A213547 Antidiagonal sums of the convolution array A213505.
1, 12, 68, 260, 777, 1960, 4368, 8856, 16665, 29524, 49764, 80444, 125489, 189840, 279616, 402288, 566865, 784092, 1066660, 1429428, 1889657, 2467256, 3185040, 4069000, 5148585, 6456996, 8031492, 9913708, 12149985, 14791712, 17895680, 21524448, 25746721, 30637740
Offset: 1
Examples
From _Henri Mühle_, Jan 23 2013: (Start) For n=2, let S=({s0,s1,s2},{(s0,s0),(s0,s1),(s0,s2),(s1,s1),(s2,s2)}) be a 2-star, and let C=({c},{(c,c)}) be a 1-chain. The a(2)=12 proper mergings of S and C are: ({s0,s1,s2,c},{(s0,s0),(s0,s1),(s0,s2),(s1,s1),(s2,s2),(c,c)}) ({s0,s1,s2,c},{(c,s0),(c,s1),(c,s2),(s0,s0),(s0,s1),(s0,s2),(s1,s1),(s2,s2),(c,c)}) ({s0,s1,s2,c},{(c,s1),(s0,s0),(s0,s1),(s0,s2),(s1,s1),(s2,s2),(c,c)}) ({s0,s1,s2,c},{(c,s2),(s0,s1),(s0,s2),(s1,s1),(s2,s2),(c,c)}) ({s0,s1,s2,c},{(c,s1),(c,s2),(s0,s1),(s0,s2),(s1,s1),(s2,s2),(c,c)}) ({s0,s1,s2,c},{(s0,c),(s0,s0),(s0,s1),(s0,s2),(s1,s1),(s2,s2),(c,c)}) ({s0,s1,s2,c},{(s0,c),(c,s1),(s0,s0),(s0,s1),(s0,s2),(s1,s1),(s2,s2),(c,c)}) ({s0,s1,s2,c},{(s0,c),(c,s2),(s0,s0),(s0,s1),(s0,s2),(s1,s1),(s2,s2),(c,c)}) ({s0,s1,s2,c},{(s0,c),(c,s1),(c,s2),(s0,s0),(s0,s1),(s0,s2),(s1,s1),(s2,s2),(c,c)}) ({s0,s1,s2,c},{(s1,c),(s0,s0),(s0,s1),(s0,s2),(s1,s1),(s2,s2),(c,c)}) ({s0,s1,s2,c},{(s2,c),(s0,s0),(s0,s1),(s0,s2),(s1,s1),(s2,s2),(c,c)}) ({s0,s1,s2,c},{(s1,c),(s2,c),(s0,s0),(s0,s1),(s0,s2),(s1,s1),(s2,s2),(c,c)}) (End)
Links
- Clark Kimberling, Table of n, a(n) for n = 1..1000
- Henri Muehle, Proper Mergings of Stars and Chains are Counted by Sums of Antidiagonals in Certain Convolution Arrays -- The Details, arXiv preprint arXiv:1301.1654 [math.CO], 2013.
- Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).
Programs
-
Mathematica
(See A213505.)
-
PARI
{a(n) = n++; (n^6 - n^2) / 60}; /* Michael Somos, Oct 08 2017 */
Formula
a(n) = (n^6 + 6*n^5 + 15*n^4 + 20*n^3 + 14*n^2 + 4*n)/60.
a(n) = 7*a(n-1) - 21*a(n-2) + 35*a(n-3) - 35*a(n-4) + 21*a(n-5) - 7*a(n-6) + a(n-7).
G.f.: x*(1+x)*(1+4*x+x^2)/(1-x)^7.
a(n) = a(-2-n) and a(n-1) = (n^6 - n^2) / 60 for all n in Z. - Michael Somos, Oct 08 2017
E.g.f.: exp(x)*x*(60 + 300*x + 350*x^2 + 140*x^3 + 21*x^4 + x^5)/60. - Stefano Spezia, Apr 07 2023
Comments