A343525 If n = Product (p_j^k_j) then a(n) = Product (2*p_j^k_j + 1), with a(1) = 1.
1, 5, 7, 9, 11, 35, 15, 17, 19, 55, 23, 63, 27, 75, 77, 33, 35, 95, 39, 99, 105, 115, 47, 119, 51, 135, 55, 135, 59, 385, 63, 65, 161, 175, 165, 171, 75, 195, 189, 187, 83, 525, 87, 207, 209, 235, 95, 231, 99, 255, 245, 243, 107, 275, 253, 255, 273, 295, 119, 693, 123, 315, 285, 129, 297, 805
Offset: 1
Links
- Alois P. Heinz, Table of n, a(n) for n = 1..20000
Programs
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Maple
a:= n-> mul(2*i[1]^i[2]+1, i=ifactors(n)[2]): seq(a(n), n=1..80); # Alois P. Heinz, Apr 18 2021
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Mathematica
a[1] = 1; a[n_] := Times @@ ((2 #[[1]]^#[[2]] + 1) & /@ FactorInteger[n]); Table[a[n], {n, 66}]
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PARI
a(n) = my(f=factor(n)); for (k=1, #f~, f[k,1] = 2*f[k,1]^f[k,2]+1; f[k,2]=1); factorback(f); \\ Michel Marcus, Apr 18 2021
Formula
a(n) = Sum_{d|n, gcd(d, n/d) = 1} d * usigma(n/d).
a(n) = Sum_{d|n, gcd(d, n/d) = 1} d * 2^omega(d).
Dirichlet g.f.: zeta(s-1) * zeta(s) * Product_{p prime} (1 + 1/p^(s-1) - 2/p^(2*s-1)). - Amiram Eldar, Jul 24 2024
Comments