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Search: a100033 -id:a100033
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A002458
a(n) = binomial(4*n+1, 2*n).
+10
20
1, 10, 126, 1716, 24310, 352716, 5200300, 77558760, 1166803110, 17672631900, 269128937220, 4116715363800, 63205303218876, 973469712824056, 15033633249770520, 232714176627630544, 3609714217008132870, 56093138908331422716, 873065282167813104916
(list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
REFERENCES
The right-hand side of a binomial coefficient identity in H. W. Gould, Combinatorial Identities, Morgantown, 1982, (3.109), page 35.
LINKS
T. D. Noe, Table of n, a(n) for n = 0..100
FORMULA
a(n) = Sum_{k=0..n} 4^k * binomial( n + k, n) * binomial( 2*n - 2*k, n - k). - Michael Somos, Feb 25 2012
a(n) = A001700(2*n) = (n+1)*A000108(2*n+1).
G.f.: (4 - (1+4*y)*c(y) - (1-4*y)*c(-y))/(2*(1 - (4*y)^2)) with y^2 = x, c(y) = g.f. for A000108 (Catalan). - Wolfdieter Lang, Dec 13 2001
a(n) ~ 2^(1/2)*Pi^(-1/2)*n^(-1/2)*2^(4*n)*{1 - 5/16*n^-1 + ...}. - Joe Keane (jgk(AT)jgk.org), Jun 11 2002
a(n) = A024492(n)*(n+1). - R. J. Mathar, Aug 10 2015
G.f.: 2F1(3/4,5/4; 3/2; 16*x). - R. J. Mathar, Aug 10 2015
D-finite with recurrence n*(2*n + 1)*a(n) - 2*(4*n - 1)*(4*n + 1)*a(n-1) = 0. - R. J. Mathar, Aug 10 2015
From Peter Bala, Nov 04 2015: (Start)
a(n) = 4^n*binomial(2*n + 1/2, n).
O.g.f.: sqrt(c(4*x)/(1 - 16*x)) = sqrt(2/(1 - 16*x)/(1 + sqrt(1 - 16*x))), where
c(y) = g.f. for A000108 (Catalan). In general, c(x)^k/sqrt(1 - 4*x) is the o.g.f. for the sequence binomial(2*n + k, n). (End) [Edited by Petros Hadjicostas, May 25 2020]
From Ilya Gutkovskiy, Jan 17 2017: (Start)
E.g.f.: 2F2(3/4,5/4; 1,3/2; 16*x).
Sum_{n>=0} 1/a(n) = 3F2(1,1,3/2; 3/4,5/4; 1/16) = 1.108563435104316693... (End)
From Peter Bala, Mar 16 2018: (Start)
The right-hand side of the binomial coefficient identity Sum_{k = 0..n} 4^(n-k) * C(2*n+1, 2*k) * C(2*k, k) = a(n).
a(n) = 4^n*hypergeom([-n, -n-1/2], [1], 1). (End)
From Peter Bala, Mar 20 2023: (Start)
a(n) = Sum_{k = 0..n} binomial(2*n+1,k)^2.
a(n) = (1/2)*hypergeom([-1 - 2*n, -1 - 2*n], [1], 1). (End)
EXAMPLE
1 + 10*x + 126*x^2 + 1716*x^3 + 24310*x^4 + 352716*x^5 + 5200300*x^6 + ...
MAPLE
A002458:=n->binomial(4*n+1, 2*n): seq(A002458(n), n=0..30); # Wesley Ivan Hurt, Jan 17 2017
MATHEMATICA
Table[Binomial[4n+1, 2n], {n, 0, 30}] (* Harvey P. Dale, Apr 04 2011 *)
4^Range[0, 22] Simplify[ CoefficientList[ Series[ Sqrt[2]/(((Sqrt[1 - 4 x] + 1)^(1/2))*Sqrt[1 - 4 x]), {x, 0, 22}], x]] (* Robert G. Wilson v, Aug 08 2011 *)
PROG
(PARI) a(n) = binomial( 4*n + 1, 2*n)
CROSSREFS
Cf. A000984, A001448, A001700, A024492, A100033, A187364, A187365.
Row sums of A067001.
KEYWORD
nonn,easy,nice
AUTHOR
N. J. A. Sloane
STATUS
approved
A187364
Trisection of A000984 (central binomial coefficients): binomial(2(3n+1),3n+1)/2, n>=0.
+10
10
1, 35, 1716, 92378, 5200300, 300540195, 17672631900, 1052049481860, 63205303218876, 3824345300380220, 232714176627630544, 14226520737620288370, 873065282167813104916, 53753604366668088230810, 3318776542511877736535400, 205397724721029574666088520
(list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
COMMENTS
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).
LINKS
Seiichi Manyama, Table of n, a(n) for n = 0..554
FORMULA
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)
MATHEMATICA
Table[c=3n+1; Binomial[2c, c]/2, {n, 0, 20}] (* Harvey P. Dale, May 10 2012 *)
CROSSREFS
A066802 binomial(6n,3n), A187365 binomial(2(3n+2),3n+2)/3!.
Cf. A002458, A100033.
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Mar 10 2011
STATUS
approved
A187365
Trisection of A000984 (central binomial coefficients): binomial(2(3n+2),3n+2)/3!, n>=0.
+10
8
1, 42, 2145, 117572, 6686100, 388934370, 22974421470, 1372238454600, 82653088824684, 5011211083256840, 305437356823765089, 18697712969443807572, 1148770108115543559100, 70797430141465286938140, 4374750896947475198160300, 270950190057528375091435920
(list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
COMMENTS
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.
LINKS
Seiichi Manyama, Table of n, a(n) for n = 0..554
FORMULA
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)
CROSSREFS
Cf. A066802 binomial(6n,3n), A187364 binomial(2*(3n+1),3n+1)/2, A002458, A100033.
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Mar 10 2011
STATUS
approved
A259557
a(n) = binomial(4*n-1, 2*n).
+10
1
1, 3, 35, 462, 6435, 92378, 1352078, 20058300, 300540195, 4537567650, 68923264410, 1052049481860, 16123801841550, 247959266474052, 3824345300380220, 59132290782430712, 916312070471295267, 14226520737620288370
(list; graph; refs; listen; history; text; internal format)
OFFSET
0,2
COMMENTS
Essentially the same as A100033.
LINKS
Table of n, a(n) for n=0..17.
V. V. Kruchinin and D. V. Kruchinin, A Generating Function for the Diagonal T_{2n,n} in Triangles, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.6.
FORMULA
G.f. A(x)=1+x*B(x)'/B(x), where B(x) is g.f. of A079489.
a(n) = A100033(n-1) for n>0.
D-finite with recurrence n*(2*n-1)*a(n) -2*(4*n-1)*(4*n-3)*a(n-1)=0. - R. J. Mathar, Jul 06 2015
a(n) = [x^(2*n)] 1/(1 - x)^(2*n). - Ilya Gutkovskiy, Oct 10 2017
From Peter Bala, Jun 11 2023: (Start)
a(n) = (1/2) * [x^n] ( (1 + x)^2/( 1 - x) )^(2*n) for n >= 1.
Right-hand side of the identity (1/2)*Sum_{k = 0..n} binomial(4*n,k)*binomial(3*n-k-1,n-k) = binomial(4*n-1,2*n) for n >= 1.
a(n) = [x^n] E(x)^n, where E(x) = exp( Sum_{k >= 1} A119259(k)*x^k/k ). (End)
a(n) = Sum_{k = 0..2*n} (-1)^k*binomial(-n, k)*binomial(-3*n-k, 2*n-k) = Sum_{k = 0..2*n} (-1)^k*binomial(n+k-1, k)*binomial(5*n-1, 2*n-k). - Peter Bala, Jun 08 2024
MATHEMATICA
Table[Binomial[4 n - 1, 2 n], {n, 0, 30}] (* Vincenzo Librandi, Jul 01 2015 *)
PROG
(PARI) vector(20, n, n--; binomial(4*n-1, 2*n)) \\ Michel Marcus, Jul 01 2015
(Magma) [Binomial(4*n-1, 2*n): n in [0..20]]; // Vincenzo Librandi, Jul 01 2015
CROSSREFS
Cf. A001700, A079489, A100033.
KEYWORD
nonn,easy
AUTHOR
Vladimir Kruchinin, Jun 30 2015
STATUS
approved
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