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%I M2811
%S 1,3,9,29,99,351,1275,4707,17577,66197,250953,956385,3660541,14061141,
%T 54177741,209295261,810375651,3143981871,12219117171,47564380971,
%U 185410909791,723668784231,2827767747951,11061198475551,43308802158651
%N a(n) = Sum_{ k=0..n } binomial(2*k,k).
%C The expression a(n) = B^n*Sum_{ k=0..n } binomial(2*k,k)/B^k gives A006134 for B=1, A082590 (B=2), A132310 (B=3), A002457 (B=4), A144635 (B=5). - _N. J. A. Sloane_, Jan 21 2009
%C T(n+1,1) from table A045912 of characteristic polynomial of negative Pascal matrix. - Michael Somos, Jul 24 2002
%C p divides a((p-3)/2) for p=11, 13, 23, 37, 47, 59, 61, 71, 73, 83, 97, 107, 109, 131, 157, 167..=A097933. Also primes congruent to {1, 2, 3, 11} mod 12 or primes p such that 3 is a square mod p (excluding 2 and 3) A038874. - _Alexander Adamchuk_, Jul 05 2006
%C Partial sums of the even central binomial coefficients. For p prime >=5, a(p-1) = 1 or -1 (mod p) according as p = 1 or -1 (mod 3) (see Pan and Sun link). - _David Callan_, Nov 29 2007
%C a(n)=Sum_{ k=0..n } b(k)*binomial(n+k,k), where b(k)=0 for n-k == 2 (mod 3), b(k)=1 for n-k == 0 or 1 (mod 6), and b(k)=-1 for n-k== 3 or 4 (mod 6). - Alzhekeyev Ascar M. - Jan 19 2012
%C a(n)=Sum_{ k=0..n-1 } c(k)*binomial(2n,k) + binomial(2n,n), where c(k)=0 for n-k == 0 (mod 3), c(k)=1 for n-k== 1 (mod 3), and c(k)=-1 for n-k==2 (mod 3). - Alzhekeyev Ascar M. - Jan 19 2012
%D W. F. Lunnon, "The Pascal matrix", Fib. Quart. vol. 15 (1977) pp. 201-204.
%D Paule, Peter. A proof of a conjecture of Knuth. Experiment. Math. 5(1996), no. 2, 83--89. MR1418955 (97k:33004)
%D M. Petkovsek et al., A=B, Peters, 1996, p. 22.
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H Vincenzo Librandi, <a href="/A006134/b006134.txt">Table of n, a(n) for n = 0..1000</a>
%H Hao Pan and Zhi-Wei Sun, <a href="http://front.math.ucdavis.edu/math.CO/0509648">A combinatorial identity with application to Catalan numbers </a> .
%F a(n) = Sum[((2k)!/(k!)^2),{k,0,n}]. a(n) = A066796(n) + 1, n>0. - _Alexander Adamchuk_, Jul 05 2006
%F G.f.: 1/((1-x)*sqrt(1-4*x)).
%F Recurrence: (n+2)*a(n+2) - (5*n+8)*a(n+1) + 2*(2*n+3)*a(n) = 0 [Emanuele Munarini, Mar 15 2011]
%F a(n) = C(2n,n) * Sum_{k=0..2n} (-1)^k*trinomial(n,k)/C(2n,k) where trinomial(n,k) = [x^k] (1 + x + x^2)^n. E.g. a(2) = C(4,2)*(1/1 - 2/4 + 3/6 - 2/4 + 1/1) = 6*(3/2) = 9 ; a(3) = C(6,3)*(1/1 - 3/6 + 6/15 - 7/20 + 6/15 - 3/6 + 1/1) = 20*(29/20) = 29. - _Paul D. Hanna_, Aug 21 2007
%F a(n) ~ 2^(2*n+2)/(3*sqrt(Pi*n)). - _Vaclav Kotesovec_, Nov 06 2012
%e 1 + 3*x + 9*x^2 + 29*x^3 + 99*x^4 + 351*x^5 + 1275*x^6 + 4707*x^7 + 17577*x^8 + ...
%p A006134 := proc(n) sum(binomial(2*k,k),k=0..n); end;
%t Table[Sum[((2k)!/(k!)^2),{k,0,n}], {n,0,50}] - _Alexander Adamchuk_, Jul 05 2006
%t a[ n_] := (4/3) Binomial[ 2 n, n] Hypergeometric2F1[ 1/2, 1, -n + 1/2, -1/3] (* Michael Somos, Jun 20 2012 *)
%o (MATLAB) n=10; x=pascal(n); trace(x)
%o (PARI) {a(n) = if( n<0, 0, polcoeff( charpoly( matrix( n+1, n+1, i, j, -binomial( i+j-2, i-1))), 1))} /* Michael Somos, Jul 10 2002 */
%o (PARI) {a(n)=binomial(2*n,n)*sum(k=0,2*n,(-1)^k*polcoeff((1+x+x^2)^n,k)/binomial(2*n,k))} - _Paul D. Hanna_, Aug 21 2007
%o (Maxima) makelist(sum(binomial(2*k,k),k,0,n),n,0,12); [Emanuele Munarini, Mar 15 2011]
%Y Cf. A000984, A066796, A097933, A038874.
%Y Cf. A132310.
%Y Cf. A006135, A006136, A045912. Differences give A000984.
%K nonn
%O 0,2
%A _N. J. A. Sloane_.
%E Simpler definition from _Alexander Adamchuk_, Jul 05 2006
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