

A095800


Triangle T(n,k) = abs( k *( (2*n+1)*(1)^(n+k)+2*k1) /4 ) read by rows, 1<=k<=n.


2



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, 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, 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
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OFFSET

1,3


COMMENTS

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.
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.
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...


REFERENCES

J. H. Conway and R. K. Guy, The Book of Numbers, SpringerVerlag New York, 1996, p. 83.


LINKS

Indranil Ghosh, Rows 1..125, flattened


FORMULA

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.


EXAMPLE

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:
1;
1, 4;
2, 2, 9;
2, 6, 3, 16;


MAPLE

A095800 := proc(n, k) k/4*( (2*n+1)*(1)^(n+k)+2*k1) ; abs(%) ; end proc:
seq(seq(A095800(n, k), k=1..n), n=1..16) ; # R. J. Mathar, Apr 17 2011


PROG

(PARI)
T(n, k) = abs( k *( (2*n+1)*(1)^(n+k)+2*k1) /4 );
for(n=1, 20, for(m=1, n, print1(T(n, m), ", ")));
\\ Joerg Arndt, Mar 05 2014
(Python)
#Generates the bfile
i=1
for n in range(1, 126):
....for k in range(1, n+1):
........print str(i)+" "+str(abs(k*((2*n+1)*(1)**(n+k)+2*k1)/4))
........i+=1 # Indranil Ghosh, Feb 17 2017


CROSSREFS

Cf. A002717 (row sums), A049778.
Sequence in context: A098134 A079191 A079184 * A055630 A182700 A136202
Adjacent sequences: A095797 A095798 A095799 * A095801 A095802 A095803


KEYWORD

nonn,tabl,changed


AUTHOR

Gary W. Adamson, Jun 07 2004


EXTENSIONS

Replaced NAME by closed form and inserted a missing row.  R. J. Mathar, Apr 17 2011


STATUS

approved



