A107980 Triangle read by rows: T(n,k) = (n+2)*(k+1)*(k+2)*(2*n-k+2)*(2*n-k+3)/24.
1, 5, 9, 14, 30, 40, 30, 70, 105, 125, 55, 135, 216, 280, 315, 91, 231, 385, 525, 630, 686, 140, 364, 624, 880, 1100, 1260, 1344, 204, 540, 945, 1365, 1755, 2079, 2310, 2430, 285, 765, 1360, 2000, 2625, 3185, 3640, 3960, 4125, 385, 1045, 1881, 2805, 3740, 4620, 5390, 6006, 6435, 6655
Offset: 0
Examples
Triangle begins: 1; 5, 9; 14, 30, 40; 30, 70, 105, 125; 55, 135, 216, 280, 315; 91, 231, 385, 525, 630, 686; 140, 364, 624, 880, 1100, 1260, 1344; 204, 540, 945, 1365, 1755, 2079, 2310, 2430; 285, 765, 1360, 2000, 2625, 3185, 3640, 3960, 4125; 385, 1045, 1881, 2805, 3740, 4620, 5390, 6006, 6435, 6655; 506, 1386, 2520, 3800, 5130, 6426, 7616, 8640, 9450, 10010, 10296;
References
- S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p. 237, K{B(n,3,l)}).
Links
- G. C. Greubel, Rows n = 0..50 of the triangle, flattened
Programs
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Maple
T:=proc(n,k) if k<=n then 1/24*(n+2)*(k+1)*(k+2)*(2*n-k+2)*(2*n-k+3) else 0 fi end: for n from 0 to 9 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form
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Mathematica
T[n_, k_]:= (1/6)*(n+2)*Binomial[k+2, 2]*Binomial[2*n-k+3, 2]; Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Dec 14 2021 *)
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Sage
def A107980(n,k): return (n+2)*(k+1)*(k+2)*(2*n-k+2)*(2*n-k+3)/24 flatten([[A107980(n,k) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, Dec 14 2021
Formula
T(n, k) = (n+2)*(k+1)*(k+2)*(2*n-k+2)*(2*n-k+3)/24.
T(n, 0) = A000330(n+1).
T(n, n) = A006414(n).
Sum_{k=0..n} T(n, k) = A006858(n+1).
T(n, n-1) = 5*binomial(n+4, 5) = 5*A000389(n+4). - G. C. Greubel, Dec 14 2021
Comments