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A096038 Triangle T(n,m) = (3*n^2-3*m^2+5*m-4+n)/2 read by rows.

%I A096038 #13 May 16 2021 12:16:42
%S A096038 1,6,4,14,12,7,25,23,18,10,39,37,32,24,13,56,54,49,41,30,16,76,74,69,
%T A096038 61,50,36,19,99,97,92,84,73,59,42,22,125,123,118,110,99,85,68,48,25,
%U A096038 154,152,147,139,128,114,97,77,54,28,186,184,179,171,160,146,129,109,86,60,31
%N A096038 Triangle T(n,m) = (3*n^2-3*m^2+5*m-4+n)/2 read by rows.
%C A096038 The triangle is obtained by subtracting the triangle A094930 from
%C A096038 its square root (also described in A094930) and then dividing each element of column m through 3*m-1.
%C A096038 For the first three rows n=1 to 3 this yields for example:
%C A096038 4;.................2;............2......................1;
%C A096038 14,25;......minus..2,5;.......=..12,20;......->.divide..6,4;
%C A096038 30,65,64;..........2,5,8;........28,60,56;..............14;12,7;
%F A096038 T(n,1) = A095794(n).
%F A096038 T(n,n) = 3*n-2.
%F A096038 T(n,m) = A094930(n,m)/(3*m-1)-1.
%o A096038 (Python)
%o A096038 def A096038(n,m):
%o A096038     return (3*n**2-3*m**2+5*m-4+n)//2
%o A096038 print( [A096038(n,m) for n in range(20) for m in range(1,n+1)] )
%o A096038 # _R. J. Mathar_, Oct 11 2009
%Y A096038 Cf. A095794, A011379, A002411, A096037, A096036, A024212.
%K A096038 nonn,tabl,easy
%O A096038 1,2
%A A096038 _Gary W. Adamson_, Jun 17 2004
%E A096038 Edited, T(3,2) corrected, and extended by _R. J. Mathar_, Oct 11 2009