A164137 Number of binary strings of length n with equal numbers of 000 and 001 substrings.

1, 2, 4, 6, 11, 19, 35, 61, 111, 200, 369, 676, 1256, 2337, 4392, 8273, 15686, 29837, 57038, 109362, 210448, 406029, 785573, 1523217, 2959853, 5761671, 11234619, 21937768, 42894822, 83969696, 164552423, 322773812, 633679446, 1245032098, 2447951456, 4816241573

Offset: 0

Author

R. H. Hardin, Aug 11 2009

Keywords

• nonn

Examples

			From _Robert P. P. McKone_, Apr 03 2024: (Start)
a(3) = 6: 010, 011, 100, 101, 110, 111.
a(4) = 11: 0001, 0100, 0101, 0110, 0111, 1010, 1011, 1100, 1101, 1110, 1111.
a(5) = 19: 00010, 00011, 01010, 01011, 01100, 01101, 01110, 01111, 10001, 10100, 10101, 10110, 10111, 11010, 11011, 11100, 11101, 11110, 11111.
(End)
		

Links

• R. H. Hardin, Table of n, a(n) for n=0..500
• Shalosh B. Ekhad and Doron Zeilberger, Automatic Solution of Richard Stanley's Amer. Math. Monthly Problem #11610 and ANY Problem of That Type, arXiv preprint arXiv:1112.6207 [math.CO], 2011. See subpages for rigorous derivations of g.f., recurrence, asymptotics for this sequence. [From _N. J. A. Sloane_, Apr 07 2012]

Crossrefs

Cf. A371662 (more 000 than 001), A371682 (more 001 than 000).

Cf. A163493 (equal 00 and 01).

Programs

Mathematica

tup[n_] := Tuples[{0, 1}, n];
cou[lst_List] := Count[lst, {0, 0, 0}] == Count[lst, {0, 0, 1}];
par[lst_List] := Partition[lst, 3, 1];
a[n_] := 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}]}] (* Robert P. P. McKone, Apr 03 2024 *)

Formula

From Robert P. P. McKone, Apr 03 2024: (Start)

a(n) = 2^n - A371662(n) - A371682(n).

Conjecture: a(n) = ((8*n-72)*a(n-10) + (20*n-160)*a(n-9) + (6*n-26)*a(n-8) + (46-5*n)*a(n-7) - 16*a(n-6) + (56-11*n)*a(n-5) + (12-n)*a(n-4) + (n-18)*a(n-3) + n*a(n-2) + 2*n*a(n-1))/n for n>=10.

(End)