A117109 Moebius transform of binomial(n+3, 4).

1, 4, 14, 30, 69, 107, 209, 295, 480, 641, 1000, 1209, 1819, 2166, 2976, 3546, 4844, 5379, 7314, 8110, 10402, 11645, 14949, 15890, 20405, 21927, 26910, 29055, 35959, 37108, 46375, 48484, 57890, 61196, 73536, 75027, 91389, 93951, 110096, 114260

Offset: 1

Author

Steve Butler, Apr 18 2006

Keywords

• nonn

Comments

Partial sums of a(n) give A015650(n).

Examples

			a(2)=4 because of the quadruples (1,1,1,1), (1,1,1,2), (1,1,2,2), (1,2,2,2).
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A007438, A015650, A117108.

Programs

Maple

b34:= unapply(expand(binomial(n+3,4)),n):
f:= proc(n) local k; uses numtheory;
add(b34(k)*mobius(n/k),k=divisors(n))
end proc:
map(f, [$1..100]); # Robert Israel, May 24 2019

Mathematica

a[n_] := Sum[Binomial[k+3, 4] MoebiusMu[n/k], {k, Divisors[n]}];
Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Feb 01 2023 *)

PARI

a(n) = sumdiv(n, k, binomial(k+3, 4)*moebius(n/k)); \\ Michel Marcus, Nov 04 2018

Formula

a(n) = |{(x,y,z,w) : 1 <= x <= y <= z <= w <= n, gcd(x,y,z,w,n) = 1}|.

G.f.: Sum_{k>=1} mu(k) * x^k / (1 - x^k)^5. - Ilya Gutkovskiy, Feb 13 2020