OFFSET
0,6
COMMENTS
It is easy to prove that this sequence is exactly one-half of sequence A360309. The connection to the number of binary n-sequences ending in 1 with exactly one more occurrence of 11 pairs than 10 pairs appears new and is related to a coin-tossing problem posed by Daniel Litt. In particular, the cumulative sum a(0)+…+a(n-1) gives the number of all binary sequences of length n with strictly more 10 pairs than 11 pairs minus the number of all binary sequences of length n with strictly more 11 pairs than 10 pairs. These results are demonstrated in the link below.
LINKS
Alois P. Heinz, Table of n, a(n) for n = 0..3330
Bruce Levin, Note on a coin-tossing problem posed by Daniel Litt, arXiv:2409.13087 [math.CO], 2024.
FORMULA
a(n) = Sum_{k=1..floor((n+1)/3)} binomial((2*k)-1,k) * binomial(n-2*k,k-1).
a(n) = A360309(n+1)/2.
EXAMPLE
There are a(5)=4 binary 5-sequences ending in 1 with one more occurrence of 11 than 10. They are 11101, 11011, and 10111 (each with two occurrences of 11 and one occurrence of 10); and 00011 (with one occurrence of 11 and zero occurrences of 10).
MAPLE
a:= proc(n) option remember; `if`(n<4, iquo(n, 2), (2*n*a(n-1)
-(n-1)*a(n-2)+(4*n-2)*a(n-3)-4*(n-2)*a(n-4))/(n+1))
end:
seq(a(n), n=0..36); # Alois P. Heinz, Oct 08 2024
CROSSREFS
KEYWORD
nonn
AUTHOR
Bruce Levin, Oct 07 2024
STATUS
approved