A384789 The number of ways in which the n-th cubefull number can be expressed as b^3 * c^4 * d^5, with b, c and d >= 1.
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 4, 1, 1
Offset: 1
Examples
a(12) = 2 since A036966(12) = 256 = 2^8 has 2 representations as b^3*c^4*d^5: 2^3 * 2^5 (b = d = 2, c = 1) and 4^4 (b = d = 1, c = 4). a(38) = 3 since A036966(38) = 4096 = 2^12 has 3 representations as b^3*c^4*d^5: 2^3 * 2^4 * 2^5 (b = c = d = 2), 8^4 (b = d = 1, c = 8) and 16^3 (b = 16, c = d = 1).
Links
- Amiram Eldar, Table of n, a(n) for n = 1..10000
Programs
-
Mathematica
f[p_, e_] := Floor[(1+(-1)^e)*(-1)^Floor[e/2]/8 + (e^2 + 12*e + 90)/120]; s[1] = 1; s[n_] := Times @@ f @@@ FactorInteger[n]; Select[Array[s, 10000], # > 0 &]
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PARI
f(e) = floor((1+(-1)^e)*(-1)^floor(e/2)/8 + (e^2 + 12*e + 90)/120); list(kmax) = {my(e); for(k = 1, kmax, e = factor(k)[, 2]; if(k == 1 || vecmin(e) > 2, print1(vecprod(apply(x -> f(x), e)), ", ")));}
Formula
a(n) >= 1.
Comments