A376884 - OEIS

%I #42 Jan 07 2025 18:00:12

%S 0,0,1,1,1,4,7,10,23,46,79,157,315,588,1137,2249,4337,8402,16495,

%T 32179,62707,122916,240837,471456,925061,1816610,3566865,7010347,

%U 13789477,27130956,53409503,105205514,207309743,408672454,805989367,1590166915,3138371715

%N Number of binary n-sequences ending in 1 with exactly one more occurrence of 11 than 10.

%C 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 Levin link below, where more formulas are given for the number of binary n-sequences ending in 1 (or 0) with exactly s more occurrences of 11 pairs than 10 pairs for any integer s.

%H Alois P. Heinz, <a href="/A376884/b376884.txt">Table of n, a(n) for n = 0..3330</a>

%H Bruce Levin, <a href="https://arxiv.org/abs/2409.13087">Note on a coin-tossing problem posed by Daniel Litt</a>, arXiv:2409.13087 [math.CO], 2024.

%F a(n) = Sum_{k=1..floor((n+1)/3)} binomial((2*k)-1,k) * binomial(n-2*k,k-1).

%F a(n) = A360309(n+1)/2.

%e 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).

%p a:= proc(n) option remember; `if`(n<4, iquo(n, 2), (2*n*a(n-1)

%p       -(n-1)*a(n-2)+(4*n-2)*a(n-3)-4*(n-2)*a(n-4))/(n+1))

%p     end:

%p seq(a(n), n=0..36);  # _Alois P. Heinz_, Oct 08 2024

%t a[n_] := Sum[Binomial[n - 2 k, n + 1 - 3 k]*Binomial[2 k, k]/2, {k, 0, Floor[(n + 1)/3]}];

%t Table[a[n], {n, 0, 36}] (* _Robert P. P. McKone_, Dec 28 2024 *)

%Y One-half of A360309(n+1).

%Y Cf. A217615.

%K nonn

%O 0,6

%A _Bruce Levin_, Oct 07 2024