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 A007773 For any circular arrangement of 0..n-1, let S = sum of squares of every sum of two contiguous numbers; then a(n) = # of distinct values of S. 4
 1, 1, 1, 3, 8, 21, 43, 69, 102, 145, 197, 261, 336, 425, 527, 645, 778, 929, 1097, 1285, 1492, 1721, 1971, 2245, 2542, 2865, 3213, 3589, 3992, 4425, 4887, 5381, 5906, 6465, 7057, 7685, 8348, 9049, 9787, 10565, 11382, 12241, 13141, 14085, 15072, 16105 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,4 LINKS Alois P. Heinz, Table of n, a(n) for n = 1..10000 K. S. Brown, The Dartboard Sequence H. L. Montgomery, Kevin Brown's enumeration problem, Manuscript, 4 Oct. 1994. (Annotated scanned copy) FORMULA For n >= 7, a(n) = (n^3-16*n+27)/6 (n odd); (n^3-16*n+30)/6 (n even). G.f.: x*(1-2*x+4*x^3+3*x^5-10*x^7+2*x^8+8*x^9-4*x^10)/((1-x)^3*(1-x^2)). - Michael Somos, May 03, 2002 a(2n) = A322594(n-4), n>=4. - R. J. Mathar, Mar 18 2019 MATHEMATICA Drop[CoefficientList[Series[x*(1-2*x+4*x^3+3*x^5-10*x^7+2*x^8+8*x^9 -4*x^10)/((1-x)^3*(1-x^2)), {x, 0, 60}], x], 1] (* G. C. Greubel, Mar 15 2019 *) PROG (PARI) a(n)=polcoeff(x*(1-2*x+4*x^3+3*x^5-10*x^7+2*x^8+8*x^9-4*x^10+O(x^n))/(1-x)^3/(1-x^2), n) (PARI) A007773(n)=if(n>5, (n^3-max(16*n, 116)+31)\6, n>3, 5*n-17, 1) \\ M. F. Hasler, Mar 15 2019 (MAGMA) R:=PowerSeriesRing(Integers(), 60); Coefficients(R!( x*(1-2*x +4*x^3+3*x^5-10*x^7+2*x^8+8*x^9-4*x^10)/((1-x)^3*(1-x^2)) )); // G. C. Greubel, Mar 15 2019 (Sage) a=(x*(1-2*x+4*x^3+3*x^5-10*x^7+2*x^8+8*x^9-4*x^10)/((1-x)^3*(1-x^2))).series(x, 60).coefficients(x, sparse=False); a[1:] # G. C. Greubel, Mar 15 2019 CROSSREFS Sequence in context: A103736 A172243 A101332 * A071078 A179903 A193045 Adjacent sequences:  A007770 A007771 A007772 * A007774 A007775 A007776 KEYWORD nonn,easy AUTHOR K. S. Brown (kevin2003(AT)delphi.com), Hugh L. Montgomery EXTENSIONS More terms from David W. Wilson, Oct 27 2000 STATUS approved

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Last modified April 15 07:40 EDT 2021. Contains 342975 sequences. (Running on oeis4.)