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A168597
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Squares of the central trinomial coefficients (A002426).
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6
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1, 1, 9, 49, 361, 2601, 19881, 154449, 1225449, 9853321, 80156209, 658076409, 5444816521, 45343869481, 379735715529, 3195538786449, 27004932177129, 229066136374761, 1949470542590481, 16640188083903609, 142415188146838161
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OFFSET
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0,3
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COMMENTS
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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
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LINKS
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FORMULA
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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) = 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)
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MAPLE
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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
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MATHEMATICA
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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 *)
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PROG
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(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)}
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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