%I #57 Mar 03 2023 16:51:48
%S 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,
%T 8,35,83,125,125,83,35,8,9,44,119,209,251,209,119,44,9,10,54,164,329,
%U 461,461,329,164,54,10,11,65,219,494,791,923,791,494,219,65,11
%N Subtract 1 from Pascal's triangle, read by rows.
%C 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
%C 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
%C From _R. J. Mathar_, Apr 25 2016: (Start)
%C If regarded as a symmetric array of the form
%C 1 2 3 4 5 ...
%C 2 5 9 14 20 ...
%C 3 9 19 34 55 ...
%C 4 14 34 69 125 ...
%C 5 20 55 125 251 ...
%C 6 27 83 209 461 ...
%C 7 35 119 329 791 ...
%C 8 44 164 494 1286 ...
%C 9 54 219 714 2001 ...
%C 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)
%C 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
%H Reinhard Zumkeller, <a href="/A014430/b014430.txt">Rows n=0..100 of triangle, flattened</a>
%F T(n, k) = T(n-1, k) + T(n-1, k-1) + 1, T(0, 0)=1. - _Ralf Stephan_, Jan 23 2005
%F G.f.: 1 / ((1-x)*(1-x*y)*(1-x*(1+y))). - _Ralf Stephan_, Jan 24 2005
%F 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
%F T(n, k) = binomial(n+2, k+1) - 1. - _G. C. Greubel_, Feb 25 2023
%e Triangle begins:
%e 1;
%e 2, 2;
%e 3, 5, 3;
%e 4, 9, 9, 4;
%e 5, 14, 19, 14, 5;
%e 6, 20, 34, 34, 20, 6;
%e 7, 27, 55, 69, 55, 27, 7;
%e 8, 35, 83, 125, 125, 83, 35, 8;
%t 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 *)
%t Table[Binomial[n+2,k+1] -1, {n,0,13}, {k,0,n}]//Flatten (* _G. C. Greubel_, Feb 25 2023 *)
%o (Haskell)
%o a014430 n k = a014430_tabl !! n !! k
%o a014430_row n = a014430_tabl !! n
%o a014430_tabl = map (init . tail) $ drop 2 a014473_tabl
%o -- _Reinhard Zumkeller_, Apr 10 2012
%o (Magma) [Binomial(n+2,k+1)-1: k in [0..n], n in [0..13]]; // _G. C. Greubel_, Feb 25 2023
%o (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
%Y Cf. A000096, A007318, A014430, A026642, A027641, A027642, A035927.
%Y Cf. A062748, A063258, A062988, A074909, A124089, A124326, A129696.
%Y Triangle with zeros: A014473.
%Y Cf. A000295 (row sums).
%K nonn,tabl,easy
%O 0,2
%A _N. J. A. Sloane_
%E More terms from _Erich Friedman_
%E Offset fixed by _Reinhard Zumkeller_, Apr 10 2012