A371358 Number of binary strings of length n which have more 00 than 01 substrings.

0, 0, 1, 2, 4, 10, 21, 42, 89, 184, 371, 758, 1546, 3122, 6315, 12782, 25780, 51962, 104759, 210934, 424404, 853806, 1716759, 3450158, 6932169, 13924260, 27959805, 56130762, 112662414, 226080318, 453595341, 909925794, 1825052601, 3660020992, 7339006091

Offset: 0

Author

Robert P. P. McKone, Mar 19 2024

Keywords

• nonn

Examples

			a(4) = 4: 0000, 0001, 1000, 1100.
a(5) = 10: 00000, 00001, 00010, 00011, 00100, 01000, 10000, 10001, 11000, 11100.
		

Links

• Robert P. P. McKone, Table of n, a(n) for n = 0..1698

Crossrefs

Cf. A163493 (equal 00 and 01), A371564 (more 01 than 00), A090129 (equal 01 and 10), A182027 (equal 00 and 11), A370048 (one more 00 than 01).

Cf. A000079(n-2) (more 01 than 10, for n>=2).

Programs

Maple

b:= proc(n, l, t) option remember; `if`(n+t<1, 0, `if`(n=0, 1,
      add(b(n-1, i, t+`if`(l=0, (-1)^i, 0)), i=0..1)))
    end:
a:= n-> b(n, 2, 0):
seq(a(n), n=0..34);  # Alois P. Heinz, Mar 20 2024

Mathematica

tup[n_] := Tuples[{0, 1}, n];
cou[lst_List] := Count[lst, {0, 0}] > Count[lst, {0, 1}];
par[lst_List] := Partition[lst, 2, 1];
a[n_] := Map[cou, Map[par, tup[n]]] // Boole // Total;
Monitor[Table[a[n], {n, 0, 18}], {n, Table[a[m], {m, 0, n - 1}]}]

PARI

{ a371358(n) = 2^(n-1) - sum(k=0, n\3, binomial(2*k,k) * (2*binomial(n-2*k,n-3*k) - binomial(n-2*k-1,n-3*k))) / 2; } \\ Max Alekseyev, May 01 2024

Formula

a(n) = 2^n - A163493(n) - A371564(n).

a(n) = ((4*n^2-15*n+7)*a(n-1) -(5*n^2-22*n+14)*a(n-2) +2*(3*n^2-14*n+10)*a(n-3) -4*(3*n^2-16*n+18)*a(n-4) +8*(n-2)*(n-4)*a(n-5)) / (n*(n-3)) for n>=5. - Alois P. Heinz, Mar 20 2024

For n >= 2, a(n) = 2*a(n-1) + A163493(n-1) - A163493(n-2) - A370048(n-2). - Max Alekseyev, Apr 30 2024

a(n) = 2^(n-1) - (1/2) * Sum_{k=0..floor(n/3)} binomial(2*k,k) * (2*binomial(n-2*k,n-3*k) - binomial(n-2*k-1,n-3*k)). - Max Alekseyev, May 01 2024

G.f. 1/(1-2*x)/2 - (1+x)/(2*sqrt(1-2*x+x^2-4*x^3+4*x^4)). - Max Alekseyev, Apr 30 2024