A104712 Pascal's triangle, with the first two columns removed.

1, 3, 1, 6, 4, 1, 10, 10, 5, 1, 15, 20, 15, 6, 1, 21, 35, 35, 21, 7, 1, 28, 56, 70, 56, 28, 8, 1, 36, 84, 126, 126, 84, 36, 9, 1, 45, 120, 210, 252, 210, 120, 45, 10, 1, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1, 78, 286, 715

Offset: 2

Author

Gary W. Adamson, Mar 19 2005

Keywords

• nonn
• tabl
• easy

Comments

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.

From Peter Luschny, Apr 10 2011: (Start)

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)

Diagonal sums are A001924(n+1). - Philippe Deléham, Jan 11 2014

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

Examples

			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
... reformatted. - _Wolfdieter Lang_, Mar 20 2015
		

Links

• G. C. Greubel, Rows n=2..100 of triangle, flattened
• D. Dugger, A Geometric Introduction to K-Theory
• Candice A. Marshall, Construction of Pseudo-Involutions in the Riordan Group, Dissertation, Morgan State University, 2017.
• T. Saito, The discriminant and the determinant of a hypersurface of even dimension (p. 4), arXiv:1110.1717 [math.AG], 2011-2012.

Crossrefs

Cf. A000295, A007318, A008292, A104713, A027641/A027642 (first Bernoulli numbers B-), A164555/A027642 (second Bernoulli numbers B+), A176327/A176289.

Programs

Magma

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

Mathematica

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

Formula

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

From Peter Bala, Jul 16 2013: (Start)

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

From Tom Copeland, Apr 11 2014: (Start)

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

Extensions

Edited and extended by David Wasserman, Jul 03 2007