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

0,2

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]

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)

Example

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)

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 *)

Crossrefs

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

Cf. A163493 (equal 00 and 01).

Sequence in context: A000786 A289080 A000694 * A018170 A113913 A002097

Adjacent sequences: A164134 A164135 A164136 * A164138 A164139 A164140

Keyword

nonn

Author

R. H. Hardin, Aug 11 2009

Status

approved