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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A247497 Triangle read by rows, T(n,k) (n>=0, 0<=k<=n) coefficients of the partial fraction decomposition of rational functions generating the columns of A247495 (the Motzkin polynomials evaluated at nonnegative integers).

Original entry on oeis.org

1, 1, 1, 2, 3, 2, 4, 10, 12, 6, 9, 33, 62, 60, 24, 21, 111, 300, 450, 360, 120, 51, 378, 1412, 3000, 3720, 2520, 720, 127, 1303, 6552, 18816, 32760, 34440, 20160, 5040, 323, 4539, 30186, 113820, 264264, 388080, 352800, 181440, 40320
Offset: 0

Views

Author

Peter Luschny, Dec 13 2014

Keywords

Examples

			Triangle starts:
[  1],
[  1,    1],
[  2,    3,    2],
[  4,   10,   12,     6],
[  9,   33,   62,    60,    24],
[ 21,  111,  300,   450,   360,   120],
[ 51,  378, 1412,  3000,  3720,  2520,   720],
[127, 1303, 6552, 18816, 32760, 34440, 20160, 5040].
.
[n=3] -> [4,10,12,6] -> 4/(x-1)+10/(x-1)^2+12/(x-1)^3+6/(x-1)^4 = 2*x*(-x+2*x^2+2)/(x-1)^4; generating function of A247495[n,3] = 0,4,14, 36,...
[n=4] -> [9,33,62,60,24] -> -9/(x-1)-33/(x-1)^2-62/(x-1)^3-60/(x-1)^4-24/(x-1)^5 = -(2-x-3*x^3+17*x^2+9*x^4)/(x-1)^5; generating function of A247495[n,4] = 2,9,42,137,...

Crossrefs

Cf. A247495, A097610, A001006, A058987, A001710.

Programs

  • Maple
    A247497_row := proc(n) local A, M, p;
A := (n,k) -> `if`(type(n-k, odd),0,n!/(k!*((n-k)/2)!^2*((n-k)/2+1))):
M := (k,x) -> add(A(k,j)*x^j,j=0..k): # Motzkin polynomial
p := expand(sum(x^k*M(n,k),k=0..infinity));
[seq((-1)^(n+1)*coeff(convert(p,parfrac),(x-1)^(-j)),j=1..n+1)] end:
seq(print(A247497_row(n)),n=0..7);

Formula

Let M_{n}(x) = sum_{k=0..n} A097610(n,k)*x^k denote the Motzkin polynomials. The T(n,k) are implicitly defined by:
sum_{k=0..n} (-1)^(n+1)*T(n,k)/(x-1)^(k+1) = sum_{k>=0} x^k*M_n(k).
T(n, 0) = A001006(n) (Motzkin numbers).
T(n, n) = A000142(n) = n!.
T(n, 1) = A058987(n+1) for n>=1.
T(n,n-1)= A001710(n+1) for n>=1.

A264766 Irregular symmetric triangle of coefficients T(n,k) of the polynomials p(n,x) = Sum_{k=0..n} binomial(n+1,k)*(1+x)^(2*k)*(-x)^(n-k) for 0 <= k <= 2*n.

Original entry on oeis.org

1, 2, 3, 2, 3, 9, 13, 9, 3, 4, 18, 40, 51, 40, 18, 4, 5, 30, 90, 165, 201, 165, 90, 30, 5, 6, 45, 170, 405, 666, 783, 666, 405, 170, 45, 6, 7, 63, 287, 840, 1736, 2646, 3039, 2646, 1736, 840, 287, 63, 7, 8, 84, 448, 1554, 3864, 7224, 10424, 11763, 10424, 7224, 3864, 1554, 448, 84, 8, 9, 108, 660, 2646, 7686, 17010, 29520, 40851, 45481, 40851, 29520, 17010, 7686, 2646, 660, 108, 9, 10
Offset: 0

Views

Author

Werner Schulte, Nov 23 2015

Keywords

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:  2   3    2
  2:  3   9   13     9     3
  3:  4  18   40    51    40    18      4
  4:  5  30   90   165   201   165     90     30      5
  5:  6  45  170   405   666   783    666    405    170    45     6
  6:  7  63  287   840  1736  2646   3039   2646   1736   840   287  63   7
  etc.
The polynomial corresponding to row 2 is p(2,x) = 3 + 9*x + 13*x^2 + 9*x^3 + 3*x^4.

Links

Crossrefs

Cf. A000108, A000984, A001006, A002426, A003462, A005061, A027907, A058987, A163774, A260056.

Programs

  • Mathematica
    T[n_, k_] := Sum[(-1)^j*Binomial[n + 1, j + 1]*Binomial[2*n - 2*j, k - j], {j, 0, n - Abs[k - n]}]; Table[T[n, k], {n,0,10}, {k,0,2*n}] // Flatten (* G. C. Greubel, Aug 12 2017 *)
  • PARI
    T(n,k) = sum(j=0, n-abs(k-n), (-1)^j*binomial(n+1,j+1)*binomial(2*n-2*j,k-j));
tabf(nn) = for (n=0, nn, for (k=0, 2*n, print1(T(n, k), ", ");); print();); \\ Michel Marcus, Nov 24 2015

Formula

T(n,k) = Sum_{j=0..n-d} (-1)^j*binomial(n+1,j+1)*binomial(2*n-2*j,k-j) if d = 0 or better d = abs(k-n), and 0 <= k <= 2*n.
Recurrence: T(n,0) = n+1, and T(n,k) = 0 for k < 0 or k > 2*n, and T(n+1,k) = T(n,k-2) + T(n,k-1) + T(n,k) + binomial(2*n+2,k) for k > 0 and n >= 0.
T(n,k) = T(n,2*n-k) for 0 <= k <= 2*n.
p(n,x) = Sum_{k=0..2*n} T(n,k)*x^k = Sum_{k=0..n} (1+x)^(2*k)*(1+x+x^2)^(n-k) = Sum_{k=0..n} binomial(n+1,k)*(1+x+x^2)^k*x^(n-k) for n >= 0.
Recurrence: p(0,x) = 1, and p(n+1,x) = (1+x+x^2)*p(n,x)+(1+x)^(2*n+2), n >= 0.
T(n,n) = Sum_{j=0..n} (-1)^(n-j)*binomial(n+1,j)*binomial(2*j,j) = A000984(n+1)-A002426(n+1) for n >= 0 (see also A163774).
Sum_{n>=0} T(n,n)*x^(n+1) = 1/sqrt(1-4*x) - 1/sqrt(1-2*x-3*x^2) for abs(x) < 1/4.
T(n,n-1) = binomial(2*n+2,n) - A027907(n+1,n) for n > 0.
T(n+1,n)/(n+2) = A000108(n+2) - A001006(n+1) for n >= 0 (see also A058987).
Row sums: p(n,1) = A005061(n+1) for n >= 0.
Alternating row sums: p(n,-1) = 1 for n >= 0.
p(n,-2) = Sum_{k=0..2*n} T(n,k)*(-2)^k = A003462(n+1) for n >= 0.
T(n,k) = Sum_{j=0..k} (-1)^j*A260056(n,j)*binomial(2*n-j,k-j) for 0 <= k <= 2*n.
A260056(n,k) = Sum_{j=0..k} (-1)^j*T(n,j)*binomial(2*n-j,k-j) for 0 <= k <= 2*n.
p(n,-1-x) = Sum{k=0..2*n} A260056(n,k)*x^(2*n-k) for n >= 0.
p(n,-x/(1+x))*(1+x)^(2*n) = Sum_{k=0..2*n} A260056(n,k)*x^k for n >= 0.
Sum_{n>=0} p(n,x)*t^n = 1/((1-t*(1+x)^2)*(1-t*(1+x+x^2))).
p(n,x)*x = (1+x)^(2*n+2) - (1+x+x^2)^(n+1), n >= 0.
T(n,k) = binomial(2*n+2,k+1) - A027907(n+1,k+1) for 0 <= k <= 2*n.
Showing 1-2 of 2 results.

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