0,1

a(n) is the trace of the n-th power of the adjacency matrix of order 8 whose rows are (up to simultaneous permutations of the rows and columns): 10111010 01111001 01111001 10111010 00111000 11111111 11111111 11111111.

For n>2, this is also the number of ways to mark one edge at every vertex of a regular n-gonal prism so that no edge is marked at both extremities.

Remarkably, for n>1, a(n)=A141221(n)+2. The fourth-order linear recurrence established by Max Alekseyev for A141221, based on the characteristic polynomial of the above (singular) matrix, namely x^4(x-1)(x^3-7x^2+9x-1) = x^4(x^4-8x^3+16x^2-10x+1). Because of the x^4 factor, the recurrence is guaranteed a priori to hold for corresponding elements of the successive powers (or sums thereof, including matrix traces) only if n is 4 or more. It happens to have a greater validity downward the case of this sequence (and A141221 as well). The recurrence would be valid down to n=0 if we had a(0)=4, which is not the case.

G. P. Michon, A screaming game for short-sighted people.

G. P. Michon, Silent circles, enumerated by Max Alekseyev.

G. P. Michon, Brocoum's Screaming Circles.

For positive values of n: a(n) = (5.3538557854308)^n + (1.5235479602692)^n + 1 + (0.1225962542999)^n. The dominant term in the above is the n-th power of (7+2*sqrt(22)*cos(atan(sqrt(5319)/73)/3))/3. For n>0 we have: a(n+4) = 8*a(n+3)-16*a(n+2)+10*a(n+1)-a(n)

a(0) = 8 because the trace of the order-8 identity matrix is 8.

a(1) = 8 because all diagonal elements of the adjacency matrix are 1 (there's a loop at each vertex).

Cf. A141221.

Adjacent sequences: A141381 A141382 A141383 this_sequence A141385 A141386 A141387

Sequence in context: A143336 A122858 A053596 this_sequence A111218 A103744 A038286

easy,nonn

Gerard P. Michon (g.michon(AT)att.net), Jun 29 2008