[go: up one dir, main page]

login
A370048
Number of binary strings of length n in which the number of substrings 00 is one more than that of substrings 01.
3
0, 0, 1, 1, 2, 6, 10, 18, 40, 76, 141, 285, 558, 1066, 2097, 4121, 8000, 15660, 30763, 60171, 117918, 231690, 454816, 893208, 1756688, 3455580, 6799195, 13388587, 26375466, 51974798, 102470402, 202108730, 398756664, 787025260, 1553900235, 3068937675, 6062944710, 11981429394, 23683822694, 46828287038
OFFSET
0,5
FORMULA
For n >= 2, a(n) = Sum_{m=0..floor((n-1)/3)} binomial(2*m,m+1) * binomial(n-1-2*m,m) + binomial(2*m+1,m) * binomial(n-2-2*m,m).
For n >= 4, a(n) = ( (n-2)*(2*n-1)*(n^2-n-4)*a(n-1) - (n^2-5*n+2)*(n^2+n-4)*a(n-2) + 2*(n-3)*n^2*(2*n-3)*a(n-3) - 4*(n-3)*(n-1)^2*n*a(n-4) ) / (n-2)^2 / (n-1) / (n+2).
a(n) = 2*A371358(n+1) - A371358(n+2) + A163493(n+1) - A163493(n).
G.f. ((1-x^2-2*x^3)*(1-2*x+x^2-4*x^3+4*x^4)^(-1/2) - 1 - x)/x^2/2, which can be expressed in terms of g.f. C(x) = (1-sqrt(1-4*x))/x/2 for Catalan number (A000108) as x*((x+1)*C(x^3/(1-x))-1)/(1-x-2*x^3*C(x^3/(1-x))).
PROG
(PARI) { a370048(n) = (n > 1) * sum(m=0, (n-1)\3, binomial(2*m, m+1) * binomial(n-1-2*m, m) + binomial(2*m+1, m) * binomial(n-2-2*m, m) ); }
(Python)
from math import comb
def A370048(n): return 0 if n<2 else 1+sum((x:=comb((k:=m<<1), m+1)*comb(n-1-k, m))+x*(k+1)*(n-1-3*m)//(m*(n-1-k)) for m in range(1, (n+2)//3)) # Chai Wah Wu, May 01 2024
CROSSREFS
KEYWORD
nonn
AUTHOR
Max Alekseyev, Apr 30 2024
STATUS
approved