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Squares of the central trinomial coefficients (A002426).
6

%I #45 Feb 10 2022 06:20:41

%S 1,1,9,49,361,2601,19881,154449,1225449,9853321,80156209,658076409,

%T 5444816521,45343869481,379735715529,3195538786449,27004932177129,

%U 229066136374761,1949470542590481,16640188083903609,142415188146838161

%N Squares of the central trinomial coefficients (A002426).

%C Ignoring initial term, a(n) equals the logarithmic derivative of A168598.

%C Partial sums of A007987. Hence, a(n) is the number of irreducible words of length at most 2n in the free group with generators x,y such that the total degree of x and the total degree of y both equal zero. - _Max Alekseyev_, Jun 05 2011

%C The number of ways a king, starting at the origin of an infinite chessboard, can return to the origin in n moves, where leaving the king where it is counts as a move. Cf. A094061. - _Peter Bala_, Feb 14 2017

%H G. C. Greubel, <a href="/A168597/b168597.txt">Table of n, a(n) for n = 0..1000</a>

%F a(n) = A002426(n)^2.

%F G.f.: hypergeom([1/12, 5/12],[1],1728*x^4*(x-1)*(9*x-1)*(3*x+1)^2/(81*x^4-36*x^3-26*x^2-4*x+1)^3)/(81*x^4-36*x^3-26*x^2-4*x+1)^(1/4). - _Mark van Hoeij_, May 07 2013

%F G.f.: 1 / AGM(1+3*x, sqrt((1-x)*(1-9*x))), where AGM(x,y) = AGM((x+y)/2,sqrt(x*y)) is the arithmetic-geometric mean. - _Paul D. Hanna_, Sep 04 2014

%F G.f.: 1 / AGM((1-x)*(1-3*x), (1+x)*(1+3*x)) = Sum_{n>=0} a(n)*x^(2*n). - _Paul D. Hanna_, Oct 04 2014

%F a(n) = (-1)^n*hypergeom([1/2,-n],[1],4)*hypergeom([(1-n)/2,-n/2],[1],4). - _Peter Luschny_, Nov 10 2014

%F a(n) ~ 3^(2*n+1) / (4*Pi*n). - _Vaclav Kotesovec_, Sep 28 2019

%F From _Peter Bala_, Feb 08 2022: (Start)

%F a(n) = Sum_{k = 0..n} (-3)^(n-k)*binomial(2*k,k)*binomial(n,k)* binomial(n+k,k).

%F n^2*(2*n-3)*a(n)= (7*n^2-14*n+6)*(2*n-1)*a(n-1) + 3*(7*n^2-14*n+6)*(2*n-3)*a(n-2) - 27*(2*n-1)*(n-2)^2*a(n-3) with a(0) = 1, a(1) = 1 and a(2) = 9.

%F G.f.: A(x) = Sum_{n >= 0} binomial(2*n,n)^2*x^n/(1 + 3*x)^(2*n+1).

%F The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all prime p and positive integers n and k.

%F Conjecture: The stronger congruences a(n*p^k) == a(n*p^(k-1)) (mod p^(2*k)) hold for all primes p >= 5 and positive integers n and k. (End)

%p a := n -> (-1)^n*hypergeom([1/2,-n],[1],4)*hypergeom([1/2-n/2,-n/2],[1], 4): seq(simplify(a(n)),n=0..20); # _Peter Luschny_, Nov 10 2014

%t Table[(-1)^n*Hypergeometric2F1[1/2, -n, 1, 4] * Hypergeometric2F1[(1 - n)/2, -n/2, 1, 4], {n, 0, 50}] (* _G. C. Greubel_, Feb 26 2017 *)

%o (PARI) {a(n)=polcoeff((1+x+x^2 +x*O(x^n))^n,n)^2}

%o for(n=0, 20, print1(a(n), ", "))

%o (PARI) /* Using AGM: */

%o {a(n)=polcoeff( 1 / agm(1+3*x, sqrt((1+3*x)^2 - 16*x +x*O(x^n))), n)}

%o for(n=0, 20, print1(a(n), ", ")) \\ _Paul D. Hanna_, Sep 04 2014

%Y Cf. A002426, A133053, A168598, A243949, A094061.

%K nonn,easy

%O 0,3

%A _Paul D. Hanna_, Dec 01 2009