A243946 -id:A243946 - cp's OEIS frontend

A245926 Expansion of g.f. sqrt( (1-x + sqrt(1-14x+x^2)) / (2(1-14*x+x^2)) ).

1, 5, 51, 587, 7123, 89055, 1135005, 14660805, 191253843, 2513963567, 33244446601, 441772827105, 5894323986301, 78912561223553, 1059543126891027, 14261959492731387, 192392702881384275, 2600355510685245087, 35206018016510388345, 477377227987055971905

Offset: 0

Paul D. Hanna, Aug 15 2014

Square each term to form a bisection of A245925.

Limit a(n+1)/a(n) = 7 + 4*sqrt(3).

			G.f.: A(x) = 1 + 5*x + 51*x^2 + 587*x^3 + 7123*x^4 + 89055*x^5 +...
where
A(x)^2 = (1-x + sqrt(1-14*x+x^2)) / (2*(1-14*x+x^2)).
Explicitly,
A(x)^2 = 1 + 10*x + 127*x^2 + 1684*x^3 + 22717*x^4 + 309214*x^5 + 4231675*x^6 + 58117672*x^7 + 800173945*x^8 +...+ A245923(n)*x^n +...

G. C. Greubel, Table of n, a(n) for n = 0..870

Column k=3 of A337389.

Cf. A245925, A245927, A245923, A243946, A337422.

A245926 := n -> sqrt(add(binomial(4*n-2*k, 2*n-k)*binomial(4*n-k, k)^2, k=0..2*n)); seq(A245926(n), n=0..20); # Peter Luschny, Aug 17 2014

CoefficientList[Series[Sqrt[(1 - x + Sqrt[1 - 14*x + x^2])/(2*(1 - 14*x + x^2))], {x,0,50}], x] (* G. C. Greubel, Jan 29 2017 *)
a[n_] := (-1)^n Hypergeometric2F1[-n, n + 1/2, 1, 4];
Table[a[n], {n, 0, 19}] (* Peter Luschny, Mar 16 2018 *)

/* From definition: */
{a(n)=polcoeff( sqrt( (1-x + sqrt(1-14*x+x^2 +x*O(x^n))) / (2*(1-14*x+x^2 +x*O(x^n))) ), n)}
for(n=0, 20, print1(a(n), ", "))

/* From formula for a(n)^2: */
{a(n)=sqrtint(sum(k=0, 2*n, sum(j=0, 4*n-2*k, (-1)^(j+k)*binomial(4*n-k,j+k)^2*binomial(j+k, k)^2)))}
for(n=0, 20, print1(a(n), ", "))

/* From formula for a(n)^2: */
{a(n) = sqrtint( sum(k=0, 2*n, binomial(2*k, k)^2*binomial(2*n+k, 2*n-k)*(-1)^k) )}
for(n=0, 20, print1(a(n), ", "))

a(n)^2 = Sum_{k=0..2*n} Sum_{j=0..4*n-2*k} (-1)^(j+k) * C(4*n-k,j+k)^2 * C(j+k,k)^2.
a(n) ~ (3*sqrt(3)-5) * (7+4*sqrt(3))^(n+1) / (4*sqrt(Pi*n)). - Vaclav Kotesovec, Aug 16 2014
a(n)^2 = C(4*n,2*n)*hyper4F3([-2*n,-2*n,-2*n,-2*n+1/2],[1,-4*n,-4*n],4). - Peter Luschny, Aug 17 2014
a(n)^2 = Sum_{k=0..2*n} binomial(4*n-2*k, 2*n-k)*binomial(4*n-k, k)^2. - Peter Luschny, Aug 17 2014
a(n)^2 = Sum_{k=0..2*n} (-1)^k * C(2*k, k)^2 * C(2*n+k, 2*n-k). - Paul D. Hanna, Aug 17 2014
From Peter Bala, Mar 14 2018: (Start)
a(n) = (-1)^n*P(2*n,sqrt(-3)), where P(n,x) denotes the n-th Legendre polynomial. See A008316.
a(n) = (1/C(2*n,n))*Sum_{k = 0..n} (-1)^(n+k)*C(n,k)*C(n+k,k)* C(2*n+2*k,n+k) = Sum_{k = 0..n} (-1)^(n+k)*C(2*k,k)*C(n,k) *C(2*n+2*k,2*n)/C(n+k,n). In general, P(2*n,sqrt(1+4*x)) = (1/C(2*n,n))*Sum_{k = 0..n} C(n,k)*C(n+k,k)*C(2*n+2*k,n+k) *x^k.
a(n) = Sum_{k = 0..2*n} C(2*n,k)^2 * u^(n-k), where u = (1 - sqrt(-3))/2 is a primitive sixth root of unity.
a(n) = (-1)^n*Sum_{k = 0..2*n} C(2*n,k)*C(2*n+k,k)*u^(2*k).
(End)
a(n) = (-1)^n*hypergeom([-n, n + 1/2], [1], 4). - Peter Luschny, Mar 16 2018
a(0) = 1, a(1) = 5 and n * (2*n-1) * (4*n-5) * a(n) = (4*n-3) * (28*n^2-42*n+9) * a(n-1) - (n-1) * (2*n-3) * (4*n-1) * a(n-2) for n > 1. - Seiichi Manyama, Aug 28 2020
From Peter Bala, May 03 2022: (Start)
Conjecture: a(n) = [x^n] ( (1 + x + x^2)*(1 + x)^2/(1 - x)^2 )^n. Equivalently, a(n) = Sum_{k = 0..n} Sum_{i = 0..k} Sum_{j = 0..n-k-i} C(n, k)*C(k, i)*C(2*n, j)*C(3*n-k-i-j-1, n-k-i-j).
If the conjecture is true then the Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and positive integers n and k. Calculation suggests that the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(2*k)) hold for primes p >= 5 and positive integers n and k. (End)

A243947 Expansion of g.f. sqrt( (1+x - sqrt(1-18x+x^2)) / (10x(1-18x+x^2)) ).

Original entry on oeis.org

1, 11, 155, 2365, 37555, 610897, 10098997, 168894355, 2849270515, 48395044705, 826479148001, 14177519463191, 244109912494525, 4216385987238575, 73024851218517275, 1267712063327871245, 22052786911315216595, 384321597582115655825, 6708530714274563938225

Offset: 0

Paul D. Hanna, Aug 17 2014

Multiply the square of each term by 5 to form a bisection of A243945.

Limit_{n->oo} a(n+1)/a(n) = 9 + 4*sqrt(5).

			G.f.: A(x) = 1 + 11*x + 155*x^2 + 2365*x^3 + 37555*x^4 + 610897*x^5 + ...
where
A(x)^2 = (1+x - sqrt(1-18*x+x^2)) / (10*x*(1-18*x+x^2)).

Seiichi Manyama, Table of n, a(n) for n = 0..500

Cf. A243945, A243946, A084769, A245927, A008316.

seq(add(1/2*binomial(2*k+1,k)*binomial(n,k)*binomial(2*n+2*k+2,2*n+1)/binomial(n+k+1,n), k = 0..n), n = 0..20); # Peter Bala, Mar 15 2018

CoefficientList[Series[Sqrt[((1+x-Sqrt[1-18x+x^2]))/(10x(1-18x+x^2))],{x,0,20}],x] (* Harvey P. Dale, Jul 31 2016 *)
a[n_] := Hypergeometric2F1[-n, n + 3/2, 1, -4];
Table[a[n], {n, 0, 18}] (* Peter Luschny, Mar 16 2018 *)

/* From definition: */
{a(n)=polcoeff( sqrt( (1+x - sqrt(1-18*x+x^2 +x^2*O(x^n))) / (10*x*(1-18*x+x^2 +x*O(x^n))) ), n)}
for(n=0, 20, print1(a(n), ", "))

/* From a(n) = sqrt( A243945(2*n+1)/5 ): */
{a(n)=sqrtint( (1/5)*sum(k=0, 2*n+1, binomial(2*k, k)^2*binomial(2*n+k+1, 2*n-k+1)) )}
for(n=0, 20, print1(a(n), ", "))

from math import comb
def A243947(n): return sum(5**(n-k)*comb(m:=k<<1,k)*comb((n<<1)+1,m) for k in range(n+1)) # Chai Wah Wu, Mar 23 2023

a(n)^2 = (1/5) * Sum_{k=0..2*n+1} C(2*k, k)^2 * C(2*n+k+1, 2*n-k+1).
a(n) ~ (9+4*sqrt(5))^(n+1) / (2*5^(1/4)*sqrt(2*Pi*n) * sqrt(5+2*sqrt(5))). - Vaclav Kotesovec, Aug 18 2014. Equivalently, a(n) ~ phi^(6*n + 9/2) / (2^(3/2) * sqrt(5*Pi*n)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Dec 08 2021
From Peter Bala, Mar 15 2018: (Start)
a(n) = (1/sqrt(5))*P(2*n+1,sqrt(5)), where P(n,x) denotes the n-th Legendre polynomial. See A008316.
a(n) = Sum_{k = 0..n} (1/2)*C(2*k+1,k)*C(n,k)*C(2*n+2*k+2,2*n+1)/C(n+k+1,n). In general, (1/sqrt(1 + 4*x))*P(2*n+1,sqrt(1+4*x)) = (1/(2*C(2*n+1,n))) * Sum_{k = 0..n} C(n,k)*C(n+k+1,k)*C(2*n+2*k+2,n+k+1)*x^k.
a(n) = (1/sqrt(5))*Sum_{k = 0..2*n+1} C(2*n+1,k)^2 * phi^(2*n-2*k+1), where phi = (sqrt(5) + 1)/2.
a(n) = (1/sqrt(5))*Sum_{k = 0..2*n+1} C(2*n+1,k)*C(2*n+1+k,k) * Phi^k, where Phi = (sqrt(5) - 1)/2. (End)
a(n) = hypergeom([-n, n + 3/2], [1], -4). - Peter Luschny, Mar 16 2018
From Peter Bala, Mar 17 2018: (Start)
a(n) = Sum_{k = 0..n} C(2*n+1,2*k)*C(2*k,k)*5^(n-k).
D-finite with recurrence: n*(4*n-3)*(2*n+1)*a(n) = (4*n-1)*(36*n^2-18*n-7)*a(n-1) - (n-1)*(2*n-1)*(4*n+1)*a(n-2). (End)

A337389 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of sqrt((1+(k-4)x+sqrt(1-2(k+4)x+((k-4)x)^2)) / (2 * (1-2(k+4)x+((k-4)*x)^2))).

Original entry on oeis.org

1, 1, 2, 1, 3, 6, 1, 4, 19, 20, 1, 5, 34, 141, 70, 1, 6, 51, 328, 1107, 252, 1, 7, 70, 587, 3334, 8953, 924, 1, 8, 91, 924, 7123, 34904, 73789, 3432, 1, 9, 114, 1345, 12870, 89055, 372436, 616227, 12870, 1, 10, 139, 1856, 20995, 184756, 1135005, 4027216, 5196627, 48620

Offset: 0

Seiichi Manyama, Aug 25 2020

			Square array begins:
    1,    1,     1,     1,      1,      1, ...
    2,    3,     4,     5,      6,      7, ...
    6,   19,    34,    51,     70,     91, ...
   20,  141,   328,   587,    924,   1345, ...
   70, 1107,  3334,  7123,  12870,  20995, ...
  252, 8953, 34904, 89055, 184756, 337877, ...

Seiichi Manyama, Antidiagonals n = 0..139, flattened

Columns k=0..5 give A000984, A082758, A337390, A245926, A001448, A243946.
Main diagonal gives A337388.
Cf. A337369, A337419.

T[n_, k_] := Sum[If[k == 0, Boole[n == j], k^(n - j)] * Binomial[2*j, j] * Binomial[2*n, 2*j], {j, 0, n}]; Table[T[k, n - k], {n, 0, 9}, {k, 0, n}] // Flatten (* Amiram Eldar, Aug 25 2020 *)

{T(n, k) = sum(j=0, n, k^(n-j)*binomial(2*j, j)*binomial(2*n, 2*j))}

T(n,k) = Sum_{j=0..n} k^(n-j) * binomial(2*j,j) * binomial(2*n,2*j).
T(0,k) = 1, T(1,k) = k+2 and n * (2*n-1) * (4*n-5) * T(n,k) = (4*n-3) * (4*(k+4)*n^2-6*(k+4)*n+k+6) * T(n-1,k) - (k-4)^2 * (n-1) * (2*n-3) * (4*n-1) * T(n-2,k) for n > 1. - Seiichi Manyama, Aug 28 2020
For fixed k > 0, T(n,k) ~ (2 + sqrt(k))^(2*n + 1/2) / sqrt(8*Pi*n). - Vaclav Kotesovec, Aug 31 2020
Conjecture: the k-th column entries, k >= 0, are given by [x^n] ( (1 + (k-2)*x + x^2)*(1 + x)^2/(1 - x)^2 )^n. This is true for k = 0 and k = 4. - Peter Bala, May 03 2022

A243945 a(n) = Sum_{k=0..n} C(2k, k)^2 C(n+k, n-k).

Original entry on oeis.org

1, 5, 49, 605, 8281, 120125, 1809025, 27966125, 440790025, 7051890125, 114160867129, 1865975723045, 30743797894681, 509948702030045, 8507207970913729, 142626515754330125, 2401552098016698025, 40591712338241826125, 688413807606268692025, 11710401759994742685125

Offset: 0

Paul D. Hanna, Aug 17 2014

nonn

The g.f.s formed from a(2*n)^(1/2) and (a(2*n+1)/5)^(1/2) are:
A243946: sqrt( (1+x + sqrt(1-18*x+x^2)) / (2*(1-18*x+x^2)) );
A243947: sqrt( (1+x - sqrt(1-18*x+x^2)) / (10*x*(1-18*x+x^2)) ).
Lim_{n->infinity} a(n+1)/a(n) = 9 + 4*sqrt(5).
Diagonal of rational function 1/(1 - (x + y + x*z + y*z + x*y*z)). - Gheorghe Coserea, Aug 24 2018

			G.f.: A(x) = 1 + 5*x + 49*x^2 + 605*x^3 + 8281*x^4 + 120125*x^5 + ... where
A(x) = 1/(1-x) + 2^2*x/(1-x)^3 + 6^2*x^2/(1-x)^5 + 20^2*x^3/(1-x)^7 + 70^2*x^4/(1-x)^9 + 252^2*x^5/(1-x)^11 + 924^2*x^6/(1-x)^13 + ... + A000984(n)^2*x^n/(1-x)^(2*n+1) + ...

Seiichi Manyama, Table of n, a(n) for n = 0..800
A. Bostan, S. Boukraa, J.-M. Maillard, J.-A. Weil, Diagonals of rational functions and selected differential Galois groups, arXiv preprint arXiv:1507.03227 [math-ph], 2015.

Cf. A243946, A243947, A245925.

&cat[ [&+[ Binomial(2*k, k)^2 * Binomial(n+k, n-k): k in [0..n]]]: n in [0..30]]; // Vincenzo Librandi, Aug 25 2018

Table[Sum[Binomial[2*k, k]^2 * Binomial[n + k, n - k],{k, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Aug 18 2014 *)
a[n_] := HypergeometricPFQ[{1/2, -n, n + 1}, {1, 1}, -4];
Table[a[n], {n, 0, 19}] (* Peter Luschny, Mar 14 2018 *)

{a(n)=sum(k=0, n, binomial(2*k, k)^2*binomial(n+k, n-k))}
for(n=0, 20, print1(a(n), ", "))

{a(n)=local(A=1); A=sum(m=0, n, binomial(2*m, m)^2 * x^m/(1-x +x*O(x^n))^(2*m+1)); polcoeff(A, n)}
for(n=0, 20, print1(a(n), ", "))

{a(n)=polcoeff( 1 / agm(1-x, sqrt((1-x)^2 - 16*x +x*O(x^n))), n)}
for(n=0,20,print1(a(n),", "))

G.f.: Sum_{n>=0} binomial(2*n, n)^2 * x^n / (1-x)^(2*n+1).
G.f.: 1 / AGM(1-x, sqrt(1-18*x+x^2)), where AGM(x,y) = AGM((x+y)/2,sqrt(x*y)) is the arithmetic-geometric mean.
a(2*n) = A243946(n)^2.
a(2*n+1) = 5 * A243947(n)^2.
Recurrence: n^2*(2*n-3)*a(n) = (2*n-1)*(19*n^2 - 38*n + 14)*a(n-1) - (2*n-3)*(19*n^2 - 38*n + 14)*a(n-2) + (n-2)^2*(2*n-1)*a(n-3). - Vaclav Kotesovec, Aug 18 2014
a(n) ~ (2+sqrt(5)) * (9+4*sqrt(5))^n / (4*Pi*n). - Vaclav Kotesovec, Aug 18 2014. Equivalently, a(n) ~ phi^(6*n + 3) / (4*Pi*n), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Dec 08 2021
a(n) = hypergeom([1/2, -n, n + 1], [1, 1], -4). - Peter Luschny, Mar 14 2018
G.f. y=A(x) satisfies: 0 = x*(x^2 - 1)*(x^2 - 18*x + 1)*y'' + (3*x^4 - 34*x^3 - 38*x^2 + 38*x - 1)*y' + (x^3 - 3*x^2 - 19*x + 5)*y. - Gheorghe Coserea, Aug 29 2018
From Peter Bala, Feb 07 2022: (Start)
a(n) = P(n,sqrt(5))^2, where P(n,x) denotes the n-th Legendre polynomial.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) hold for all primes p and positive integers n and k. (End)

A300945 Rectangular array A(n, k) = hypergeom([-k, k + n/2 - 1], [1], -4) with row n >= 0 and k >= 0, read by ascending antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 3, 25, 1, 5, 43, 425, 1, 7, 65, 661, 7025, 1, 9, 91, 965, 10515, 116625, 1, 11, 121, 1345, 15105, 171097, 1951625, 1, 13, 155, 1809, 20995, 243525, 2828101, 32903225, 1, 15, 193, 2365, 28401, 337877, 4001345, 47284251, 558265825

Offset: 0

Peter Luschny, Mar 16 2018

			[0] 1,  1,  25,  425,  7025, 116625,  1951625,  32903225, ... [A299845]
[1] 1,  3,  43,  661, 10515, 171097,  2828101,  47284251, ... [A299506]
[2] 1,  5,  65,  965, 15105, 243525,  4001345,  66622085, ...
[3] 1,  7,  91, 1345, 20995, 337877,  5544709,  92234527, ... [A243946]
[4] 1,  9, 121, 1809, 28401, 458649,  7544041, 125700129, ... [A084769]
[5] 1, 11, 155, 2365, 37555, 610897, 10098997, 168894355, ... [A243947]
[6] 1, 13, 193, 3021, 48705, 800269, 13324417, 224028877, ...

Cf. A299845, A299506, A243946, A084769, A243947.

Cf. A300946.

Arow[n_, len_] := Table[Hypergeometric2F1[-k, k + n/2 - 1, 1, -4], {k, 0, len}];
Table[Print[Arow[n, 7]], {n, 0, 6}];
T[n_, k_] := If[k==0, 1, 4^k*Sum[(5/4)^j*Binomial[k, j]*Binomial[k - 2 + ((n - k)/2), j - 2 + ((n - k)/2)] ,{j, 0, n}]]; Flatten[Table[T[n, k],{n, 0, 8}, {k, 0, n}]] (* Detlef Meya, May 28 2024 *)

T(n, k) = if k = 0 then 1, otherwise 4^k*Sum_{j=0..n} (5/4)^j * binomial(k, j) * binomial(k - 2 + ((n - k)/2), j - 2 + ((n - k)/2)). - Detlef Meya, May 28 2024

A299845 a(n) = hypergeom([-n, n - 1], [1], -4).

Original entry on oeis.org

1, 1, 25, 425, 7025, 116625, 1951625, 32903225, 558265825, 9522632225, 163160773625, 2806202183625, 48420275891025, 837813745045425, 14531896733426025, 252593595973313625, 4398859688478578625, 76733590756134492225, 1340547988367851940825, 23451231922182584693225

Offset: 0

Peter Luschny, Mar 16 2018

nonn

Robert Israel, Table of n, a(n) for n = 0..798

Cf. A299506, A243946, A084769, A243947.

Cf. A300945, A300946.

f:= gfun:-rectoproc({4*n*(n-2)^2*a(n)+4*(n-1)^2*(n-3)*a(n-2)-4*(2*n-3)*(9*n^2-27*n+17)*a(n-1)=0,
a(0)=1,a(1)=1,a(2)=25},a(n),remember):
map(f, [$0..100]); # Robert Israel, Mar 21 2018

a[n_] := Hypergeometric2F1[-n, n - 1, 1, -4]; Table[a[n], {n, 0, 19}]
a[0]:=1; a[1]:=1; a[n_] := 4^n*Sum[(5/4)^k*(Gamma[n + 1]*Gamma[n - 1])/(Gamma[k + 1]*Gamma[n - k + 1]^2*Gamma[k - 1]),{k,0,n}]; Flatten[Table[a[n],{n,0,19}]] (* Detlef Meya, May 22 2024 *)

4*n*(n-2)^2*a(n) + 4*(n-1)^2*(n-3)*a(n-2) - 4*(2*n-3)*(9*n^2-27*n+17)*a(n-1) = 0.  - Robert Israel, Mar 21 2018
a(n) ~ 2^(-3/2) * 5^(3/4) * phi^(6*n - 3) / sqrt(Pi*n), where phi = A001622 = (1 + sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Jul 05 2018
a(n) = 4^n*Sum_{k=0..n} (5/4)^k*Gamma(n + 1)*Gamma(n - 1)/(Gamma(k + 1)*Gamma(n - k + 1)^2*Gamma(k - 1)) for n >= 2. - Detlef Meya, May 22 2024

A299506 a(n) = hypergeom([-n, n - 1/2], [1], -4).

Original entry on oeis.org

1, 3, 43, 661, 10515, 171097, 2828101, 47284251, 797456947, 13540982665, 231188344401, 3964874384863, 68252711769373, 1178662654873191, 20409993947488075, 354260920943874245, 6161735337225790035, 107368528677807960185, 1873946997372948997345, 32754419073618998202975

Offset: 0

Peter Luschny, Mar 16 2018

nonn

Cf. A299845, A243946, A084769, A243947.

Cf. A300945, A300946.

a[n_] := Hypergeometric2F1[-n, n - 1/2, 1, -4]; Table[a[n], {n, 0, 19}]
a[n_] := 4^n*Sum[(5/4)^k*(Gamma[n + 1]*Gamma[n - 1/2])/(Gamma[k + 1]*Gamma[n - k + 1]^2*Gamma[k - 1/2]),{k,0,n}]; Flatten[Table[a[n],{n,0,19}]] (* Detlef Meya, May 22 2024 *)

From Vaclav Kotesovec, Jul 05 2018: (Start)
Recurrence: n*(2*n - 3)*(4*n - 7)*a(n) = 9*(4*n - 5)*(4*n^2 - 10*n + 5)*a(n-1) - (n-1)*(2*n - 5)*(4*n - 3)*a(n-2).
a(n) ~ 2^(-3/2) * sqrt(5) * phi^(6*n - 3/2) / sqrt(Pi*n), where phi = A001622 = (1 + sqrt(5))/2 is the golden ratio. (End)
a(n) = 4^n*Sum_{k=0..n} (5/4)^k*(Gamma(n + 1)*Gamma(n - 1/2))/(Gamma(k + 1)*Gamma(n - k + 1)^2*Gamma(k - 1/2)). - Detlef Meya, May 22 2024

cp's OEIS Frontend

A245926 Expansion of g.f. sqrt( (1-x + sqrt(1-14*x+x^2)) / (2*(1-14*x+x^2)) ).

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A243947 Expansion of g.f. sqrt( (1+x - sqrt(1-18*x+x^2)) / (10*x*(1-18*x+x^2)) ).

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Maple

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A337389 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of sqrt((1+(k-4)*x+sqrt(1-2*(k+4)*x+((k-4)*x)^2)) / (2 * (1-2*(k+4)*x+((k-4)*x)^2))).

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A243945 a(n) = Sum_{k=0..n} C(2*k, k)^2 * C(n+k, n-k).

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Magma

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A300945 Rectangular array A(n, k) = hypergeom([-k, k + n/2 - 1], [1], -4) with row n >= 0 and k >= 0, read by ascending antidiagonals.

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A299845 a(n) = hypergeom([-n, n - 1], [1], -4).

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Maple

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A299506 a(n) = hypergeom([-n, n - 1/2], [1], -4).

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A245926 Expansion of g.f. sqrt( (1-x + sqrt(1-14x+x^2)) / (2(1-14*x+x^2)) ).

A243947 Expansion of g.f. sqrt( (1+x - sqrt(1-18x+x^2)) / (10x(1-18x+x^2)) ).

A337389 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of sqrt((1+(k-4)x+sqrt(1-2(k+4)x+((k-4)x)^2)) / (2 * (1-2(k+4)x+((k-4)*x)^2))).

A243945 a(n) = Sum_{k=0..n} C(2k, k)^2 C(n+k, n-k).