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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A116382 Riordan array (1/sqrt(1-4*x^2), (1-2*x^2*c(x^2))*(x^2*c(x^2))/(x*(1-x-x^2*c(x^2)))) where c(x) is the g.f. of A000108.

Original entry on oeis.org

1, 0, 1, 2, 1, 1, 0, 3, 2, 1, 6, 4, 5, 3, 1, 0, 10, 10, 8, 4, 1, 20, 15, 21, 19, 12, 5, 1, 0, 35, 42, 42, 32, 17, 6, 1, 70, 56, 84, 92, 77, 50, 23, 7, 1, 0, 126, 168, 192, 180, 131, 74, 30, 8, 1, 252, 210, 330, 405, 400, 326, 210, 105, 38, 9, 1
Offset: 0

Views

Author

Paul Barry, Feb 12 2006

Keywords

Comments

Row sums are A116383. Diagonal sums are A116384.
First column has e.g.f. Bessel_I(0,2*x) (A000984 with interpolated zeros).
Second column has e.g.f. Bessel_I(1,2*x) + Bessel_I(2,2*x) (A037952).
Third column has e.g.f. Bessel_I(2,2*x) + 2*Bessel_I(3,2*x) + Bessel_I(4,2*x) (A116385).
A binomial-Bessel triangle: column k has e.g.f. Sum_{j=0..k} C(k,j) * Bessel_I(k+j,2*x).

Examples

			Triangle begins
    1;
    0,   1;
    2,   1,   1;
    0,   3,   2,   1;
    6,   4,   5,   3,   1;
    0,  10,  10,   8,   4,   1;
   20,  15,  21,  19,  12,   5,   1;
    0,  35,  42,  42,  32,  17,   6,   1;
   70,  56,  84,  92,  77,  50,  23,   7,  1;
    0, 126, 168, 192, 180, 131,  74,  30,  8, 1;
  252, 210, 330, 405, 400, 326, 210, 105, 38, 9, 1;
		

Programs

  • GAP
    Flat(List([0..10], n-> List([0..n], k-> Sum([0..n], j-> (-1)^(n-j)*Binomial(n,j)*Sum([0..j], m-> Binomial(j,m-k)*Binomial(m,j-m)  ))))); # G. C. Greubel, May 22 2019
  • Magma
    T:= func< n,k | (&+[(-1)^(n-j)*Binomial(n,j)*(&+[Binomial(j,m-k)* Binomial(m,j-m): m in [0..j]]): j in [0..n]]) >;
    [[T(n,k): k in [0..n]]: n in [0..10]]; // G. C. Greubel, May 22 2019
    
  • Mathematica
    T[n_, k_] := Sum[(-1)^(n-j)*Binomial[n, j]*Sum[Binomial[j, i-k]* Binomial[i, j-i], {i, 0, j}], {j, 0, n}];
    Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jan 24 2018 *)
  • PARI
    {T(n,k) = sum(j=0,n, (-1)^(n-j)*binomial(n,j)*sum(m=0,j, binomial(j,m-k)*binomial(m,j-m) ))}; \\ G. C. Greubel, May 22 2019
    
  • Sage
    def T(n, k): return sum((-1)^(n-j)*binomial(n,j)*sum(binomial(j,m-k)*binomial(m,j-m) for m in (0..j)) for j in (0..n))
    [[T(n, k) for k in (0..n)] for n in (0..10)] # G. C. Greubel, May 22 2019
    

Formula

Riordan array (1/sqrt(1-4*x^2), sqrt(1-4*x^2)*(1-sqrt(1-4*x^2))/(x-2*x^2 + x*sqrt(1-4*x^2))).
Number triangle T(n,k) = Sum{j=0..n} (-1)^(n-j)* C(n,j)*Sum_{i=0..j} C(j,i-k)*C(i,j-i).

A116387 Expansion of 1/(sqrt(1-2*x-3*x^2)*(2-M(x))), where M(x) is the g.f. of the Motzkin numbers A001006.

Original entry on oeis.org

1, 2, 7, 22, 72, 234, 763, 2486, 8099, 26372, 85833, 279226, 907946, 2951066, 9587981, 31140034, 101104048, 328162170, 1064856217, 3454513274, 11204337056, 36332719182, 117795920249, 381848062066, 1237615088203, 4010710218384
Offset: 0

Views

Author

Paul Barry, Feb 12 2006

Keywords

Comments

Binomial transform of A116383.
The substitution x-> x/(1+x+x^2) in the g.f. (this might be called an inverse Motzkin transform) yields the g.f. of A074331. - R. J. Mathar, Nov 10 2008

Programs

  • GAP
    List([0..30], n-> Sum([0..n], k-> Sum([0..n], j-> Binomial(n, j-k)*Binomial(j, n-j) ))); # G. C. Greubel, May 23 2019
  • Magma
    [(&+[ (&+[Binomial(n, j-k)*Binomial(j, n-j): j in [0..n]]) : k in [0..n]]): n in [0..30]]; // G. C. Greubel, May 23 2019
    
  • Mathematica
    Table[Sum[Binomial[n,j-k]Binomial[j,n-j],{k,0,n},{j,0,n}],{n,0,30}] (* Harvey P. Dale, Feb 08 2012 *)
  • PARI
    {a(n) = sum(k=0,n, sum(j=0,n, binomial(n, j-k)*binomial(j,n-j)))}; \\ G. C. Greubel, May 23 2019
    
  • Sage
    [sum( sum(binomial(n, j-k)*binomial(j,n-j) for j in (0..n)) for k in (0..n)) for n in (0..30)] # G. C. Greubel, May 23 2019
    

Formula

a(n) = Sum_{k=0..n} Sum_{j=0..n} C(n,j-k)*C(j,n-j).
Conjecture: n*(17*n-142)*a(n) + (17*n^2 + 95*n + 138)*a(n-1) + (-391*n^2 + 2488*n - 2908)*a(n-2) + (-17*n^2 - 603*n + 1892)*a(n-3) + 2*(697*n-2021)*(n-4)*a(n-4) + 60*(17*n-47)*(n-4)*a(n-5) = 0. - R. J. Mathar, Nov 15 2011
a(n) ~ (1+sqrt(5))^n * (5+sqrt(5)) / 10. - Vaclav Kotesovec, Feb 08 2014
Showing 1-2 of 2 results.