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%I #23 Feb 22 2019 01:55:14
%S 1,4,5,9,13,14,16,25,29,30,25,41,50,54,55,36,61,77,86,90,91,49,85,110,
%T 126,135,139,140,64,113,149,174,190,199,203,204,81,145,194,230,255,
%U 271,280,284,285,100,181,245,294,330,355,371,380,384,385,121,221,302
%N Table of truncated square pyramid numbers, read by antidiagonals.
%C The n-th row contains n numbers: n^2, n^2 + (n-1)^2, ..., n^2 + (n-1)^2 + ... + 1^2.
%C All numbers that appear in the table are listed in ascending order at A034705.
%C All numbers that appear twice or more are listed at A130052.
%C The left column is A000290 (the squares).
%C The top row is A000330 (the square pyramidal numbers).
%C 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, ...
%C 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.
%C The main diagonal is A050410.
%F 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)
%F 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
%e The 17th term is entry 2 on antidiagonal 6, so we sum two terms: 6^2 + 5^2 = 61.
%e Table begins:
%e 1 5 14 30 55 91 140 204 ...
%e 4 13 29 54 90 139 203 ...
%e 9 25 50 86 135 199 ...
%e 16 41 77 126 190 ...
%e 25 61 110 174 ...
%e 36 85 149 ...
%e 49 113 ...
%e 64 ...
%e ...
%t 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 *)
%t 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,
%t i + 1 - j}, {y, 0, j}], {i, n, n}, {j, 1, n}]; Flatten[Array[f, 10]] (* _Stefano Spezia_, Nov 28 2018 *)
%Y See comments; also cf. A000330, A059255.
%K nonn,easy,tabl
%O 1,2
%A _Allan C. Wechsler_, Nov 27 2018
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