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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.

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