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%I #64 May 20 2023 07:48:38
%S 1,1,1,1,1,5,1,15,8,1,35,84,1,70,469,180,1,126,1869,3044,1,210,5985,
%T 26060,8064,1,330,16401,152900,193248,1,495,39963,696905,2286636,
%U 604800,1,715,88803,2641925,18128396,19056960,1,1001,183183,8691683,109425316,292271616,68428800,1,1365,355355,25537655,539651112,2961802480,2699672832
%N 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.
%C Row n contains floor(n/2) + 1 terms.
%H Gheorghe Coserea, <a href="/A185263/b185263.txt">Rows n = 0..202, flattened</a>
%H Nikita Alexeev and Peter Zograf, <a href="http://arxiv.org/abs/1111.3061">Hultman numbers, polygon gluings and matrix integrals</a>, arXiv preprint arXiv:1111.3061 [math.PR], 2011.
%F From _Gheorghe Coserea_, Jan 29 2018: (Start)
%F 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))
%F 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))
%F T(n,k) = -Stirling1(n+2, n+1-2*k)/binomial(n+2,2), where Stirling1(n,k) is defined by A048994.
%F A000142(n) = p(n)(1), A052572(n) = p(n)(2) for n > 0, A060593(n) = T(2*n, n) for n > 0. (End)
%F 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
%e Triangle begins:
%e n\k| 0 1 2 3 4 5 6
%e -----+---------------------------------------------------
%e 0 | 1
%e 1 | 1
%e 2 | 1 1
%e 3 | 1 5
%e 4 | 1 15 8
%e 5 | 1 35 84
%e 6 | 1 70 469 180
%e 7 | 1 126 1869 3044
%e 8 | 1 210 5985 26060 8064
%e 9 | 1 330 16401 152900 193248
%e 10 | 1 495 39963 696905 2286636 604800
%e 11 | 1 715 88803 2641925 18128396 19056960
%e 12 | 1 1001 183183 8691683 109425316 292271616 68428800
%e ...
%e Polynomials p_n(t):
%e p_0 = t;
%e p_1 = t^2;
%e p_2 = t^3 + t;
%e p_3 = t^4 + 5*t^2;
%e p_4 = t^5 + 15*t^3 + 8*t;
%e p_5 = t^6 + 35*t^4 + 84*t^2;
%e p_6 = t^7 + 70*t^5 + 469*t^3 + 180*t;
%e p_7 = t^8 + 126*t^6 + 1869*t^4 + 3044*t^2;
%e ...
%e A(x;t) = t + t^2*x/1! + (t^3 + t)*x^2/2! + (t^4 + 5*t^2)*x^3/3! + ...
%t T[n_, k_] := Abs[StirlingS1[n+2, n-2k+1]]/Binomial[n+2, 2];
%t Table[T[n, k], {n, 0, 13}, {k, 0, n/2}] // Flatten (* _Jean-François Alcover_, Aug 12 2018 *)
%o (PARI)
%o seq(N) = {
%o my(p=vector(N), t='t, v); p[1] = t^2; p[2] = t^3 + t;
%o for (n=3, N,
%o p[n] = ((2*n+1)*t*p[n-1] + (n-1)*(n^2-t^2)*p[n-2])/(n+2));
%o v = vector(#p, n, vector(1+n\2, k, polcoeff(p[n], n+1-2*(k-1))));
%o concat([[1]], v);
%o };
%o concat(seq(13))
%o (PARI)
%o N=14; x='x+O('x^(N+1));
%o concat(apply(p->select(a->a!=0, Vec(p)), Vec(serlaplace(((1-x)^(-t) - (1+x)^t)/x^2))))
%o (PARI)
%o T(n,k) = -stirling(n+2, n+1-2*k, 1)/binomial(n+2,2);
%o concat(1, concat(vector(13, n, vector(1+n\2, k, T(n, k-1)))))
%o \\ _Gheorghe Coserea_, Jan 29 2018
%Y Cf. A038720, A048994, A060593, A185259.
%Y For uncompressed form of polynomial coefficients, in order of increasing powers, see A164652.
%K nonn,tabf
%O 0,6
%A _N. J. A. Sloane_, Jan 21 2012
%E More terms from _Gheorghe Coserea_, Jan 29 2018
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