A008977 -id:A008977 - cp's OEIS frontend

A184898 a(n) = C(2n,n) * (8^n/n!^2) * Product_{k=0..n-1} (8k+1)*(8k+7).

1, 112, 90720, 105100800, 142542960000, 211337613527040, 331831362513530880, 542307255307827609600, 912855634598629193472000, 1571864775032876891607040000, 2755743023914838714304931102720

Offset: 0

Paul D. Hanna, Jan 25 2011

nonn

			G.f.: A(x) = 1 + 112*x + 90720*x^2 + 105100800*x^3 +...
A(x)^(1/2) = 1 + 56*x + 43792*x^2 + 50098048*x^3 +...+ A184897(n)*x^n +...

Cf. A184897; variants: A184423, A008977, A184892, A001421, A184896.

{a(n)=(2*n)!/n!^2*(8^n/n!^2)*prod(k=0,n-1,(8*k+1)*(8*k+7))}

Self-convolution of A184897, where A184897(n) = (8^n/n!^2) * Product_{k=0..n-1} (16k+1)*(16k+7).

a(n) ~ sqrt(2-sqrt(2)) * 2^(11*n - 1) / (Pi^(3/2) * n^(3/2)). - Vaclav Kotesovec, Oct 05 2020

A370294 G.f.: exp(Sum_{k>=1} (4k)!/(4!k!^4) * x^k/k).

Original entry on oeis.org

1, 1, 53, 5186, 663444, 98703235, 16179000550, 2837251240021, 522937525075783, 100134345595461824, 19762585810520535829, 3997199042964419204924, 825055790810846248226675, 173231819660726985218760834, 36906136513918240767383588700, 7962139696794640558535530147729

Offset: 0

Vaclav Kotesovec, Feb 14 2024

nonn

Cf. A008977, A082368, A229452, A370295, A333042.

CoefficientList[Series[Exp[Sum[(4*k)!/(4!*k!^4)*x^k/k, {k, 1, 20}]], {x, 0, 20}], x]
CoefficientList[Series[Exp[x*HypergeometricPFQ[{1, 1, 5/4, 3/2, 7/4}, {2, 2, 2, 2}, 256*x]], {x, 0, 20}], x]

G.f. A(x) = G(x)^(1/24), where G(x) is the g.f. for A333042.

a(n) ~ c * 4^(4*n)/n^(5/2), where c = exp(HypergeometricPFQ[{1, 1, 5/4, 3/2, 7/4}, {2, 2, 2, 2}, 1] / 256) / (24*sqrt(2)*Pi^(3/2)) = 0.005320414767134132512371690902604699480645296829596277834542636529157577...

A333042 G.f.: exp(Sum_{k>=1} (4k)!/k!^4 x^k/k).

Original entry on oeis.org

1, 24, 1548, 155744, 19893054, 2937661200, 477691374152, 83161733788992, 15230338934722749, 2900395347525785464, 569718535329796732476, 114759815105897160007392, 23602808330272138320592494, 4940203531008336735249385488, 1049571237547858314991495867848

Offset: 0

Vaclav Kotesovec, Mar 06 2020

From Peter Bala, Feb 08 2023: (Start)
Let A(x) denote the o.g.f. of the sequence. The sequence defined by b(n) := [x^n] A(x)^n for n >= 1 begins [24, 3672, 703968, 149835864, 33911355024, 7993981771488, 1940145241321920, ...]. We conjecture that b(n) satisfies the supercongruences b(n*p^r) == b(n*p^(r-1)) ( mod p^(3*r) ) for prime p >= 5 and all positive integers n and r.
More generally, for a positive integer m, set A_m(x) = exp( Sum_{n >= 1} (m*n)!/(n!^m) * x^n/n ) and define a sequence {b_m(n): n >= 1} by b_m(n) := [x^n] A_m(x)^n. Then we conjecture that b_m(n) is an integer sequence satisfying the same supercongruences. (End)

Cf. A000108, A008977, A229451, A229452, A333043, A370294.

CoefficientList[Series[Exp[Sum[(4*k)!/k!^4*x^k/k, {k, 1, 20}]], {x, 0, 20}], x]
CoefficientList[Series[Exp[24*x*HypergeometricPFQ[{1, 1, 5/4, 3/2, 7/4}, {2, 2, 2, 2}, 256*x]], {x, 0, 20}], x] (* Vaclav Kotesovec, Feb 09 2024 *)

a(n) ~ c * 4^(4*n)/n^(5/2), where c = exp(3*HypergeometricPFQ[{1, 1, 5/4, 3/2, 7/4}, {2, 2, 2, 2}, 1] / 32) / (sqrt(2)*Pi^(3/2)) = 0.14496966... - Vaclav Kotesovec, Mar 06 2020, updated Feb 16 2024

a(0) = 1; a(n) = (1/n) * Sum_{k=1..n} A008977(k) * a(n-k). - Seiichi Manyama, Feb 09 2024

A361032 Square array read by ascending antidiagonals: T(n,k) = F(n) * (4k)!/(k!(k + n + 1)!^3), where F(n) = (1/8)(4n + 4)!/(n + 1)!; n, k >= 0.

Original entry on oeis.org

3, 315, 9, 46200, 280, 280, 7882875, 17325, 3675, 17325, 1466593128, 1513512, 116424, 116424, 1513512, 288592936632, 162954792, 5885880, 2134440, 5885880, 162954792, 59064793444800, 20193091776, 399072960, 67953600, 67953600, 399072960, 20193091776, 12445136556298875

Offset: 0

Peter Bala, Mar 04 2023

The Catalan numbers A000108 are given by the formula Catalan(k) = (2*k)!/(k!*(k + 1)!). Gessel (1992) considered generalized Catalan numbers defined by Catalan(n,k) = J(n) * (2*k)!/(k!*(k + n + 1)!), where J(n) = (2*n + 2)!/(2*(n + 1)!) = (2^n)*Product_{j = 0..n} (2*j + 1) is chosen so that these numbers are always integers. Gessel's generalized Catalan numbers are particular cases of super ballot numbers. See A135573 for a table of these generalized Catalan numbers.
For this table we carry out an analogous construction using the numbers B(k) = (4*k)!/k!^4 = A008977(k) in place of the central binomial numbers (2*k)!/k!^2. We define sequences {B(n,k) : k >= 0}, n = 0, 1, 2, ..., by B(n,k) = F(n) * (4*k)!/(k!*(k + n + 1)!^3), where choosing F(n) = (4*n + 4)!/(8*(n + 1)!) = (1/2)*(4^n)*Product_{j = 0..n} (4*j + 1)*(4*j + 2)*(4*j + 3) appears to produce integer values for these quantities. The rows of the square array below are the sequences  {B(0,k)}, {B(1,k)}, {B(2,k)}, ....
An alternative expression is B(n,k) =  G(n,k) * B(n+k+1), where G(n,k) = (1/8)*Product_{j = 0..n} ( (4*j + 1)*(4*j + 2)*(4*j + 3)/((4*k + 4*j + 1)*(4*k + 4*j + 2)*(4*k + 4*j + 3)) ).

			The square array with rows n >= 0 and columns k >= 0 begins:
  n\k|          0          1          2           3           4 ...
  ----------------------------------------------------------------------
   0 |          3          9        280       17325     1513512 ...
   1 |        315        280       3675      116424     5885880 ...
   2 |      46200      17325     116424     2134440    67953600 ...
   3 |    7882875    1513512    5885880    67953600  1449322875 ...
   4 | 1466593128  162954792  399072960  3086579925 46235189000 ...
   5 |  ...
  ...
As a triangle:
 Row
  0 |             3
  1 |           315          9
  2 |         46200        280      280
  3 |       7882875      17325     3675    17325
  4 |    1466593128    1513512   116424   116424  1513512
  5 |  288592936632  162954792  5885880  2134440  5885880  162954792

Andrew Howroyd, Table of n, a(n) for n = 0..1325 (first 51 antidiagonals)
Ira M. Gessel, Super ballot numbers, J. Symbolic Comp., 14 (1992), 179-194.

A361033 (row 0), A361034 (row 2), A361035 (row 3).

Cf. A008977, A135573, A361027.

# as a square array
T := proc (n, k) (-1)^k*(1/8)*256^(n+1+k)*binomial(n+1/4, n+1+k)*binomial(n+2/4, n+1+k)* binomial(n+3/4, n+1+k); end proc:
for n from 0 to 10 do seq(T(n, k), k = 0..10); end do;
# as a triangle
T := proc (n, k) (-1)^k*(1/8)*256^(n+1+k)*binomial(n+1/4, n+1+k)*binomial(n+2/4, n+1+k)* binomial(n+3/4, n+1+k); end proc:
for n from 0 to 10 do seq(T(n-k, k), k = 0..n); end do;

T(n,k) = (-1)^k*(1/8)*256^(n+k+1)*binomial(n+1/4, n+k+1)*binomial(n+1/2, n+k+1)* binomial(n+3/4, n+k+1) \\ Andrew Howroyd, Jan 05 2024

T(n,k) = (-1)^k*(1/8)*256^(n+k+1)*binomial(n+1/4, n+k+1)*binomial(n+1/2, n+k+1)* binomial(n+3/4, n+k+1).
P-recursive: (n + k + 1)^3*T(n,k) = 4*(4*k - 1)*(4*k - 2)*(4*k - 3)*T(n,k-1) with T(n,0) = (1/8)*(4*n + 4)!/(n + 1)!^4 = (1/8)*A008977(n+1).
(n + k + 1)^3*T(n,k) = 4*(4*n + 1)*(4*n + 2)*(4*n + 3)*T(n-1,k) with T(0,k) = 3*(4*k)!/(k!*(k+1)!^3) = A361033(k).
T(n,k) = (1/2) * (1/(2*Pi))^3 * 256^(n+k+1) * Integral_{x = 0..1} (1 - x)^(n+1/4)*x^(k-1/4) dx * Integral_{x = 0..1} (1 - x)^(n+1/2)*x^(k-1/2) dx * Integral_{x = 0..1} (1 - x)^(n+3/4)*x^(k-3/4) dx.

A361033 a(n) = 3(4n)!/(n!*(n+1)!^3).

Original entry on oeis.org

3, 9, 280, 17325, 1513512, 162954792, 20193091776, 2768662192725, 409716429837000, 64358256798795960, 10605621798062141760, 1817833036248401270280, 321997225483126007438400, 58649494641569379926280000, 10941649720331183519046796800, 2084191938036600263793119045925

Offset: 0

Peter Bala, Mar 01 2023

Row 0 of A361032.
The central binomial numbers A000984(n) = (2*n)!/n!^2 have the property that A000984(n) is divisible by n + 1 and the result (2*n)!/(n!*(n+1)!) is the n-th Catalan number A000108(n). Similarly, the  numbers A008977(n) = (4*n)!/n!^4 appear to have the property that 3*A008977(n) is divisible by (n + 1)^3, leading to the present sequence. Cf. A361028. Do these numbers have a combinatorial interpretation?
Conjecture: a(n) is odd iff n = 2^k - 1 for some k >= 0.

Cf. A000108, A007226, A007228, A008977, A361028, A361032, A361034, A361035.

Maple
```
seq(3*(4*n)!/(n!*(n+1)!^3), n = 0..20);
```

Table[3 (4n)!/(n! ((n+1)!)^3),{n,0,15}] (* Harvey P. Dale, Jul 30 2024 *)

a(n) = 3*A008977(n)/(n+1)^3.
a(n) = (3/4)*A008977(n+1)/((4*n+1)*(4*n+2)*(4*n+3)).
a(n) = (1/2)*A007228(n)*A007226(n)*A000108(n).
P-recursive: a(n) = 4*(4*n-1)*(4*n-2)*(4*n-3)/(n+1)^3 * a(n-1) with a(0) = 3.
The o.g.f. A(x) satisfies the differential equation
x^3*(1 - 256*x)*A(x)''' + x^2*(6 - 1152*x)*A(x)'' + x*(7 - 816*x)*A(x)' + (1 - 24*x)*A(x) - 3 = 0 with A(0) = 3, A'(0) = 9 and A''(0) = 560.
a(n) ~ 3*sqrt(1/(2*Pi^3)) * 2^(8*n)/n^(9/2).

A367568 a(n) = Product_{k=0..n} (4*k)! / k!^4.

Original entry on oeis.org

1, 24, 60480, 22353408000, 1409672968704000000, 16539333509029163728896000000, 38185078618454141182825889242546176000000, 18043150250179542387558306410182977707728856678400000000, 1796395750154420920494206475343190362781863323574704301041254400000000000

Offset: 0

Vaclav Kotesovec, Nov 23 2023

nonn

Cf. A000178, A008977, A268505, A168488, A268196, A262261.

Cf. A007685, A367567, A367569, A367570, A367571.

Table[Product[(4*k)!/k!^4, {k, 0, n}], {n, 0, 10}]
Table[Product[Binomial[4*k,k] * Binomial[3*k,k] * Binomial[2*k,k], {k, 0, n}], {n, 0, 10}]

a(n) = Product_{k=0..n} binomial(4*k,k) * binomial(3*k,k) * binomial(2*k,k).
a(n) = A268505(n) / A000178(n)^4.
a(n) = A268505(n) / A168488(n).
a(n) = A007685(n) * A268196(n) * A262261(n).
a(n) ~ A^(15/4) * sqrt(Gamma(1/4)) * 2^(4*n^2 + 7*n/2 - 7/6) * exp(3*n/2 - 5/16) / (n^(3*n/2 + 17/16) * Pi^(3*n/2 + 7/4)), where A is the Glaisher-Kinkelin constant A074962.

A378130 Decimal expansion of 24L^2/(5^(7/4)Pi^2), where L is the lemniscate constant (A062539).

Original entry on oeis.org

9, 9, 9, 9, 9, 6, 3, 8, 3, 1, 5, 9, 0, 8, 4, 1, 2, 7, 7, 7, 2, 7, 6, 3, 4, 9, 9, 1, 8, 4, 7, 0, 6, 1, 1, 2, 8, 0, 8, 9, 4, 3, 4, 8, 8, 7, 7, 0, 3, 5, 9, 6, 6, 1, 3, 2, 9, 0, 9, 5, 9, 5, 0, 4, 9, 2, 6, 8, 1, 5, 2, 7, 3, 9, 9, 2, 1, 6, 4, 9, 2, 2, 9, 9, 3, 7, 4, 7, 9, 1

Offset: 0

Paolo Xausa, Nov 18 2024

			0.999996383159084127772763499184706112808943488770...

Wikipedia, Lemniscate constant (includes this constant).
Index to sequences related to the Gamma function.

Cf. A008977, A062539, A091670.

Cf. A377999, A378128, A378129, A378129, A378131, A378132.

First[RealDigits[12*Pi/(5^(7/4)*Gamma[3/4]^4), 10, 100]] (* or *)
First[RealDigits[Sum[((-1)^k/6635520^k)*(4*k)!/k!^4, {k, 0, Infinity}], 10, 100]] (* or *)
First[RealDigits[HypergeometricPFQ[{1/4, 1/2, 3/4}, {1, 1}, -1/25920], 10, 100]]

Equals 12*Pi/(5^(7/4)*Gamma(3/4)^4) = 12*A091670/5^(7/4).
Equals Sum_{k >= 0} ((-1)^k/6635520^k)*(4*k)!/(k!)^4 = Sum_{k >= 0} ((-1)^k/6635520^k)*A008977(k).
Equals pFq(1/4, 1/2, 3/4; 1, 1; -1/25920), where pFq is the generalized hypergeometric function.

A361034 a(n) = 2520(4n)!/(n!*(n+2)!^3).

Original entry on oeis.org

315, 280, 3675, 116424, 5885880, 399072960, 33129291195, 3190228041000, 344161801063080, 40616781150254400, 5155510596280207800, 695029472211496161600, 98570579229528369624000, 14597207555235045670540800, 2243893009052293495117018875, 356344642367340570239409729000

Offset: 0

Peter Bala, Mar 01 2023

Row 1 of A361032.
The central binomial numbers A000984(n) = (2*n)!/n!^2 have the property that 6*A000984(n) is divisible by (n + 1)*(n + 2) and the result (2*n)!/(n!*(n+2)!) is the super ballot number A007054(n). Similarly, the  numbers A008977(n) = (4*n)!/n!^4 appear to have the property that 2520*A008977(n) is divisible by ((n + 1)*(n + 2))^3, leading to the present sequence. Cf. A361029.
Conjecture: a(n) is odd iff n = 2^k - 2 for some k >= 1.

Cf. A007054, A008977, A361029, A361032, A361033, A361035.

seq(2520*(4*n)!/(n!*(n+2)!^3), n = 0..20);

a(n) = 2520*A008977(n)/((n+1)*(n+2))^3.
a(n) = (315/2)*A008977(n+2)/((4*n+1)*(4*n+2)*(4*n+3)*(4*n+5)*(4*n+6)*(4*n+7)).
P-recursive: a(n) = 4*(4*n-1)*(4*n-2)*(4*n-3)/(n+2)^3 * a(n-1) with a(0) = 315.
The o.g.f. A(x) satisfies the differential equation
x^3*(1 - 256*x)*A(x)''' + x^2*(9 - 1152*x)*A(x)'' + x*(19 - 816*x)*A(x)' + (8 - 24*x)*A(x) - 2520 = 0 with A(0) = 315, A'(0) = 280 and A''(0) = 7350.
a(n) ~ 630*sqrt(8/Pi^3) * 2^(8*n)/n^(15/2).

A361035 a(n) = 9979200 * (4n)!/(n!(n+3)!^3).

Original entry on oeis.org

46200, 17325, 116424, 2134440, 67953600, 3086579925, 179961581800, 12633303042360, 1023952465972800, 93080123469333000, 9292590788015304000, 1003030870975774344000, 115656146295979953692160, 14112534648127632044761125, 1808633485822731984665865000

Offset: 0

Peter Bala, Mar 01 2023

Row 2 of A361032.
The central binomial numbers A000984(n) = (2*n)!/n!^2 have the property that 60*A000984(n) is divisible by (n + 1)*(n + 2)*(n + 3) and the result (2*n)!/(n!*(n+3)!) is the super ballot number A007272(n). Similarly, the  numbers A008977(n) = (4*n)!/n!^4 appear to have the property that 9979200*A008977(n) is divisible by ((n + 1)*(n + 2)*(n + 3))^3, leading to the present sequence. Cf. A361030.
Conjecture: a(n) is odd iff n = 2^k - 3 for some k >= 2.

Cf. A007272, A008977, A361030, A361032, A361033, A361034.

seq( 9979200 * (4*n)!/(n!*(n+3)!^3 ), n = 0..20);

a(n) = 9979200 * A008977(n)/((n+1)*(n+2)*(n+3))^3.
a(n) = (15925)*A008977(n+3)/((4*n+1)*(4*n+2)*(4*n+3)*(4*n+5)*(4*n+6)*(4*n+7)*(4*n+9)*(4*n+10)*(4*n+11)).
P-recursive: a(n) = 4*(4*n-1)*(4*n-2)*(4*n-3)/(n+3)^3 * a(n-1) with a(0) = 46200.
The o.g.f. A(x) satisfies the differential equation
x^3*(1 - 256*x)*A(x)''' + x^2*(12 - 1152*x)*A(x)'' + x*(37 - 816*x)*A(x)' + (27 - 24*x)*A(x) - 1247400 = 0 with A(0) = 46200, A'(0) = 17325 and A''(0) = 232848.
a(n) ~ 2494800*sqrt(8/Pi^3) * 2^(8*n)/n^(21/2).

A262014 Triangle in which the g.f. for row n is (1-x)^(4n+1) Sum_{j>=0} binomial(n+j-1,j)^4 * x^j, read by rows of k=0..3*n terms.

Original entry on oeis.org

1, 1, 11, 11, 1, 1, 72, 603, 1168, 603, 72, 1, 1, 243, 6750, 49682, 128124, 128124, 49682, 6750, 243, 1, 1, 608, 40136, 724320, 4961755, 15018688, 21571984, 15018688, 4961755, 724320, 40136, 608, 1, 1, 1275, 167475, 6021225, 84646275, 554083761, 1858142825, 3363309675, 3363309675, 1858142825, 554083761, 84646275, 6021225, 167475, 1275, 1, 1, 2376, 554931, 35138736, 879018750, 10490842656, 66555527346, 239677178256, 509723668476, 654019630000, 509723668476

Offset: 0

Paul D. Hanna, Sep 10 2015

			Triangle begins:
 1;
 1, 11, 11, 1;
 1, 72, 603, 1168, 603, 72, 1;
 1, 243, 6750, 49682, 128124, 128124, 49682, 6750, 243, 1;
 1, 608, 40136, 724320, 4961755, 15018688, 21571984, 15018688, 4961755, 724320, 40136, 608, 1;
 1, 1275, 167475, 6021225, 84646275, 554083761, 1858142825, 3363309675, 3363309675, 1858142825, 554083761, 84646275, 6021225, 167475, 1275, 1;
 1, 2376, 554931, 35138736, 879018750, 10490842656, 66555527346, 239677178256, 509723668476, 654019630000, 509723668476, 239677178256, 66555527346, 10490842656, 879018750, 35138736, 554931, 2376, 1;
 ...
Row g.f.s begin:
 n=0: (1) = (1-x) * (1 + x + x^2 + x^3 + x^4 +...);
 n=1: (1 + 11*x + 11*x^2 + x^3)  =  (1-x)^5 * (1 + 2^4*x + 3^4*x^2 + 4^4*x^3 + 5^4*x^4 + 6^4*x^5 +...);
 n=2: (1 + 72*x + 603*x^2 + 1168*x^3 + 603*x^4 + 72*x^5 + x^6)  =  (1-x)^9 * (1 + 3^4*x + 6^4*x^2 + 10^4*x^3 + 15^4*x^5 + 21^4*x^6 +...);
 n=3: (1 + 243*x + 6750*x^2 + 49682*x^3 + 128124*x^4 + 128124*x^5 + 49682*x^6 + 6750*x^7 + 243*x^8 + x^9)  =  (1-x)^13 * (1 + 4^4*x + 10^4*x^2 + 20^4*x^3 + 35^4*x^4 + 56^4*x^5 + 84^4*x^6 +...);
 ...

Paul D. Hanna, Table of n, a(n) for n = 0..1425, as a flattened triangle of rows 0..30
Ilia Gaiur, Vladimir Rubtsov, and Duco van Straten, Product formulas for the Higher Bessel functions, arXiv:2405.03015 [math.AG], 2024. See p. 18.
Feihu Liu, Guoce Xin, and Chen Zhang, Ehrhart Polynomials of Order Polytopes: Interpreting Combinatorial Sequences on the OEIS, arXiv:2412.18744 [math.CO], 2024. See p. 14.

Cf. A008977 (row sums), A262015 (diagonal), A202750, A258402.

Cf. A181544 (triangle variant).

{T(n, k)=polcoeff(sum(j=0, n+k, binomial(n+j, j)^4*x^j)*(1-x)^(4*n+1), k)}
for(n=0, 10, for(k=0, 3*n, print1(T(n, k), ", ")); print(""))

Row sums form A008977(n) = (4*n)!/(n!)^4.
T(n,1) = A258402(n) = (n^2 + 4*n + 6) * n^2.
From Sergii Voloshyn, Dec 17 2024: (Start)
Let E be the operator D*x*D*x*D*x*D, where D denotes the derivative operator d/dx. Then (1/(n)!^4)  * E^n(1/(1 - x)) = (row n generating polynomial)/(1 - x)^(4*n+1) = Sum_{j>=0} binomial(n+j,j)^4 * x^j.
For example, when n = 2 we have (1/2!)^4*E^3(1/(1 - x)) = (1 + 243 x + 6750 x^2 + 49682 x^3 + 128124 x^4 + 128124 x^5 + 49682 x^6 + 6750 x^7 + 243 x^8 + x^9)/(1 - x)^13.  (End)

cp's OEIS Frontend

A184898 a(n) = C(2n,n) * (8^n/n!^2) * Product_{k=0..n-1} (8k+1)*(8k+7).

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A370294 G.f.: exp(Sum_{k>=1} (4*k)!/(4!*k!^4) * x^k/k).

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A333042 G.f.: exp(Sum_{k>=1} (4*k)!/k!^4 * x^k/k).

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A361032 Square array read by ascending antidiagonals: T(n,k) = F(n) * (4*k)!/(k!*(k + n + 1)!^3), where F(n) = (1/8)*(4*n + 4)!/(n + 1)!; n, k >= 0.

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A361033 a(n) = 3*(4*n)!/(n!*(n+1)!^3).

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A367568 a(n) = Product_{k=0..n} (4*k)! / k!^4.

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A378130 Decimal expansion of 24*L^2/(5^(7/4)*Pi^2), where L is the lemniscate constant (A062539).

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A361034 a(n) = 2520*(4*n)!/(n!*(n+2)!^3).

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A361035 a(n) = 9979200 * (4*n)!/(n!*(n+3)!^3).

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A370294 G.f.: exp(Sum_{k>=1} (4k)!/(4!k!^4) * x^k/k).

A333042 G.f.: exp(Sum_{k>=1} (4k)!/k!^4 x^k/k).

A361032 Square array read by ascending antidiagonals: T(n,k) = F(n) * (4k)!/(k!(k + n + 1)!^3), where F(n) = (1/8)(4n + 4)!/(n + 1)!; n, k >= 0.

A361033 a(n) = 3(4n)!/(n!*(n+1)!^3).

A378130 Decimal expansion of 24L^2/(5^(7/4)Pi^2), where L is the lemniscate constant (A062539).

A361034 a(n) = 2520(4n)!/(n!*(n+2)!^3).

A361035 a(n) = 9979200 * (4n)!/(n!(n+3)!^3).

A262014 Triangle in which the g.f. for row n is (1-x)^(4n+1) Sum_{j>=0} binomial(n+j-1,j)^4 * x^j, read by rows of k=0..3*n terms.