



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
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OFFSET

0,11


COMMENTS

A combinatorial interpretation is given in the Edgar link.


LINKS

Table of n, a(n) for n=0..63.
G. E. Andrews, A. S. Fraenkel, and J. A. Sellers, Characterizing the number of mary partitions modulo m, The American Mathematical Monthly, Vol. 122, No. 9 (November 2015), pp. 880885.
G. E. Andrews, A. S. Fraenkel, and J. A. Sellers, Characterizing the number of mary partitions modulo m.
Tom Edgar, The distribution of the number of parts of mary partitions modulo m., arXiv:1603.00085 [math.CO], 2016.


FORMULA

Let b(0) = 1 and b(n) = b(n1) + 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)).


MATHEMATICA

b[0] = 1; b[n_] := b[n] = b[n1] + b[Floor[n/5]];
c[n_] := If[OddQ[n], 2 Count[Table[Binomial[n, k], {k, 0, (n1)/2}], c_ /; !Divisible[c, 5]], 2 Count[Table[Binomial[n, k], {k, 0, (n2)/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}] (* JeanFrançois Alcover, Feb 15 2019 *)


PROG

(Sage)
def b(n):
A=[1]
for i in [1..n]:
A.append(A[i1] + A[i//5])
return A[n]
print([(b(n)prod(x+1 for x in n.digits(5)))/5 for n in [0..63]])


CROSSREFS

Cf. A005706, A194459, A268127, A268128, A268443.
Sequence in context: A105591 A130497 A178154 * A263144 A126715 A158805
Adjacent sequences: A270771 A270772 A270773 * A270775 A270776 A270777


KEYWORD

nonn


AUTHOR

Tom Edgar, Mar 22 2016


STATUS

approved



