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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A092371 Triangle read by rows: T(n, k) = binomial(n, k) * binomial(n+k, n-k).

Original entry on oeis.org

1, 6, 1, 18, 15, 1, 40, 90, 28, 1, 75, 350, 280, 45, 1, 126, 1050, 1680, 675, 66, 1, 196, 2646, 7350, 5775, 1386, 91, 1, 288, 5880, 25872, 34650, 16016, 2548, 120, 1, 405, 11880, 77616, 162162, 126126, 38220, 4320, 153, 1, 550, 22275, 205920, 630630
Offset: 1

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Author

Benoit Cloitre, Mar 20 2004

Keywords

Comments

Related to the coefficients of x^k y^k in the n-th power of x^2 + x*y + 2*x + y + 1. - F. Chapoton, Jan 04 2025

Examples

			Triangle starts:
  [1]   1;
  [2]   6,    1;
  [3]  18,   15,     1;
  [4]  40,   90,    28,     1;
  [5]  75,  350,   280,    45,     1;
  [6] 126, 1050,  1680,   675,    66,    1;
  [7] 196, 2646,  7350,  5775,  1386,   91,   1;
  [8] 288, 5880, 25872, 34650, 16016, 2548, 120, 1;

Crossrefs

First column = A002411, second column = A001297, third column = A107418, fourth column = A105251, fifth column = A104673.
Main diagonal = 1, second diagonal = A000384.
Cf. A063007, A006480 (central terms), A082759 (row sums + 1).
Cf. A104684.

Programs

  • Maple
    T := (n, k) -> binomial(n, k) * binomial(n+k, n-k):  # Peter Luschny, Jan 04 2025
  • PARI
    T(n,k) = binomial(n,k)*binomial(n+k,n-k)

Formula

T(n, k) = [x^(n-k)] F(-n, -n-k; 1; x). - Paul Barry, Sep 04 2008

A290310 Irregular triangle read by rows. Row n gives the coefficients of the polynomial multiplying the exponential function in the e.g.f. of the (n+1)-th diagonal sequences of triangle A008459 (Pascal squares). T(n,k) for n >= 0 and k = 0..2*n.

Original entry on oeis.org

1, 1, 3, 2, 1, 8, 19, 18, 6, 1, 15, 69, 147, 162, 90, 20, 1, 24, 176, 624, 1251, 1500, 1070, 420, 70, 1, 35, 370, 1920, 5835, 11253, 14240, 11830, 6230, 1890, 252, 1, 48, 687, 4850, 20385, 55908, 104959, 137886, 127050, 80640, 33642, 8316, 924, 1, 63, 1169, 10703, 58821, 214123, 545629, 1004307, 1356194, 1347318, 974862, 500346, 172788, 36036, 3432
Offset: 0

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Author

Wolfdieter Lang, Jul 27 2017

Keywords

Comments

The length of row n of this irregular triangle is 2*n+1.
A008459 gives the squares of the entries of Pascal's triangle A007318.
The e.g.f. of the (n+1)-th diagonal sequence of the squares of Pascal's triangle (A008459) is EDP2(n, t) = Sum_{m=0..n} binomial(n+m, m)^2*t^m/m!, for n >= 0. It turns out to be EDP2(n, t) = exp(t)*Sum_{k=0..2*n} T(n, k)*t^k/k!.
This has been computed from the corresponding o.g.f.: GDP2(n, x) = Sum_{m=0..n} binomial(n+m, m)^2*x^m, which is GDP2(n, x) = Sum_{m=0..n} binomial(n,m)^2*x^m / (1 - x)^(2*n+1) (see the triangle A008459, and comments in A288876 on how to compute these o.g.f.s). To obtain the e.g.f.s from the o.g.f.s the formulas (23) - (25) of the W. Lang link given in A060187 have been used.

Examples

			The irregular triangle T(n, k) begins:
  n\k 0  1   2    3     4     5      6      7      8     9    10   11  12 ..,
  0:  1
  1:  1  3   2
  2:  1  8  19   18     6
  3:  1 15  69  147   162    90     20
  4:  1 24 176  624  1251  1500   1070    420     70
  5:  1 35 370 1920  5835 11253  14240  11830   6230  1890   252
  6:  1 48 687 4850 20385 55908 104959 137886 127050 80640 33642 8316 924
  ...
  7: 1 63 1169 10703 58821 214123 545629 1004307 1356194 1347318 974862 500346 172788 36036 3432
  ...
n = 3: The e.g.f. of the fourth diagonal sequence of A008459 is A001249 = [1, 16, 100, ...] is exp(t)*(1 + 15*t + 69*t^2/2! + 147*t^3/3! + 162*t^4/4! + 90*t^5/5! + 20*t^6/6!). The corresponding o.g.f. from which the e.g.f. has been computed is (1 + x)*(1 + 8*x + x^2)/(1 - x)^7 = (1 + 9*x + 9*x^2 + x^3)/(1 - x)^7.

Links

Crossrefs

T(n, n) = A005258(n). The squares of the first diagonals are in A000012, A000290(n+1), A000537(n+1), A001249, A288876 (for d = 0..4).
Cf. A007318, A008459, A063007, A000984, A002457.

Programs

  • Maple
    T := (n,k) -> binomial(2*n, k)*hypergeom([-k, -n, -n], [1, -2*n], 1):
seq(seq(simplify(T(n,k)),k=0..2*n),n=0..7); # Peter Luschny, Feb 10 2018
  • Mathematica
    Table[Sum[Binomial[2 n - i, k - i] Binomial[n, i]^2, {i, 0, k}], {n, 0, 7}, {k, 0, 2 n}] // Flatten (* Michael De Vlieger, Jul 30 2017 *)

Formula

T(n, k) = Sum_{i = 0..k} binomial(2*n - i, k-i)*binomial(n, i)^2.
From Peter Bala, Feb 06 2018: (Start)
T(n,k) = Sum_{i = 0..k} (-1)^(k-i)*binomial(k,i)*binomial(n+i,i)^2.
T(n,k) = Sum_{i = 0..k} binomial(n,i)*binomial(n,k-i)*binomial(n+k-i,k-i).
T(n,2*n) = binomial(2*n,n) = A000984(n); T(n+1,2*n+1) = 3*(2*n+1)!/n!^2 = 3*A002457(n).
Recurrence: (2*n-k)*(2*n-k-1)T(n,k) = (5*n^2+2*n*k+1-4*n-k)*T(n-1,k) - (n-1)^2*T(n-2,k).
n-th row polynomial R(n,x) = (1 + x)^n * P(n,2*x + 1) = (1 + x)^n * the n-th row polynomial of A063007, where P(n,x) is the n-th Legendre polynomial.
R(n,x) = Sum_{k >= 0} binomial(n+k,k)^2 * x^k/(1 + x)^(k+1).
(x - 1)^(2*n)/x^n * R(n,1/(x - 1)) = Sum_{k = 0..n} binomial(n,k)^2*x^k, the n-th row polynomial of A008459.
R(n,x) = (1 + x)^n o (1 + x)^n where o denotes the black diamond product of power series as defined in Dukes and White.
R(n,x) = coefficient of u^n*v^n in the expansion of the rational function 1/((1+x)*(1+u)(1+v) - x). (End)
T(n,k) = binomial(2*n, k)*hypergeom([-k, -n, -n],[1, -2*n], 1). - Peter Luschny, Feb 10 2018

A370258 Triangle read by rows: T(n, k) = binomial(n, k)*binomial(2*n+k, k), 0 <= k <= n.

Original entry on oeis.org

1, 1, 3, 1, 10, 15, 1, 21, 84, 84, 1, 36, 270, 660, 495, 1, 55, 660, 2860, 5005, 3003, 1, 78, 1365, 9100, 27300, 37128, 18564, 1, 105, 2520, 23800, 107100, 244188, 271320, 116280, 1, 136, 4284, 54264, 339150, 1139544, 2089164, 1961256, 735471, 1, 171, 6840, 111720, 921690, 4239774, 11306064
Offset: 0

Views

Author

Peter Bala, Feb 13 2024

Keywords

Comments

Compare with A063007(n, k) = binomial(n, k)*binomial(n+k, k), the table of coefficients of the shifted Legendre polynomials P(n, 2*x + 1).

Examples

			Triangle begins
n\k| 0    1     2      3       4       5       6       7
- - - - - - - - - - - - - - - - - - - - - - - - - - - - -
 0 | 1
 1 | 1    3
 2 | 1   10    15
 3 | 1   21    84     84
 4 | 1   36   270    660     495
 5 | 1   55   660   2860    5005    3003
 6 | 1   78  1365   9100   27300   37128   18564
 7 | 1  105  2520  23800  107100  244188  271320  116280
 ...

Crossrefs

A114496 (row sums), A000984 (alt. row sums unsigned), A005809 (main diagonal), A090763 (first subdiagonal), A014105 (column 1).
Cf. A026000, A063007, A102537, A110608, A339710.

Programs

  • Maple
    seq(print(seq(binomial(n, k)*binomial(2*n+k, k), k = 0..n)), n = 0..10);

Formula

n-th row polynomial R(n, x) = Sum_{k = 0..n} binomial(n, k)*binomial(2*n+k, k)*x^k = (1 + x)^n * Sum_{k = 0..n} binomial(n, k)*binomial(2*n, k)*(x/(1 + x))^k = Sum_{k = 0..n} A110608(n, n-k)*x^k*(1 + x)^(n-k).
(x - 1)^n * R(n, 1/(x - 1)) = Sum_{k = 0..n} binomial(n,k)*binomial(2*n, n-k)*x^k = the n-th row polynomial of A110608.
R(n, x) = hypergeom([-n, 2*n + 1], [1], -x).
Second-order differential equation: ( (1 + x)^n * (x + x^2)*R(n, x)' )' = n*(2*n + 1)*(1 + x)^n * R(n, x), where the prime indicates differentiation w.r.t. x.
Equivalently, x*(1 + x)*R(n, x)'' + ((n + 2)*x + 1)*R(n, x)' - n*(2*n + 1)*R(n, x)' = 0.
Analog of Rodrigues' formula for the shifted Legendre polynomials:
R(n, x) = 1/(1 + x)^n * 1/n! * (d/dx)^n (x*(1 + x)^2)^n.
Analog of Rodrigues' formula for the Legendre polynomials:
R(n, (x-1)/2) = 1/(n!*2^n) * 1/(1 + x)^n *(d/dx)^n ((x - 1)*(x + 1)^2)^n.
Orthogonality properties:
Integral_{x = -1..0} (1 + x)^n * R(n, x) * R(m, x) dx = 0 for n > m.
Integral_{x = -1..0} (1 + x)^n * R(n, x)^2 dx = 1/(3*n + 1).
Integral_{x = -1..0} (1 + x)^(n+m) * R(n, x) * R(m, x) dx = 0 for m >= 2*n + 1 or m <= (n - 1)/2.
Integral_{x = -1..0} (1 + x)^k * R(n, x) dx = 0 for n <= k <= 2*n - 1;
Integral_{x = -1..0} (1 + x)^(2*n) * R(n, x) dx = (2*n)!*n!/(3*n+1)! = 1/A090816(n).
Recurrence for row polynomials:
2*n*(2*n - 1)*((9*n - 12)*x + 8*n - 11)*(1 + x)*R(n, x) = (9*(3*n - 1)*(3*n - 2)*(3*n - 4)*x^3 + 3*(3*n - 1)*(3*n - 2)*(20*n - 27)*x^2 + 6*(3*n - 2)*(20*n^2 - 34*n + 9)*x + 2*(32*n^3 - 76*n^2 + 50*n - 9))*R(n-1, x) - 2*(n - 1)*(2*n - 3)*((9*n - 3)*x + 8*n - 3)*R(n-2, x), with R(0, x) = 1, R(1, x) = 1 + 3*x.
Conjecture: exp( Sum_{n >= 1} R(n,t)*x^n/n ) = 1 + (1 + 3*t)*x + (1 + 8*t + 12*t^2)*x^2 + ... is the o.g.f. for A102537. If true, then it would follows that, for each integer t, the sequence u = {R(n,t) : n >= 0} satisfies the Gauss congruences u(m*p^r) == u(m*p^(r-1)) (mod p^r) for all primes p and positive integers m and r.
R(n, 1) = A114496(n); R(n, -1) = (-1)^n * A000984(n).
R(n, 2) = A339710(n); R(n, -2) = (-1)^n * A026000(n).
(2^n)*R(n, -1/2) = A234839(n).
Previous Showing 61-63 of 63 results.

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