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%I #27 Sep 08 2022 08:46:16
%S 1,1,1,1,4,1,1,15,9,1,1,64,66,16,1,1,325,490,190,25,1,1,1956,3915,
%T 2120,435,36,1,1,13699,34251,23975,6755,861,49,1,1,109600,328804,
%U 283136,101990,17696,1540,64,1,1,986409,3452436,3534636,1554966,342846,40404,2556,81,1
%N Triangle read by rows, T(n,k) = Sum_{j=0..n} (-1)^(n-j)*C(-j-1,-n-1)*L(j,k), L the unsigned Lah numbers A271703, for n>=0 and 0<=k<=n.
%C This is the Sheffer (aka exponential Riordan) matrix T = P*L = A007318*A271703 = (exp(x), x/(1-x)). Note that P = A007318 is Sheffer (exp(t), t) (of the Appell type). The Sheffer a-sequence is [1,1,repeat(0)] and the z-sequence has e.g.f. (x/(1+x))*(1 - exp(-x/(1+x)) given in A288869 / A000027. Because the column k=0 has only entries 1, the z-sequence gives fractional representations of 1. See A288869. - _Wolfdieter Lang_, Jun 20 2017
%H G. C. Greubel, <a href="/A271705/b271705.txt">Rows n = 0..50 of the triangle, flattened</a>
%H Marin Knežević, Vedran Krčadinac, and Lucija Relić, <a href="https://arxiv.org/abs/2012.15307">Matrix products of binomial coefficients and unsigned Stirling numbers</a>, arXiv:2012.15307 [math.CO], 2020.
%F From _Wolfdieter Lang_, Jun 20 2017: (Start)
%F T(n, k) = Sum_{m=k..n} A007318(n, m)*A271703(m, k), n >= k >= 0, and 0 for k < m. See also the name.
%F E.g.f. of column k: exp(x)*(x/(1-x))^k/k! (Sheffer property), k >= 0.
%F E.g.f. of triangle (or row polynomials in x): exp(z)*exp((x*z/(1-z)).
%F Recurrence for T(n, k), k >= 1, with T(n, 0) = 1, T(n, k) = 0 if n < k: T(n, k) = (n/k)*T(n-1, k-1) + n*T(n-1, k), n >= 1, k = 1..n. (From the a-sequence with column k=0 as input.) (End)
%F T(n, k) = Sum_{j=0..n-k} j!*binomial(n, j+k)*binomial(j+k, k)*binomial(j+k-1, k-1) with T(n, 0) = 1. - _G. C. Greubel_, Jan 09 2022
%e Triangle starts:
%e 1;
%e 1, 1;
%e 1, 4, 1;
%e 1, 15, 9, 1;
%e 1, 64, 66, 16, 1;
%e 1, 325, 490, 190, 25, 1;
%e 1, 1956, 3915, 2120, 435, 36, 1;
%e ...
%e Recurrence: T(3, 2) = (3/2)*4 + 3*1 = 9. - _Wolfdieter Lang_, Jun 20 2017
%p L := (n,k) -> `if`(k<0 or k>n,0,(n-k)!*binomial(n,n-k)*binomial(n-1,n-k)):
%p T := (n,k) -> add(L(j,k)*binomial(-j-1,-n-1)*(-1)^(n-j), j=0..n):
%p seq(seq(T(n,k), k=0..n), n=0..9);
%t T[n_, k_]:= If[k==0, 1, Sum[((k*j!)/(j+k))*Binomial[n, j+k]*Binomial[j+k, k]^2, {j,0,n-k}]];
%t Table[T[n, k], {n,0,12}, {k,0,n}]//Flatten (* _G. C. Greubel_, Jan 09 2022 *)
%o (Magma)
%o B:=Binomial;
%o A271705:= func< n,k | k eq 0 select 1 else (&+[B(n, j+k)*B(j+k, k)*B(j+k-1, k-1)*Factorial(j): j in [0..n-k]]) >;
%o [A271705(n,k): k in [0..n], n in [0..12]]; // _G. C. Greubel_, Jan 09 2022
%o (Sage)
%o b=binomial
%o def A271705(n,k): return 1 if (k==0) else sum(factorial(j-k)*b(n, j)*b(j, k)*b(j-1, k-1) for j in (k..n))
%o flatten([[A271705(n,k) for k in (0..n)] for n in (0..12)]) # _G. C. Greubel_, Jan 09 2022
%Y Cf. A000290 (diag n, n-1), A062392 (diag n, n-2).
%Y Cf. A007526 (col. 1), A134432 (col. 2).
%Y Cf. A052844 (row sums), A059110 (matrix inverse).
%Y Cf. A007318, A271703, A288869.
%K nonn,easy,tabl
%O 0,5
%A _Peter Luschny_, Apr 14 2016
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