This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A163493 #59 May 02 2024 10:06:23
%S A163493 1,2,2,3,6,9,15,30,54,97,189,360,675,1304,2522,4835,9358,18193,35269,
%T A163493 68568,133737,260802,509132,995801,1948931,3816904,7483636,14683721,
%U A163493 28827798,56637969,111347879,219019294,431043814,848764585,1672056525,3295390800,6497536449
%N A163493 Number of binary strings of length n which have the same number of 00 and 01 substrings.
%C A163493 A variation of problem 11424 in the American Mathematical Monthly. Terms were brute-force calculated using Maple 10.
%C A163493 Proposed Problem 11610 in the Dec 2011 A.M.M.
%C A163493 From _Gus Wiseman_, Jul 27 2021: (Start)
%C A163493 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:
%C A163493 (1) (2) (3) (4) (5) (6) (7)
%C A163493 (11) (21) (31) (41) (51) (61)
%C A163493 (121) (122) (123) (124)
%C A163493 (221) (222) (223)
%C A163493 (1112) (321) (322)
%C A163493 (1211) (1122) (421)
%C A163493 (1221) (1132)
%C A163493 (2112) (1231)
%C A163493 (2211) (2122)
%C A163493 (2221)
%C A163493 (3112)
%C A163493 (3211)
%C A163493 (11131)
%C A163493 (12121)
%C A163493 (13111)
%C A163493 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.
%C A163493 (End)
%H A163493 Alois P. Heinz, <a href="/A163493/b163493.txt">Table of n, a(n) for n = 0..3328</a> (first 501 terms from R. H. Hardin)
%H A163493 Shalosh B. Ekhad and Doron Zeilberger, <a href="http://arxiv.org/abs/1112.6207">Automatic Solution of Richard Stanley's Amer. Math. Monthly Problem #11610 and ANY Problem of That Type</a>, 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]
%H A163493 R. Stanley, <a href="https://www.jstor.org/stable/10.4169/amer.math.monthly.118.10.936">Problem 11610</a>, Amer. Math. Monthly, 118 (2011), 937; 120 (2013), 943-944.
%F A163493 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
%F A163493 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
%F A163493 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
%F A163493 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
%F A163493 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
%F A163493 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
%F A163493 a(n) = (1 + 3*hypergeom([1/2, 1-3*n/8, (1-n)/3, (2-n)/3, -n/3],[1, (1-n)/2, 1-n/2, -3*n/8],-27))/2 for n > 0. - _Stefano Spezia_, Apr 26 2024
%F A163493 a(n) ~ 2^n / sqrt(Pi*n). - _Vaclav Kotesovec_, Apr 26 2024
%e A163493 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 + ...
%e A163493 From _Gus Wiseman_, Jul 27 2021: (Start)
%e A163493 The a(0) = 1 though a(6) = 15 binary strings:
%e A163493 () (0) (1,0) (0,0,1) (0,0,1,0) (0,0,1,1,0) (0,0,0,1,0,1)
%e A163493 (1) (1,1) (1,1,0) (0,0,1,1) (0,0,1,1,1) (0,0,1,0,0,1)
%e A163493 (1,1,1) (0,1,0,0) (0,1,1,0,0) (0,0,1,1,1,0)
%e A163493 (1,0,0,1) (1,0,0,1,0) (0,0,1,1,1,1)
%e A163493 (1,1,1,0) (1,0,0,1,1) (0,1,0,0,0,1)
%e A163493 (1,1,1,1) (1,0,1,0,0) (0,1,1,1,0,0)
%e A163493 (1,1,0,0,1) (1,0,0,1,1,0)
%e A163493 (1,1,1,1,0) (1,0,0,1,1,1)
%e A163493 (1,1,1,1,1) (1,0,1,1,0,0)
%e A163493 (1,1,0,0,1,0)
%e A163493 (1,1,0,0,1,1)
%e A163493 (1,1,0,1,0,0)
%e A163493 (1,1,1,0,0,1)
%e A163493 (1,1,1,1,1,0)
%e A163493 (1,1,1,1,1,1)
%e A163493 (End)
%p A163493 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:
%p A163493 # second Maple program:
%p A163493 b:= proc(n, l, t) option remember; `if`(n-abs(t)<0, 0, `if`(n=0, 1,
%p A163493 add(b(n-1, i, t+`if`(l=0, (-1)^i, 0)), i=0..1)))
%p A163493 end:
%p A163493 a:= n-> b(n, 1, 0):
%p A163493 seq(a(n), n=0..36); # _Alois P. Heinz_, Mar 20 2024
%t A163493 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}];
%t A163493 Table[a[n], {n, 0, 40}] (* _Jean-François Alcover_, Nov 28 2017, after _Joel B. Lewis_ *)
%t A163493 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 *)
%t A163493 a[0]:=1; a[n_]:=(1 + 3*HypergeometricPFQ[{1/2, 1-3*n/8, (1-n)/3, (2-n)/3, -n/3},{1, (1-n)/2, 1-n/2, -3*n/8}, -27])/2; Array[a,37,0] (* _Stefano Spezia_, Apr 26 2024 *)
%o A163493 (Python)
%o A163493 from math import comb
%o A163493 def A163493(n): return 2+sum((x:=comb((k:=m<<1)-1,m)*comb(n-k,m))+(x*(n-3*m)<<1)//(n-k) for m in range(1,n//3+1)) if n else 1 # _Chai Wah Wu_, May 01 2024
%Y A163493 Antidiagonal sums of the matrices A345197.
%Y A163493 Row sums of A345907.
%Y A163493 Taking diagonal instead of antidiagonal sums gives A345908.
%Y A163493 A011782 counts compositions (or binary strings).
%Y A163493 A097805 counts compositions by alternating (or reverse-alternating) sum.
%Y A163493 A103919 counts partitions by sum and alternating sum (reverse: A344612).
%Y A163493 A316524 gives the alternating sum of prime indices (reverse: A344616).
%Y A163493 Compositions of n, 2n, or 2n+1 with alternating/reverse-alternating sum k:
%Y A163493 - k = 0: counted by A088218, ranked by A344619/A344619.
%Y A163493 - k = 1: counted by A000984, ranked by A345909/A345911.
%Y A163493 - k = -1: counted by A001791, ranked by A345910/A345912.
%Y A163493 - k = 2: counted by A088218, ranked by A345925/A345922.
%Y A163493 - k = -2: counted by A002054, ranked by A345924/A345923.
%Y A163493 - k >= 0: counted by A116406, ranked by A345913/A345914.
%Y A163493 - k <= 0: counted by A058622(n-1), ranked by A345915/A345916.
%Y A163493 - k > 0: counted by A027306, ranked by A345917/A345918.
%Y A163493 - k < 0: counted by A294175, ranked by A345919/A345920.
%Y A163493 - k != 0: counted by A058622, ranked by A345921/A345921.
%Y A163493 - k even: counted by A081294, ranked by A053754/A053754.
%Y A163493 - k odd: counted by A000302, ranked by A053738/A053738.
%Y A163493 Cf. A000041, A000070, A000096, A000097, A000124, A000346, A007318, A008549, A025047, A131577, A238279.
%K A163493 nonn
%O A163493 0,2
%A A163493 _Christopher Carl Heckman_, Jul 29 2009