A108578 Number of 3 X 3 magic squares with magic sum 3n.
0, 0, 0, 0, 8, 24, 32, 56, 80, 104, 136, 176, 208, 256, 304, 352, 408, 472, 528, 600, 672, 744, 824, 912, 992, 1088, 1184, 1280, 1384, 1496, 1600, 1720, 1840, 1960, 2088, 2224, 2352, 2496, 2640, 2784, 2936, 3096, 3248, 3416, 3584, 3752, 3928, 4112, 4288
Offset: 1
Keywords
Examples
a(5) = 8 because there are 8 3 X 3 magic squares with entries 1,...,9 and magic sum 15.
Links
- T. Zaslavsky, Table of n, a(n) for n = 1..10000.
- M. Beck and T. Zaslavsky, An enumerative geometry for magic and magilatin labellings, Ann. Combinatorics, 10 (2006), no. 4, 395-413. MR 2007m:05010. Zbl 1116.05071. - _Thomas Zaslavsky_, Jan 29 2010
- Matthias Beck and Thomas Zaslavsky, Six Little Squares and How their Numbers Grow, Journal of Integer Sequences, 13 (2010), Article 10.6.2.
- Index entries for linear recurrences with constant coefficients, signature (1,1,0,-1, -1,1).
Programs
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Magma
I:=[0,0,0,0,8,24]; [n le 6 select I[n] else Self(n-1)+Self(n-2)-Self(n-4)-Self(n-5)+Self(n-6): n in [1..60]]; // Vincenzo Librandi, Sep 01 2018
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Mathematica
LinearRecurrence[{1, 1, 0, -1, -1, 1}, {0, 0, 0, 0, 8, 24}, 50] (* Jean-François Alcover, Sep 01 2018 *) CoefficientList[Series[8 x^4 (1 + 2 x) / ((1 - x) (1 - x^2) (1 - x^3)), {x, 0, 50}], x] (* Vincenzo Librandi, Sep 01 2018 *) -
PARI
a(n)=(1/9)*(2*n^2-32*n+[144,78,120,126,96,102][(n%18)/3+1])
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PARI
x='x+O('x^99); concat(vector(4), Vec(8*x^5*(1+2*x)/((1-x)*(1-x^2)*(1-x^3)))) \\ Altug Alkan, Sep 01 2018
Formula
G.f.: [8*x^5*(1+2*x)] / [(1-x)*(1-x^2)*(1-x^3)].
a(n) = a(n-1) + a(n-2) - a(n-4) - a(n-5) + a(n-6). - Vincenzo Librandi, Sep 01 2018
Extensions
Edited by N. J. A. Sloane, Feb 05 2010
Corrected g.f. to account for previous change in parameter n from magic sum s to s/3; by Thomas Zaslavsky, Mar 12 2010
Comments