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
Links
- Daniel Forgues, Table of n, a(n) for n = 1..100000 (first 1000 terms from T. D. Noe)
- Index to divisibility sequences
Crossrefs
Apart from initial terms, same as A064478.
Programs
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Haskell
a003959 1 = 1 a003959 n = product $ map (+ 1) $ a027746_row n -- Reinhard Zumkeller, Apr 09 2012
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Maple
a:= n-> mul((i[1]+1)^i[2], i=ifactors(n)[2]): seq(a(n), n=1..80); # Alois P. Heinz, Sep 13 2017
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Mathematica
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 *)
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PARI
a(n)=if(n<1,0,direuler(p=2,n,1/(1-X-p*X))[n]) /* Ralf Stephan */
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)
Dirichlet g.f.: zeta(s-1) * Product_{primes p} (1 + 1/(p^s - p - 1)). - Vaclav Kotesovec, Aug 22 2021
Extensions
Definition reedited (with formula) by Daniel Forgues, Nov 17 2009
Comments