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

%I A270774 #18 Dec 18 2023 08:26:02
%S A270774 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,
%T A270774 20,23,24,25,26,27,32,33,34,35,36,43,44,45,46,47,56,57,58,59,60,73,76,
%U A270774 79,82,85,91,94,97,100,103,112,115,118,121
%N A270774 a(n) = (A005706(n) - A194459(n))/5.
%C A270774 A combinatorial interpretation is given in the Edgar link.
%H A270774 G. E. Andrews, A. S. Fraenkel, and J. A. Sellers, <a href="http://www.jstor.org/stable/10.4169/amer.math.monthly.122.9.880">Characterizing the number of m-ary partitions modulo m</a>, The American Mathematical Monthly, Vol. 122, No. 9 (November 2015), pp. 880-885.
%H A270774 G. E. Andrews, A. S. Fraenkel, and J. A. Sellers, <a href="https://georgeandrews1.github.io/pdf/300.pdf">Characterizing the number of m-ary partitions modulo m</a>.
%H A270774 Tom Edgar, <a href="http://arxiv.org/abs/1603.00085">The distribution of the number of parts of m-ary partitions modulo m.</a>, arXiv:1603.00085 [math.CO], 2016.
%F A270774 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)).
%t A270774 b[0] = 1; b[n_] := b[n] = b[n-1] + b[Floor[n/5]];
%t A270774 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]]];
%t A270774 a[n_] := (b[n] - c[n])/5;
%t A270774 Table[a[n], {n, 0, 63}] (* _Jean-François Alcover_, Feb 15 2019 *)
%o A270774 (Sage)
%o A270774 def b(n):
%o A270774     A=[1]
%o A270774     for i in [1..n]:
%o A270774         A.append(A[i-1] + A[i//5])
%o A270774     return A[n]
%o A270774 print([(b(n)-prod(x+1 for x in n.digits(5)))/5 for n in [0..63]])
%Y A270774 Cf. A005706, A194459, A268127, A268128, A268443.
%K A270774 nonn
%O A270774 0,11
%A A270774 _Tom Edgar_, Mar 22 2016