2024 03:09:51

%S 1,9,25,27,121,225,289,243,125,1089,961,675,1681,2601,3025,2187,3481,

%T 1125,4489,3267,7225,8649,6889,6075,1331,15129,3125,7803,11881,27225,

%U 16129,177147,24025,31329,34969,3375,24649,40401,42025,29403,32041,65025,36481,25947,15125

%N If n = Product (p_j^k_j) then a(n) = Product (prime(p_j)^prime(k_j)).

%H <a href="/index/Pri#prime_indices">Index entries for sequences computed from indices in prime factorization</a>.

%F Multiplicative with a(p^e) = prime(p)^prime(e). - _M. F. Hasler_, Nov 20 2018

%F Sum_{n>=1} 1/a(n) = Product_{m>=1} (1 + Sum_{k>=1} 1/prime(m)^prime(k)) = 1.22718741... . - _Amiram Eldar_, Jan 20 2024

%e a(12) = a(2^2 * 3^1) = prime(2)^prime(2) * prime(3)^prime(1) = 3^3 * 5^2 = 675.

%t a[n_] := Times @@ (Prime[#[[1]]]^Prime[#[[2]]] & /@ FactorInteger[n]); a[1] = 1; Table[a[n], {n, 45}]

%o (PARI) a(n) = my(f=factor(n)); prod(k=1, #f~, prime(f[k,1])^prime(f[k,2])); \\ _Michel Marcus_, Nov 20 2018

%o (PARI) apply( A321874(n)=factorback(apply(prime,factor(n))), [1..49]) \\ _M. F. Hasler_, Nov 20 2018

%Y Cf. A000040, A064988, A304203.

%K nonn,mult

%O 1,2

%A _Ilya Gutkovskiy_, Nov 20 2018

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Last modified August 23 23:54 EDT 2024. Contains 375396 sequences. (Running on oeis4.)