A376888 The sum of divisors of n that are products of factors of the form p^(k!) with multiplicities not larger than their multiplicity in n, where p is a prime and k >= 1, when the factorization of n is uniquely done using the factorial-base representation of the exponents in the prime factorization of n.
1, 3, 4, 5, 6, 12, 8, 15, 10, 18, 12, 20, 14, 24, 24, 21, 18, 30, 20, 30, 32, 36, 24, 60, 26, 42, 40, 40, 30, 72, 32, 63, 48, 54, 48, 50, 38, 60, 56, 90, 42, 96, 44, 60, 60, 72, 48, 84, 50, 78, 72, 70, 54, 120, 72, 120, 80, 90, 60, 120, 62, 96, 80, 65, 84, 144
Offset: 1
Examples
For n = 12 = 2^2 * 3^1, the representation of 2 in factorial base is 10, i.e., 2 = 2!, so 12 = (2^(2!))^1 * (3^(1!))^1 and a(12) is the sum of the 4 divisors 1 + 3 + 4 + 12 = 20.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
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Mathematica
ff[q_, s_] := (q^(s + 1) - 1)/(q - 1); f[p_, e_] := Module[{k = e, m = 2, r, s = {}}, While[{k, r} = QuotientRemainder[k, m]; k != 0 || r != 0, If[r > 0, AppendTo[s, {p^(m - 1)!, r}];]; m++]; Times @@ ff @@@ s]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] -
PARI
fdigits(n) = {my(k = n, m = 2, r, s = []); while([k, r] = divrem(k, m); k != 0 || r != 0, s = concat(s, r); m++); s;} a(n) = {my(f = factor(n), p = f[, 1], e = f[, 2], d); prod(i = 1, #p, prod(j = 1, #d=fdigits(e[i]), (p[i]^(j!*(d[j]+1)) - 1)/(p[i]^j! - 1)));}
Formula
Multiplicative: if e = Sum_{k>=1} d_k * k! (factorial base representation), then a(p^e) = Product_{k>=1} (p^(k!*{d_k+1}) - 1)/(p^(k!) - 1).
Comments