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.

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A387369 a(n) = Sum_{k=0..n} 2^k * 3^(n-k) * binomial(n+2,k) * binomial(n+2,n-k).

Original entry on oeis.org

1, 15, 174, 1850, 18915, 189525, 1877596, 18476820, 181083285, 1770245675, 17278828842, 168496597230, 1642259489143, 16002398658225, 155919866646840, 1519307275471400, 14806582620440553, 144329229195062535, 1407215890063071910, 13724133021646678050, 133885448856624266571
Offset: 0

Views

Author

Seiichi Manyama, Aug 27 2025

Keywords

Links

Crossrefs

Cf. A006442, A387368.
Cf. A387339.

Programs

  • Magma
    [&+[2^k * 3^(n-k) * Binomial(n+2,k) * Binomial(n+2,n-k): k in [0..n]]: n in [0..25]]; // Vincenzo Librandi, Aug 29 2025
  • Mathematica
    Table[Sum[2^k * 3^(n-k)*Binomial[n+2,k]*Binomial[n+2, n-k],{k,0,n}],{n,0,25}] (* Vincenzo Librandi, Aug 29 2025 *)
  • PARI
    a(n) = sum(k=0, n, 2^k*3^(n-k)*binomial(n+2, k)*binomial(n+2, n-k));

Formula

a(n) = Sum_{k=0..n} 3^k * 2^(n-k) * binomial(n+2,k) * binomial(n+2,n-k).
n*(n+4)*a(n) = (n+2) * (5*(2*n+3)*a(n-1) - (n+1)*a(n-2)) for n > 1.
a(n) = Sum_{k=0..floor(n/2)} 6^k * 5^(n-2*k) * binomial(n+2,n-2*k) * binomial(2*k+2,k).
a(n) = [x^n] (1+5*x+6*x^2)^(n+2).
E.g.f.: exp(5*x) * BesselI(2, 2*sqrt(6)*x) / 6, with offset 2.

A084774 Coefficients of 1/sqrt(1-14*x+9*x^2); also, a(n) is the central coefficient of (1+7x+10x^2)^n.

Original entry on oeis.org

1, 7, 69, 763, 8881, 106407, 1298949, 16065483, 200630241, 2524253767, 31947470149, 406281388443, 5187375332881, 66454791792487, 853788052488069, 10996378059281643, 141934540736139201, 1835494145265388167, 23776671158743933509, 308463567293772941883
Offset: 0

Views

Author

Paul D. Hanna, Jun 11 2003

Keywords

Comments

G.f.: 1/sqrt(1-2*b*x+(b^2-4*c)*x^2) yields central coefficients of (1+b*x+c*x^2)^n.
Diagonal of rational functions 1/(1 - x - 2*y - 3*x*y), 1/(1 - x - 2*y*z - 3*x*y*z). - Gheorghe Coserea, Jul 06 2018

Links

Crossrefs

Cf. A006442, A084771.

Programs

  • GAP
    List([0..20],n->Sum([0..n],k->Binomial(n,k)^2*2^k*5^(n-k))); # Muniru A Asiru, Jul 29 2018
  • Magma
    [3^n*Evaluate(LegendrePolynomial(n), 7/3) : n in [0..40]]; // G. C. Greubel, May 31 2023
  • Mathematica
    Table[Sum[Binomial[n,k]^2*2^k*5^(n-k),{k,0,n}],{n,0,20}] (* Vaclav Kotesovec, Oct 14 2012 *)
Table[n! SeriesCoefficient[E^(7 x) BesselI[0, 2 Sqrt[10] x], {x,0,n}], {n,0,20}] (* Vincenzo Librandi, May 10 2013 *)
Table[3^n*LegendreP[n, 7/3], {n,0,40}] (* G. C. Greubel, May 31 2023 *)
a[n_] := 3^n*HypergeometricPFQ[{-n, n + 1}, {1}, -2/3]; Flatten[Table[a[n], {n,0,19}]] (* Detlef Meya, May 22 2024 *)
  • PARI
    for(n=0,30,t=polcoeff((1+7*x+10*x^2)^n,n,x); print1(t","))
  • PARI
    {a(n)=sum(k=0, n, binomial(n, k)^2*2^k*5^(n-k))} \\ Paul D. Hanna, Sep 28 2012
  • SageMath
    [3^n*gen_legendre_P(n, 0, 7/3) for n in range(41)] # G. C. Greubel, May 31 2023

Formula

a(n) = Sum_{k=0..n} binomial(n,k)^2 * 2^k * 5^(n-k). - Paul D. Hanna, Sep 28 2012
D-finite with recurrence: n*a(n) = 7*(2*n-1)*a(n-1) - 9*(n-1)*a(n-2). - Vaclav Kotesovec, Oct 14 2012
a(n) ~ sqrt(200 + 70*sqrt(10))*(7 + 2*sqrt(10))^n/(20*sqrt(Pi*n)) = (sqrt(2) + sqrt(5))^(2*n+1)/(2*10^(1/4)*sqrt(Pi*n)). - Vaclav Kotesovec, Oct 14 2012
a(n) = 3^n * LegendreP(n, 7/3). - G. C. Greubel, May 31 2023
a(n) = 3^n*hypergeom([-n, n + 1], [1], -2/3). - Detlef Meya, May 22 2024

A376467 Triangular array read by rows: A063007 * A007318.

Original entry on oeis.org

1, 3, 2, 13, 18, 6, 63, 132, 90, 20, 321, 900, 930, 420, 70, 1683, 5910, 8190, 5600, 1890, 252, 8989, 37926, 65940, 60480, 30870, 8316, 924, 48639, 239624, 501228, 577080, 395010, 160776, 36036, 3432, 265729, 1497096, 3660300, 5072760, 4358970, 2378376, 804804, 154440, 12870, 1462563, 9274410, 25951860, 42060480, 43513470, 29801772, 13513500, 3912480, 656370, 48620
Offset: 0

Views

Author

Peter Bala, Sep 30 2024

Keywords

Comments

Note that the n-th row generating polynomial of A063007 is equal to P(n,2*x + 1), where P(n,x) denotes the n-th Legendre polynomial.
The matrix product A063007 * A007318^(-1) is equal to a signed version of A063007 and A007318^(-1) * A063007 = A115951.

Examples

			Triangle begins
 n\k|     0       1       2       3       4       5      6     7
 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
  0 |     1
  1 |     3       2
  2 |    13      18       6
  3 |    63     132      90      20
  4 |   321     900     930     420      70
  5 |  1683    5910    8190    5600    1890     252
  6 |  8989   37926   65940   60480   30870    8316    924
  7 | 48639  239624  501228  577080  395010  160776  36036  3432
  ...

Crossrefs

A000984 (main diagonal), A002457( (1/3)*first subdiagonal ), A001850 (Column 0), A002695 ( (1/2)*Column 1 ), A277660 ( (1/3)*Column 2 ), A006442 (row sums).
Cf. A063007, A007318, A115951, A118384.

Programs

  • Maple
    A376467 := proc(n, k); add(binomial(n, j)*binomial(n+j, j)*binomial(j, k), j = k..n) end:
seq(print(seq(A376467(n, k) , k = 0..n)), n = 0..10);

Formula

T(n, k) = Sum_{j = k..n} binomial(n, j)*binomial(n+j, j)*binomial(j, k).
(n - k)*T(n, k) = 3*(2*n - 1)*T(n-1, k) - (n + k - 1)*T(n-2, k).
T(n, k) = (1/k!) * (d/dx)^k (P(n, 2*x+1)) evaluated at x = 1, where P(n,x) denotes the n-th Legendre polynomial.
G.f. for triangle: 1/sqrt(1 - 6*t + t^2 - 4*t*x) = 1 + (3 + 2*x)*t + (13 + 18*x + 6*x^2)*t^2 + ....
G.f. for column k: binomial(2*k, k) * x^k/(1 - 6*x + x^2)^(k+1/2).
T(n, k) is divisible by binomial(2*k, k) and the array ( T(n, k)/binomial(2*k, k) )n,k >= 0 is the Riordan array (1/sqrt(1 - 6*x + x^2), x/(1 - 6*x + x^2)).
T(n, k) is divisible by binomial(n+k, k) and the array ( T(n, k)/binomial(n+k, k) )n,k >= 0 is the Riordan array A118384.
T(n, n) = binomial(2*n, n); T(n, n-1) = 3*(2*n-1)!/(n-1)!^2 = 3 * A002457(n-1) for n >= 1.
The n-th row polynomial R(n, x) = Sum_{k = 0..n} binomial(n, k)*binomial(n+k, k)*(1 + x)^k = P(n, 2*x+3) = hypergeom([-n, n+1], [1], -1-x).
Recurrence: n*R(n, x) = (2*x + 3)*(2*n - 1)*R(n-1, x) - (n - 1)*R(n-2, x) with R(0, x) = 1.
If we set R(-1,x) = 1, we can run the recurrence backwards to give R(-n, x) = Sum_{k = 0..n} binomial(-n, k)*binomial(-n+k, k)*(1 + x)^k = R(n-1, x).
R(n, x) = (-1)^n * R(n, -x-3).
R(n, x) = 1/n! * (d/dx)^n( ((1 + x)*(2 + x))^n ).
R(1, x) = 3 + 2*x divides R(2*n+1, x) in the polynomial ring Z[x].

A387366 Expansion of 1/(1 - 10*x + x^2)^(3/2).

Original entry on oeis.org

1, 15, 186, 2150, 23955, 260925, 2798740, 29688300, 312289605, 3263403275, 33922215822, 351081270930, 3620347505047, 37217828876025, 381591426746280, 3903412392243800, 39848499404096265, 406072116038615175, 4131456665470332130, 41974347760312761150, 425899035044461953051
Offset: 0

Views

Author

Seiichi Manyama, Aug 27 2025

Keywords

Links

Crossrefs

Cf. A006442, A387367.
Cf. A387368.

Programs

  • Magma
    R := PowerSeriesRing(Rationals(), 34); f := 1/(1 - 10*x + x^2)^(3/2); coeffs := [ Coefficient(f, n) : n in [0..33] ]; coeffs; // Vincenzo Librandi, Aug 29 2025
  • Mathematica
    CoefficientList[Series[1/(1-10*x+x^2)^(3/2),{x,0,33}],x] (* Vincenzo Librandi, Aug 29 2025 *)
  • PARI
    my(N=30, x='x+O('x^N)); Vec(1/(1-10*x+x^2)^(3/2))

Formula

n*a(n) = 5*(2*n+1)*a(n-1) - (n+1)*a(n-2) for n > 1.
a(n) = ((n+2)/2) * A387368(n).
a(n) = (-1)^n * Sum_{k=0..n} (1/10)^(n-2*k) * binomial(-3/2,k) * binomial(k,n-k).
Previous Showing 11-14 of 14 results.

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