A195806 Number of triangular of a 5 X 5 X 5 0..n arrays with all rows and diagonals having the same length having the same sum, with corners zero.

16, 105, 496, 1759, 5052, 12469, 27412, 55059, 102952, 181543, 304908, 491563, 765184, 1155567, 1699684, 2442553, 3438468, 4752283, 6460432, 8652429, 11432392, 14920189, 19253232, 24588229, 31102456, 38995845, 48492976, 59844451, 73329300

Offset: 1

Author

R. H. Hardin, Sep 23 2011

Keywords

• nonn

Examples

			Some solutions for n=4:
      0          0          0          0          0          0          0
     0 1        2 2        1 1        1 4        4 2        4 1        0 0
    2 0 2      1 0 4      0 3 0      4 2 0      2 4 2      1 0 4      3 2 3
   1 0 0 0    3 3 0 0    1 3 3 1    2 0 4 3    2 4 4 4    2 3 0 2    0 2 2 0
  0 0 2 1 0  0 1 1 4 0  0 1 0 1 0  0 3 2 2 0  0 4 2 2 0  0 3 1 3 0  0 0 3 0 0
		

Links

• R. H. Hardin, Table of n, a(n) for n = 1..32
• M. Kauers and C. Koutschan, Some D-finite and some possibly D-finite sequences in the OEIS, arXiv:2303.02793 [cs.SC], 2023.

Crossrefs

Row 5 of A195805.

Formula

From Manuel Kauers and Christoph Koutschan, Mar 01 2023: (Start)

Conjectured recurrence: a(n) - 3*a(n+1) + 2*a(n+2) - a(n+3) + 6*a(n+4) - 5*a(n+5) - 3*a(n+6) + 3*a(n+8) + 5*a(n+9) - 6*a(n+10) + a(n+11) - 2*a(n+12) + 3*a(n+13) - a(n+14) = 0.

Conjectured closed form as a quasi-polynomial:

a(6*n) = 1 + 25*n + 158*n^2 + 650*n^3 + 2275*n^4 + 4680*n^5 + 4680*n^6.

a(6*n+1) = 16 + 198*n + 1133*n^2 + 3900*n^3 + 8125*n^4 + 9360*n^5 + 4680*n^6.

a(6*n+2) = 105 + 1087*n + 4922*n^2 + 12350*n^3 + 17875*n^4 + 14040*n^5 + 4680*n^6.

a(6*n+3) = 496 + 4148*n + 14783*n^2 + 28600*n^3 + 31525*n^4 + 18720*n^5 + 4680*n^6.

a(6*n+4) = 1759 + 12121*n + 35258*n^2 + 55250*n^3 + 49075*n^4 + 23400*n^5 + 4680*n^6.

a(6*n+5) = (1+n)^2*(5052 + 19370*n + 28405*n^2 + 18720*n^3 + 4680*n^4). (End)