A059329 Number of 3 X 3 matrices, with elements from {0,...,n}, having the property that the middle element of each of the eight 3-element horizontal, vertical and diagonal lines equals the average of the two end elements.

1, 2, 7, 12, 25, 38, 63, 88, 129, 170, 231, 292, 377, 462, 575, 688, 833, 978, 1159, 1340, 1561, 1782, 2047, 2312, 2625, 2938, 3303, 3668, 4089, 4510, 4991, 5472, 6017, 6562, 7175, 7788, 8473, 9158, 9919, 10680, 11521, 12362, 13287, 14212

Offset

0,2

Comments

The bisections of the first differences of {a(n)} give A001844 (the centered triangular numbers n^2+(n-1)^2).

Quasipolynomial of order 2. - Charles R Greathouse IV, Dec 07 2011

Also, the number of 3 X 3 magic squares with elements from {0,...,n} and with duplicate elements allowed. If [[a,b,c], [d,e,f], [g,h,i]] satisfies the property in the description of this sequence, then [[h,a,f], [c,e,g], [d,i,b]] is a magic square, and conversely. - David Radcliffe, Apr 13 2020

Links

G. C. Greubel, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (2,1,-4,1,2,-1).

Formula

From Frank Ellermann: (Start)

even: a(2*n) = (4*n^3 + 6*n^2 + 8*n + 3)/3.

odd: a(2*n-1) = (4*n^3 + 2*n)/3. (End)

a(n) = Sum_{k=0..n} A109613(k)*A109613(n-k). - Reinhard Zumkeller, Dec 05 2009

From Colin Barker, Mar 29 2013: (Start)

a(n) = ((1+n)*(9+3*(-1)^n+4*n+2*n^2))/12.

G.f.: (x^2+1)^2 / ((x-1)^4*(x+1)^2). (End)

E.g.f.: (1/12)*(3*(1 - x)*exp(-x) + (9 + 21*x + 12*x^2 + 2*x^3)*exp(x)). - G. C. Greubel, Jan 07 2017

Mathematica

LinearRecurrence[{2, 1, -4, 1, 2, -1}, {1, 2, 7, 12, 25, 38}, 50] (* G. C. Greubel, Jan 07 2017 *)

Prog

(PARI) a(n)=if(n%2, 4*n^3+2*n, 4*n^3+6*n^2+8*n+3)/3 \\ Charles R Greathouse IV, Dec 07 2011

Crossrefs

Sequence in context: A122264 A379790 A079824 * A242201 A350093 A177747

Adjacent sequences: A059326 A059327 A059328 * A059330 A059331 A059332

Keyword

nonn,easy

Author

John W. Layman, Jan 26 2001

Status

approved