A322135 Table of truncated square pyramid numbers, read by antidiagonals.

1, 4, 5, 9, 13, 14, 16, 25, 29, 30, 25, 41, 50, 54, 55, 36, 61, 77, 86, 90, 91, 49, 85, 110, 126, 135, 139, 140, 64, 113, 149, 174, 190, 199, 203, 204, 81, 145, 194, 230, 255, 271, 280, 284, 285, 100, 181, 245, 294, 330, 355, 371, 380, 384, 385, 121, 221, 302

Offset: 1

Author

Allan C. Wechsler, Nov 27 2018

Keywords

• nonn
• easy
• tabl

Comments

The n-th row contains n numbers: n^2, n^2 + (n-1)^2, ..., n^2 + (n-1)^2 + ... + 1^2.

All numbers that appear in the table are listed in ascending order at A034705.

All numbers that appear twice or more are listed at A130052.

The left column is A000290 (the squares).

The top row is A000330 (the square pyramidal numbers).

The columns are A000290, A099776 (or a tail of A001844), a tail of A005918 or A120328, a tail of A027575, a tail of A027578, a tail of A027865, ...

The first two rows are A000330 and a tail of A168599, but subsequent rows are not currently in the OEIS, and are all tails of A000330 minus various constants.

The main diagonal is A050410.

Examples

			The 17th term is entry 2 on antidiagonal 6, so we sum two terms: 6^2 + 5^2 = 61.
Table begins:
   1   5  14  30  55  91 140 204 ...
   4  13  29  54  90 139 203 ...
   9  25  50  86 135 199 ...
  16  41  77 126 190 ...
  25  61 110 174 ...
  36  85 149 ...
  49 113 ...
  64 ...
  ...
		

Crossrefs

See comments; also cf. A000330, A059255.

Programs

Mathematica

T[n_,k_] = Sum[(n+i)^2, {i,0,k-1}]; Table[T[n-k+1, k], {n,1,10},  {k,1,n}] // Flatten (* Amiram Eldar, Nov 28 2018 *)
f[n_] := Table[SeriesCoefficient[-((y (y (1 + y) + x (1 - 2 y - 3 y^2) + x^2 (1 - 3 y + 4 y^2)))/((-1 + x)^3 (-1 + y)^4)) , {x, 0,
i + 1 - j}, {y, 0, j}], {i, n, n}, {j, 1, n}]; Flatten[Array[f, 10]] (* Stefano Spezia, Nov 28 2018 *)

Formula

T(n,k) = n^2 + (n+1)^2 + ... + (n+k-1)^2 = A000330(n + k - 1) - A000330(n - 1) = T(n, k) = k*n^2 + (k^2 - k)*n + (1/3*k^3 - 1/2*k^2 + 1/6*k)

G.f.: -y*(y*(1 + y) + x*(1 - 2*y - 3*y^2) + x^2*(1 - 3*y + 4*y^2))/((- 1 + x)^3*(- 1 + y)^4). - Stefano Spezia, Nov 28 2018