cp's OEIS Frontend

From Colin Barker, Nov 19 2017: (Start)

G.f.: (1 + 4*x + 41*x^2 + 127*x^3 + 41*x^4 - 4*x^5 + x^6 - x^7) / ((1 - 16*x^2 + x^4)*(1 + 16*x^2 + x^4)).

a(n) = 254*a(n-4) - a(n-8) for n > 7.

(End)

The proof of the g.f. runs like the one given for the denominators in A294973. The recurrence for a(n) is the same but the input is now a(0) = b(0) = 1 and a(-1) = 1, (a(-2) = 0). - Wolfdieter Lang, Nov 19 2017

From Colin Barker, Nov 19 2017: (Start)

G.f.: (1 + 3*x - x^2)*(1 + 32*x^2 + x^4) / ((1 - 16*x^2 + x^4)*(1 + 16*x^2 + x^4)).

a(n) = 254*a(n-4) - a(n-8) for n > 7.

(End)

The g.f. is correct: denominator recurrence a(n) = b(n)*a(n-1) + a(n-2), a(-1) = 0, a(0) = 1, (a(-2) = a(0) = 1) with b(n) modulo 4 from the continued fraction given above: b(0) = 1, b(4*(k+1)) = 2, b(4*k+1) = 3 = b(4*k+3) and b(4*k+2) = 10, for k >= 0. The 4-section is G(x) = Sum_{k>=0} a(k)*x^k = G_0(x^4) + x*G_1(x^4) + x^2*G_2(x^4) + x^3*G_3(x^4) with G_j(x) = Sum_{k>=0} a(4*k+j)*x^k, for j=0..3. The recurrence leads to four equations (omit the x here): G_1 = 3*G_0 + x*G_3, G_2 = 10*G_1 + G_4, G_3 = 3*G_2 + G_1, G_0 = 2*x*G_3 +1 + x*G_2 (using a(-2) = 1). This can be solved to obtain for G(x) = (1 + 3*x + 31*x^2 + 96*x^3 - 31*x^4 + 3*x^5 - x^6)/(1 - 254*x^4 + x^8), and the numerator and denominator factorize like given in the above conjecture. - Wolfdieter Lang, Nov 19 2017