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A014430 Subtract 1 from Pascal's triangle, read by rows. 9
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 (list; table; graph; refs; listen; history; text; internal format)
OFFSET
0,2
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
LINKS
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
EXAMPLE
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;
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 *)
PROG
(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
(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
CROSSREFS
Triangle with zeros: A014473.
Cf. A000295 (row sums).
Sequence in context: A115262 A128141 A252829 * A360196 A225622 A196436
KEYWORD
nonn,tabl,easy
AUTHOR
EXTENSIONS
More terms from Erich Friedman
Offset fixed by Reinhard Zumkeller, Apr 10 2012
STATUS
approved

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Last modified June 22 15:06 EDT 2024. Contains 373587 sequences. (Running on oeis4.)