OFFSET
0,3
COMMENTS
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
KEYWORD
nonn,base
AUTHOR
Peter Kagey, Mar 16 2019
STATUS
approved