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A306921 Number of ways of breaking the binary expansion of n into consecutive blocks with no leading zeros.

%I A306921 #30 May 09 2019 10:10:17
%S A306921 1,1,2,2,3,3,4,4,4,4,6,6,6,6,8,8,5,5,8,8,9,9,12,12,8,8,12,12,12,12,16,
%T A306921 16,6,6,10,10,12,12,16,16,12,12,18,18,18,18,24,24,10,10,16,16,18,18,
%U A306921 24,24,16,16,24,24,24,24,32,32,7,7,12,12,15,15,20,20,16
%N A306921 Number of ways of breaking the binary expansion of n into consecutive blocks with no leading zeros.
%C A306921 The number 0 is not considered to have a leading zero.
%C A306921 a(n) >= 2^(A000120(n) - 1).
%C A306921 a(2^n - 1) = 2^(n-1) for n > 0.
%C A306921 a(2^n) = n+1.
%C A306921 Conjecture: n appears A067824(n) times for n > 1.
%H A306921 Peter Kagey, <a href="/A306921/b306921.txt">Table of n, a(n) for n = 0..8192</a>
%F A306921 From _Charlie Neder_, May 08 2019: (Start)
%F A306921 If n = k*2^e + {0,1} with k odd and e > 0, then a(n) = a(k)*(e+1).
%F A306921 Proof: Each partition of n is uniquely determined by a partition of k (call it K) and a choice of some number, from 0 to e, of trailing digits to append to the final part in K, since any remaining digits must appear as singletons. The conjecture follows, since each ordered factorization of a number m produces two numbers n such that a(n) = m, one of each parity, and A067824(n) = 2*A074206(n).
%F A306921 Corollary: For n >= 1, a(2n) = a(2n+1) = Product{k+1 | k in row n of A066099}. (End)
%e A306921 For n = 13 the a(13) = 6 partitions are [1101], [1, 101], [110, 1], [1, 10, 1], [11, 0, 1], and [1, 1, 0, 1].
%e A306921 Notice that [1, 1, 01] and [11, 01] are not valid partitions because the last part has a leading zero.
%Y A306921 Cf. A000120, A067824, A074206, A066099.
%Y A306921 A321318 gives number of distinct sums of such partitions.
%K A306921 nonn,base
%O A306921 0,3
%A A306921 _Peter Kagey_, Mar 16 2019