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A306921 Number of ways of breaking the binary expansion of n into consecutive blocks with no leading zeros. 2
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, 16, 6, 6, 10, 10, 12, 12, 16, 16, 12, 12, 18, 18, 18, 18, 24, 24, 10, 10, 16, 16, 18, 18, 24, 24, 16, 16, 24, 24, 24, 24, 32, 32, 7, 7, 12, 12, 15, 15, 20, 20, 16 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,3

COMMENTS

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

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

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

a(2^n) = n+1.

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

LINKS

Peter Kagey, Table of n, a(n) for n = 0..8192

FORMULA

From Charlie Neder, May 08 2019: (Start)

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

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).

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

EXAMPLE

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].

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

CROSSREFS

Cf. A000120, A067824, A074206, A066099.

A321318 gives number of distinct sums of such partitions.

Sequence in context: A117695 A074794 A234309 * A048686 A278959 A090501

Adjacent sequences:  A306918 A306919 A306920 * A306922 A306923 A306924

KEYWORD

nonn,base

AUTHOR

Peter Kagey, Mar 16 2019

STATUS

approved

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Last modified April 9 10:32 EDT 2020. Contains 333348 sequences. (Running on oeis4.)