A348734 Numerator of Product((p+1)^e / ((p^e)+1)), when n = Product(p^e), with p primes, and e their exponents.
1, 1, 1, 9, 1, 1, 1, 3, 8, 1, 1, 9, 1, 1, 1, 81, 1, 8, 1, 9, 1, 1, 1, 3, 18, 1, 16, 9, 1, 1, 1, 81, 1, 1, 1, 72, 1, 1, 1, 3, 1, 1, 1, 9, 8, 1, 1, 81, 32, 18, 1, 9, 1, 16, 1, 3, 1, 1, 1, 9, 1, 1, 8, 729, 1, 1, 1, 9, 1, 1, 1, 24, 1, 1, 18, 9, 1, 1, 1, 81, 128, 1, 1, 9, 1, 1, 1, 3, 1, 8, 1, 9, 1, 1, 1, 81, 1, 32, 8
Offset: 1
Links
- Antti Karttunen, Table of n, a(n) for n = 1..21125
Programs
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Mathematica
f[p_, e_] := (p + 1)^e/(p^e + 1); a[1] = 1; a[n_] := Numerator[Times @@ f @@@ FactorInteger[n]]; Array[a, 100] (* Amiram Eldar, Nov 05 2021 *)
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PARI
A003959(n) = { my(f = factor(n)); for(i=1, #f~, f[i, 1]++); factorback(f); }; A034448(n) = { my(f = factor(n)); prod(k=1, #f~, 1+(f[k, 1]^f[k, 2])); }; \\ After code in A034448 A348734(n) = { my(u=A003959(n)); (u/gcd(u, A034448(n))); };
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PARI
A348734(n) = { my(f = factor(n)); numerator(prod(k=1, #f~, ((1+f[k, 1])^f[k, 2])/(1+(f[k, 1]^f[k, 2])))); };
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