A279497 Number of pentagonal numbers dividing n.
1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 1, 2, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 3, 2, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 2, 1, 2, 2, 1, 1, 1, 2, 1, 1, 1, 1, 3, 1, 1, 1, 1, 2, 2, 1, 1, 1, 4, 1, 2, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 2, 1, 2, 1, 2, 1, 1, 2, 2, 1, 1, 1, 2, 1, 2, 1, 1, 3, 1, 1, 2, 1, 3, 1, 1, 1, 1, 2, 1, 2, 1, 1, 3
Offset: 1
Examples
a(12) = 2 because 12 has 6 divisors {1,2,3,4,6,12} among which 2 divisors {1,12} are pentagonal numbers.
Links
- Antti Karttunen, Table of n, a(n) for n = 1..101270
- Eric Weisstein's World of Mathematics, Pentagonal Number.
- Index to sequences related to polygonal numbers.
Programs
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Mathematica
Rest[CoefficientList[Series[Sum[x^(k (3k -1)/2)/(1 - x^(k (3k -1)/2)), {k, 120}], {x, 0, 120}], x]] Table[Count[Divisors[n],?(IntegerQ[(1+Sqrt[1+24#])/6]&)],{n,120}] (* _Harvey P. Dale, Jan 05 2022 *) -
PARI
a(n) = sumdiv(n, d, ispolygonal(d, 5)); \\ Michel Marcus, Jul 27 2022
Formula
G.f.: Sum_{k>=1} x^(k*(3*k-1)/2)/(1 - x^(k*(3*k-1)/2)).
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 3*log(3) - Pi/sqrt(3) = 1.482037... (A244641). - Amiram Eldar, Jan 02 2024
a(n) = Sum_{d|n} A255849(d). - Antti Karttunen, Jan 14 2025