A084569 Partial sums of A084570.

1, 3, 9, 21, 44, 82, 142, 230, 355, 525, 751, 1043, 1414, 1876, 2444, 3132, 3957, 4935, 6085, 7425, 8976, 10758, 12794, 15106, 17719, 20657, 23947, 27615, 31690, 36200, 41176, 46648, 52649, 59211, 66369, 74157, 82612, 91770, 101670, 112350, 123851

Offset

0,2

Comments

Conjecture: a(n) is the number of perimeter-magic (hollow) squares of order 3 with magic sum n+3. Order 3 means each of the 4 edges has 3 elements >=1; the square has 8 elements. The elements do not need to be distinct, and squares obtained by rotations are counted only once. The square (read ccw) for magic sum 3 has elements 1 1 1 1 1 1 1 1. The 3 squares with magic sum 4 are 1 1 2 1 1 1 2 1, 1 1 2 1 1 2 1 2 and 1 2 1 2 1 2 1 2. - R. J. Mathar, Mar 08 2025

Links

Table of n, a(n) for n=0..40.

Index entries for linear recurrences with constant coefficients, signature (4,-5,0,5,-4,1).

Formula

a(n) = (-1)^n/8 + (n^4 + 6*n^3 + 17*n^2 + 30*n + 21)/24.

a(n) = Sum_{k=0..n} Sum_{j=0..k} Sum_{i=0..j} (i + (-1)^i).

G.f.: ( -1+x-2*x^2 ) / ( (1+x)*(x-1)^5 ). - R. J. Mathar, Mar 08 2025

a(n)+a(n+1) = A116701(n+3)-1. - R. J. Mathar, Mar 08 2025

Mathematica

Accumulate[LinearRecurrence[{3, -2, -2, 3, -1}, {1, 2, 6, 12, 23}, 50]] (* or *) LinearRecurrence[{4, -5, 0, 5, -4, 1}, {1, 3, 9, 21, 44, 82}, 50] (* Harvey P. Dale, Nov 12 2014 *)

Crossrefs

Cf. A116701.

Sequence in context: A011843 A080549 A175006 * A192970 A110964 A107351

Adjacent sequences: A084566 A084567 A084568 * A084570 A084571 A084572

Keyword

easy,nonn,changed

Author

Paul Barry, May 31 2003

Status

approved