A133146 Antidiagonal sums of the triangle A133128.

2, 5, 7, 14, 18, 29, 35, 50, 58, 77, 87, 110, 122, 149, 163, 194, 210, 245, 263, 302, 322, 365, 387, 434, 458, 509, 535, 590, 618, 677, 707, 770, 802, 869, 903, 974, 1010, 1085, 1123, 1202, 1242, 1325, 1367, 1454, 1498, 1589, 1635, 1730, 1778, 1877, 1927, 2030

Offset: 0

Author

Paul Curtz, Aug 27 2008

Keywords

• nonn
• less

Examples

			a(2) = A133128(2,0) + A133128(1,1) = 10 - 3 = 7.
a(3) = A133128(3,0) + A133128(2,1) = 17 - 3 = 14.
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..10000
• Index entries for linear recurrences with constant coefficients, signature (1, 2, -2, -1, 1).

Programs

Magma

[19/8 +5*n/4 +3*n^2/4 -(-1)^n*(n/4+3/8): n in [0..60]]; // Vincenzo Librandi, Aug 10 2011

Mathematica

CoefficientList[Series[(1+2x)(2-x+x^3)/((1-x)^3(1+x)^2),{x,0,60}],x] (* or *) LinearRecurrence[{1,2,-2,-1,1},{2,5,7,14,18},60] (* Harvey P. Dale, Aug 26 2013 *)

Formula

First differences: a(n+1) - a(n) = A059029(n+1).

Bisections: a(2n+1) = A005918(n+1). a(2n) = A141631(n+1).

G.f.: (1+2*x)(2 - x + x^3)/((1-x)^3*(1+x)^2). - R. J. Mathar, Oct 15 2008

a(n) = 19/8 + 5*n/4 + 3*n^2/4 - (-1)^n*(n/4 + 3/8). - R. J. Mathar, Oct 15 2008

From Harvey P. Dale, Aug 26 2013: (Start)

a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5); a(0)=2, a(1)=5, a(2)=7, a(3)=14, a(4)=18. (End)

Extensions

Edited and extended by R. J. Mathar, Oct 15 2008