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A173261
Array T(n,k) read by antidiagonals: T(n,2k)=1, T(n,2k+1)=n, n>=2, k>=0.
1
1, 1, 2, 1, 3, 1, 1, 4, 1, 2, 1, 5, 1, 3, 1, 1, 6, 1, 4, 1, 2, 1, 7, 1, 5, 1, 3, 1, 1, 8, 1, 6, 1, 4, 1, 2, 1, 9, 1, 7, 1, 5, 1, 3, 1, 1, 10, 1, 8, 1, 6, 1, 4, 1, 2, 1, 11, 1, 9, 1, 7, 1, 5, 1, 3, 1, 1, 12, 1, 10, 1, 8, 1, 6, 1, 4, 1, 2, 1, 13, 1, 11, 1, 9, 1, 7, 1, 5, 1, 3, 1, 1, 14, 1, 12, 1, 10, 1, 8, 1, 6, 1, 4, 1, 2
OFFSET
2,3
COMMENTS
One may define another array B(n,0) = -1, B(n,k) = T(n,k-1) + 2*B(n,k-1), n>=2, which also starts in columns k>=0, as follows:
-1, -1, 0, 1, 4, 9, 20, 41, 84, 169, 340, 681, 1364 ...: A084639;
-1, -1, 1, 3, 9, 19, 41, 83, 169, 339, 681, 1363, 2729;
-1, -1, 2, 5, 14, 29, 62, 125, 254, 509, 1022, 2045, 4094;
-1, -1, 3, 7, 19, 39, 83, 167, 339, 679, 1363, 2727, 5459 ...: -A173114;
B(n,k) = (n-1)*A001045(k) - T(n,k).
First differences are B(n,k+1) - B(n,k) = (n-1)*A001045(k).
FORMULA
From G. C. Greubel, Dec 03 2021: (Start)
T(n, k) = (1/2)*((n+3) - (n+1)*(-1)^k).
Sum_{k=0..n} T(n-k, k) = A024206(n).
Sum_{k=0..floor((n+2)/2)} T(n-2*k+2, k) = (1/16)*(2*n^2 4*n -5*(1 +(-1)^n) + 4*sin(n*Pi/2)) (diagonal sums).
T(2*n-2, n) = A093178(n). (End)
EXAMPLE
The array T(n,k) starts in row n=2 with columns k>=0 as:
1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2 ... A000034;
1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3 ... A010684;
1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4 ... A010685;
1, 5, 1, 5, 1, 5, 1, 5, 1, 5, 1, 5 ... A010686;
1, 6, 1, 6, 1, 6, 1, 6, 1, 6, 1, 6 ... A010687;
1, 7, 1, 7, 1, 7, 1, 7, 1, 7, 1, 7 ... A010688;
1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8 ... A010689;
1, 9, 1, 9, 1, 9, 1, 9, 1, 9, 1, 9 ... A010690;
1, 10, 1, 10, 1, 10, 1, 10, 1, 10, 1, 10 ... A010691.
Antidiagonal triangle begins as:
1;
1, 2;
1, 3, 1;
1, 4, 1, 2;
1, 5, 1, 3, 1;
1, 6, 1, 4, 1, 2;
1, 7, 1, 5, 1, 3, 1;
1, 8, 1, 6, 1, 4, 1, 2;
1, 9, 1, 7, 1, 5, 1, 3, 1;
1, 10, 1, 8, 1, 6, 1, 4, 1, 2;
1, 11, 1, 9, 1, 7, 1, 5, 1, 3, 1;
1, 12, 1, 10, 1, 8, 1, 6, 1, 4, 1, 2;
1, 13, 1, 11, 1, 9, 1, 7, 1, 5, 1, 3, 1;
1, 14, 1, 12, 1, 10, 1, 8, 1, 6, 1, 4, 1, 2;
MATHEMATICA
T[n_, k_]:= (1/2)*((n+3) - (n+1)*(-1)^k);
Table[T[n-k, k], {n, 2, 17}, {k, 2, n}]//Flatten (* G. C. Greubel, Dec 03 2021 *)
PROG
(Sage) flatten([[(1/2)*((n-k+3) - (n-k+1)*(-1)^k) for k in (2..n)] for n in (2..17)]) # G. C. Greubel, Dec 03 2021
KEYWORD
nonn,tabl,easy
AUTHOR
Paul Curtz, Feb 14 2010
STATUS
approved