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%I #23 Aug 02 2023 00:32:18
%S 0,0,1,0,4,4,0,10,20,10,0,20,60,60,20,0,35,140,210,140,35,0,56,280,
%T 560,560,280,56,0,84,504,1260,1680,1260,504,84,0,120,840,2520,4200,
%U 4200,2520,840,120,0,165,1320,4620,9240,11550,9240,4620,1320,165,0,220,1980,7920,18480,27720,27720,18480,7920,1980,220
%N Fourth slice along the 1-2-plane in the cube a(m,n,o) = a(m-1,n,o) + a(m,n-1,o) + a(m,n,o-1) for which the first slice is Pascal's triangle (slice read by antidiagonals).
%C Essentially the same as triangle A178820 with an additional column of zeros at the left border. - _Georg Fischer_, Jul 31 2023
%F a(m-1,n,o) + a(m,n-1,o) + a(m,n,o-1) with initialization values a(1,0,0) = 1 and a(m<>1=0, n>=0, 0>=o) = 0.
%e The second row is 1, 4, 10, 20, 35, 56, 84, 120, 165, 220 = A000292, i.e., Tetrahedral (or pyramidal) numbers: binomial(n+2,3) = n(n+1)(n+2)/6 (core).
%e The third row is 4, 20, 60, 140, 280, 504, 840, 1320, 1980, 2860 = A033488 = n*(n+1)*(n+2)*(n+3)/6.
%e The main diagonal is 0, 4, 60, 560, 4200, 27720, 168168, 960960, 5250960, 27713400 = {0} U A002803*4.
%e Triangle starts:
%e 0
%e 0, 1
%e 0, 4, 4
%e 0, 10, 20, 10
%e 0, 20, 60, 60, 20
%e 0, 35, 140, 210, 140, 35
%e 0, 56, 280, 560, 560, 280, 56
%e 0, 84, 504, 1260, 1680, 1260, 504, 84
%p T:=(n, k)->binomial(n+3, 3)*binomial(n, k): seq(print(seq(T(n-1, k-1), k=0..n)), n=0..10); # _Georg Fischer_, Jul 31 2023
%Y Cf. A000292, A003506, A007318, A094305, A121306, A119800, A178820.
%K nonn,tabl
%O 0,5
%A _Thomas Wieder_, Aug 06 2006
%E a(55)-a(56) corrected and more terms from _Georg Fischer_, Jul 31 2023
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