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 A243213 Number of ways to place 4 points on a triangular grid of side length n so that no three of them are vertices of an equilateral triangle with sides parallel to the grid. 5
 3, 128, 1062, 5160, 18591, 55113, 142005, 329045, 701160, 1395975, 2626953, 4713723, 8120322, 13503350, 21770766, 34153758, 52292385, 78337890, 115072320, 166048850, 235753353, 329791143, 455099307, 620189115, 835418766, 1113301553, 1468849515, 1919958285 (list; graph; refs; listen; history; text; internal format)
 OFFSET 3,1 LINKS Heinrich Ludwig, Table of n, a(n) for n = 3..1000 Index entries for linear recurrences with constant coefficients, signature (6,-12,2,27,-36,0,36,-27,-2,12,-6,1) FORMULA a(n) = (n^8 + 4*n^7 - 6*n^6 - 80*n^5 - 15*n^4 + 532*n^3 - 244*n^2 - 432*n)/384 + IF(MOD(n, 2) = 1)*(-n^2 - n + 12)/16. G.f.: x^3*(7*x^7-33*x^6-15*x^5-38*x^4-318*x^3-330*x^2-110*x-3) / ((x-1)^9*(x+1)^3). - Colin Barker, Jun 09 2014 EXAMPLE There are exactly a(3) = 3 ways to place 4 points (x) on a 3X3X3 grid, no three of them being vertices of an equilateral triangle: . x x x x . x x . x . x x x . . x x PROG (PARI) Vec(x^3*(7*x^7-33*x^6-15*x^5-38*x^4-318*x^3-330*x^2-110*x-3)/((x-1)^9*(x+1)^3) + O(x^100)) \\ Colin Barker, Jun 09 2014 CROSSREFS Cf. A243211, A243209, A000217, A050534, A243212, A243214. Sequence in context: A134711 A163850 A258671 * A264582 A108713 A319222 Adjacent sequences: A243210 A243211 A243212 * A243214 A243215 A243216 KEYWORD nonn,easy AUTHOR Heinrich Ludwig, Jun 09 2014 STATUS approved

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Last modified August 13 04:15 EDT 2024. Contains 375113 sequences. (Running on oeis4.)