%I
%S 0,0,4,0,3,1,0,11,8,12,0,8,3,0,4,0,24,16,27,24,28,0,19,5,3,8,5,9,
%T 0,43,24,48,40,51,48,52,0,36,7,12,12,4,15,12,16,0,68,32,75,56,80,72,
%U 83,80,84,0,59,9,27
%N Square array read by antidiagonals downwards giving the first differences A261327(n+p)  A261327(n), with p >= 0.
%C Successive rows:
%C p
%C 0: 0, 0, 0, 0, 0, 0, 0, ...
%C 1: 4, 3, 11, 8, 24, 19, 43, ...
%C 2: 1, 8, 3, 16, 5, 24, 7, ...
%C 3: 12, 0, 27, 3, 48, 12, 75, ...
%C 4: 4, 24, 8, 40, 12, 56, 16, ...
%C 5: 28, 5, 51, 4, 80, 3, 115, ...
%C 6: 9, 48, 15, 72, 21, 96, 27, ...
%C ... .
%C Main diagonal: alternate 3*n^2, 3.
%C From p>0, the rows are multiples of 1, 1, 3, 4, 1, 3, 1, 8, 3, 5, 1, 12, 1, 7, 3, 16, 1, ... . Sequences appearing after division: shifted A144433 or A195161, A064680. For p=3, we have (n+2)^2, n^2.
%C First column: alternate n^2, 4*(n^2 + n + 1). Its first differences (4, 3, 11, 8, 24, ...) is the sequence of the square array for p=1.
%C Third column: 0, 3, 8, 15, ... is A005563(n).
%C Fifth column: 5, 21, 45, 77, ... is a bisection of A061037(n).
%C Seventh column: 7, 16, 40, 55, 91, 112, ... is a subsequence of A061039(n).
%C Etc. From the Rydberg spectra of the hydrogen atom (mentioned in A261327).
%C Starting for instance from p=3,at the main antidiagonal,yields:
%C 3: 12, 0, 27, 3, ... see p=3
%C 2: 1, 8, 3, 16, 5, ... p=2
%C 1: 4, 3, 11, 8, 24, 19, ... p=1.
%Y Cf. A000290, A002061, A005563, A061037, A061039, A064680, A112087, A144433, A195161, A261327.
%K sign,tabl
%O 0,3
%A _Paul Curtz_, Jan 05 2017
