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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.

A318107 Triangle read by rows: T(n,k) = (3*n - 2*k)!/((n-k)!^3*k!).

Original entry on oeis.org

1, 6, 1, 90, 24, 1, 1680, 630, 60, 1, 34650, 16800, 2520, 120, 1, 756756, 450450, 92400, 7560, 210, 1, 17153136, 12108096, 3153150, 369600, 18900, 336, 1, 399072960, 325909584, 102918816, 15765750, 1201200, 41580, 504, 1, 9465511770, 8779605120, 3259095840, 617512896, 63063000, 3363360, 83160, 720, 1
Offset: 0

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Author

Gheorghe Coserea, Sep 18 2018

Keywords

Comments

Diagonal of rational function R(x,y,z,t) = 1/(1 - (x + y + z + t*x*y*z)) with respect to x,y,z, i.e., T(n,k) = [(xyz)^n*t^k] R(x,y,z,t).
Annihilating differential operator: x*(2*t*x + 1)*((t*x - 1)^3 + 27*x)*Dx^2 + (6*t^4*x^4 - 8*t^3*x^3 - 3*t*(t - 18)*x^2 + 6*(t + 9)*x - 1)*Dx + (t*x - 1)*(t*(2*t^2*x^2 + 2*t*x - 1) - 6).

Examples

			A(x;t) = 1 + (6 + t)*x + (90 + 24*t + t^2)*x^2 + (1680 + 630*t + 60*t^2 + t^3)*x^3 + ...
Triangle starts:
n\k [0]        [1]        [2]        [3]       [4]      [5]    [6]  [7]
[0] 1;
[1] 6,         1;
[2] 90,        24,        1;
[3] 1680,      630,       60,        1;
[4] 34650,     16800,     2520,      120,      1;
[5] 756756,    450450,    92400,     7560,     210,     1;
[6] 17153136,  12108096,  3153150,   369600,   18900,   336,   1;
[7] 399072960, 325909584, 102918816, 15765750, 1201200, 41580, 504, 1;
[8] ...
		

Crossrefs

Programs

  • PARI
    T(n,k) = (3*n - 2*k)!/((n-k)!^3*k!);
    concat(vector(10, n, vector(n, k, T(n-1, k-1))))
    /* test:
    P(n, v='t) = subst(Polrev(vector(n+1, k, T(n, k-1)), 't), 't, v);
    diag(expr, N=22, var=variables(expr)) = {
      my(a = vector(N));
      for (k = 1, #var, expr = taylor(expr, var[#var - k + 1], N));
      for (n = 1, N, a[n] = expr;
        for (k = 1, #var, a[n] = polcoef(a[n], n-1)));
      return(a);
    };
    apply_diffop(p, s) = { \\ apply diffop p (encoded as Pol in Dx) to Ser s
      s=intformal(s);
      sum(n=0, poldegree(p, 'Dx), s=s'; polcoef(p, n, 'Dx) * s);
    };
    \\ diagonal property:
    x='x; y='y; z='z; t='t;
    diag(1/(1 - (x+y+z + t*x*y*z)), 11, [x,y,z]) == vector(11, n, P(n-1))
    \\ annihilating diffop:
    y = Ser(vector(101, n, P(n-1)), 'x);
    p=x*(2*t*x + 1)*((t*x - 1)^3 + 27*x)*Dx^2 + (6*t^4*x^4 - 8*t^3*x^3 - 3*t*(t - 18)*x^2 + 6*(t + 9)*x  - 1)*Dx + (t*x - 1)*(t*(2*t^2*x^2 + 2*t*x - 1) - 6);
    0 == apply_diffop(p, y)
    */

Formula

Let P_n(t) = Sum_{k=0..n} T(n,k)*t^k. Then A000172(n) = P_n(-4), A318108(n) = P_n(-3), A318109(n) = P_n(-2), A124435(n) = P_n(-1), A006480(n) = P_n(0), A081798(n) = P_n(1).
G.f. y = Sum_{n>=0} P_n(t)*x^n satisfies:
0 = x*(2*t*x + 1)*((t*x - 1)^3 + 27*x)*y'' + (6*t^4*x^4 - 8*t^3*x^3 - 3*t*(t - 18)*x^2 + 6*(t + 9)*x - 1)*y' + (t*x - 1)*(t*(2*t^2*x^2 + 2*t*x - 1) - 6)*y.