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Irregular triangle by columns derived from (1, 2, 3, ...) * (1, 2, 3, ...).
4

%I #27 May 13 2022 09:59:38

%S 1,2,3,2,4,4,5,6,3,6,8,6,7,10,9,4,8,12,12,8,9,14,15,12,5,10,16,18,16,

%T 10,11,18,21,20,15,6,12,20,24,24,20,12,13,22,27,28,25,18,7,14,24,30,

%U 32,30,24,14,15,26,33,36,35,30,21,8,16,28,36,40,40,36,28,16

%N Irregular triangle by columns derived from (1, 2, 3, ...) * (1, 2, 3, ...).

%C Given a (1, 2, 3, ...) * (1, 2, 3, ...) multiplication table; leftmost column of the triangle = (1, 2, 3, ...). Then shift down each successive column of the array twice to get this irregular triangle.

%H Stefano Spezia, <a href="/A173997/b173997.txt">First 200 rows of the triangle, flattened</a>

%F T(n, k) = k*(2 - 2*k + n), with 1 <= k <= floor((n + 1)/2). - _Stefano Spezia_, Apr 19 2022

%e Given:

%e 1, 2, 3, 4, 5, ...

%e 2, 4, 6, 8, 10, ...

%e 3, 6, 9, 12, 15, ...

%e 4, 8, 12, 16, 20, ...

%e ...

%e After the shift twice operation, we obtain:

%e 1;

%e 2;

%e 3, 2;

%e 4, 4;

%e 5, 6, 3;

%e 6, 8, 6;

%e 7, 10, 9, 4;

%e 8, 12, 12, 8;

%e 9, 14, 15, 12, 5;

%e 10, 16, 18, 16, 10;

%e 11, 18, 21, 20, 15, 6;

%e 12, 20, 24, 24, 20, 12;

%e ...

%t Flatten[Table[k(2-2k+n),{n,16},{k,Floor[(n+1)/2]}]] (* _Stefano Spezia_, Apr 19 2022 *)

%Y Cf. A003991, A006918 (row sums).

%K nonn,easy,tabf

%O 1,2

%A _Gary W. Adamson_, Mar 05 2010