A321331 - OEIS

%I #39 Nov 10 2023 09:36:05

%S 1,1,2,1,6,3,1,14,18,4,1,30,75,40,5,1,62,270,260,75,6,1,126,903,1400,

%T 700,126,7,1,254,2898,6804,5250,1596,196,8,1,510,9075,31080,34755,

%U 15876,3234,288,9,1,1022,27990,136420,212625,136962,41160,6000,405,10,1,2046,85503,583000,1233650,1076922,447909,95040,10395,550,11

%N Triangle read by rows: T(n, k) = (k+1)*S2(n+1, k+1), for n >= k >= 0, and S2 = A048993 (Stirling2).

%C This lower triangular matrix T is the inverse of the triangular matrix with elements Narumi[-1](n,m)/(m+1) = S1(n+1, m+1)/(n+1), with the Narumi triangle for parameter a = -1, and S1 = A048994 (Stirling1), i.e., Sum_{k=m..n} T(n, k) * S1(k+1, m+1)/(k+1) = delta_{n,m} (Kronecker symbol).

%C This triangle arises from the inverse of the rational Sheffer matrix Narumi[-1] = (log(1+x)/x, log(1+x) (such special Sheffer matrices (g(x), x*g(x)) define elements of the Narumi subgroup). The inverse matrix is (Narumi[-1])^(-1) = ((exp(x) - 1)/x, exp(x) - 1).

%C In order to have an integer matrix one takes T(n, k) := (n+1)*(Narumi[-1])^(-1)(n, k) = (k+1)*S2(n+1, k+1). The connection to S2 = A048993 results from the general relation between each Narumi-type matrix N = (g(x), x*g(x)) and its associated Sheffer matrix J = (1, x*g(x)) (this is of the Jabotinsky-type), i.e., N(n, m) = (m+1)*J(n+1, m+1)/(n+1), or with the row polynomials Npol(n, x) = (1/(n+1))*(d/dx)Jpol(n+1, x).

%C The signed triangle (-1)^(n-k)*A028421(n, k) (with upper diagonals filled with zeros) gives the integer matrix Nscaled with elements (n+1)*Narumi[-1](n,k). This inverse of Nscaled has the rational elements (Narumi[-1])^(-1)(k, m)/(m+1) = (1/(k+1))*S2(k+1, m+1).

%C The a- and z- sequence for the Sheffer matrix (Narumi[-1])^(-1) (see A006232 for a link on these sequences) have e.g.f.s Ea(x) = x/log(1 + x) and Ez(x) = 1/log(1 + x) - 1/x, hence a(n) = A006232(n)/A006233(n) and z(n) = A006232(n+1)/A075178(n), for n >= 0. This leads to the recurrence for T(n, k) given in the formula section.

%C The Boas-Buck-type column recurrence (see the link, also for references) uses the sequence with o.g.f. GBB(y) = exp(y)/(exp(y) - 1) - 1/y, with BB(n) = (-1)^(n+1)*A060054(n+1 ) / A227830(n+1), for n >= 1. For the recurrence see the formula section.

%C The Meixner-type identity (see the Meixner link) for the row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k, derived from the one for the Narumi[-1]^(-1) row polynomials is Sum_{k=1..n} (-1)^{k+1}*(1/k)*(d/dx)^k R(n, x)/(n+1) = R(n-1, x), for n >= 1, and R(0, x) = 1. Here d/dx is a differentiation operator.

%C The Roman-type recurrence for the row polynomials (see the reference, Corollary 3.7.2. p. 50) becomes, with the z-sequence from above: R(n, x) = ((n+1)/n)*{(x + 1/2)*1 + (x - z(1))*d/dx - Sum_{k=2..n-1} (1/k!)*z(k)*(d/dx)^k}*R(n-1, x), for n >= 1, and R(0, x) = 1.

%C The triangle is the exponential Riordan square (cf. A321620) of exp(x)-1 with an additional main diagonal of zeros. - _Peter Luschny_, Jan 03 2019

%D Steven Roman, The umbral calculus, Academic Press, 1984.

%H Wolfdieter Lang, <a href="/A321331/a321331_1.pdf">On the Generating Functions of the Boas-Buck Sequences for the Inverse of Riordan and Sheffer Matrices</a>

%H J. Meixner, <a href="https://doi.org/10.1112/jlms/s1-9.1.6">Orthogonale Polynomsysteme mit einer besonderen Gestalt der erzeugenden Funktion</a>, J. Lond. Math. Soc. 9 (1934), 6-13.

%F T(n, k) = (k+1)*A048993(n+1, k+1), with A048993 = Stirling2, for n >= k >= 0, and 0 otherwise.

%F T(n, k) = (n+1)*(Narumi[a=-1])^(-1)(n, k), with the Narumi[a=-1] matrix with entries (-1)^(n-k)*A028421(n, k)/(n+1).

%F E.g.f. for column k sequence: E(k, x) = (x*d/dx + 1)*EN(k, x), where EN(k, x) = exp(x) - 1)^(k+1)/(x*k!) is the e.g.f. for the (Narumi[a=-1])^(-1) columns. Hence E(k, x) = exp(x)*(exp(x) - 1)*(k+1)/k!, for k >= 0.

%F E.g.f. for (ordinary) row polynomials R(n, x): Epol(z, x) = exp(z)*exp(x*(exp(z) - 1))*(1 + x*(exp(z) - 1)).

%F Recurrence (from Stirling2): T(n, k) = 0 for n < k; T(n, 0) = (k + 1)*T(n-1, k), for n <= 1, T(0, 0) = 1; T(n, k) = (k+1)*(T(n-1, k) + T(n-1, k-1)/k), for n >= 1, k >= 1.

%F Recurrence (from a- and z-sequence, see above): a = {1, 1/2, -1/6, 1/4, -19/30, 9/4, ...}, z = {1/2, -1/12, 1/12, -19/120, 9/20, -863/504, ...}.

%F T(n, k) = 0, for n < k; T(n, 0) = (n+1)*Sum_{j=0..n-1} z(j)*T(n-1, j), for n >= 1, with T(0, 0) = 1; T(n, k) = ((n+1)/k)*Sum_{j=0..n-m} binomial(k-1+j, j)*a(j)*T(n-1, k-1+j).

%F Recurrence for column k, from the Boas-Buck-type sequence BB(n) = (-1)^(n+1)*A060054(n+1)/A227830(n+1), for n >= 0; BB = {1/2, 1/12, 0, -1/720, 0, 1/30240, 0, -1/1209600, ...}: T(n, k) = 0, for  n < k; T(n, n) = n+1, for n >= 0; T(n, k) = ((n+1)!*(k+1)/(n-k))*Sum_{j=k..n-1} (1/(j+1)!)*BB(n-(j+1))*T(j, k), for n >= 0 and k = 0, 1, ..., n-1.

%F T(n, k) = Stirling2(n+2, k+1) - Stirling2(n+1, k). - _Peter Luschny_, May 26 2020

%e The triangle T(n, k) begins:

%e n\k  0    1     2      3       4       5      6     7     8   9 10 ...

%e ----------------------------------------------------------------------

%e 0:   1

%e 1:   1    2

%e 2:   1    6     3

%e 3:   1   14    18      4

%e 4:   1   30    75     40       5

%e 5:   1   62   270    260      75       6

%e 6:   1  126   903   1400     700     126      7

%e 7:   1  254  2898   6804    5250    1596    196     8

%e 8:   1  510  9075  31080   34755   15876   3234   288     9

%e 9:   1 1022 27990 136420  212625  136962  41160  6000   405  10

%e 10:  1 2046 85503 583000 1233650 1076922 447909 95040 10395 550 11

%e ...

%e Recurrence (from Stirling2): T(4, 2) = 3*(T(3, 2) + T(3, 1)/2) = 3*(18 + 14/2) = 75.

%e Recurrence (from a- and z-sequence): T(4, 0) = 5*((1/2)*T(3, 0) - (1/12)*T(3, 1) + (1/12)*T(3, 2) - (19/120)*T(3, 3)) = 5*(1/2 - 14/12 + 18/12 - 4*19/120) = 1; T(4,2) = (5/2)*(1*1*T(3, 1) + 2*(1/2)*T(3, 2) + 3*(-1/6)* T(3, 3)) = (5/2)*(14 + 18 - 2) = 75.

%e Recurrence for column k=2 (Boas-Buck-type): T(4, 2) = (5!*3/2)*((1/3!)*(1/12)*T(2, 2) + (1/4!)*(1/2)*T(3, 2)) = (5!*3/2)*((1/72)*3 + (1/48)*18) = 75.

%e Meixner identity for the row polynomials, for n = 3: {d/dx  - (1/2)*(d/dx)^2 + (1/3)*(d/dx)^3)}*R(3, x)/4) = ((14 - 36/2 + 24/3) + (36 - 24/2)*x + 12*x^2)/4 = (1 + 6*x + 3*x^2) = R(2, x).

%e Roman type recurrence for row polynomials: R(n, 3) = (3/2)*{(x + 1/12)*(1 + 6*x + 3*x^2) + (x - (-1/2))*(6 + 6*x) - (1/2!)*(1/12)*6} = 1 + 14*x + 18*x^2 + 4*x^3.

%p T:=(n,k)->(k+1)*Stirling2(n+1,k+1): seq(seq(T(n,k),k=0..n),n=0..10); # _Muniru A Asiru_, Dec 03 2018

%t T[n_, k_] := (k+1) * StirlingS2[n+1, k+1];  Table[T[n, k], {n,0,10}, {k, 0, n}] //Flatten (* _Amiram Eldar_, Dec 03 2018 *)

%o (GAP) Flat(List([0..10],n->List([0..n],k->(k+1)*Stirling2(n+1,k+1)))); # _Muniru A Asiru_, Dec 03 2018

%o (Sage) # uses[riordan_square from A321620]

%o riordan_square(exp(x) - 1, 10, True) # _Peter Luschny_, Jan 03 2019

%o (PARI) T(n, k) = (k+1)*stirling(n+1, k+1, 2) \\ _Thomas Scheuerle_, Nov 10 2023

%Y Cf. A006232, A006233, A060054, A028421, A048993, A048994, A075178, A227830.

%K nonn,tabl,easy

%O 0,3

%A _Wolfdieter Lang_, Dec 03 2018