A185263 Triangle T(n,k) read by rows: coefficients (in compressed forms) in order of decreasing exponents of polynomials p_n(t) related to Hultman numbers.

1, 1, 1, 1, 1, 5, 1, 15, 8, 1, 35, 84, 1, 70, 469, 180, 1, 126, 1869, 3044, 1, 210, 5985, 26060, 8064, 1, 330, 16401, 152900, 193248, 1, 495, 39963, 696905, 2286636, 604800, 1, 715, 88803, 2641925, 18128396, 19056960, 1, 1001, 183183, 8691683, 109425316, 292271616, 68428800, 1, 1365, 355355, 25537655, 539651112, 2961802480, 2699672832

Offset: 0

Author

N. J. A. Sloane, Jan 21 2012

Keywords

• nonn
• tabf

Comments

Row n contains floor(n/2) + 1 terms.

Examples

			Triangle begins:
  n\k| 0    1      2       3         4         5        6
-----+---------------------------------------------------
   0 | 1
   1 | 1
   2 | 1    1
   3 | 1    5
   4 | 1   15      8
   5 | 1   35     84
   6 | 1   70    469     180
   7 | 1  126   1869    3044
   8 | 1  210   5985   26060      8064
   9 | 1  330  16401  152900    193248
  10 | 1  495  39963  696905   2286636    604800
  11 | 1  715  88803 2641925  18128396  19056960
  12 | 1 1001 183183 8691683 109425316 292271616 68428800
  ...
Polynomials p_n(t):
  p_0 = t;
  p_1 = t^2;
  p_2 = t^3 +     t;
  p_3 = t^4 +   5*t^2;
  p_4 = t^5 +  15*t^3 +    8*t;
  p_5 = t^6 +  35*t^4 +   84*t^2;
  p_6 = t^7 +  70*t^5 +  469*t^3 +  180*t;
  p_7 = t^8 + 126*t^6 + 1869*t^4 + 3044*t^2;
  ...
A(x;t) = t + t^2*x/1! + (t^3 + t)*x^2/2! + (t^4 + 5*t^2)*x^3/3! + ...
		

Links

• Gheorghe Coserea, Rows n = 0..202, flattened
• Nikita Alexeev and Peter Zograf, Hultman numbers, polygon gluings and matrix integrals, arXiv preprint arXiv:1111.3061 [math.PR], 2011.

Crossrefs

Cf. A038720, A048994, A060593, A185259.

For uncompressed form of polynomial coefficients, in order of increasing powers, see A164652.

Programs

Mathematica

T[n_, k_] := Abs[StirlingS1[n+2, n-2k+1]]/Binomial[n+2, 2];
Table[T[n, k], {n, 0, 13}, {k, 0, n/2}] // Flatten (* Jean-François Alcover, Aug 12 2018 *)

PARI

seq(N) = {
  my(p=vector(N), t='t, v); p[1] = t^2; p[2] = t^3 + t;
  for (n=3, N,
    p[n] = ((2*n+1)*t*p[n-1] + (n-1)*(n^2-t^2)*p[n-2])/(n+2));
  v = vector(#p, n, vector(1+n\2, k, polcoeff(p[n], n+1-2*(k-1))));
  concat([[1]], v);
};
concat(seq(13))

PARI

N=14; x='x+O('x^(N+1));
concat(apply(p->select(a->a!=0, Vec(p)), Vec(serlaplace(((1-x)^(-t) - (1+x)^t)/x^2))))

PARI

T(n,k) = -stirling(n+2, n+1-2*k, 1)/binomial(n+2,2);
concat(1, concat(vector(13, n, vector(1+n\2, k, T(n, k-1)))))
\\ Gheorghe Coserea, Jan 29 2018

Formula

From Gheorghe Coserea, Jan 29 2018: (Start)

p(n) = Sum_{k=0..floor(n/2)} T(n,k)*t^(n+1-2*k) satisfies (n+2)*p(n) = (2*n+1)*t*p(n-1) + (n-1)*(n^2-t^2)*p(n-2), n >= 2. (th. 3, (iii))

E.g.f. A(x;t) = Sum_{n>=0} p(n)*x^n/n! = ((1-x)^(-t) - (1+x)^t)/x^2. (th. 3, (i))

T(n,k) = -Stirling1(n+2, n+1-2*k)/binomial(n+2,2), where Stirling1(n,k) is defined by A048994.

A000142(n) = p(n)(1), A052572(n) = p(n)(2) for n > 0, A060593(n) = T(2*n, n) for n > 0. (End)

n-th row polynomial R(n,x) satisfies x*R(n,x^2) = (1/2)*( P(n+1,x) - P(n+1,-x) )/ binomial(n+2,2), where P(k,x) = (1 + x)*(1 + 2*x) * ... *(1 + k*x). - Peter Bala, May 14 2023

Extensions

More terms from Gheorghe Coserea, Jan 29 2018