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
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
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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
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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).
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