A005810 -id:A005810 - cp's OEIS frontend

A257633 a(n) = binomial(4*n + 2,n).

1, 6, 45, 364, 3060, 26334, 230230, 2035800, 18156204, 163011640, 1471442973, 13340783196, 121399651100, 1108176102180, 10142940735900, 93052749919920, 855420636763836, 7877932561061640, 72667580816130436, 671262558647881200, 6208770443303347920

Offset: 0

Peter Bala, Nov 04 2015

Michael De Vlieger, Table of n, a(n) for n = 0..1025
N. J. Wildberger and Dean Rubine, A Hyper-Catalan Series Solution to Polynomial Equations, and the Geode, Amer. Math. Monthly (2025). See section 12.

Cf. A002293, A004331, A005810, A052203, A224274, A262977.

#A257633
seq(binomial(4*n + 2,n), n = 0..20);

Table[Binomial[4*n + 2, n], {n, 0, 120}] (* Michael De Vlieger, Apr 11 2025 *)

vector(30, n, n--; binomial(4*n+2, n)) \\ Altug Alkan, Nov 05 2015

The o.g.f. equals f(x)*g(x)^2, where f(x) is the o.g.f. for A005810 and g(x) is the o.g.f. for A002293. More generally, f(x)*g(x)^k is the o.g.f. for the sequence binomial(4*n + k,n). Cf. A262977 (k = -1), A005810 (k = 0), A052203 (k = 1), A224274 (k = 3) and A004331 (k = 4).

A226761 G.f.: 1 / (1 + 12xG(x)^2 - 13xG(x)^3) where G(x) = 1 + x*G(x)^4 is the g.f. of A002293.

Original entry on oeis.org

1, 1, 16, 118, 1004, 8601, 75076, 662796, 5903676, 52949332, 477533356, 4326309406, 39343725716, 358943047438, 3283745710968, 30112624408488, 276715616909148, 2547523969430508, 23491659440021920, 216942761366305144, 2006084011596742384, 18572529488934397689

Offset: 0

Paul D. Hanna, Jun 16 2013

nonn

			G.f.: A(x) = 1 + x + 16*x^2 + 118*x^3 + 1004*x^4 + 8601*x^5 +...
A related series is G(x) = 1 + x*G(x)^4, where
G(x) = 1 + x + 4*x^2 + 22*x^3 + 140*x^4 + 969*x^5 + 7084*x^6 +...
G(x)^2 = 1 + 2*x + 9*x^2 + 52*x^3 + 340*x^4 + 2394*x^5 + 17710*x^6 +...
G(x)^3 = 1 + 3*x + 15*x^2 + 91*x^3 + 612*x^4 + 4389*x^5 + 32890*x^6 +...
such that A(x) = 1/(1 + 12*x*G(x)^2 - 13*x*G(x)^3).

Vincenzo Librandi and Joerg Arndt, Table of n, a(n) for n = 0..200

Cf. A002293.

Cf. A005810, A078995, A147855, A226733, A385605, A386811.

Table[Sum[Binomial[2*n+3*k,n-k]*Binomial[2*n-3*k,k],{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Jun 17 2013 *)

{a(n)=local(G=1+x); for(i=0, n, G=1+x*G^4+x*O(x^n)); polcoeff(1/(1+12*x*G^2-13*x*G^3), n)}
for(n=0, 30, print1(a(n), ", "))

{a(n)=local(G=1+x); for(i=0, n, G=1+x*G^4+x*O(x^n)); polcoeff(1/(1-x*G^2-13*x^2*G^6), n)}
for(n=0, 30, print1(a(n), ", "))

{a(n)=sum(k=0, n, binomial(2*n+3*k, n-k)*binomial(2*n-3*k, k))}
for(n=0, 30, print1(a(n), ", "))

{a(n)=sum(k=0, n, binomial(3*k, n-k)*binomial(4*n-3*k, k))}
for(n=0, 30, print1(a(n), ", "))

{a(n)=sum(k=0, n, binomial(4*n+3*k, n-k)*binomial(-3*k, k))}
for(n=0, 30, print1(a(n), ", "))

a(n) = Sum_{k=0..n} C(3*k, n-k) * C(4*n-3*k, k).
a(n) = Sum_{k=0..n} C(n+3*k, n-k) * C(3*n-3*k, k).
a(n) = Sum_{k=0..n} C(2*n+3*k, n-k) * C(2*n-3*k, k).
a(n) = Sum_{k=0..n} C(3*n+3*k, n-k) * C(n-3*k, k).
a(n) = Sum_{k=0..n} C(4*n+3*k, n-k) * C(-3*k, k).
G.f.: 1 / (1 - x*G(x)^2 - 13*x^2*G(x)^6) where G(x) = 1 + x*G(x)^4 is the g.f. of A002293.
a(n) ~ 2^(8*n+5/2)/(7*3^(3*n+1/2)*sqrt(Pi*n)). - Vaclav Kotesovec, Jun 17 2013
From Seiichi Manyama, Aug 05 2025: (Start)
a(n) = [x^n] 1/((1+3*x) * (1-x)^(3*n+1)).
a(n) = Sum_{k=0..n} (-4)^(n-k) * binomial(4*n+1,k).
a(n) = Sum_{k=0..n} (-3)^(n-k) * binomial(3*n+k,k). (End)
From Seiichi Manyama, Aug 14 2025: (Start)
a(n) = Sum_{k=0..n} (-3)^k * 4^(n-k) * binomial(4*n+1,k) * binomial(4*n-k,n-k).
G.f.: G(x)^2/((-3+4*G(x)) * (4-3*G(x))) where G(x) = 1+x*G(x)^4 is the g.f. of A002293. (End)
G.f.: B(x)^2/(1 + 7*(B(x)-1)/4), where B(x) is the g.f. of A005810. - Seiichi Manyama, Aug 15 2025

A346578 a(n) = (1/(4n)) Sum_{d|n} mu(n/d) * binomial(4*d,d).

Original entry on oeis.org

1, 3, 18, 112, 775, 5598, 42287, 328640, 2615085, 21191125, 174303162, 1451424960, 12211799223, 103655906781, 886568152950, 7633233227520, 66105170315083, 575445689879247, 5032380942945321, 44191451767056400, 389514699012969936, 3444925385161998518, 30561576846316109863

Offset: 1

Ilya Gutkovskiy, Jul 24 2021

nonn

Inverse Euler transform of A002293.

Moebius transform of A261497.

Cf. A002293, A005810, A008683, A022553, A261497, A346577, A346579, A346580, A346581, A346582.

Table[(1/(4 n)) Sum[MoebiusMu[n/d] Binomial[4 d, d], {d, Divisors[n]}], {n, 23}]

a(n) = sumdiv(n, d, moebius(n/d)*binomial(4*d,d))/(4*n); \\ Michel Marcus, Jul 24 2021

A371771 a(n) = Sum_{k=0..floor(n/3)} binomial(4n-3k-1,n-3*k).

Original entry on oeis.org

1, 3, 21, 166, 1377, 11748, 102088, 898677, 7987305, 71517307, 644134026, 5829345492, 52964836184, 482846377185, 4414405051413, 40458397722306, 371605426607673, 3419639400458316, 31521758873514301, 291000881055737811, 2690082750919841442

Offset: 0

Seiichi Manyama, Apr 05 2024

nonn

Robert Israel, Table of n, a(n) for n = 0..1000

Cf. A005810, A147855, A262977.

Cf. A371758, A371770, A371772.

f:= proc(n) local k; add(binomial(4*n-3*k-1,n-3*k),k=0..n/3) end proc:
map(f, [$0..30]); # Robert Israel, Feb 28 2025

a(n) = sum(k=0, n\3, binomial(4*n-3*k-1, n-3*k));

a(n) = [x^n] 1/((1-x^3) * (1-x)^(3*n)).
a(n) = binomial(4*n-1, n)*hypergeom([1, (1-n)/3, (2-n)/3, -n/3], [(1-4*n)/3, 2*(1-2*n)/3, 1-4*n/3], 1). - Stefano Spezia, Apr 06 2024
From Vaclav Kotesovec, Apr 08 2024: (Start)
Recurrence: 81*n*(3*n - 2)*(3*n - 1)*(9037*n^4 - 61391*n^3 + 154035*n^2 - 169317*n + 68836)*a(n) = 27*(2394805*n^7 - 19820156*n^6 + 66654684*n^5 - 117198990*n^4 + 115250735*n^3 - 62650734*n^2 + 17209736*n - 1814400)*a(n-1) - 3*(7021749*n^7 - 58192764*n^6 + 196050236*n^5 - 345531070*n^4 + 340849311*n^3 - 186035886*n^2 + 51353864*n - 5443200)*a(n-2) + 8*(2*n - 3)*(4*n - 9)*(4*n - 7)*(9037*n^4 - 25243*n^3 + 24084*n^2 - 9272*n + 1200)*a(n-3).
a(n) ~ 2^(8*n + 9/2) / (7 * sqrt(Pi*n) * 3^(3*n + 3/2)). (End)

A378484 Expansion of (Sum_{k>=0} binomial(4k,k) x^k)^4.

Original entry on oeis.org

1, 16, 208, 2480, 28176, 310336, 3344688, 35472672, 371570320, 3853862080, 39650662720, 405221752112, 4117879215472, 41643345090240, 419362920305952, 4207604570770752, 42079232716865424, 419609034657373120, 4173470598366784960, 41413032430984848832, 410071444666659404352

Offset: 0

Seiichi Manyama, Nov 28 2024

nonn

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

Cf. A000302, A378483.

Cf. A005810, A078995.

nmax = 20; CoefficientList[Series[Sum[Binomial[4*k,k] * x^k, {k, 0, nmax}]^4, {x, 0, nmax}], x] (* Vaclav Kotesovec, Jul 19 2025 *)

my(N=30, x='x+O('x^N)); Vec(sum(k=0, N, binomial(4*k, k)*x^k)^4)

a(n) = Sum_{i+j+k+l=n, i,j,k,l >= 0} binomial(4*i,i) * binomial(4*j,j) * binomial(4*k,k) * binomial(4*l,l).
G.f.: B(x)^4 where B(x) is the g.f. of A005810.
27*a(n) - 256*a(n-1) = 18*A078995(n) + 8*A005810(n) for n > 0.
a(n) ~ n * 2^(8*n + 2) / 3^(3*n + 2) * (1 + 2^(7/2)/(3^(3/2)*sqrt(Pi*n))). - Vaclav Kotesovec, Jul 19 2025

A378802 a(n) = n * binomial(4*n, n).

Original entry on oeis.org

0, 4, 56, 660, 7280, 77520, 807576, 8288280, 84146400, 847289520, 8476605280, 84362730452, 836022413616, 8255176274800, 81266247493200, 797911337890800, 7816430993273280, 76417576884236016, 745777615780501920, 7266758081613043600, 70706322844243486400, 687103929058903836480

Offset: 0

Amiram Eldar, Dec 07 2024

Amiram Eldar, Table of n, a(n) for n = 0..500
Necdet Batir and Anthony Sofo, On some series involving reciprocals of binomial coefficients, Appl. Math. Comp., Vol. 220 (2013), pp. 331-338.

Cf. A005810, A225847, A229703, A374522.

a[n_] := n * Binomial[4*n, n]; Array[a, 25, 0]

PARI
```
a(n) = n * binomial(4*n, n);
```

a(n) = n * A005810(n).
a(n) = A374522(n) + n.
a(n) == 0 (mod 4).
Sum_{n>=1} 1/a(n) = A225847.
Sum_{n>=1} (-1)^n/a(n) = A229703.

A385498 a(n) = Sum_{k=0..n} 2^(n-k) * binomial(4*n,k).

Original entry on oeis.org

1, 6, 48, 408, 3564, 31626, 283548, 2560872, 23255964, 212101176, 1941110628, 17815257048, 163896843300, 1510891524252, 13952756564424, 129048895061208, 1195191116753436, 11082661017288264, 102877353868090080, 955912961224763232, 8889969049985302464

Offset: 0

Seiichi Manyama, Jul 30 2025

Cf. A100192, A385004, A386699.

Cf. A005810, A066381, A371814, A386701.

Table[(81/8)^n - Binomial[4*n, n]*(-1 + Hypergeometric2F1[1, -3*n, 1 + n, -1/2]), {n,0,25}] (* Vaclav Kotesovec, Jul 30 2025 *)

a(n) = sum(k=0, n, 2^(n-k)*binomial(4*n, k));

a(n) = [x^n] 1/((1-3*x) * (1-x)^(3*n)).
a(n) = Sum_{k=0..n} 3^(n-k) * binomial(3*n+k-1,k).
From Vaclav Kotesovec, Jul 30 2025: (Start)
Recurrence: 24*n*(3*n - 2)*(3*n - 1)*(139*n^3 - 366*n^2 + 143*n + 132)*a(n) = (588665*n^6 - 2281011*n^5 + 2262209*n^4 + 1245939*n^3 - 3359986*n^2 + 1877400*n - 322560)*a(n-1) - 648*(2*n - 3)*(4*n - 7)*(4*n - 5)*(139*n^3 + 51*n^2 - 172*n + 48)*a(n-2).
a(n) ~ 2^(8*n + 1/2) / (sqrt(Pi*n) * 3^(3*n - 1/2)). (End)
G.f.: g/((3-2*g) * (4-3*g)) where g = 1+x*g^4 is the g.f. of A002293. - Seiichi Manyama, Aug 13 2025
a(n) = Sum_{k=0..n} 3^k * (-2)^(n-k) * binomial(4*n,k) * binomial(4*n-k-1,n-k). - Seiichi Manyama, Aug 15 2025
G.f.: 1/(1 - x*g^3*(12-6*g)) where g = 1+x*g^4 is the g.f. of A002293. - Seiichi Manyama, Aug 17 2025

A096130 Triangle read by rows: T(n,k) = binomial(k*n,n), 1 <= k <= n.

Original entry on oeis.org

1, 1, 6, 1, 20, 84, 1, 70, 495, 1820, 1, 252, 3003, 15504, 53130, 1, 924, 18564, 134596, 593775, 1947792, 1, 3432, 116280, 1184040, 6724520, 26978328, 85900584, 1, 12870, 735471, 10518300, 76904685, 377348994, 1420494075, 4426165368, 1, 48620, 4686825, 94143280, 886163135, 5317936260, 23667689815, 85113005120, 260887834350

Offset: 1

Amarnath Murthy, Jul 04 2004

			Triangle begins:
  1;
  1,   6;
  1,  20,   84;
  1,  70,  495,  1820;
  1, 252, 3003, 15504, 53130;
  ...

Seiichi Manyama, Rows n = 1..140, flattened

Row-sums give A096131. The leading diagonal is A014062. Cf. A096131.

Cf. A007318.

Flat(List([1..10],n->List([1..n],k->Binomial(k*n,n)))); # Muniru A Asiru, Aug 12 2018

a:=(n,k)->binomial(k*n,n): seq(seq(a(n,k),k=1..n),n=1..10); # Muniru A Asiru, Aug 12 2018

tabl(nrows) = {for (n=1, nrows, for (k=1, n, print1(binomial(k*n, n), ", ");); print(););} \\ Michel Marcus, May 14 2013

T(n, 1) = 1;
T(n, 2) = A000984(n) for n > 1;
T(n, 3) = A005809(n) for n > 2;
T(n, 4) = A005810(n) for n > 3;
T(n, n) = A014062(n).

Corrected and extended by Reinhard Zumkeller, Jan 09 2005

A357508 a(n) = binomial(4n,2n) - 2binomial(4n,n).

Original entry on oeis.org

-1, -2, 14, 484, 9230, 153748, 2434964, 37748520, 580043790, 8886848740, 136151207764, 2088760285456, 32108266614164, 494648505828904, 7637081136832840, 118158193386475984, 1831647087068431374, 28444051172077725444, 442429676097305612324

Offset: 0

Peter Bala, Oct 01 2022

Sun and Wan's supercongruence stated below apparently generalizes as follows:
Let m be an integer and k a positive integer. Define u(n) = binomial((m+2)*n,(k+1)*n) - binomial(m,k)*binomial((m+2)*n,n). We conjecture that u(n) == u(1) (mod p^5) for all primes p >= 7. [added 22 Oct 2022: the conjecture is true: apply Helou and Terjanian, Section 3, Proposition 2.]
Conjecture: for r >= 2, u(p^r) == u(p^(r-1)) ( mod p^(3*r+3) ) for all primes p >= 5. - Peter Bala, Oct 13 2022

C. Helou and G. Terjanian, On Wolstenholme’s theorem and its converse, J. Number Theory 128 (2008), 475-499.
Z.-W. Sun and D. Wan, On Fleck quotients, arXiv:math/0603462 [math.NT], 2006-2007.

Cf. A001448, A005810, A357509.

seq(binomial(4*n,2*n) - 2*binomial(4*n,n), n = 0..20);

a(n) = A001448(n) - 2*A005810(n).

a(p) == -2 (mod p^5) for all primes p >= 7. (Sun and Wan, Corollary 1.5.)

A364507 a(n) = (5n)!(4n)! / ((3n)!^2 * (2n)! n!).

Original entry on oeis.org

1, 40, 5880, 1101100, 229265400, 50678855040, 11641642112100, 2746924727976000, 661097260785195000, 161538994454795003200, 39949572934939198410880, 9976687616280042928424700, 2511716999955421326631644900, 636662322699394050738883008000

Offset: 0

Peter Bala, Jul 27 2023

Row 2 of A364506.

			Examples of supercongruences:
a(7) - a(1) = 2746924727976000 - 40 = (2^3)*5*(7^4)*28601881799 == 0 (mod 7^4).
a(11) - a(1) = 9976687616280042928424700 - 40 = (2^2)*5*(11^3)*18397*3568463* 5708869513 == 0 (mod 11^3).

Paolo Xausa, Table of n, a(n) for n = 0..400
Romeo Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862--2012), arXiv:1111.3057 [math.NT], 2011.
Wikipedia, Dixon's identity

Cf. A001450, A005810, A006480, A364506, A364508.

seq( (5*n)!*(4*n)!*(2*n)! / ((3*n)!^2 * (2*n)!^2 * n!), n = 0..15);

A364507[n_]:=(5n)!(4n)!/((3n)!^2(2n)!n!);Array[A364507,15,0] (* Paolo Xausa, Oct 06 2023 *)

a(n) = Sum_{k = -n..n} (-1)^k * binomial(2*n, n + k) * binomial(4*n, 2*n + k)^2 (showing a(n) to be integral). Compare with Dixon's identity, Sum_{k = -n..n} (-1)^k * binomial(2*n, n + k)^3 = (3*n)!/n!^3 = A006480(n).
P-recursive: a(n) = (20/9)*(4*n-1)*(4*n-3)*(5*n-1)*(5*n-2)*(5*n-3)*(5*n-4)/((3*n-1)^2*(3*n-2)^2*n^2) * a(n-1) with a(0) = 1.
a(n) ~ c^n * sqrt(10)/(6*Pi*n), where c = (2^6)*(5^5)/(3^6).
a(n) = [x^n] G(x)^(20*n), where the power series G(x) = 1 + 2*x + 69*x^2 + 5647*x^3 + 618860*x^4 + 79241349*x^5 + 11177111981*x^6 + 1684171189810*x^7 + 266238907746252*x^8 + ... appears to have integer coefficients.
exp( Sum_{n > = 1} a(n)*x^n/n ) = F(x)^20, where the power series F(x) = 1 + 2*x + 149*x^2 + 18647*x^3 + 2913620*x^4 + 515276389*x^5 + 98628630997*x^6 + 19944410220744*x^7 + 4199273746072180*x^8 + ... appears to have integer coefficients.
Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and all positive integers n and r [added Aug 04 2023: the conjecture follows from Meštrović. equation 39].
a(n) = binomial(4*n,n)*binomial(5*n,2*n). - Christian Krause, Aug 03 2023

cp's OEIS Frontend

A257633 a(n) = binomial(4*n + 2,n).

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A226761 G.f.: 1 / (1 + 12*x*G(x)^2 - 13*x*G(x)^3) where G(x) = 1 + x*G(x)^4 is the g.f. of A002293.

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A346578 a(n) = (1/(4*n)) * Sum_{d|n} mu(n/d) * binomial(4*d,d).

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A371771 a(n) = Sum_{k=0..floor(n/3)} binomial(4*n-3*k-1,n-3*k).

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A378484 Expansion of (Sum_{k>=0} binomial(4*k,k) * x^k)^4.

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A378802 a(n) = n * binomial(4*n, n).

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A385498 a(n) = Sum_{k=0..n} 2^(n-k) * binomial(4*n,k).

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A096130 Triangle read by rows: T(n,k) = binomial(k*n,n), 1 <= k <= n.

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A226761 G.f.: 1 / (1 + 12xG(x)^2 - 13xG(x)^3) where G(x) = 1 + x*G(x)^4 is the g.f. of A002293.

A346578 a(n) = (1/(4n)) Sum_{d|n} mu(n/d) * binomial(4*d,d).

A371771 a(n) = Sum_{k=0..floor(n/3)} binomial(4n-3k-1,n-3*k).

A378484 Expansion of (Sum_{k>=0} binomial(4k,k) x^k)^4.

A357508 a(n) = binomial(4n,2n) - 2binomial(4n,n).

A364507 a(n) = (5n)!(4n)! / ((3n)!^2 * (2n)! n!).