OFFSET
2,3
COMMENTS
Let F(r,n,d) = binomial(r+d-2,d-1)* ((r-1)*(n-2)+d) /d be the d-dimensional pyramidal numbers. Then A(n,k) = F(k,n,5).
Sum of the n-th antidiagonal is binomial(n+4,7) + binomial(n+4,5) = A055797(n-1). - Mathew Englander, Oct 27 2020
LINKS
Michael De Vlieger, Table of n, a(n) for n = 2..11176 (rows 2 <= n <= 150, flattened)
FORMULA
A(n+2,k) = Sum_{j=0..k-1} A080852(n,j).
A(n,k) = binomial(k+3,4) + (n-2)*binomial(k+3,5). - Mathew Englander, Oct 27 2020
EXAMPLE
The array starts in rows n>=2 and columns k>=1 as
1 5 15 35 70 126 210 330 495 715 1001 1365 1820
1 6 21 56 126 252 462 792 1287 2002 3003 4368 6188
1 7 27 77 182 378 714 1254 2079 3289 5005 7371 10556
1 8 33 98 238 504 966 1716 2871 4576 7007 10374 14924
1 9 39 119 294 630 1218 2178 3663 5863 9009 13377 19292
1 10 45 140 350 756 1470 2640 4455 7150 11011 16380 23660
1 11 51 161 406 882 1722 3102 5247 8437 13013 19383 28028
1 12 57 182 462 1008 1974 3564 6039 9724 15015 22386 32396
1 13 63 203 518 1134 2226 4026 6831 11011 17017 25389 36764
MATHEMATICA
Table[Binomial[k + 3, 4] + (# - 2)*Binomial[k + 3, 5] &[m - k + 1], {m, 2, 12}, {k, m - 1}] // Flatten (* Michael De Vlieger, Nov 05 2020 *)
CROSSREFS
KEYWORD
AUTHOR
R. J. Mathar, Jul 30 2016
STATUS
approved