A163407 -id:A163407 - cp's OEIS frontend

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

Offset: 1

Marc LeBrun

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

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

Apart from initial terms, same as A064478.
Cf. A003958, A027746, A036987, A061142, A163407.
Cf. A063441, A165824, A165825, A165826, A165827.
Cf. A165828, A165829, A165831, A167350, A166590.
Cf. A166642, A166643, A166644, A166645, A166646.
Cf. A166647, A166649, A166650, A166586, A165830.
Cf. A167351, A168065, A168066.

a003959 1 = 1
a003959 n = product $ map (+ 1) $ a027746_row n
-- Reinhard Zumkeller, Apr 09 2012

a:= n-> mul((i[1]+1)^i[2], i=ifactors(n)[2]):
seq(a(n), n=1..80);  # Alois P. Heinz, Sep 13 2017

a[1] = 1; a[n_] := (fi = FactorInteger[n]; Times @@ ((fi[[All, 1]]+1)^fi[[All, 2]])); a /@ Range[67] (* Jean-François Alcover, Apr 22 2011 *)

a(n)=if(n<1,0,direuler(p=2,n,1/(1-X-p*X))[n]) /* Ralf Stephan */

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

Definition reedited (with formula) by Daniel Forgues, Nov 17 2009

A180411 Sum of the semiprime divisors (with repetition) of the n-th number with two or more distinct semiprime divisors.

Original entry on oeis.org

16, 21, 24, 30, 32, 31, 37, 42, 41, 48, 39, 48, 45, 56, 45, 54, 51, 51, 61, 72, 59, 57, 55, 80, 71, 64, 65, 78, 61, 96, 70, 77, 75, 69, 91, 90, 71, 67, 87, 80, 101, 120, 87, 75, 128, 77, 101, 93, 72, 114, 121, 87, 81, 91, 152, 81, 126, 111, 113, 107, 90, 78, 168, 103, 93, 129, 123, 176

Offset: 1

Jonathan Vos Post, Sep 02 2010

nonn

This is to A164865 [Sum of the distinct semiprime divisors of the n-th number with two or more distinct semiprime divisors], as bigomega [A001222, Number of prime divisors of n (counted with multiplicity)] is to omega [A001221, Number of distinct primes dividing n].
The sum of semiprime divisors (with multiplicity) of all k such that A086971(k) > 1.
This is to A001414 [Integer log of n: sum of primes dividing n (with repetition)], as semiprimes A001358 are to primes A000040.

			a(1) = 16 because the first number (greater than 1) such that the sum of numbers of prime factors with and without repetitions does not equal the number of divisors, is a(2) = 12 = (2^2)*3 whose semiprime factors are (2^2 = 4) once and (2*3) with multiplicity two hence (4*1)*1 + (3*3)*2 = 4 + 12 = 16.
a(6) = 31 because 30 = 2*3*5 has multiplicity one semiprime factors (2*3), (2*5), (3*5), which sum to 6+10+15 = 31.

Cf. A001221, A001222, A001358, A086971, A164865.

a(n) = A163407(A102467(n+1)).

Formula, edits, and more terms from Charles R Greathouse IV, Sep 03 2010

cp's OEIS Frontend

A003959 If n = Product p(k)^e(k) then a(n) = Product (p(k)+1)^e(k), a(1) = 1.

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