A080852 Square array of 4D pyramidal numbers, read by antidiagonals.
1, 1, 4, 1, 5, 10, 1, 6, 15, 20, 1, 7, 20, 35, 35, 1, 8, 25, 50, 70, 56, 1, 9, 30, 65, 105, 126, 84, 1, 10, 35, 80, 140, 196, 210, 120, 1, 11, 40, 95, 175, 266, 336, 330, 165, 1, 12, 45, 110, 210, 336, 462, 540, 495, 220, 1, 13, 50, 125, 245, 406, 588, 750, 825, 715, 286
Offset: 0
Examples
Array, n >= 0, k >= 0, begins 1 4 10 20 35 56 ... 1 5 15 35 70 126 ... 1 6 20 50 105 196 ... 1 7 25 65 140 266 ... 1 8 30 80 175 336 ...
Crossrefs
Programs
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Derive
vector(vector(poly_coeff(Taylor((1+kx)/(1-x)^5,x,11),x,n),n,0,11),k,-1,10)
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Derive
VECTOR(VECTOR(comb(k+3,3)+comb(k+3,4)n, k, 0, 11), n, 0, 11)
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Maple
A080852 := proc(n,k) binomial(k+4,4)+(n-1)*binomial(k+3,4) ; end proc: seq( seq(A080852(d-k,k),k=0..d),d=0..12) ; # R. J. Mathar, Oct 01 2021
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Mathematica
T[n_, k_] := Binomial[k+3, 3] + Binomial[k+3, 4]n; Table[T[n-k, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, May 05 2023 *)
Formula
T(n, k) = binomial(k + 4, 4) + (n-1)*binomial(k + 3, 4), corrected Oct 01 2021.
T(n, k) = T(n - 1, k) + C(k + 3, 4) = T(n - 1, k) + k(k + 1)(k + 2)(k + 3)/24.
G.f. for rows: (1 + nx)/(1 - x)^5, n >= -1.
T(n,k) = sum_{j=0..k} A080851(n,j). - R. J. Mathar, Jul 28 2016
Comments