A074635 -id:A074635 - cp's OEIS frontend

A074649 a(0) = 1; for n >= 1, a(n) = sum(binomial(n,k)^3*binomial(n+k,k+1)^2,k = 0..n)/n^2.

1, 2, 23, 434, 10897, 327270, 11076235, 408850370, 16119036965, 669177449258, 28960814900899, 1297024187184478, 59777126587203937, 2822927389747980806, 136132927766691327651, 6685542830954666301218, 333618582889745741654221

Offset: 0

Karol A. Penson, Aug 28 2002

nonn

Vaclav Kotesovec, Recurrence (of order 5)

Cf. A074635.

a[0] = 1; a[n_] := Sum[Binomial[n, k]^3*Binomial[n + k, k + 1]^2, {k, 0, n}]/n^2; Table[a[n], {n, 0, 16}]

Special values of the hypergeometric function 5F4, in Maple notation : a(n)= hypergeom([n+1, n+1, -n, -n, -n], [1, 1, 2, 2], -1), n=0, 1...

a(n) ~ c * d^n / n^4, where d = 63.74669588201779948... is the root of the equation -2 - 247*d - 11666*d^2 - 502*d^3 - 53*d^4 + d^5 = 0 and c = 0.3611541... - Vaclav Kotesovec, Mar 02 2014

Edited by Robert G. Wilson v, Aug 29 2002

A073530 a(n) = (1/n)Sum_{k=0..n} binomial(n, k)binomial(n+k, k+1)*binomial(n+k, k) with a(0) = 1.

Original entry on oeis.org

1, 3, 22, 225, 2706, 35861, 507060, 7510005, 115175530, 1815002145, 29231242206, 479251119815, 7975209124260, 134398986236625, 2289535943534920, 39370761619959165, 682603570436824602, 11921040322642855193

Offset: 0

Karol A. Penson, Aug 29 2002

nonn

G. C. Greubel, Table of n, a(n) for n = 0..770

Cf. A074635, A074649.

A073530:= func< n | n eq 0 select 1 else (1/n)*(&+[Binomial(n,j)* Binomial(n+j,j+1)*Binomial(n+j,j): j in [0..n]]) >;
[A073530(n): n in [0..30]]; // G. C. Greubel, Dec 27 2022

p3 := x^3+5*x^2+39*x-2; p4 := x^4+4*x^3+30*x^2-20*x+1;
y := hypergeom([1/12, 5/12], [1], -1728*p3*x^4/p4^3)/p4^(1/4);
a1 := p3/(5*x^2+8*x); a2 := (13*x^3-197*x^2-60*x+16)/(5*x^2+8*x)^2;
ogf := a1*y - Int(a2*y,x) - 89/32;
series(ogf,x=0,20);  # Mark van Hoeij, Apr 03 2013

Table[ HypergeometricPFQ[{n+1,n+1,-n}, {1,2}, -1], {n,0,20}] (* Robert G. Wilson v *)

def A073530(n):
    if (n==0): return 1
    else: return sum(binomial(n,j)*binomial(n+j,j+1)*binomial(n+j,j) for j in range(n+1))/n
[A073530(n) for n in range(31)] # G. C. Greubel, Dec 27 2022

a(n) = hypergeometric3F2([n+1, n+1, -n], [1, 2], -1).
Recurrence: 2*(n-1)*n*(n+1)*(59*n^2 - 235*n + 216)*a(n) = 3*(n-1)*(767*n^4 - 3822*n^3 + 6065*n^2 - 3602*n + 720)*a(n-1) + (n-2)*(n-1)*(295*n^3 - 1470*n^2 + 2051*n - 792)*a(n-2) + (n-3)^2*(n-2)*(59*n^2 - 117*n + 40)*a(n-3). - Vaclav Kotesovec, Mar 02 2014
a(n) ~ c * d^n/n^2, where d = 1/6*(39 + (61128 - 177*sqrt(177))^(1/3) + (3*(20376 + 59*sqrt(177)))^(1/3)) = 19.62866250831184052... is the root of the equation 2*d^3 - 39*d^2 - 5*d = 1 and c = sqrt(1/2 + sqrt(231/59) * cosh(arccosh(51 * sqrt(177/77)/77)/3)/2)/Pi = 0.38852216850573971010943128486103656013508... - Vaclav Kotesovec, Mar 02 2014, updated Mar 17 2024

A075132 a(0)=1; for n > 0, a(n)=sum(binomial(n,k)binomial(n+k,k+1)binomial(n+k+1,k),k=0..n)/n.

Original entry on oeis.org

1, 4, 33, 356, 4415, 59690, 855925, 12809620, 198034395, 3140695856, 50845026815, 837207042270, 13982565667915, 236366796137230, 4037392483733629, 69589308151724084, 1209013441314791891

Offset: 0

Karol A. Penson, Sep 04 2002

nonn

Cf. A075514, A074635, A074649, A073530.

p3 := x^3+5*x^2+39*x-2; p4 := x^4+4*x^3+30*x^2-20*x+1;
y := hypergeom([1/12, 5/12], [1], -1728*p3*x^4/p4^3)/p4^(1/4);
a1 := p3/(5*x+8); a2 := (5*x^3+21*x^2-42*x+178)/(5*x+8)^2;
ogf := (a1*y - Int(a2*y,x) + 1/4)/x;
series(ogf, x=0, 20);  # Mark van Hoeij, Apr 05 2013

Table[ HypergeometricPFQ[{n + 1, n + 2, -n}, {1, 2}, -1], {n, 0, 16}] (* Robert G. Wilson v *)

Special values of the hypergeometric function 3F2, in Maple notation: a(n)=hypergeom([ n+1, n+2, -n], [1, 2], -1)
Recurrence: 2*n*(n+1)^2*(59*n^2 - 121*n + 48)*a(n) = n*(2301*n^4 - 4719*n^3 + 1411*n^2 + 943*n - 384)*a(n-1) + (n-1)*(295*n^4 - 900*n^3 + 631*n^2 + 170*n - 168)*a(n-2) + (n-3)*(n-2)*(n-1)*(59*n^2 - 3*n - 14)*a(n-3). - Vaclav Kotesovec, Mar 02 2014
a(n) ~ c * r^n/n^2, where r = 1/6*(39 + (61128 - 177*sqrt(177))^(1/3) + (3*(20376 + 59*sqrt(177)))^(1/3)) = 19.62866250831184052... is the root of the equation 2*r^3 - 39*r^2 - 5*r = 1 and c = sqrt(5/3 + (6647924463616 - 1376397312 * sqrt(177))^(1/3) / 11328 + ((55017913 + 11391*sqrt(177))/2)^(1/3) / (12*59^(2/3)))/Pi = 0.7107949125787545435298127032606174925961199835... - Vaclav Kotesovec, Mar 02 2014, updated Mar 17 2024

A075514 a(0)=1; for n > 0, a(n)=sum(binomial(n,k)(binomial(n+k,k+1)^2) binomial(n+k,k),k=0..n).

Original entry on oeis.org

1, 3, 154, 7941, 429036, 24123105, 1399303662, 83176893681, 5041470373624, 310449199290489, 19369215839341710, 1221826010437625703, 77798300823672280164, 4993767938470070592261, 322795606469564782029126

Offset: 0

Karol A. Penson, Sep 02 2002

nonn

Cf. A074635, A074649, A073530.

Flatten[{1,Table[Sum[Binomial[n,k]*(Binomial[n+k,k+1]^2)* Binomial[n+k,k],{k,0,n}],{n,1,20}]}] (* Vaclav Kotesovec, Mar 02 2014 *)

Special values of the hypergeometric function 4F3, in Maple notation: a(n)=n^2*hypergeom([n+1, n+1, n+1, -n], [1, 2, 2], -1), n=1, 2....
Recurrence: 4*(n-1)^2*(n+1)^2*(29412*n^8 - 523944*n^7 + 3995715*n^6 - 17035566*n^5 + 44400751*n^4 - 72401280*n^3 + 72040928*n^2 - 39898368*n + 9379584)*a(n) = 2*(4176504*n^12 - 80664804*n^11 + 681787110*n^10 - 3320158377*n^9 + 10317109990*n^8 - 21386158690*n^7 + 29990437762*n^6 - 28191974977*n^5 + 17142077578*n^4 - 6206039632*n^3 + 1063468848*n^2 + 4760064*n - 18759168)*a(n-1) + 4*(823536*n^12 - 17141040*n^11 + 157530180*n^10 - 841802850*n^9 + 2898036925*n^8 - 6724482767*n^7 + 10676838689*n^6 - 11506383284*n^5 + 8122517663*n^4 - 3444627899*n^3 + 682922191*n^2 + 15710352*n - 19910592)*a(n-2) + 2*(n-3)*(235296*n^11 - 4544496*n^10 + 38143266*n^9 - 182713089*n^8 + 551659187*n^7 - 1094072109*n^6 + 1439496807*n^5 - 1235496354*n^4 + 653944224*n^3 - 183118980*n^2 + 13441600*n + 2965248)*a(n-3) - (n-4)^2*(n-3)^2*(29412*n^8 - 288648*n^7 + 1151643*n^6 - 2417028*n^5 + 2879446*n^4 - 1930604*n^3 + 642371*n^2 - 53200*n - 12768)*a(n-4). - Vaclav Kotesovec, Mar 02 2014
a(n) ~ c * d^n / n^(3/2), where d = 71.39297952064022156... is the root of the equation 1 - 16*d - 112*d^2 - 284*d^3 + 4*d^4 = 0, and c = 0.2216473197208166381284001749414... - Vaclav Kotesovec, Mar 02 2014

A277060 a(n) = (1/2) * Sum_{k=0..n} (binomial(n,k) * binomial(n+k,k+1))^2 for n >= 0.

Original entry on oeis.org

0, 1, 28, 729, 19376, 529575, 14835780, 424231465, 12338211520, 363931754949, 10862528888300, 327501958094003, 9959845931792784, 305175084350065267, 9412306255856822388, 291982561878565118025, 9104382992541189221120

Offset: 0

Seiichi Manyama, Nov 07 2016

Conjecture: the supercongruences a(p-1) == 1 (mod p^4) holds for all primes p >= 5 and a(p^2-1) == 1 (mod p^5) holds for all primes p >= 3. - Peter Bala, Mar 22 2023

Seiichi Manyama, Table of n, a(n) for n = 0..656

Cf. 1/2 * Sum_{k=0..n} (binomial(n,k) * binomial(n+k,k+1))^m: A050151 (m=1), this sequence (m=2).

Cf. A005259, A074635.

a := proc(n) option remember; if n = 0 then 0 elif n = 1 then 1 else ( (2*n-1)*(51*n^4-102*n^3+19*n^2+ 32*n-14)*n^2*a(n-1) - n^2*(n-2)*(3*n^2-1)*(n-1)^2*a(n-2) )/( (n-1)^2*(3*n^2-6*n+2)*(n+1)^3 ) end if; end:
seq(a(n), n = 0..20); # Peter Bala, Mar 22 2023

a(n)=my(t=n); if(n<2, return(n)); sum(k=1,n, t*=(n-k+1)*(n+k)/k/(k+1); t^2, n^2)/2 \\ Charles R Greathouse IV, Nov 07 2016

a(n) = n^2 * A074635(n)/2.
From Peter Bala, Mar 22 2023: (Start)
a(n) = Sum_{k = 0..n-1} binomial(n+1,k)*binomial(n-1,k)*binomial(n+k,k)^2.
P-recursive: (n-1)^2*(3*n^2-6*n+2)*(n+1)^3*a(n) = (2*n-1)*(51*n^4-102*n^3+19*n^2+ 32*n-14)*n^2*a(n-1) - n^2*(n-2)*(3*n^2-1)*(n-1)^2*a(n-2) with a(0) = 0 and a(1) = 1.
a(n) ~ sqrt(12 + 17*sqrt(2)/2)*(17 + 12*sqrt(2))^n/(4*n^(3/2)*Pi^(3/2)). (End)

cp's OEIS Frontend

A074649 a(0) = 1; for n >= 1, a(n) = sum(binomial(n,k)^3*binomial(n+k,k+1)^2,k = 0..n)/n^2.

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A073530 a(n) = (1/n)*Sum_{k=0..n} binomial(n, k)*binomial(n+k, k+1)*binomial(n+k, k) with a(0) = 1.

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A075132 a(0)=1; for n > 0, a(n)=sum(binomial(n,k)*binomial(n+k,k+1)*binomial(n+k+1,k),k=0..n)/n.

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A075514 a(0)=1; for n > 0, a(n)=sum(binomial(n,k)*(binomial(n+k,k+1)^2)* binomial(n+k,k),k=0..n).

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A277060 a(n) = (1/2) * Sum_{k=0..n} (binomial(n,k) * binomial(n+k,k+1))^2 for n >= 0.

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A073530 a(n) = (1/n)Sum_{k=0..n} binomial(n, k)binomial(n+k, k+1)*binomial(n+k, k) with a(0) = 1.

A075132 a(0)=1; for n > 0, a(n)=sum(binomial(n,k)binomial(n+k,k+1)binomial(n+k+1,k),k=0..n)/n.

A075514 a(0)=1; for n > 0, a(n)=sum(binomial(n,k)(binomial(n+k,k+1)^2) binomial(n+k,k),k=0..n).