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A168065 If n = Product p(k)^e(k) then a(n) = (Product (p(k)+1)^e(k) + Product (p(k)-1)^e(k))/2, a(1) = 1. 6

%I #12 Jul 08 2023 20:15:43

%S 1,2,3,5,5,7,7,14,10,11,11,19,13,15,16,41,17,26,19,29,22,23,23,55,26,

%T 27,36,39,29,40,31,122,34,35,36,74,37,39,40,83,41,54,43,59,56,47,47,

%U 163,50,62,52,69,53,100,56,111,58,59,59,112,61,63,76,365,66,82,67,89,70,84,71

%N If n = Product p(k)^e(k) then a(n) = (Product (p(k)+1)^e(k) + Product (p(k)-1)^e(k))/2, a(1) = 1.

%C a(n) = n iff n is 1 or a prime;

%C a(n) = n+1 iff n is a biprime, i.e., n = p*q, p <= q primes;

%C a(n) = n+(p+q+r) iff n is a triprime, i.e., n = p*q*r, p <= q <= r primes;

%C a(n) = n + (p*q + p*r + p*s + q*r + q*s + r*s) + 1 iff n is a quadprime, i.e., n = p*q*r*s, p <= q <= r <= s primes;

%C ...

%H Daniel Forgues, <a href="/A168065/b168065.txt">Table of n, a(n) for n = 1..100000</a>

%F a(n) = (A003959(n) + A003958(n))/2.

%o (PARI) a(n) = {f = factor(n); return ((prod(k=1, #f~, (f[k, 1]+1)^f[k, 2]) + prod(k=1, #f~, (f[k, 1]-1)^f[k, 2]))/2);} \\ _Michel Marcus_, Jun 13 2013

%Y Cf. A003958, A003959, A168066.

%K nonn

%O 1,2

%A _Daniel Forgues_, Nov 18 2009

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