A135225 Pascal's triangle A007318 augmented with a leftmost border column of 1's.

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 3, 3, 1, 1, 1, 4, 6, 4, 1, 1, 1, 5, 10, 10, 5, 1, 1, 1, 6, 15, 20, 15, 6, 1, 1, 1, 7, 21, 35, 35, 21, 7, 1, 1, 1, 8, 28, 56, 70, 56, 28, 8, 1

Offset: 0

Author

Gary W. Adamson, Nov 23 2007

Keywords

• nonn
• easy
• tabl

Comments

Row sums give A094373.

From Peter Bala, Sep 08 2011: (Start)

This augmented Pascal array, call it P, has interesting connections with the Bernoulli polynomials B(n,x). The infinitesimal generator S of P is the array such that exp(S) = P. The array S is obtained by augmenting the infinitesimal generator A132440 of the Pascal triangle with an initial column [0, 0, 1/2, 1/6, 0, -1/30, ...] on the left. The entries in this column, after the first two zeros, are the Bernoulli values B(n,1), n>=1.

The array P is also connected with the problem of summing powers of consecutive integers. In the array P^n, the entry in position p+1 of the first column is equal to sum {k = 1..n} k^p - see the Example section below.

For similar results for the square of Pascal's triangle see A062715.

Note: If we augment Pascal's triangle with the column [1, 1, x, x^2, x^3, ...] on the left, the resulting lower unit triangular array has the Bernoulli polynomials B(n,x) in the first column of its infinitesimal generator. The present case is when x = 1.

(End)

Examples

			First few rows of the triangle:
  1;
  1, 1;
  1, 1, 1;
  1, 1, 2, 1;
  1, 1, 3, 3, 1;
  1, 1, 4, 6, 4, 1;
...
The infinitesimal generator for P begins:
  /0
  |0.......0
  |1/2.....1...0
  |1/6.....0...2....0
  |0.......0...0....3....0
  |-1/30...0...0....0....4....0
  |0.......0...0....0....0....5....0
  |1/42....0...0....0....0....0....6....0
  |...
  \
The array P^n begins:
  /1
  |1+1+...+1........1
  |1+2+...+n........n.........1
  |1+2^2+...+n^2....n^2.....2*n........1
  |1+2^3+...+n^3....n^3.....3*n^2....3*n.......1
  |...
  \
More generally, the array P^t, defined as exp(t*S) for complex t, begins:
  /1
  |B(1,1+t)-B(1,1)..........1
  |1/2*(B(2,1+t)-B(2,1))....t.........1
  |1/3*(B(3,1+t)-B(3,1))....t^2.....2*t........1
  |1/4*(B(4,1+t)-B(4,1))....t^3.....3*t^2....3*t.......1
  |...
  \
		

Links

• G. C. Greubel, Rows n = 0..100 of triangle, flattened

Crossrefs

Cf. A007318, A027641, A027642, A062715, A094373, A103438, A132440.

Programs

GAP

T:= function(n,k)
    if k=0 then return 1;
    else return Binomial(n-1,k-1);
    fi; end;
Flat(List([0..10], n-> List([0..n], k-> T(n,k) ))); # G. C. Greubel, Nov 19 2019

Magma

T:= func< n, k | k eq 0 select 1 else Binomial(n-1, k-1) >;
[T(n,k): k in [0..n], n in [0..10]]; // G. C. Greubel, Nov 19 2019

Maple

T:= proc(n, k) option remember;
      if k=0 then 1
    else binomial(n-1, k-1)
      fi; end:
seq(seq(T(n, k), k=0..n), n=0..10); # G. C. Greubel, Nov 19 2019

Mathematica

T[n_, k_]:= T[n, k]= If[k==0, 1, Binomial[n-1, k-1]]; Table[T[n, k], {n, 0, 10}, {k, 0, n}]//Flatten (* G. C. Greubel, Nov 19 2019 *)

PARI

T(n,k) = if(k==0, 1, binomial(n-1, k-1)); \\ G. C. Greubel, Nov 19 2019

Sage

@CachedFunction
def T(n,k):
    if (k==0): return 1
    else: return binomial(n-1, k-1)
[[T(n,k) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Nov 19 2019

Formula

A103451 * A007318 * A000012(signed), where A000012(signed) = (1; -1,1; 1,-1,1; ...); as infinite lower triangular matrices.

Given A007318, binomial(n,k) is shifted to T(n+1,k+1) and a leftmost border of 1's is added.

Extensions

Corrected by R. J. Mathar, Apr 16 2013