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%I #28 May 16 2021 12:17:21
%S 1,1,4,2,2,9,2,6,3,16,3,4,12,4,25,3,8,6,20,5,36,4,6,15,8,30,6,49,4,10,
%T 9,24,10,42,7,64,5,8,18,12,35,12,56,8,81,5,12,12,28,15,48,14,72,9,100,
%U 6,10,21,16,40,18,63,16,90,10,121,6,14,15,32,20,54,21,80,18,110,11,144,7,12,24,20,45,24,70,24,99,20,132
%N Triangle T(n,k) = abs( k *( (2*n+1)*(-1)^(n+k)+2*k-1) /4 ) read by rows, 1<=k<=n.
%C 1. Triangles of increasing sizes are subdivided using a triangular array. Then as shown on p. 83 of Conway and Guy, the series A002717 (1, 5, 13, 27, 48, 78, 118...) denotes the total number of triangles in each figure.
%C 2. As a conjecture, each row of A095800 could be a distribution governing distinct subsets of types of triangles having the sum in the "How Many Triangles" series A002717. Thus 1 = 1; 5 = (1 + 4), 13 = (2 + 2 + 9)...etc.
%C 3. Powers of the matrices have alternating signs such that odd rows begin with (+) and even rows begin with (-), as: 1; -1, 4; 2, -2, 9; -2, 6, -3, 16; 3, -4, 12, -4, 25;... Signed row sums = A049778: 1, 3, 9, 17, 32, 48...
%D J. H. Conway and R. K. Guy, The Book of Numbers, Springer-Verlag New York, 1996, p. 83.
%H Indranil Ghosh, <a href="/A095800/b095800.txt">Rows 1..125, flattened</a>
%F Let M(n,k) = (-1)^(k+1)*k, 1<=k<=n be the infinite lower triangular matrix with 1, -2, 3,.. up to the diagonal, and the upper triangular part all zeros. The 3x3 submatrix would be [1 0 0 / 1 -2 0 / 1 -2 3]. The current triangle contains the absolute values of the matrix square M^2.
%e 1. [1 0 0 / 1 -2 0 / 1 -2 3]^2 = [1 0 0 / 1 -4 0 / 2 -2 9]. Then change the (-) signs to (+) getting the first 3 rows of the triangle:
%e 1;
%e 1, 4;
%e 2, 2, 9;
%e 2, 6, 3, 16;
%p A095800 := proc(n,k) k/4*( (2*n+1)*(-1)^(n+k)+2*k-1) ; abs(%) ; end proc:
%p seq(seq(A095800(n,k),k=1..n),n=1..16) ; # _R. J. Mathar_, Apr 17 2011
%o (PARI)
%o T(n,k) = abs( k *( (2*n+1)*(-1)^(n+k)+2*k-1) /4 );
%o for(n=1,20,for(m=1,n,print1(T(n,m),", ")));
%o \\ _Joerg Arndt_, Mar 05 2014
%o (Python)
%o # Generates the b-file
%o i=1
%o for n in range(1,126):
%o for k in range(1,n+1):
%o print(str(i)+" "+str(abs(k*((2*n+1)*(-1)**(n+k)+2*k-1)//4)))
%o i+=1 # _Indranil Ghosh_, Feb 17 2017
%Y Cf. A002717 (row sums), A049778.
%K nonn,tabl
%O 1,3
%A _Gary W. Adamson_, Jun 07 2004
%E Replaced NAME by closed form and inserted a missing row. - _R. J. Mathar_, Apr 17 2011
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