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%I #6 Apr 18 2020 22:14:15
%S 0,0,2,0,2,2,0,2,4,2,2,4,2,2,8,2,4,6,2,6,6,4,8,6,4,6,10,4,10,10,0,10,
%T 12,6,10,10,6,8,16,8,8,12,10,10,14,12,8,14,12,10,18,10,14,16,8,14,20,
%U 14,10,20,10,10,32,10,12,22,10,18,22,16,16,20,16,16
%N Number of solutions to n = i+j, 0 <= i,j <= n for which A010060(i)=A010060(j)=0, i != j.
%C This is the same as the number of solutions to n = i+j, 0 <= i,j <= n for which A010060(i)=A010060(j)=1, i != j.
%H J. Lambek and L. Moser, <a href="https://doi.org/10.4153/CMB-1959-013-x">On some two way classifications of integers</a>, Can. Math. Bull 2 (1959) 85-89.
%F We have
%F a(4n+1) = -a(2n)+a(2n+1)+a(4n)
%F a(4n+2) = a(2n)+a(2n+1)
%F a(8n) = 3a(2n)+a(2n+1)
%F a(8n+3) = 3a(2n)+a(2n+1)-a(4n)+a(4n+3)
%F a(8n+4) = a(2n+1)+a(4n)
%F a(8n+7) = 4a(2n+1),
%F which provides a fast algorithm to compute a(n).
%e For n = 19 the only solutions are (i,j) in {(9,10), (10,9)}, so a(19) = 2.
%Y Cf. A010060.
%K nonn
%O 1,3
%A _Jeffrey Shallit_, Apr 18 2020
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