cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

A288470 a(n) = Sum_{k=0..n} binomial(n,k)*binomial(2*n,2*k).

Original entry on oeis.org

1, 2, 14, 92, 646, 4652, 34124, 253528, 1901638, 14368844, 109208164, 833981128, 6394017436, 49185717752, 379438594136, 2934361958192, 22741538394694, 176582855512588, 1373431963785332, 10698376362421096, 83447762846703796, 651690159076273192, 5095051571420324264, 39874449115469939152, 312350761370734541596
Offset: 0

Views

Author

Robert Israel, Jun 09 2017

Keywords

Comments

Row sums of A155495.
a(n) is the constant term in the expansion of (-1 + (1 + x + 1/x)^2)^n. - Seiichi Manyama, Nov 21 2019

Crossrefs

Programs

  • Maple
    f:= gfun:-rectoproc({n*(2*n-1)*a(n) = (32*(n-2))*(2*n-5)*a(n-3)+(8*(9*n^2-31*n+28))*a(n-2)+(2*(3*n^2+7*n-9))*a(n-1), a(0)=1,a(1)=2, a(2)=14},a(n),remember):
    map(f, [$0..30]);
  • Mathematica
    Table[Sum[Binomial[n, k] Binomial[2 n, 2 k], {k, 0, n}], {n, 0, 24}] (* Michael De Vlieger, Jun 09 2017 *)
  • PARI
    {a(n) = polcoef((-1+(1+x+1/x)^2)^n, 0)} \\ Seiichi Manyama, Nov 21 2019

Formula

a(n) = hypergeom([-n,-n,1/2-n],[1/2,1],-1).
n*(2*n-1)*a(n) = (32*(n-2))*(2*n-5)*a(n-3)+(8*(9*n^2-31*n+28))*a(n-2)+(2*(3*n^2+7*n-9))*a(n-1).
G.f.: sqrt((1-2*x+sqrt(1-8*x))/(2*(1-7*x-8*x^2))).
a(n) ~ 8^n / sqrt(3*Pi*n). - Vaclav Kotesovec, Nov 27 2017
a(n) = Sum_{k=0..floor(n/2)} binomial(2*n,k) * binomial(3*n-2*k-1,n-2*k). - Seiichi Manyama, Feb 13 2024
From Peter Bala, Aug 30 2025: (Start)
n*(2*n - 1)*(3*n - 4)*a(n) = 2*(21*n^3 - 49*n^2 + 33*n - 6)*a(n-1) + 8*(n - 1)*(3*n - 1)*(2*n - 3)*a(n-2) with a(0) = 1 and a(1) = 2.
a(n) = Sum_{0 <= i, j <= n/2} binomial(2*n, j)*binomial(2*n+i-1, i)*binomial(2*n, n- 2*i-2*j) (verified to satisfy the above second-order recurrence using the MulZeil procedure in Doron Zeilberger's MultiZeilberger Maple package).
Equivalently, a(n) = [x^n] ( (1 + x + x^2 + x^3)/(1 - x^2) )^(2*n). Cf. A240688.
The Gauss congruences hold: a(n*p^k) == a(n*p^(k-1)) (mod p^k) for all primes p and positive integers n and k.
Conjecture: the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(2*k)) hold for primes p >= 5 and positive integers n and k. (End)

A168518 Expansion of g.f. (1/2)*( a*(1+x)^n + b*(1-x)^(n+2)*LerchPhi(x, -n-1, 1) + c*2^(n+1)*(1-x)^(n+1)*LerchPhi(x, -n, 1/2) ), where a = -4, b = 2, and c = 2, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 12, 1, 1, 51, 51, 1, 1, 170, 514, 170, 1, 1, 521, 3646, 3646, 521, 1, 1, 1552, 22247, 49472, 22247, 1552, 1, 1, 4591, 125565, 534995, 534995, 125565, 4591, 1, 1, 13590, 677776, 5058698, 9506078, 5058698, 677776, 13590, 1, 1, 40341, 3560448, 43870968, 140136690, 140136690, 43870968, 3560448, 40341, 1
Offset: 0

Views

Author

Roger L. Bagula, Nov 28 2009

Keywords

Examples

			Triangle begins as:
  1;
  1,     1;
  1,    12,       1;
  1,    51,      51,        1;
  1,   170,     514,      170,         1;
  1,   521,    3646,     3646,       521,         1;
  1,  1552,   22247,    49472,     22247,      1552,        1;
  1,  4591,  125565,   534995,    534995,    125565,     4591,       1;
  1, 13590,  677776,  5058698,   9506078,   5058698,   677776,   13590,     1;
  1, 40341, 3560448, 43870968, 140136690, 140136690, 43870968, 3560448, 40341, 1;
		

Crossrefs

Programs

  • Mathematica
    p[x_, n_, a_, b_, c_]= (1/2)*(a*(1+x)^n + b*(1-x)^(n+2)*LerchPhi[x,-n-1,1] + c*2^(n+1)*(1-x)^(n+1)*LerchPhi[x,-n,1/2]);
    Table[CoefficientList[p[x,n,-4,2,2], x], {n,0,10}]//Flatten (* modified by G. C. Greubel, Mar 31 2022 *)
  • Sage
    def A168518(n,k,a,b,c): return (1/2)*( a*binomial(n,k) + sum( (-1)^(k-j)*(b*binomial(n+2, k-j)*(j+1)^(n+1) + 2*c*binomial(n+1,k-j)*(2*j+1)^n) for j in (0..k)) )
    flatten([[A168518(n,k,-4,2,2) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Mar 31 2022

Formula

G.f.: (1/2)*( a*(1+x)^n + b*(1-x)^(n+2)*LerchPhi(x, -n-1, 1) + c*2^(n+1)*(1 - x)^(n+1)*LerchPhi(x, -n, 1/2) ), where a = -4, b = 2, and c = 2.
From G. C. Greubel, Mar 31 2022: (Start)
T(n, k) = (1/2)*( a*binomial(n,k) + sum( (-1)^(k-j)*(b*binomial(n+2, k-j)*(j+1)^(n+1) + 2*c*binomial(n+1,k-j)*(2*j+1)^n) for j in (0..k)) ), with a = -4, b = 2, and c = 2.
T(n, n-k) = T(n, k). (End)

Extensions

Edited by G. C. Greubel, Mar 31 2022

A155497 Triangle T(n, k) = binomial(n, k)*binomial(n+1, k+1)*binomial(2*n, 2*k)/(n-k+1), read by rows.

Original entry on oeis.org

1, 1, 1, 1, 18, 1, 1, 90, 90, 1, 1, 280, 1400, 280, 1, 1, 675, 10500, 10500, 675, 1, 1, 1386, 51975, 161700, 51975, 1386, 1, 1, 2548, 196196, 1471470, 1471470, 196196, 2548, 1, 1, 4320, 611520, 9417408, 22702680, 9417408, 611520, 4320, 1, 1, 6885, 1652400, 46781280, 231567336, 231567336, 46781280, 1652400, 6885, 1
Offset: 0

Views

Author

Roger L. Bagula, Jan 23 2009

Keywords

Examples

			Triangle begins as:
  1;
  1,    1;
  1,   18,       1;
  1,   90,      90,        1;
  1,  280,    1400,      280,         1;
  1,  675,   10500,    10500,       675,         1;
  1, 1386,   51975,   161700,     51975,      1386,        1;
  1, 2548,  196196,  1471470,   1471470,    196196,     2548,       1;
  1, 4320,  611520,  9417408,  22702680,   9417408,   611520,    4320,    1;
  1, 6885, 1652400, 46781280, 231567336, 231567336, 46781280, 1652400, 6885, 1;
		

Crossrefs

Programs

  • Magma
    [Binomial(n, k)*Binomial(n+1, k+1)*Binomial(2*n, 2*k)/(n-k+1): k in [0..n], n in [0..12]]; // G. C. Greubel, May 29 2021
    
  • Mathematica
    T[n_, k_]:= Binomial[n,k]*Binomial[n+1,k+1]*Binomial[2*n,2*k]/(n-k+1);
    Table[T[n, k], {n,0,10}, {k,0,n}]//Flatten (* modified by G. C. Greubel, May 29 2021 *)
  • Sage
    flatten([[binomial(n, k)*binomial(n+1, k+1)*binomial(2*n, 2*k)/(n-k+1) for k in (0..12)] for n in (0..12)]) # G. C. Greubel, May 29 2021

Formula

T(n, k) = binomial(n, k)*binomial(2*n, 2*k)*f(n)/(f(k)*f(n-k)), where f(n) = (n+1)!.
T(n, k) = binomial(n, k)*binomial(n+1, k+1)*binomial(2*n, 2*k)/(n-k+1).
From G. C. Greubel, May 29 2021: (Start)
Sum_{k=0..n} T(n, k) = Hypergeometric4F3([-n, -n, -n-1, -n-1/2], [1/2, 1, 2], 1).
T(n, k) = binomial(n, k)*A155495(n, k)/(n-k+1).
T(n, k) = binomial(2*n, 2*k)*A103371(n, k)/(n-k+1). (End)

Extensions

Edited by G. C. Greubel, May 29 2021

A155516 Triangle T(n, k) = binomial(2*n, 2*k)*binomial(2*n+1, 2*k+1)/(2*n-2*k+1), read by rows.

Original entry on oeis.org

1, 1, 1, 1, 20, 1, 1, 105, 105, 1, 1, 336, 1764, 336, 1, 1, 825, 13860, 13860, 825, 1, 1, 1716, 70785, 226512, 70785, 1716, 1, 1, 3185, 273273, 2147145, 2147145, 273273, 3185, 1, 1, 5440, 866320, 14158144, 34763300, 14158144, 866320, 5440, 1, 1, 8721, 2372112, 71954064, 367479684, 367479684, 71954064, 2372112, 8721, 1
Offset: 0

Views

Author

Roger L. Bagula, Jan 23 2009

Keywords

Examples

			Triangle begins as:
  1;
  1,    1;
  1,   20,       1;
  1,  105,     105,        1;
  1,  336,    1764,      336,         1;
  1,  825,   13860,    13860,       825,         1;
  1, 1716,   70785,   226512,     70785,      1716,        1;
  1, 3185,  273273,  2147145,   2147145,    273273,     3185,       1;
  1, 5440,  866320, 14158144,  34763300,  14158144,   866320,    5440,    1;
  1, 8721, 2372112, 71954064, 367479684, 367479684, 71954064, 2372112, 8721, 1;
		

Crossrefs

Cf. A155495.

Programs

  • Magma
    [Binomial(2*n, 2*k)*Binomial(2*n+1, 2*k+1)/(2*n-2*k+1): k in [0..n], n in [0..12]]; // G. C. Greubel, May 29 2021
    
  • Mathematica
    T[n_, k_]:= Binomial[2*n, 2*k]*Binomial[2*n+1, 2*k+1]/(2*n-2*k+1);
    Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* modified by G. C. Greubel, May 29 2021 *)
  • Sage
    flatten([[binomial(2*n, 2*k)*binomial(2*n+1, 2*k+1)/(2*n-2*k+1) for k in (0..12)] for n in (0..12)]) # G. C. Greubel, May 29 2021

Formula

T(n, k) = (2*n + 1)!!/((2*k + 1)!!*(2*(n-k) + 1)!!)*binomial(2*n, 2*k)*binomial(n, k).
From G. C. Greubel, May 29 2021: (Start)
T(n, k) = (2*n + 1)!!/((2*k + 1)!!*(2*(n-k) + 1)!!)*A155495(n, k).
T(n, k) = binomial(2*n, 2*k)*binomial(2*n+1, 2*k+1)/(2*n-2*k+1).
Sum_{k=0..n} T(n, k) = Hypergeometric4F3([-n,-n,-n-1/2,-n+1/2], [1/2,1,3/2], 1). (End)

Extensions

Edited by G. C. Greubel, May 29 2021
Showing 1-4 of 4 results.