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Number of ways of breaking the binary expansion of n into consecutive blocks with no leading zeros.
2

%I #30 May 09 2019 10:10:17

%S 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 16,6,6,10,10,12,12,16,16,12,12,18,18,18,18,24,24,10,10,16,16,18,18,

%U 24,24,16,16,24,24,24,24,32,32,7,7,12,12,15,15,20,20,16

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

%C The number 0 is not considered to have a leading zero.

%C a(n) >= 2^(A000120(n) - 1).

%C a(2^n - 1) = 2^(n-1) for n > 0.

%C a(2^n) = n+1.

%C Conjecture: n appears A067824(n) times for n > 1.

%H Peter Kagey, <a href="/A306921/b306921.txt">Table of n, a(n) for n = 0..8192</a>

%F From _Charlie Neder_, May 08 2019: (Start)

%F If n = k*2^e + {0,1} with k odd and e > 0, then a(n) = a(k)*(e+1).

%F 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 Corollary: For n >= 1, a(2n) = a(2n+1) = Product{k+1 | k in row n of A066099}. (End)

%e 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 Notice that [1, 1, 01] and [11, 01] are not valid partitions because the last part has a leading zero.

%Y Cf. A000120, A067824, A074206, A066099.

%Y A321318 gives number of distinct sums of such partitions.

%K nonn,base

%O 0,3

%A _Peter Kagey_, Mar 16 2019