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 A008233 a(n) = floor(n/4)*floor((n+1)/4)*floor((n+2)/4)*floor((n+3)/4). 4
 0, 0, 0, 0, 1, 2, 4, 8, 16, 24, 36, 54, 81, 108, 144, 192, 256, 320, 400, 500, 625, 750, 900, 1080, 1296, 1512, 1764, 2058, 2401, 2744, 3136, 3584, 4096, 4608, 5184, 5832, 6561, 7290, 8100, 9000, 10000, 11000, 12100, 13310, 14641, 15972, 17424, 19008, 20736 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,6 REFERENCES Mubayi, Dhruv. Counting substructures II: Hypergraphs, Combinatorica 33 (2013), no. 5, 591--612. MR3132928. LINKS Vincenzo Librandi, Table of n, a(n) for n = 0..3000 Index entries for linear recurrences with constant coefficients, signature (2, -1, 0, 3, -6, 3, 0, -3, 6, -3, 0, 1, -2, 1). FORMULA Let b(n) = A002620(n), the quarter-squares. Then this sequence is b(0)*b(0), b(0)*b(1), b(1)*b(1), b(1)*b(2), b(2)*b(2), b(2)*b(3), ... a(n) = +2*a(n-1) -a(n-2) +3*a(n-4) -6*a(n-5) +3*a(n-6) -3*a(n-8) +6*a(n-9) -3*a(n-10) +a(n-12) -2*a(n-13) +a(n-14). G.f. -x^4*(1+x^6+x^2+2*x^3+x^4) / ( (1+x)^3*(x^2+1)^3*(x-1)^5 ). - R. J. Mathar, Feb 20 2011 MAPLE A008233:=n->floor(n/4)*floor((n+1)/4)*floor((n+2)/4)*floor((n+3)/4); seq(A008233(n), n=0..50); # Wesley Ivan Hurt, Dec 31 2013 MATHEMATICA Table[Floor[n/4]*Floor[(n + 1)/4]*Floor[(n + 2)/4]*Floor[(n + 3)/4], {n, 0, 50}] (* Stefan Steinerberger, Apr 03 2006 *) PROG (Haskell) a008233 n = product \$ map (`div` 4) [n..n+3] -- Reinhard Zumkeller, Jun 08 2011 (MAGMA) [Floor(n/4)*Floor((n+1)/4)*Floor((n+2)/4)*Floor((n+3)/4): n in [0..50]]; // Vincenzo Librandi, Jun 09 2011 CROSSREFS Cf. A002620. Sequence in context: A160158 A010072 A005943 * A224815 A031923 A225549 Adjacent sequences:  A008230 A008231 A008232 * A008234 A008235 A008236 KEYWORD nonn,nice,easy AUTHOR EXTENSIONS More terms from Stefan Steinerberger, Apr 03 2006 STATUS approved

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