A014430 Subtract 1 from Pascal's triangle, read by rows.

1, 2, 2, 3, 5, 3, 4, 9, 9, 4, 5, 14, 19, 14, 5, 6, 20, 34, 34, 20, 6, 7, 27, 55, 69, 55, 27, 7, 8, 35, 83, 125, 125, 83, 35, 8, 9, 44, 119, 209, 251, 209, 119, 44, 9, 10, 54, 164, 329, 461, 461, 329, 164, 54, 10, 11, 65, 219, 494, 791, 923, 791, 494, 219, 65, 11

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• tabl
• easy

Comments

Each value of the sequence (T(x,y)) is equal to the sum of all values in Pascal's Triangle that are in the rectangle defined by the tip (0,0) and the position (x,y). - Florian Kleedorfer (florian.kleedorfer(AT)austria.fm), May 23 2005

To clarify T(n,k) and A129696: We subtract I = Identity matrix from Pascal's triangle to obtain the beheaded variant, A074909. Then take column sums starting from the top of A074909 to get triangle A014430. Row sums of the inverse of triangle T(n,k) gives the Bernoulli numbers, A027641/A026642. Alternatively, triangle T(n,k) as an infinite lower triangular matrix * [the Bernoulli numbers as a vector] = [1, 1, 1, ...]. Given the B_n version starting (1, 1/2, 1/6, ...) triangle T(n,k) * the B_n vector [1, 1/2, 1/6, 0, -1/30, ...] = the triangular numbers. - Gary W. Adamson, Mar 13 2012

From R. J. Mathar, Apr 25 2016: (Start)

If regarded as a symmetric array of the form

1 2 3 4 5 ...

2 5 9 14 20 ...

3 9 19 34 55 ...

4 14 34 69 125 ...

5 20 55 125 251 ...

6 27 83 209 461 ...

7 35 119 329 791 ...

8 44 164 494 1286 ...

9 54 219 714 2001 ...

it contains the rows (and columns) A000096, A062748, A063258, A062988, A124089, ..., A035927 and so on and counts the multisets of digits of numbers in base b>=2 with d>=1 digits (equivalent to the comment in A035927). (End)

Proof of Florian Kleedorfer's formula: Take sums of the columns of the rectangle - these are all binomial coefficients by the Hockey Stick Identity. Note the locations of these coefficients: They form a row going almost all the way to the edge, only missing the 1 - apply the Hockey Stick Identity again. - James East, Jul 03 2020

Examples

			Triangle begins:
  1;
  2,  2;
  3,  5,  3;
  4,  9,  9,   4;
  5, 14, 19,  14,   5;
  6, 20, 34,  34,  20,  6;
  7, 27, 55,  69,  55, 27,  7;
  8, 35, 83, 125, 125, 83, 35, 8;
		

Links

• Reinhard Zumkeller, Rows n=0..100 of triangle, flattened

Crossrefs

Cf. A000096, A007318, A014430, A026642, A027641, A027642, A035927.

Cf. A062748, A063258, A062988, A074909, A124089, A124326, A129696.

Triangle with zeros: A014473.

Cf. A000295 (row sums).

Programs

Haskell

a014430 n k = a014430_tabl !! n !! k
a014430_row n = a014430_tabl !! n
a014430_tabl = map (init . tail) $ drop 2 a014473_tabl
-- Reinhard Zumkeller, Apr 10 2012

Magma

[Binomial(n+2,k+1)-1: k in [0..n], n in [0..13]]; // G. C. Greubel, Feb 25 2023

Mathematica

Table[Sum[Sum[Binomial[m, j], {m, j, j+(n-k)}], {j,0,k}], {n,0,10}, {k, 0,n}]//Flatten (* Michael De Vlieger, Sep 01 2020 *)
Table[Binomial[n+2,k+1] -1, {n,0,13}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 25 2023 *)

SageMath

flatten([[binomial(n+2,k+1)-1 for k in range(n+1)] for n in range(14)]) # G. C. Greubel, Feb 25 2023

Formula

T(n, k) = T(n-1, k) + T(n-1, k-1) + 1, T(0, 0)=1. - Ralf Stephan, Jan 23 2005

G.f.: 1 / ((1-x)*(1-x*y)*(1-x*(1+y))). - Ralf Stephan, Jan 24 2005

T(n, k) = Sum_{j=0..k} Sum_{m=j..j+(n-k)} binomial(m, j). - Florian Kleedorfer (florian.kleedorfer(AT)austria.fm), May 23 2005

T(n, k) = binomial(n+2, k+1) - 1. - G. C. Greubel, Feb 25 2023

Extensions

More terms from Erich Friedman

Offset fixed by Reinhard Zumkeller, Apr 10 2012