A377264 Consider the recurrence d(k) = (d(k-3)*d(k-2) + 1)/(d(k-5)*d(k-4)*d(k-3)^2*d(k-2)^2*d(k-1)), with d(0..4) = {1,1,1,2,1}. a(n) = numerator(d(2*n+1)).

1, 2, 3, 14, 69, 413, 7222, 90211, 2577626, 127385577, 5092018073, 655664812074, 78618294607139, 13276948495989478, 5995083279193033837, 1895278734817024984181, 1542333923096758721461086, 1867485777936169465836858947, 2020248742951852823878208914098, 7078136335206254534330825538868049

Offset: 0

Author

Thomas Scheuerle, Oct 22 2024

Keywords

• nonn

Comments

Consider the sequence s(k) with ordinary generating function: 1/(1-d(0)*x/(1-d(1)*x/(1-d(2)*x/(...)))), the Hankel sequence transform of s(k) is A006720 starting with the third term.

This is a special case of a more general theorem: Consider the sequence h(k) with generating function 1/(1-c(0)*x/(1-c(1)*x/(1-c(2)*x/(...)))), if c(2*k+1) = (c(2*k-2)*c(2*k-1) + t)/(c(2*k-4)*c(2*k-3)*c(2*k-2)^2*c(2*k-1)^2*c(2*k)) for all k with c(< 0) = 1, then the Hankel sequence transform of h(k) satisfies a Somos-4 A(1, t) recurrence.

If we would change the start condition into d(0..4) = {1,1,-1,-2,(5/2)}, the expansion of the continued fraction generating function would give us A171416, its Hankel sequence transform is again A006720. There exist infinitely many sequences with the same Hankel sequence transform.

Crossrefs

Cf. A006720, A028940.

The Hankel transform is directly related to A006720: A157002, A157003, A160702, A171416, A173992, A173993, A254314.

Programs

PARI

d(n) = if(n<5, [1,1,1,2,1][n+1], (d(n-3)*d(n-2)+1)/(d(n-5)*d(n-4)*d(n-3)^2*d(n-2)^2*d(n-1)))
a(n) = numerator(d(2*n+1))

PARI

a(n) = -numerator(ellmul(ellinit([0, 3, -1, 2, 0]), [-1,0], 2*n+1)[1])

Formula

a(n) = A006720(n+1)*A006720(n+3).

denominator(d(2*n+1)) = A006720(n+2)^2.

-a(n)/A006720(n+3)^2 are the x-coordinates of (2*n+1) times [-1,0] on the curve y^2 - y = x^3 + 3*x^2 + 2*x. "Times" means here the multiplication under the elliptic group law.