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.

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

Offset: 1

Author

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

Keywords

• nonn
• easy

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, Oct 04 1994. (Annotated scanned copy)

Programs

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

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 *)

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

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

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

Extensions

More terms from David W. Wilson, Oct 27 2000