A110813 A triangle of pyramidal numbers.
1, 3, 1, 5, 4, 1, 7, 9, 5, 1, 9, 16, 14, 6, 1, 11, 25, 30, 20, 7, 1, 13, 36, 55, 50, 27, 8, 1, 15, 49, 91, 105, 77, 35, 9, 1, 17, 64, 140, 196, 182, 112, 44, 10, 1, 19, 81, 204, 336, 378, 294, 156, 54, 11, 1, 21, 100, 285, 540, 714, 672, 450, 210, 65, 12, 1, 23, 121, 385, 825
Offset: 0
Examples
The number triangle T(n, k) begins n\k 0 1 2 3 4 5 6 7 8 9 10 11 0: 1 1: 3 1 2: 5 4 1 3: 7 9 5 1 4: 9 16 14 6 1 5: 11 25 30 20 7 1 6: 13 36 55 50 27 8 1 7: 15 49 91 105 77 35 9 1 8: 17 64 140 196 182 112 44 10 1 9: 19 81 204 336 378 294 156 54 11 1 10: 21 100 285 540 714 672 450 210 65 12 1 11: 23 121 385 825 1254 1386 1122 660 275 77 13 1 ... reformatted by _Wolfdieter Lang_, Mar 23 2015 As a number square S(n, k) = T(n+k, k), rows begin 1, 1, 1, 1, 1, 1, ... 3, 4, 5, 6, 7, 8, ... 5, 9, 14, 20, 27, 35, ... 7, 16, 30, 50, 77, 112, ... 9, 25, 55, 105, 182, 294, ...
Links
- G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Crossrefs
Programs
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Mathematica
Table[2*Binomial[n + 1, k + 1] - Binomial[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, Oct 19 2017 *) -
PARI
for(n=0,10, for(k=0,n, print1(2*binomial(n+1, k+1) - binomial(n,k), ", "))) \\ G. C. Greubel, Oct 19 2017
Formula
Number triangle T(n, k) = C(n, k)*(2n-k+1)/(k+1) = 2*C(n+1, k+1) - C(n, k); Riordan array ((1+x)/(1-x)^2, x/(1-x)); As a number square read by antidiagonals, T(n, k)=C(n+k, k)(2n+k+1)/(k+1).
Equals A007318 * an infinite bidiagonal matrix with 1's in the main diagonal and 2's in the subdiagonal. - Gary W. Adamson, Dec 01 2007
Binomial transform of an infinite lower triangular matrix with all 1's in the main diagonal, all 2's in the subdiagonal and the rest zeros. - Gary W. Adamson, Dec 12 2007
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k) - T(n-2,k-1), T(0,0)=T(1,1)=1, T(1,0)=3, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Nov 30 2013
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(7 + 9*x + 5*x^2/2! + x^3/3!) = 7 + 16*x + 30*x^2/2! + 50*x^3/3! + 77*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 21 2014
T(n, k) = ps(1, 2; k, n-k) with ps(a, d; k, n) = sum(ps(a, d; k-1, j), j=0..n) and input ps(a, d; 0, j) = a + d*j. See the iterated partial sums comment from Mar 23 2015 above. - Wolfdieter Lang, Mar 23 2015
From Franck Maminirina Ramaharo, May 21 2018: (Start)
T(n,k) = coefficients in the expansion of ((x + 2)*(x + 1)^n - 2)/x.
T(n,k) = A097207(n,n-k).
G.f.: (y + 1)/((y - 1)*(x*y + y - 1)).
E.g.f.: ((x + 2)*exp(x*y + y) - 2*exp(y))/x.
(End)
Comments