A168597 Squares of the central trinomial coefficients (A002426).

1, 1, 9, 49, 361, 2601, 19881, 154449, 1225449, 9853321, 80156209, 658076409, 5444816521, 45343869481, 379735715529, 3195538786449, 27004932177129, 229066136374761, 1949470542590481, 16640188083903609, 142415188146838161, 1221800234100831441, 10504959504381567729

Offset: 0

Author

Paul D. Hanna, Dec 01 2009

Keywords

• nonn
• easy

Comments

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

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

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

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000

Crossrefs

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

Programs

Maple

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

Mathematica

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 *)
CoefficientList[Series[(2 EllipticK[(16 x)/(1 + 3 x)^2])/(Pi (1 + 3 x)), {x, 0, 28}], x, 26]  (* After Mark van Hoeij, Peter Luschny, May 13 2025 *)

PARI

{a(n)=polcoeff((1+x+x^2 +x*O(x^n))^n,n)^2}
for(n=0, 20, print1(a(n), ", "))

PARI

/* Using AGM: */
{a(n)=polcoeff( 1 / agm(1+3*x, sqrt((1+3*x)^2 - 16*x +x*O(x^n))), n)}
for(n=0, 20, print1(a(n), ", ")) \\ Paul D. Hanna, Sep 04 2014

Formula

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

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

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

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

a(n) = (-1)^n*hypergeom([1/2,-n],[1],4)*hypergeom([(1-n)/2,-n/2],[1],4). - Peter Luschny, Nov 10 2014

a(n) ~ 3^(2*n+1) / (4*Pi*n). - Vaclav Kotesovec, Sep 28 2019

From Peter Bala, Feb 08 2022: (Start)

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

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.

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

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.

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)

G.f.: hypergeom([1/2, 1/2],[1],16*x/(1+3*x)^2)/(1+3*x). - Mark van Hoeij, May 13 2025