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A003959
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If n = Product p(k)^e(k) then a(n) = Product (p(k)+1)^e(k), a(1) = 1.
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86
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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
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1,2
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COMMENTS
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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
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LINKS
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FORMULA
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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
Dirichlet g.f.: zeta(s-1) * Product_{primes p} (1 + 1/(p^s - p - 1)). - Vaclav Kotesovec, Aug 22 2021
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MAPLE
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a:= n-> mul((i[1]+1)^i[2], i=ifactors(n)[2]):
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MATHEMATICA
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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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PROG
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(PARI) a(n)=if(n<1, 0, direuler(p=2, n, 1/(1-X-p*X))[n]) /* Ralf Stephan */
(Haskell)
a003959 1 = 1
a003959 n = product $ map (+ 1) $ a027746_row n
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CROSSREFS
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Apart from initial terms, same as A064478.
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KEYWORD
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nonn,easy,nice,mult
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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