A104712 - OEIS

A000295 (Eulerian numbers) gives the row sums.

Write A004736 and Pascal's triangle as infinite lower triangular matrices A and B; then A*B is this triangle.

A slight variation has a combinatorial interpretation: remove the last column and the second one from Pascal's triangle. Let P(m, k) denote the set partitions of {1,2,..,n} with the following properties:

(a) Each partition has at least one singleton block;

(c) k is the size of the largest block of the partition;

(b) m = n - k + 1 is the number of parts of the partition.

Then A000295(n) = Sum_{k=1..n} card(P(n-k+1,k)).

For instance, A000295(4) = P(4,1) + P(3,2) + P(2,3) + P(1,4) = card({1|2|3|4}) + card({1|2|34, 1|3|24,1|4|23, 2|3|14, 2|4|13, 3|4|12}) + card({1|234, 2|134, 3|124, 4|123}) = 1 + 6 + 4 = 11.

This interpretation can be superimposed on the sequence by changing the offset to 1 and adding the value 1 in front. The triangle then starts

1;

1, 3;

1, 6, 4;

1, 10, 10, 5;

1, 15, 20, 15, 6;

...

(End)

Relation to K-theory: T acting on the column vector (d,-d^2,d^3,...) generates the Euler classes for a hypersurface of degree d in CP^n. Cf. Dugger p. 168, A111492, A238363, and A135278. - Tom Copeland, Apr 11 2014

a(n,k) = binomial(n,k), for 2 <= k <= n.

The following remarks assume an offset of 0.

Riordan array (1/(1 - x)^3, x/(1 - x)).

O.g.f.: 1/(1 - t)^2*1/(1 - (1 + x)*t) = 1 + (3 + x)*t + (6 + 4*x + x^2)*t^2 + ....

E.g.f.: (1/x*d/dt)^2 (exp(t)*(exp(x*t) - 1 - x*t) = 1 + (3 + x)*t + (6 + 4*x + x^2)*t^2/2! + ....

The infinitesimal generator for this triangle has the sequence [3,4,5,...] on the main subdiagonal and 0's elsewhere. (End)

As triangle T(n,k), 0<=k<=n: T(n,k) = 3*T(n-1,k) + T(n-1,k-1) - 3*T(n-2,k) - 2*T(n-2,k-1) + T(n-3,k) + T(n-3,k-1), T(0,0)=1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Jan 11 2014

A) The infinitesimal generator for this matrix is given in A132681 with m=2. See that entry for numerous relations to differential operators and the Laguerre polynomials of order m=2, i.e., Lag(n,t,2) = Sum_{j=0..n} binomial(n+2,n-j)*(-t)^j/j!.

B) O.g.f.: 1 / { [ 1 - t * x/(1-x) ] * (1-x)^3 }

C) O.g.f. of row e.g.f.s: exp[t*x/(1-x)]/(1-x)^3 = [Sum_{n>=0} x^n * Lag(n,-t,2)] = 1 + (3 + t)*x + (6 + 4t + t^2/2!)*x^2 + (10 + 10t + 5t^2/2! + t^3/3!)*x^3 + ....

D) E.g.f. of row o.g.f.s: [(1+t)*exp((1+t)*x) - (1+t+t*x)exp(x)]/t^2. (End)

O.g.f. for m-th row (m=n-2): [(1+x)^(m+2)-(1+(m+2)*x)]/x^2. - Tom Copeland, Apr 16 2014

Reverse T = [St2]*dP*[St1]- dP = [St2]*(exp(x*M)-I)*[St1]-(exp(x*M)-I) with top two rows of zeros removed, [St1]=padded A008275 just as [St2]=A048993=padded A008277, dP= A132440, M=A238385-I, and I=identity matrix. Cf. A238363. - Tom Copeland, Apr 26 2014

O.g.f. of column k (with k leading zeros): (x^k)/(1-x)^(k+1), k >= 2. - Wolfdieter Lang, Mar 20 2015

The triangle a(n, k) begins:

n\k 2 3 4 5 6 7 8 9 10 11 12 13

2: 1

3: 3 1

4: 6 4 1

5: 10 10 5 1

6: 15 20 15 6 1

7: 21 35 35 21 7 1

8: 28 56 70 56 28 8 1

9: 36 84 126 126 84 36 9 1

10: 45 120 210 252 210 120 45 10 1

11: 55 165 330 462 462 330 165 55 11 1

12: 66 220 495 792 924 792 495 220 66 12 1

13: 78 286 715 1287 1716 1716 1287 715 286 78 13 1

t[n_, k_] := Binomial[n, k]; Table[ t[n, k], {n, 2, 13}, {k, 2, n}] // Flatten (* Robert G. Wilson v, Apr 16 2011 *)

(PARI) for(n=2, 10, for(k=2, n, print1(binomial(n, k), ", "))) \\ G. C. Greubel, May 15 2018

(Magma) /* As triangle */ [[Binomial(n, k): k in [2..n]]: n in [2..10]]; // G. C. Greubel, May 15 2018