A116091 -id:A116091 - cp's OEIS frontend

A066802 a(n) = binomial(6n,3n).

1, 20, 924, 48620, 2704156, 155117520, 9075135300, 538257874440, 32247603683100, 1946939425648112, 118264581564861424, 7219428434016265740, 442512540276836779204, 27217014869199032015600, 1678910486211891090247320, 103827421287553411369671120, 6435067013866298908421603100

Offset: 0

Benoit Cloitre, Jan 18 2002

For the trisection of a sequence (here A000984) given by its real o.g.f. see a comment and a reference under A187357.

Harry J. Smith, Table of n, a(n) for n = 0..100
R. J. Mathar, The Eggenberger-Polya urn process: Probabilities of revisited ball ratios, vixra:2502.0097 (2025) Table 1

Cf. A001450, A066798, A187364 (binomial(2(3n+1),3n+1)/2), A187365 (binomial(2(3n+2),3n+2)/3!).

[Binomial(6*n, 3*n): n in [0..15]]; // G. C. Greubel, Feb 17 2020

a := n -> hypergeom([-3*n, -3*n], [1], 1):
seq(simplify(a(n)), n=0..13); # Peter Luschny, Mar 19 2018

Table[Binomial[6n, 3n], {n,0,13}] (* Jean-François Alcover, Jun 03 2019 *)

a(n) = { binomial(6*n, 3*n) } \\ Harry J. Smith, Mar 28 2010

[binomial(6*n, 3*n) for n in (0..15)] # G. C. Greubel, Feb 17 2020

a(n) = A000984(3*n).
a(n) = Sum_{i=0..n} Sum_{j=0..n} Sum_{k=0..n} binomial(n, i)*binomial(n, j) *binomial(n, k)*binomial(3n, i+j+k). - Benoit Cloitre, Mar 08 2005
O.g.f. (with a(0):=1): (cb(x^(1/3)) + sqrt(2)*P(x^(1/3))*sqrt(1/P(x^(1/3))+1+2*x^(1/3)))/3, with cb(x):=1/sqrt(1-4*x) (o.g.f. of A000984) and P(x):=P(-1/2,4*x) = 1/sqrt(1+4*x+16*x^2) (o.g.f. of A116091, with P(x,z) the o.g.f. of the Legendre polynomials). - Wolfdieter Lang, Mar 24 2011
D-finite with recurrence n*(3n-1)*(3n-2)*a(n) = 8*(6n-5)*(6n-1)*(2n-1)*a(n-1). - R. J. Mathar, Sep 17 2012
a(n) = GegenbauerC(3*n, -3*n, -1). - Peter Luschny, May 07 2016
a(n) = hypergeom([-3*n, -3*n], [1], 1). - Peter Luschny, Mar 19 2018
a(n) ~ 2^(6*n)/sqrt(3*Pi*n). - Vaclav Kotesovec, Jun 07 2019
From Peter Bala, Feb 16 2020: (Start)
a(m*p^k) == a(m*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers m and k.
a(n) = [(x*y)^(3*n)] (1 + x + y)^(6*n). Cf. A001448. (End)
Conjecture: a(n) = [x^n] G(x)^(2*n), where G(x) = (1 + x)*(1 - 6*x + x^2)/(2*x) + (x^2 - 1)*sqrt(1 - 14*x + x^2)/(2*x) = 1 + 10*x + 81*x^2 + 720*x^3 + .... The algebraic function G(x) satisfies the quadratic equation x*G(x)^2 - (1 - 5*x - 5*x^2 + x^3)*G(x) + (1 + x)^4 = 0. Cf. A001450. - Peter Bala, Oct 27 2022
a(n) = Sum_{k = 0..3*n} binomial(3*n+k-1, k). - Peter Bala, Jun 04 2024
O.g.f: 3F2(1/6,1/2,5/6; 1/3,2/3 ; 64*x). - R. J. Mathar, Jan 11 2025

Extended to a(0)=1 by M. F. Hasler, Oct 06 2014

A098332 Expansion of 1/sqrt(1 - 2x + 9x^2).

Original entry on oeis.org

1, 1, -3, -11, 1, 81, 141, -363, -1791, -479, 13597, 29877, -54911, -353807, -223443, 2539989, 6806529, -8302527, -73999299, -73313931, 489731841, 1584548241, -1110170163, -15812965611, -21391839999, 94696016481

Offset: 0

Paul Barry, Sep 03 2004

Central coefficients of (1 + x - 2*x^2)^n.
Binomial transform of 1/sqrt(1+8*x^2), or (1,0,-4,0,24,0,...).
Binomial transform is A098336.

R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd ed., 1994, ex. 7.56, p. 575.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Hacène Belbachir and Abdelghani Mehdaoui, Recurrence relation associated with the sums of square binomial coefficients, Quaestiones Mathematicae (2021) Vol. 44, Issue 5, 615-624.
Hacène Belbachir, Abdelghani Mehdaoui, and László Szalay, Diagonal Sums in the Pascal Pyramid, II: Applications, J. Int. Seq., Vol. 22 (2019), Article 19.3.5.
Tony D. Noe, On the Divisibility of Generalized Central Trinomial Coefficients, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.7.

Cf. A126869, A012000, A116091, A098341.

a := n -> hypergeom([1/2-n/2, -n/2], [1], -8);
seq(round(evalf(a(n),99)),n=0..30); # Peter Luschny, Sep 18 2014

Table[(-3)^n*LegendreP[n,-1/3],{n,0,20}] (* Vaclav Kotesovec, Jul 23 2013 *)
CoefficientList[Series[1/Sqrt[1 - 2*x + 9*x^2], {x,0,50}], x] (* G. C. Greubel, Feb 18 2017 *)

x='x+O('x^25); Vec(1/sqrt(1 - 2*x + 9*x^2)) \\ G. C. Greubel, Feb 18 2017

E.g.f.: exp(x)*BesselI(0, 2*sqrt(-2)*x);
a(n) = Sum_{k=0..floor(n/2)} binomial(n,2*k) * binomial(2*k,k) * (-2)^k.
a(n) = Sum_{k=0..floor(n/2)} binomial(n,k) * binomial(n-k,k) * (-2)^k.
a(n) = (-1)^n * Sum_{k=0..n} binomial(n,k)^2 * (-2)^k.
G.f.: A(x) = 1/(2*T(0)+3*x-1) where T(k) = 1 - 2*x/(1 + x/T(k+1)); (continued fraction, 2-step ). - Sergei N. Gladkovskii, Aug 23 2012
D-finite with recurrence: a(n+2) = ((2*n+3)*a(n+1))/(n+2) - (9*(n+1)*a(n))/(n+2) with a(0)=1, a(1)=1. (See Graham, Knuth, and Patashnik). - Alexander R. Povolotsky, Aug 23 2012
a(n) = hypergeom([1/2-n/2, -n/2], [1], -8). - Peter Luschny, Sep 18 2014
a(n) = (3/2)*(9/2)^n*Sum_{k >= 0} (-1/2)^k*binomial(n+k,k)^2. - Peter Bala, Mar 02 2017

A187364 Trisection of A000984 (central binomial coefficients): binomial(2(3n+1),3n+1)/2, n>=0.

Original entry on oeis.org

1, 35, 1716, 92378, 5200300, 300540195, 17672631900, 1052049481860, 63205303218876, 3824345300380220, 232714176627630544, 14226520737620288370, 873065282167813104916, 53753604366668088230810, 3318776542511877736535400, 205397724721029574666088520

Offset: 0

Wolfdieter Lang, Mar 10 2011

See a comment under A187363 concerning trisection.

This appears also in the trisection of A001700 (central binomials in the odd numbered Pascal rows): binomial(2*(3*n)+1,3*n+1).

Seiichi Manyama, Table of n, a(n) for n = 0..554

Cf. A066802 (binomial(6n,3n)), A187365 (binomial(2(3n+2),3n+2)/3!).

Cf. A002458, A100033.

Table[c=3n+1;Binomial[2c,c]/2,{n,0,20}] (* Harvey P. Dale, May 10 2012 *)

a(n) = binomial(2*(3*n+1),3*n+1)/2, n>=0.
a(n) = binomial(2*(3*n)+1,3*n+1), n>=0.
O.g.f.: (cb(x^(1/3)) - sqrt(2)*P(x^(1/3))*sqrt(1/P(x^(1/3))-(1+8*x^(1/3))/2))/(6*x^(1/3)), with cb(x):=1/sqrt(1-4*x) (o.g.f. of A000984) and P(x):=P(-1/2,4*x)=1/sqrt(1+4*x+16*x^2) (o.g.f. of A116091, with P(x,z) the o.g.f. of the Legendre polynomials).
From Peter Bala, Mar 19 2023: (Start)
a(n) = (1/2)*Sum_{k = 0..3*n+1} binomial(3*n+1,k)^2.
a(n) = (1/2)*hypergeom([-1 - 3*n, -1 - 3*n], [1], 1).
a(n) = 8*(2*n - 1)*(6*n + 1)*(6*n - 1)/(n*(3*n + 1)*(3*n - 1)) * a(n-1). (End)
Right-hand side of the binomial sum identity (1/18) * Sum_{k = 0..6*n+3} (-1)^(n+k) * (k/(2*n + 1))^2 * binomial(6*n+3, k)^2 = a(n). - Peter Bala, Nov 05 2024

A307884 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals, where column k is the expansion of 1/sqrt(1 + 2(k-1)x + ((k+1)*x)^2).

Original entry on oeis.org

1, 1, 1, 1, 0, 1, 1, -1, -2, 1, 1, -2, -3, 0, 1, 1, -3, -2, 11, 6, 1, 1, -4, 1, 28, 1, 0, 1, 1, -5, 6, 45, -74, -81, -20, 1, 1, -6, 13, 56, -255, -92, 141, 0, 1, 1, -7, 22, 55, -554, 477, 1324, 363, 70, 1, 1, -8, 33, 36, -959, 2376, 2689, -3656, -1791, 0, 1

Offset: 0

Seiichi Manyama, May 02 2019

Column k is the diagonal of the rational function 1 / ((1-x)*(1-y) + k*x*y). - Seiichi Manyama, Jul 11 2020

More generally, column k is the diagonal of the rational function r / ((1-r*x)*(1-r*y) + r-1 + (k-r+1)*r*x*y) for any nonzero real number r. - Seiichi Manyama, Jul 22 2020

			Square array begins:
  1,   1,   1,    1,    1,    1,      1, ...
  1,   0,  -1,   -2,   -3,   -4,     -5, ...
  1,  -2,  -3,   -2,    1,    6,     13, ...
  1,   0,  11,   28,   45,   56,     55, ...
  1,   6,   1,  -74, -255, -554,   -959, ...
  1,   0, -81,  -92,  477, 2376,   6475, ...
  1, -20, 141, 1324, 2689, -804, -20195, ...

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

Columns k=2..4 give (-1)^n * A098332, A116091, (-1)^n * A098341.
Main diagonal gives A307885.
T(n,n-1) gives A335310.
Cf. A307883, A336179.

T[n_, k_] := Sum[If[k == j == 0, 1, (-k)^j] * Binomial[n, j]^2, {j, 0, n}]; Table[T[k, n - k], {n, 0, 10}, {k, 0, n}] // Flatten (* Amiram Eldar, May 13 2021 *)

T(n,k) is the coefficient of x^n in the expansion of (1 - (k-1)*x - k*x^2)^n.
T(n,k) = Sum_{j=0..n} (-k)^j * binomial(n,j)^2.
T(n,k) = Sum_{j=0..n} (-k-1)^(n-j) * binomial(n,j) * binomial(n+j,j).
n * T(n,k) = -(k-1) * (2*n-1) * T(n-1,k) - (k+1)^2 * (n-1) * T(n-2,k).

A187365 Trisection of A000984 (central binomial coefficients): binomial(2(3n+2),3n+2)/3!, n>=0.

Original entry on oeis.org

1, 42, 2145, 117572, 6686100, 388934370, 22974421470, 1372238454600, 82653088824684, 5011211083256840, 305437356823765089, 18697712969443807572, 1148770108115543559100, 70797430141465286938140, 4374750896947475198160300, 270950190057528375091435920

Offset: 0

Wolfdieter Lang, Mar 10 2011

See a comment under A187357 concerning trisection.

This appears also in the trisection of A001700: binomial(2*(3*n+1)+1,(3*n+1)+1)/3.

Seiichi Manyama, Table of n, a(n) for n = 0..554

Cf. A066802 binomial(6n,3n), A187364 binomial(2*(3n+1),3n+1)/2, A002458, A100033.

a(n)=binomial(2*(3*n+2),3*n+2)/3!, n>=0.
a(n)=binomial(3*(2*n+1),3*n+2)/3, n>=0.
O.g.f.:(cb(x^(1/3)) - sqrt(2)*P(x^(1/3))*sqrt(1/P(x^(1/3))-(1-4*x^(1/3))/2))/(18*x^(2/3)),
with cb(x):=1/sqrt(1-4*x) (o.g.f. of A000984) and P(x):=P(-1/2,4*x)=1/sqrt(1+4*x+16*x^2) (o.g.f. of A116091, with P(x,z)the o.g.f. of the Legendre polynomials).
From Peter Bala, Mar 19 2023: (Start)
a(n) = (1/6)*Sum_{k = 0..3*n+2} binomial(3*n+2,k)^2.
a(n) = (1/6)*hypergeom([-2 - 3*n, -2 - 3*n], [1], 1).
a(n) = 8*(2*n + 1)*(6*n + 1)*(6*n - 1)/(n*(3*n + 1)*(3*n + 2)) * a(n-1). (End)

A012000 Expansion of 1/sqrt(1 - 4x + 16x^2).

Original entry on oeis.org

1, 2, -2, -28, -74, 92, 1324, 3656, -4826, -70228, -197372, 267896, 3921724, 11126936, -15347432, -225505648, -643622906, 897078476, 13214495764, 37869162392, -53170602284, -784672445368, -2255295815192, 3183829452272, 47051201187676, 135537088268792, -192142210448216

Offset: 0

Wouter Meeussen

sign

			G.f. = 1 + 2*x - 2*x^2 - 28*x^3 - 74*x^4 + 92*x^5 + 1324*x^6 + 3656*x^7 + ...

T. D. Noe, Table of n, a(n) for n = 0..200
Hacène Belbachir and Abdelghani Mehdaoui, Recurrence relation associated with the sums of square binomial coefficients, Quaestiones Mathematicae (2021) Vol. 44, Issue 5, 615-624.
Hacène Belbachir, Abdelghani Mehdaoui, and László Szalay, Diagonal Sums in the Pascal Pyramid, II: Applications, J. Int. Seq., Vol. 22 (2019), Article 19.3.5.
Tony D. Noe, On the Divisibility of Generalized Central Trinomial Coefficients, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.7.

Cf. A116091, A098332, A126869.

a := n -> 4^n*hypergeom([-n,1+n],[1],1/4);
seq(simplify(a(n)),n=0..26); # Peter Luschny, May 09 2016

Table[ 2^(2n) LegendreP[ n, 1/2 ], {n, 12} ]

{a(n) = 2^(2*n) * subst( pollegendre(n), x, 1/2)} /* Michael Somos, Dec 03 2001 */

a(n) = polcoeff(((1 - x)*(1 + 3*x))^n, n); \\ Michel Marcus, Aug 16 2015

G.f.: 1/sqrt(1 - 4*x + 16*x^2). - Vladeta Jovovic, May 13 2003
Scaled Legendre polynomials evaluated at 1/2: 2^(2n)P(n, 1/2). - Michael Somos, Dec 03 2001
a(n) = (-1)^n*Sum_{k=0..n} C(n,k)^2*(-3)^k. - Benoit Cloitre, Oct 25 2003
a(n) = Sum_{k=0..floor(n/2)} C(n,k)*C(2(n-k),n)(-4)^k. - Paul Barry, Sep 08 2004
D-finite with recurrence: n*a(n) + 2*(1-2*n)*a(n-1) + 16*(n-1)*a(n-2) = 0. - R. J. Mathar, Nov 14 2011
G.f.: G(0), where G(k) = 1 + x*(1-x)*(4*k+1)/( 4*k+2 - x*(1-x)*(4*k+2)*(4*k+3)/(x*(1-x)*(4*k+3) + 4*(k+1)/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Jul 06 2013
a(n) ~ 2^(2*n+1)*sin(Pi/12*(4*n+5))/(3^(1/4)*sqrt(Pi*n)). - Vaclav Kotesovec, Jul 30 2013
a(n) = [x^n] ((1 - x)*(1 + 3*x))^n. - Peter Bala, Aug 16 2015
a(n) = 4^n*hypergeom([-n,1+n], [1], 1/4). - Peter Luschny, May 09 2016
From Peter Bala, Nov 28 2021: (Start)
a(n) = 4^(2*n+1)*Sum_{k >= n} (-1)^(n+k)*C(k,n)^2*(1/3)^(k+1).
a(n) = (-1)^n*hypergeom([-n,-n], [1], -3).
a(n) = (4/3)*(16/3)^n*hypergeom([n+1,n+1], [1], -1/3). (End)

More terms from Vladeta Jovovic, May 13 2003

A098341 Expansion of 1/sqrt(1 - 6x + 25x^2).

Original entry on oeis.org

1, 3, 1, -45, -255, -477, 2689, 25203, 82945, -90045, -2379519, -11581677, -12063999, 197669475, 1423716225, 3645266355, -12180238335, -156702949245, -626511576575, 51239061075, 15179398450945, 87687927568035, 151934475887745

Offset: 0

Paul Barry, Sep 03 2004

Central coefficients of (1 + 3*x - 4*x^2)^n.
(-1)^n*a(n) is the sum of squares of coefficients of (1+2*i*x)^n where i=sqrt(-1) (see PARI code). - Joerg Arndt, Jul 06 2011
Binomial transform of A098337.
Second binomial transform of A098334.

Seiichi Manyama, Table of n, a(n) for n = 0..1000
Hacène Belbachir and Abdelghani Mehdaoui, Recurrence relation associated with the sums of square binomial coefficients, Quaestiones Mathematicae (2021) Vol. 44, Issue 5, 615-624.
Hacène Belbachir, Abdelghani Mehdaoui, and László Szalay, Diagonal Sums in the Pascal Pyramid, II: Applications, J. Int. Seq., Vol. 22 (2019), Article 19.3.5.
Tony D. Noe, On the Divisibility of Generalized Central Trinomial Coefficients, Journal of Integer Sequences, Vol. 9 (2006), Article 06.2.7.

Cf. A098332, A012000, A116091, A126869.

Table[(-5)^n*LegendreP[n,-3/5],{n,0,20}] (* Vaclav Kotesovec, Jul 23 2013 *)
CoefficientList[Series[1/Sqrt[1-6x+25x^2],{x,0,30}],x] (* Harvey P. Dale, Aug 22 2014 *)

a(n)={local(v=Vec((1+2*I*x)^n)); (-1)^n*sum(k=1,#v,v[k]^2);} /* Joerg Arndt, Jul 06 2011 */

a(n)={local(v=Vec((1+2*I*x)^n)); sum(k=1,#v, real(v[k])^2-imag(v[k])^2);} /* Joerg Arndt, Jul 06 2011 */

A098341 = lambda n: (-1)^n*hypergeometric([-n,-n], [1], -4)
[Integer(A098341(n).n(100)) for n in (0..22)] # Peter Luschny, Sep 23 2014

E.g.f.: exp(3*x)*BesselI(0, 4*I*x), I=sqrt(-1).
a(n) = (-1)^n*Sum_{k=0..n} binomial(n, k)^2*(-4)^k.
a(n) = (-1)^n*hypergeometric([-n,-n], [1], -4). - Peter Luschny, Sep 23 2014
D-finite with recurrence: n*a(n) +3*(-2*n+1)*a(n-1) +25*(n-1)*a(n-2)=0. - R. J. Mathar, Nov 27 2014
From Peter Bala, Nov 28 2021: (Start)
a(n) = (5^n)*P(n,3/5), where P(n,x) is the n-th Legendre polynomial.
a(n) = [x^n] ((1 - x)*(1 + 4*x))^n.
a(n) = 5^(2*n+1)*Sum_{k >= n} (-1)^(n+k)*binomial(k,n)^2*(1/4)^(k+1).
a(n) = (5/4)*(25/4)^n*hypergeom([n+1, n+1], [1], -1/4). (End)

A335310 a(n) = Sum_{k=0..n} binomial(n,k) * binomial(n+k,k) * (-n)^(n-k).

Original entry on oeis.org

1, 1, -2, 11, -74, 477, -804, -84425, 3315334, -102211207, 3005297956, -88338323709, 2627003399164, -78764141488043, 2341929797646648, -66394419743289105, 1609460569459689286, -18001777147777896975, -1625299659961386724524, 196005371138608184827003

Offset: 0

Ilya Gutkovskiy, May 31 2020

sign

Seiichi Manyama, Table of n, a(n) for n = 0..386

Cf. A098332, A116091, A126869, A307884, A307885, A331657, A335309.

Join[{1}, Table[Sum[Binomial[n, k] Binomial[n + k, k] (-n)^(n - k), {k, 0, n}], {n, 1, 19}]]
Table[SeriesCoefficient[1/Sqrt[1 + 2 (n - 2) x + n^2 x^2], {x, 0, n}], {n, 0, 19}]
Table[n! SeriesCoefficient[Exp[(2 - n) x] BesselI[0, 2 Sqrt[1 - n] x], {x, 0, n}], {n, 0, 19}]
Table[Hypergeometric2F1[-n, -n, 1, 1 - n], {n, 0, 19}]

a(n) = sum(k=0, n, binomial(n,k)^2*(1-n)^k); \\ Michel Marcus, Jun 01 2020

a(n) = central coefficient of (1 - (n - 2)*x - (n - 1)*x^2)^n.
a(n) = [x^n] 1 / sqrt(1 + 2*(n - 2)*x + n^2*x^2).
a(n) = n! * [x^n] exp((2 - n)*x) * BesselI(0,2*sqrt(1 - n)*x).
a(n) = Sum_{k=0..n} binomial(n,k)^2 * (1-n)^k.

A336729 G.f. A(x) satisfies: A(x) = 1 + x * A(x) / (1 + 3 * x * A(x)).

Original entry on oeis.org

1, 1, -2, 1, 10, -38, 28, 289, -1262, 1054, 11044, -51302, 45604, 482068, -2319176, 2140129, 22753378, -111964106, 105927508, 1130780062, -5652760340, 5444054956, 58291068808, -294808277414, 287740874260, 3088109246572, -15758505143192, 15541351662484, 167103084713608

Offset: 0

Seiichi Manyama, Aug 02 2020

sign

Seiichi Manyama, Table of n, a(n) for n = 0..1000

Column k=3 of A336727.

Cf. A007564, A116091.

a[0] = 1; a[n_] := Sum[(-3)^(n - k) * Binomial[n, k] * Binomial[n , k - 1], {k, 1, n}] / n; Array[a, 29, 0] (* Amiram Eldar, Aug 02 2020 *)

{a(n) = local(A=1+x*O(x^n)); for(i=0, n, A=1+x*A/(1+3*x*A)); polcoef(A, n)}

{a(n) = if(n==0, 1, sum(k=1, n, (-3)^(n-k)*binomial(n, k)*binomial(n, k-1))/n)}

N=40; x='x+O('x^N); Vec(2/(1-4*x+sqrt(1+4*x+16*x^2)))

{a(n) = sum(k=0, n, (-3)^k*4^(n-k)*binomial(n, k)*binomial(n+k, n)/(k+1))}

a(0) = 1 and a(n) = (1/n) * Sum_{k=1..n} (-3)^(n-k) * binomial(n,k) * binomial(n,k-1) for n > 0.
G.f.: 2/(1 - 4*x + sqrt(1 + 4*x + 16*x^2)).
a(n) = Sum_{k=0..n} (-3)^k * 4^(n-k) * binomial(n,k) * binomial(n+k,n)/(k+1).
(n+1) * a(n) = -2 * (2*n-1) * a(n-1) - 16 * (n-2) * a(n-2) for n>1. - Seiichi Manyama, Aug 08 2020
a(n) ~ 2^(2*n - 1/2) * ((sqrt(3) + 1)*sin(2*Pi*n/3) + (sqrt(3) - 1)*cos(2*Pi*n/3)) / (3^(3/4) * sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Dec 04 2020

A116092 Expansion of 1/sqrt(1+8x+64x^2).

Original entry on oeis.org

1, -4, -8, 224, -1184, -2944, 84736, -467968, -1235456, 35956736, -202108928, -548651008, 16063381504, -91151859712, -251452325888, 7389369073664, -42180470767616, -117581870006272, 3464100777558016, -19854347412176896, -55753417460547584, 1645577388148391936

Offset: 0

Paul Barry, Feb 04 2006

8th binomial transform is expansion of 1/sqrt(1-8*x+64*x^2).

G. C. Greubel, Table of n, a(n) for n = 0..1000
Hacène Belbachir, Abdelghani Mehdaoui, László Szalay, Diagonal Sums in the Pascal Pyramid, II: Applications, J. Int. Seq., Vol. 22 (2019), Article 19.3.5.

Cf. A116091.

List([0..30], n-> 2^n*Sum([0..n], k-> (-3)^k*Binomial(n,k)* Binomial(n, n-k))); # G. C. Greubel, May 10 2019

R:=PowerSeriesRing(Rationals(), 30); Coefficients(R!( 1/Sqrt(1+8*x+64*x^2) )); // G. C. Greubel, May 10 2019

CoefficientList[Series[1/Sqrt[1+8*x+64*x^2], {x, 0, 30}], x] (* G. C. Greubel, May 10 2019 *)

my(x='x+O('x^30)); Vec(1/sqrt(1+8*x+64*x^2)) \\ G. C. Greubel, May 10 2019

(1/sqrt(1+8*x+64*x^2)).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, May 10 2019

E.g.f.: exp(-4*x)*Bessel_I(0, 2*sqrt(-12)*x).
a(n) = 2^n*Sum_{k=0..n} C(n,n-k)*C(n,k)*(-3)^k.
a(n) = 2^n*A116091(n).
D-finite with recurrence: n*a(n) +4*(2*n-1)*a(n-1) +64*(n-1)*a(n-2)=0. - R. J. Mathar, Nov 07 2012

cp's OEIS Frontend

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A116092 Expansion of 1/sqrt(1+8x+64x^2).