cp's OEIS Frontend

A268128 a(n) = (A000123(n) - A001316)/2.

%I A268128 #20 Dec 18 2023 08:26:13
%S A268128 0,0,1,1,4,5,8,9,17,21,28,33,45,53,66,75,100,117,140,161,193,221,258,
%T A268128 291,344,389,446,499,573,639,722,797,913,1013,1132,1249,1393,1533,
%U A268128 1698,1859,2060,2253,2478,2699,2965,3223,3522,3813,4173,4517,4910,5299,5753
%N A268128 a(n) = (A000123(n) - A001316)/2.
%H A268128 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 A268128 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 A268128 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 A268128 Let b(0) = 1 and b(n) = b(n-1) + b(floor(n/2)) and let c(n) = Product_{i=0..k}(n_i+1) where n = Sum_{i=0..k}n_i*2^i is the binary representation of n. Then a(n) = (1/2)*(b(n) - c(n)).
%t A268128 b[0] = 1; b[n_] := b[n] = b[Floor[n/2]] + b[n - 1];
%t A268128 c[n_] := Sum[Mod[Binomial[n, k], 2], {k, 0, n}];
%t A268128 a[n_] := (b[n] - c[n])/2;
%t A268128 Table[a[n], {n, 0, 52}] (* _Jean-François Alcover_, Dec 12 2018 *)
%o A268128 (Sage)
%o A268128 def b(n):
%o A268128     A=[1]
%o A268128     for i in [1..n]:
%o A268128         A.append(A[i-1] + A[floor(i/2)])
%o A268128     return A[n]
%o A268128 [(b(n)-prod(x+1 for x in n.digits(2)))/2 for n in [0..60]]
%Y A268128 Cf. A005704, A006047, A268127.
%K A268128 nonn
%O A268128 0,5
%A A268128 _Tom Edgar_, Jan 26 2016