|
| |
|
|
A321331
|
|
Triangle read by rows: T(n, k) = (k+1)*S2(n+1, k+1), for n >= k >= 0, and S2 = A048993 (Stirling2).
|
|
3
|
|
|
|
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, 700, 126, 7, 1, 254, 2898, 6804, 5250, 1596, 196, 8, 1, 510, 9075, 31080, 34755, 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
(list;
table;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,3
|
|
|
COMMENTS
|
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).
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).
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).
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).
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.
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.
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.
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.
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
|
|
|
REFERENCES
|
Steven Roman, The umbral calculus, Academic Press, 1984.
|
|
|
LINKS
|
|
|
|
FORMULA
|
T(n, k) = (k+1)*A048993(n+1, k+1), with A048993 = Stirling2, for n >= k >= 0, and 0 otherwise.
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).
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.
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)).
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.
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, ...}.
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).
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.
T(n, k) = Stirling2(n+2, k+1) - Stirling2(n+1, k). - Peter Luschny, May 26 2020
|
|
|
EXAMPLE
|
The triangle T(n, k) begins:
n\k 0 1 2 3 4 5 6 7 8 9 10 ...
----------------------------------------------------------------------
0: 1
1: 1 2
2: 1 6 3
3: 1 14 18 4
4: 1 30 75 40 5
5: 1 62 270 260 75 6
6: 1 126 903 1400 700 126 7
7: 1 254 2898 6804 5250 1596 196 8
8: 1 510 9075 31080 34755 15876 3234 288 9
9: 1 1022 27990 136420 212625 136962 41160 6000 405 10
10: 1 2046 85503 583000 1233650 1076922 447909 95040 10395 550 11
...
Recurrence (from Stirling2): T(4, 2) = 3*(T(3, 2) + T(3, 1)/2) = 3*(18 + 14/2) = 75.
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.
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.
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).
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.
|
|
|
MAPLE
|
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
|
|
|
MATHEMATICA
|
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 *)
|
|
|
PROG
|
(GAP) Flat(List([0..10], n->List([0..n], k->(k+1)*Stirling2(n+1, k+1)))); # Muniru A Asiru, Dec 03 2018
(Sage) # uses[riordan_square from A321620]
riordan_square(exp(x) - 1, 10, True) # Peter Luschny, Jan 03 2019
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
