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A185779
Third accumulation array of Pascal's triangle (as a rectangle), by antidiagonals.
2
1, 4, 4, 10, 17, 10, 20, 45, 45, 20, 35, 95, 126, 95, 35, 56, 175, 281, 281, 175, 56, 84, 294, 546, 662, 546, 294, 84, 120, 462, 966, 1358, 1358, 966, 462, 120, 165, 690, 1596, 2534, 2941, 2534, 1596, 690, 165, 220, 990, 2502, 4410, 5790, 5790, 4410, 2502, 990, 220, 286, 1375, 3762, 7272, 10620, 12021, 10620, 7272, 3762, 1375, 286, 364, 1859, 5467, 11484, 18432, 23229, 23229, 18432, 11484, 5467, 1859, 364, 455
OFFSET
1,2
COMMENTS
Using "Axxxxxx < Ayyyyyy" to mean that Ayyyyyy is the accumulation array of Axxxxxx, as defined at A144112:
A185779 < A144225 < A007318 < A014430 < A077023 < A185779, where each of these is formatted as a rectangle (e.g., A007318 is Pascal's triangle). See A185778.
row 1: A000292
row 2: A095667
FORMULA
T(n,k) = C(n+k+4,n+2) - (k+3)*(k+4)/2 - (k+2)*n*(k*n+n+3*k+7)/4, for k>=1, n>=1.
EXAMPLE
Northwest corner:
1....4...10...20...35
4....17..45...95...175
10...45..126..281..546
20...95..281..662..1358
MATHEMATICA
f[n_, k_] := Binomial[n + k + 4, n + 2] - (k + 3)*(k + 4)/2 - (k + 2)* n*(k*n + n + 3*k + 7)/4; TableForm[Table[f[n, k], {n, 1, 5}, {k, 1, 5}]]
Table[f[n - k + 1, k], {n, 10}, {k, n, 1, -1}] // Flatten (* G. C. Greubel, Jul 12 2017 *)
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Clark Kimberling, Feb 03 2011
STATUS
approved