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A163493 Number of binary strings of length n which have the same number of 00 and 01 substrings. 26
1, 2, 2, 3, 6, 9, 15, 30, 54, 97, 189, 360, 675, 1304, 2522, 4835, 9358, 18193, 35269, 68568, 133737, 260802, 509132, 995801, 1948931, 3816904, 7483636, 14683721, 28827798, 56637969, 111347879, 219019294, 431043814, 848764585, 1672056525, 3295390800, 6497536449 (list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
COMMENTS
A variation of problem 11424 in the American Mathematical Monthly. Terms were brute-force calculated using Maple 10.
Proposed Problem 11610 in the Dec 2011 A.M.M.
From Gus Wiseman, Jul 27 2021: (Start)
Also the antidiagonal sums of the matrices counting integer compositions by length and alternating sum (A345197). So a(n) is the number of integer compositions of n + 1 of length (n - s + 3)/2, where s is the alternating sum of the composition. For example, the a(0) = 1 through a(6) = 7 compositions are:
(1) (2) (3) (4) (5) (6) (7)
(11) (21) (31) (41) (51) (61)
(121) (122) (123) (124)
(221) (222) (223)
(1112) (321) (322)
(1211) (1122) (421)
(1221) (1132)
(2112) (1231)
(2211) (2122)
(2221)
(3112)
(3211)
(11131)
(12121)
(13111)
For a bijection with the main (binary string) interpretation, take the run-lengths of each binary string of length n + 1 that satisfies the condition and starts with 1.
(End)
LINKS
Alois P. Heinz, Table of n, a(n) for n = 0..3328 (first 501 terms from R. H. Hardin)
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]
R. Stanley, Problem 11610, Amer. Math. Monthly, 118 (2011), 937; 120 (2013), 943-944.
FORMULA
G.f.: 1/2/(1-x) + (1+2*x)/2/sqrt((1-x)*(1-2*x)*(1+x+2*x^2)). - Richard Stanley, corrected Apr 29 2011
G.f.: (1 + sqrt( 1 + 4*x / ((1 - x) * (1 - 2*x) * (1 + x + 2*x^2)))) / (2*(1 - x)). - Michael Somos, Jan 30 2012
a(n) = sum( binomial(2*k-1, k)*binomial(n-2*k,k) + binomial(2*k, k)*binomial(n-2*k-1, k), k=0..floor(n/3)). - Joel B. Lewis, May 21 2011
Conjecture: -n*a(n) +(2+n)*a(n-1) +(3n-12)*a(n-2) +(12-n)*a(n-3) +(2n-18)*a(n-4)+(56-12n)*a(n-5) +(8n-40)*a(n-6)=0. - R. J. Mathar, Nov 28 2011
G.f. y = A(x) satisfies x = (1 - x) * (1 - 2*x) * (1 + x + 2*x^2) * y * (y * (1 - x) - 1). - Michael Somos, Jan 30 2012
Sequence a(n) satisfies 0 = a(n) * (n^2-2*n) + a(n-1) * (-3*n^2+8*n-2) + a(n-2) * (3*n^2-10*n+2) + a(n-3) * (-5*n^2+18*n-6) + a(n-4) * (8*n^2-34*n+22) + a(n-5) * (-4*n^2+20*n-16) except if n=1 or n=2. - Michael Somos, Jan 30 2012
EXAMPLE
1 + 2*x + 2*x^2 + 3*x^3 + 6*x^4 + 9*x^5 + 15*x^6 + 30*x^7 + 54*x^8 + 97*x^9 + ...
From Gus Wiseman, Jul 27 2021: (Start)
The a(0) = 1 though a(6) = 15 binary strings:
() (0) (1,0) (0,0,1) (0,0,1,0) (0,0,1,1,0) (0,0,0,1,0,1)
(1) (1,1) (1,1,0) (0,0,1,1) (0,0,1,1,1) (0,0,1,0,0,1)
(1,1,1) (0,1,0,0) (0,1,1,0,0) (0,0,1,1,1,0)
(1,0,0,1) (1,0,0,1,0) (0,0,1,1,1,1)
(1,1,1,0) (1,0,0,1,1) (0,1,0,0,0,1)
(1,1,1,1) (1,0,1,0,0) (0,1,1,1,0,0)
(1,1,0,0,1) (1,0,0,1,1,0)
(1,1,1,1,0) (1,0,0,1,1,1)
(1,1,1,1,1) (1,0,1,1,0,0)
(1,1,0,0,1,0)
(1,1,0,0,1,1)
(1,1,0,1,0,0)
(1,1,1,0,0,1)
(1,1,1,1,1,0)
(1,1,1,1,1,1)
(End)
MAPLE
with(combinat): count := proc(n) local S, matches, A, k, i; S := subsets(\{seq(i, i=1..n)\}): matches := 0: while not S[finished] do A := S[nextvalue](): k := 0: for i from 1 to n-1 do: if not (i in A) and not (i+1 in A) then k := k + 1: fi: if not (i in A) and (i+1 in A) then k := k - 1: fi: od: if (k = 0) then matches := matches + 1: fi: end do; return(matches); end proc:
# second Maple program:
b:= proc(n, l, t) option remember; `if`(n-abs(t)<0, 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, 1, 0):
seq(a(n), n=0..36); # Alois P. Heinz, Mar 20 2024
MATHEMATICA
a[0] = 1; a[n_] := Sum[Binomial[2*k - 1, k]*Binomial[n - 2*k, k] + Binomial[2*k, k]*Binomial[n - 2*k - 1, k], {k, 0, n/3}];
Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Nov 28 2017, after Joel B. Lewis *)
Table[Length[Select[Tuples[{0, 1}, n], Count[Partition[#, 2, 1], {0, 0}]==Count[Partition[#, 2, 1], {0, 1}]&]], {n, 0, 10}] (* Gus Wiseman, Jul 27 2021 *)
CROSSREFS
Antidiagonal sums of the matrices A345197.
Row sums of A345907.
Taking diagonal instead of antidiagonal sums gives A345908.
A011782 counts compositions (or binary strings).
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A316524 gives the alternating sum of prime indices (reverse: A344616).
Compositions of n, 2n, or 2n+1 with alternating/reverse-alternating sum k:
- k = 0: counted by A088218, ranked by A344619/A344619.
- k = 1: counted by A000984, ranked by A345909/A345911.
- k = -1: counted by A001791, ranked by A345910/A345912.
- k = 2: counted by A088218, ranked by A345925/A345922.
- k = -2: counted by A002054, ranked by A345924/A345923.
- k >= 0: counted by A116406, ranked by A345913/A345914.
- k <= 0: counted by A058622(n-1), ranked by A345915/A345916.
- k > 0: counted by A027306, ranked by A345917/A345918.
- k < 0: counted by A294175, ranked by A345919/A345920.
- k != 0: counted by A058622, ranked by A345921/A345921.
- k even: counted by A081294, ranked by A053754/A053754.
- k odd: counted by A000302, ranked by A053738/A053738.
Sequence in context: A079069 A019465 A077074 * A321405 A054200 A137216
KEYWORD
nonn,changed
AUTHOR
STATUS
approved

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Last modified March 28 22:04 EDT 2024. Contains 371254 sequences. (Running on oeis4.)