OFFSET
1,1
COMMENTS
In the equipotential triangle with the medians carried out from vertices, it is possible to find 16 triangles of a different form, size and orientation. Adding to the original triangle 3 triangles equivalent to it (with medians carried out vertices) we can construct a second row of triangles. In the resulting triangle, you can count 104 triangles of different form, size and orientation.
Quasipolynomial of order 30. - Charles R Greathouse IV, Aug 04 2013
LINKS
Lev Volkov, Table of n, a(n) for n = 1..10000
MathPuzzle, Bill Daly
Index entries for linear recurrences with constant coefficients, signature (1,0,1,0,0,-1,-1,0,0,1,0,1,-1).
FORMULA
a(n) = (1678*n^3+3117*n^2+88*n-345*Mod[n,2]-320*Mod[n,3]-90*Mod[n,4]-288*Mod[n^3-n^2+n,5])/240. (Bill Daly)
G.f.: x*(2*x^11 +6*x^10 +68*x^9 +167*x^8 +302*x^7 +417*x^6 +479*x^5 +439*x^4 +334*x^3 +199*x^2 +88*x +16) / ((x -1)^4*(x +1)*(x^2 +1)*(x^2 +x +1)*(x^4 +x^3 +x^2 +x +1)). - Colin Barker, Sep 06 2014
MATHEMATICA
a[n_] := (1678*n^3 + 3117*n^2 + 88*n - 345*Mod[n, 2] - 320*Mod[n, 3] - 90*Mod[n, 4] - 288*Mod[n^3 - n^2 + n, 5])/240; Table[a[n], {n, 20}]
LinearRecurrence[{1, 0, 1, 0, 0, -1, -1, 0, 0, 1, 0, 1, -1}, {16, 104, 303, 653, 1196, 1978, 3032, 4410, 6148, 8292, 10875, 13955, 17556}, 40] (* Harvey P. Dale, Dec 08 2019 *)
PROG
(PARI) a(n)=(1678*n^3+3117*n^2+88*n-n%2*345-n%3*320-n%4*90-(n^3-n^2+n)%5*288)/240 \\ Charles R Greathouse IV, Aug 04 2013
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Lev Volkov, Mar 28 2012
EXTENSIONS
Offset corrected by Andrew Howroyd, Jan 17 2018
STATUS
approved