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 1-6 of 6 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

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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)

A369263 Expansion of (1/x) * Series_Reversion( x * (1-x)^2 / (1+x^2)^3 ).

Original entry on oeis.org

1, 2, 10, 54, 329, 2126, 14356, 100030, 713956, 5193064, 38354066, 286860714, 2168308302, 16537766036, 127114940840, 983657456878, 7657060437148, 59917814944376, 471062428422152, 3718952705982232, 29471640802526185, 234356062245289566, 1869405604537134116
Offset: 0

Views

Author

Seiichi Manyama, Jan 18 2024

Keywords

Crossrefs

Programs

  • PARI
    my(N=30, x='x+O('x^N)); Vec(serreverse(x*(1-x)^2/(1+x^2)^3)/x)
    
  • PARI
    a(n, s=2, t=3, u=2) = sum(k=0, n\s, binomial(t*(n+1), k)*binomial((u+1)*(n+1)-s*k-2, n-s*k))/(n+1);

Formula

a(n) = (1/(n+1)) * Sum_{k=0..floor(n/2)} binomial(3*n+3,k) * binomial(3*n-2*k+1,n-2*k).
a(n) = (1/(n+1)) * [x^n] ( 1/(1-x)^2 * (1+x^2)^3 )^(n+1). - Seiichi Manyama, Feb 14 2024

A378290 Square array T(n,k), n >= 0, k >= 0, read by antidiagonals downwards, where T(n,0) = 0^n and T(n,k) = k * Sum_{r=0..n} binomial(n+2*r+k,r) * binomial(r,n-r)/(n+2*r+k) for k > 0.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 4, 0, 1, 3, 9, 19, 0, 1, 4, 15, 46, 104, 0, 1, 5, 22, 82, 262, 614, 0, 1, 6, 30, 128, 486, 1588, 3816, 0, 1, 7, 39, 185, 789, 3027, 10053, 24595, 0, 1, 8, 49, 254, 1185, 5052, 19543, 65686, 162896, 0, 1, 9, 60, 336, 1689, 7801, 33290, 129606, 439658, 1101922, 0
Offset: 0

Views

Author

Seiichi Manyama, Nov 21 2024

Keywords

Examples

			Square array begins:
  1,    1,     1,     1,     1,     1,     1, ...
  0,    1,     2,     3,     4,     5,     6, ...
  0,    4,     9,    15,    22,    30,    39, ...
  0,   19,    46,    82,   128,   185,   254, ...
  0,  104,   262,   486,   789,  1185,  1689, ...
  0,  614,  1588,  3027,  5052,  7801, 11430, ...
  0, 3816, 10053, 19543, 33290, 52490, 78552, ...
		

Crossrefs

Columns k=0..2 give A000007, A186997, A218045(n+2).

Programs

  • PARI
    T(n, k, t=3, u=1) = if(k==0, 0^n, k*sum(r=0, n, binomial(t*r+u*(n-r)+k, r)*binomial(r, n-r)/(t*r+u*(n-r)+k)));
    matrix(7, 7, n, k, T(n-1, k-1))

Formula

G.f. A_k(x) of column k satisfies A_k(x) = ( 1 + x * A_k(x)^(3/k) * (1 + x * A_k(x)^(1/k)) )^k for k > 0.
G.f. of column k: B(x)^k where B(x) is the g.f. of A186997.
B(x)^k = B(x)^(k-1) + x * B(x)^(k+2) + x^2 * B(x)^(k+3). So T(n,k) = T(n,k-1) + T(n-1,k+2) + T(n-2,k+3) for n > 1.

A369208 Expansion of (1/x) * Series_Reversion( x * (1-x)^2 / (1+x^2) ).

Original entry on oeis.org

1, 2, 8, 38, 200, 1122, 6576, 39790, 246672, 1558658, 10001592, 64997814, 426922392, 2829624514, 18901301984, 127115260894, 859978039840, 5848754717314, 39964745880552, 274231943135686, 1888891689752680, 13055393137141282, 90517646431869328
Offset: 0

Views

Author

Seiichi Manyama, Jan 18 2024

Keywords

Crossrefs

Programs

  • PARI
    my(N=30, x='x+O('x^N)); Vec(serreverse(x*(1-x)^2/(1+x^2))/x)
    
  • PARI
    a(n, s=2, t=1, u=2) = sum(k=0, n\s, binomial(t*(n+1), k)*binomial((u+1)*(n+1)-s*k-2, n-s*k))/(n+1);

Formula

a(n) = (1/(n+1)) * Sum_{k=0..floor(n/2)} binomial(n+1,k) * binomial(3*n-2*k+1,n-2*k).
a(n) = (1/(n+1)) *[x^n] ( 1/(1-x)^2 * (1+x^2) )^(n+1). - Seiichi Manyama, Feb 14 2024

A369226 Expansion of (1/x) * Series_Reversion( x * (1-x) / (1+x^2)^2 ).

Original entry on oeis.org

1, 1, 4, 13, 53, 220, 968, 4373, 20271, 95705, 458904, 2228220, 10934524, 54143848, 270189008, 1357428997, 6860264323, 34853234867, 177900211204, 911867479717, 4691701977973, 24222505191984, 125448280976224, 651555603531308, 3392951906596708, 17711433386188300
Offset: 0

Views

Author

Seiichi Manyama, Jan 18 2024

Keywords

Crossrefs

Programs

  • PARI
    my(N=30, x='x+O('x^N)); Vec(serreverse(x*(1-x)/(1+x^2)^2)/x)
    
  • PARI
    a(n, s=2, t=2, u=1) = sum(k=0, n\s, binomial(t*(n+1), k)*binomial((u+1)*(n+1)-s*k-2, n-s*k))/(n+1);

Formula

a(n) = (1/(n+1)) * Sum_{k=0..floor(n/2)} binomial(2*n+2,k) * binomial(2*n-2*k,n-2*k).
a(n) = (1/(n+1)) * [x^n] ( 1/(1-x) * (1+x^2)^2 )^(n+1). - Seiichi Manyama, Feb 14 2024

A218182 Number of truth tables of bracketed formulas (case 1).

Original entry on oeis.org

0, 0, 1, 6, 33, 194, 1198, 7676, 50581, 340682, 2335186, 16237284, 114255994, 812107412, 5822171548, 42052209400, 305714145869, 2235262899418, 16426616425002, 121265916776148, 898878250833358, 6687497426512700, 49920590244564484
Offset: 0

Views

Author

N. J. A. Sloane, Oct 23 2012

Keywords

Crossrefs

Programs

  • PARI
    my(x='x+O('x^30)); concat([0,0], Vec(((1-8*x)^(1/2)-3)*((2+2*(1-8*x)^(1/2)+8*x)^(1/2)-2)/8)) \\ Michel Marcus, Oct 21 2020

Formula

Yildiz gives a g.f.
G.f.: ((1-8*x)^(1/2)-3)*((2+2*(1-8*x)^(1/2)+8*x)^(1/2)-2)/8. - Mark van Hoeij, May 16 2013
Showing 1-6 of 6 results.