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A185506
Accumulation array, T, of the natural number array A000027, by antidiagonals.
5
1, 3, 4, 7, 11, 10, 14, 23, 26, 20, 25, 42, 51, 50, 35, 41, 70, 88, 94, 85, 56, 63, 109, 140, 156, 155, 133, 84, 92, 161, 210, 240, 250, 237, 196, 120, 129, 228, 301, 350, 375, 374, 343, 276, 165, 175, 312, 416, 490, 535, 550, 532, 476, 375, 220, 231, 415, 558, 664, 735, 771, 770, 728, 639, 495, 286
OFFSET
1,2
COMMENTS
Suppose that R={R(n,k) : n>=1, k>=1} is a rectangular array. The accumulation array of R is given by T(n,k) = Sum_{R(i,j): 1<=i<=n, 1<=j<=k}. (See A144112.)
The formula for the integer T(n,k) has denominator 12. The 2nd, 3rd, and 4th accumulation arrays of A000027 have formulas in which the denominators are 144, 2880, and 86400, respectively; see A185507, A185508, and A185509.
FORMULA
T(n,k) = k*n*(2*n^2 + 3*(k+1)*n + 2*k^2 - 3*k + 5)/12.
EXAMPLE
The natural number array A000027 starts with
1, 2, 4, 7, ...
3, 5, 8, 12, ...
6, 9, 13, 18, ...
...
T(n,k) is the sum of numbers in the rectangle with corners at (1,1) and (n,k) of A000027, so that a corner of T is as follows:
1, 3, 7, 14, 25, 41
4, 11, 23, 42, 70, 109
10, 26, 51, 88, 140, 210
20, 50, 94, 156, 240, 350
35, 85, 155, 250, 375, 535
MATHEMATICA
f[n_, k_]:=k*n*(2n^2+3(k+1)*n+2k^2-3k+5)/12;
TableForm[Table[f[n, k], {n, 1, 10}, {k, 1, 15}]]
Table[f[n-k+1, k], {n, 14}, {k, n, 1, -1}]//Flatten
CROSSREFS
Cf. A004006 (row 1), A000292 (col 1), A051925 (col 2), A185505 (1st diagonal).
Sequence in context: A061447 A219188 A283428 * A328379 A080591 A291063
KEYWORD
nonn,tabl
AUTHOR
Clark Kimberling, Jan 29 2011
STATUS
approved