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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

%I #27 Sep 08 2022 08:44:35

%S 1,1,1,3,8,21,43,69,102,145,197,261,336,425,527,645,778,929,1097,1285,

%T 1492,1721,1971,2245,2542,2865,3213,3589,3992,4425,4887,5381,5906,

%U 6465,7057,7685,8348,9049,9787,10565,11382,12241,13141,14085,15072,16105

%N 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.

%H Alois P. Heinz, <a href="/A007773/b007773.txt">Table of n, a(n) for n = 1..10000</a>

%H K. S. Brown, <a href="http://www.mathpages.com/home/kmath025.htm">The Dartboard Sequence</a>

%H H. L. Montgomery, <a href="/A007773/a007773.pdf">Kevin Brown's enumeration problem</a>, Manuscript, 4 Oct. 1994. (Annotated scanned copy)

%F For n >= 7, a(n) = (n^3-16*n+27)/6 (n odd); (n^3-16*n+30)/6 (n even).

%F 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

%F a(2n) = A322594(n-4), n>=4. - _R. J. Mathar_, Mar 18 2019

%t 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 *)

%o (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)

%o (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

%o (Magma) R<x>:=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

%o (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

%K nonn,easy

%O 1,4

%A K. S. Brown (kevin2003(AT)delphi.com), Hugh L. Montgomery

%E More terms from _David W. Wilson_, Oct 27 2000

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Last modified March 29 09:59 EDT 2024. Contains 371268 sequences. (Running on oeis4.)