A143785 Antidiagonal sums of the triangle A120070.

3, 8, 20, 36, 63, 96, 144, 200, 275, 360, 468, 588, 735, 896, 1088, 1296, 1539, 1800, 2100, 2420, 2783, 3168, 3600, 4056, 4563, 5096, 5684, 6300, 6975, 7680, 8448, 9248, 10115, 11016, 11988, 12996, 14079, 15200, 16400, 17640, 18963, 20328, 21780, 23276

Offset: 1

Author

Paul Curtz, Sep 01 2008

Keywords

• nonn
• easy

Comments

Let b(n) be the sequence (0,0,0,3,8,20,36,...), with offset 0. Then b(n) is the number of triples (w,x,y) having all terms in {0,...,n} and w < range{w,x,y}. - Clark Kimberling, Jun 11 2012

Consider a(n) with two 0's prepended and offset 1. Call the new sequence b(n) and consider the partitions of n into two parts (p,q). Then b(n) represents the sum of all the products (p + q) * (q - p) where p <= q. - Wesley Ivan Hurt, Apr 12 2018

Examples

			First diagonal 3 = 3.
Second diagonal 8 = 8.
Third diagonal 5+15 = 20.
Fourth diagonal 24+12 = 36.
		

Links

• Iain Fox, Table of n, a(n) for n = 1..10000 (first 1000 terms from Colin Barker)
• Index entries for linear recurrences with constant coefficients, signature (2,1,-4,1,2,-1).

Crossrefs

Cf. A035006, A099721 (bisections).

Cf. A120070, A002620.

Programs

Magma

[(n+2)*(2*n^2+4*n-(-1)^n+1)/8: n in [1..50]]; // Vincenzo Librandi, Jan 22 2018

Mathematica

Rest@ CoefficientList[Series[x (3 + 2 x + x^2)/((1 + x)^2*(x - 1)^4), {x, 0, 44}], x] (* Michael De Vlieger, Dec 22 2017 *)
LinearRecurrence[{2, 1, -4, 1, 2, -1}, {3, 8, 20, 36, 63, 96}, 60] (* Vincenzo Librandi, Jan 22 2018 *)

PARI

Vec(x*(3+2*x+x^2)/((1+x)^2*(x-1)^4) + O(x^50)) \\ Colin Barker, May 07 2016

Formula

a(n+1) - a(n) = A032438(n+2).

a(n) = A006918(n-2) + 2*A006918(n-1) + 3*A006918(n). - R. J. Mathar, Jul 01 2011

G.f.: x*(3+2*x+x^2) / ( (1+x)^2*(x-1)^4 ). - R. J. Mathar, Jul 01 2011

a(n) = (n+2)*(2*n^2 + 4*n - (-1)^n + 1)/8. - Ilya Gutkovskiy, May 07 2016

From Colin Barker, May 07 2016: (Start)

a(n) = (n^3 + 4*n^2 + 4*n)/4 for n even.

a(n) = (n^3 + 4*n^2 + 5*n + 2)/4 for n odd.

a(n) = 2*a(n-1) + a(n-2) - 4*a(n-3) + a(n-4) + 2*a(n-5) - a(n-6) for n > 6. (End)

a(n) = Sum_{k=1..n+1} floor((n+1)*k/2). - Wesley Ivan Hurt, Apr 01 2017

a(n) = (n+2)*floor((n+1)^2/4) ( = (n+2)*A002620(n+1) ) for n > 0. - Heinrich Ludwig, Dec 22 2017

E.g.f.: e^(-x) * (-2 + x + e^(2*x)*(2 + 19*x + 14*x^2 + 2*x^3))/8. - Iain Fox, Dec 29 2017