A003959 - OEIS

A003959

If n = Product p(k)^e(k) then a(n) = Product (p(k)+1)^e(k), a(1) = 1.

1, 3, 4, 9, 6, 12, 8, 27, 16, 18, 12, 36, 14, 24, 24, 81, 18, 48, 20, 54, 32, 36, 24, 108, 36, 42, 64, 72, 30, 72, 32, 243, 48, 54, 48, 144, 38, 60, 56, 162, 42, 96, 44, 108, 96, 72, 48, 324, 64, 108, 72, 126, 54, 192, 72, 216, 80, 90, 60, 216, 62, 96, 128, 729, 84, 144, 68

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OFFSET

1,2

COMMENTS

Completely multiplicative.

Sum of divisors of n with multiplicity. If n = p^m, the number of ways to make p^k as a divisor of n is C(m,k); and sum(C(m,k)*p^k) = (p+1)^k. The rest follows because the function is multiplicative. - Franklin T. Adams-Watters, Jan 25 2010

LINKS

Daniel Forgues, Table of n, a(n) for n = 1..100000 (first 1000 terms from T. D. Noe)

Index to divisibility sequences

FORMULA

Multiplicative with a(p^e) = (p+1)^e. - David W. Wilson, Aug 01 2001

Sum_{n>0} a(n)/n^s = Product_{p prime} 1/(1-p^(-s)-p^(1-s)) (conjectured). - Ralf Stephan, Jul 07 2013

This follows from the absolute convergence of the sum (compare with a(n) = n^2) and the Euler product for completely multiplicative functions. Convergence occurs for at least Re(s)>3. - Thomas Anton, Jul 15 2021

Sum_{k=1..n} a(k) ~ c * n^2, where c = A065488/2 = 1/(2*A005596) = 1.3370563627850107544802059152227440187511993141988459926... - Vaclav Kotesovec, Jul 17 2021

From Thomas Scheuerle, Jul 19 2021: (Start)

a(n) = gcd(A166642(n), A166643(n)).

a(n) = A166642(n)/A061142(n).

a(n) = A166643(n)/A165824(n).

a(n) = A166644(n)/A165825(n).

a(n) = A166645(n)/A165826(n).

a(n) = A166646(n)/A165827(n).

a(n) = A166647(n)/A165828(n).

a(n) = A166649(n)/A165830(n).

a(n) = A166650(n)/A165831(n).

a(n) = A167351(n)/A166590(n). (End)

Dirichlet g.f.: zeta(s-1) * Product_{primes p} (1 + 1/(p^s - p - 1)). - Vaclav Kotesovec, Aug 22 2021