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%I #11 Apr 26 2022 13:01:41
%S 0,1,4,3,6,6,8,8,10,10,11,14,14,16,18,18,18,20
%N Let c(s) denote A109812(s). Suppose c(s) = 2^n - 1, and define m(n), p(n), r(n) by m(n) = c(s-1)/2^n, p(n) = c(s+1)/2^n, r(n) = max(m(n), p(n)); sequence gives m(n).
%C The sequences m, p, r are well-defined since every number appears in A109812, and if A109812(s) = 2^n - 1, then by definition both A109812(s-1) and A109812(s+1) must be multiples of 2^n.
%C The sequences m, p, r are discussed in A352920.
%C (We assume A109812(0)=0 in order to get m(1)=0.)
%Y Cf. A109812, A113233, A352203, A352204, A352336, A352359, A352917-A352923.
%K nonn,more
%O 1,3
%A _David Broadhurst_, Aug 17 2022 (entry created by _N. J. A. Sloane_, Apr 24 2022)