A270774 a(n) = (A005706(n) - A194459(n))/5.

0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 3, 3, 3, 3, 3, 6, 6, 6, 6, 6, 10, 10, 10, 10, 10, 16, 17, 18, 19, 20, 23, 24, 25, 26, 27, 32, 33, 34, 35, 36, 43, 44, 45, 46, 47, 56, 57, 58, 59, 60, 73, 76, 79, 82, 85, 91, 94, 97, 100, 103, 112, 115, 118, 121

Offset: 0

Author

Tom Edgar, Mar 22 2016

Keywords

• nonn

Comments

A combinatorial interpretation is given in the Edgar link.

Links

• G. E. Andrews, A. S. Fraenkel, and J. A. Sellers, Characterizing the number of m-ary partitions modulo m, The American Mathematical Monthly, Vol. 122, No. 9 (November 2015), pp. 880-885.
• G. E. Andrews, A. S. Fraenkel, and J. A. Sellers, Characterizing the number of m-ary partitions modulo m.
• Tom Edgar, The distribution of the number of parts of m-ary partitions modulo m., arXiv:1603.00085 [math.CO], 2016.

Crossrefs

Cf. A005706, A194459, A268127, A268128, A268443.

Programs

Mathematica

b[0] = 1; b[n_] := b[n] = b[n-1] + b[Floor[n/5]];
c[n_] := If[OddQ[n], 2 Count[Table[Binomial[n, k], {k, 0, (n-1)/2}], c_ /; !Divisible[c, 5]], 2 Count[Table[Binomial[n, k], {k, 0, (n-2)/2}], c_ /; !Divisible[c, 5]] + Boole[!Divisible[Binomial[n, n/2], 5]]];
a[n_] := (b[n] - c[n])/5;
Table[a[n], {n, 0, 63}] (* Jean-François Alcover, Feb 15 2019 *)

Sage

def b(n):
    A=[1]
    for i in [1..n]:
        A.append(A[i-1] + A[i//5])
    return A[n]
print([(b(n)-prod(x+1 for x in n.digits(5)))/5 for n in [0..63]])

Formula

Let b(0) = 1 and b(n) = b(n-1) + b(floor(n/5)) and let c(n) = Product_{i=0..k}(n_i+1) where n = Sum_{i=0..k}n_i*5^i is the base 5 representation of n. Then a(n) = (1/5)*(b(n) - c(n)).