A108958 Number of unordered pairs of distinct length-n binary words having the same number of 1's.
0, 1, 6, 27, 110, 430, 1652, 6307, 24054, 91866, 351692, 1350030, 5196204, 20050108, 77542376, 300507427, 1166737574, 4537436578, 17672369756, 68922740122, 269127888644, 1052047384708, 4116711169496, 16123793452942, 63205286441660, 247959232919620, 973469645715192
Offset: 1
Examples
a(3) = 6 because the pairs are {001,010}, {001,100}, {010,100}, {011,101}, {011,110}, {101,110}.
Links
- Michael De Vlieger, Table of n, a(n) for n = 1..1664
- Mircea Merca, A Special Case of the Generalized Girard-Waring Formula, J. Integer Sequences, Vol. 15 (2012), Article 12.5.7.
- Ran Pan and Jeffrey B. Remmel, Paired patterns in lattice paths, arXiv:1601.07988 [math.CO], 2016.
- Stanislav Sýkora, Magnetic Resonance on OEIS, Stan's NMR Blog (Dec 31, 2014), Retrieved Nov 12, 2019.
Programs
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Magma
[Binomial(2*n,n)-(2^n+Binomial(2*n,n))/2: n in [1..30]]; // Vincenzo Librandi, Feb 01 2016
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Maple
with(combinat) a:= proc(n) add(binomial(binomial(n,k), 2), k=0..n) end;
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Mathematica
Table[Binomial[2 n, n] - (2^n + Binomial[2 n, n])/2, {n, 30}] (* Vincenzo Librandi, Feb 01 2016 *)
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PARI
a(n)=binomial(2*n-1,n-1)-2^(n-1) \\ Charles R Greathouse IV, Feb 01 2016
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Python
from math import comb def A108958(n): return comb((n<<1)-1,n-1)-(1<
Chai Wah Wu, Sep 23 2022
Formula
a(n) = Sum_{k=0..n} binomial(binomial(n, k), 2).
From Vladeta Jovovic, Jul 24 2005: (Start)
E.g.f.: exp(2*x)*(BesselI(0, 2*x)-1)/2. (End)
a(n) = (1/2)*Sum_{i+j>n,0<=i,j<=n} binomial(i+j,i). - Benoit Cloitre, May 26 2006
Conjecture: n*(n-2)*a(n) +2*(-3*n^2+7*n-3)*a(n-1) +4*(n-1)*(2*n-3) *a(n-2)=0. - R. J. Mathar, Apr 04 2012
a(n) = Sum_{0Mircea Merca, Apr 05 2012
a(n) = 1/2*Sum_{k=1..n} binomial(n,k)^2 - binomial(n,k). - Gerry Martens, Oct 09 2022
a(n) ~ 2^(2*n-1)/sqrt(n*Pi). - Stefano Spezia, Apr 17 2024
Comments