A171416 -id:A171416 - cp's OEIS frontend

A307733 a(0) = a(1) = 1; a(n) = a(n-1) + a(n-2) + Sum_{k=0..n-1} a(k) * a(n-k-1).

1, 1, 4, 14, 54, 220, 934, 4090, 18344, 83850, 389214, 1829736, 8693962, 41685714, 201442188, 980091814, 4797070022, 23603701828, 116688837886, 579312087802, 2887020896016, 14437318756818, 72424982972862, 364366674463824, 1837954750285458

Offset: 0

Ilya Gutkovskiy, Jul 05 2020

nonn

Cf. A002212, A004148, A006318, A085139, A128720, A143330, A171416, A175934, A245734.

a[0] = a[1] = 1; a[n_] := a[n] = a[n - 1] + a[n - 2] + Sum[a[k] a[n - k - 1], {k, 0, n - 1}]; Table[a[n], {n, 0, 24}]
nmax = 24; CoefficientList[Series[(1 - x - x^2 - Sqrt[1 - 6 x + 3 x^2 + 2 x^3 + x^4])/(2 x), {x, 0, nmax}], x]

G.f. A(x) satisfies: A(x) = (1 - x + x*A(x)^2) / (1 - x - x^2).

G.f.: (1 - x - x^2 - sqrt(1 - 6*x + 3*x^2 + 2*x^3 + x^4)) / (2*x).

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

Original entry on oeis.org

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

Offset: 0

Thomas Scheuerle, Oct 22 2024

nonn

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.

Cf. A006720, A028940.

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

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))

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

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.

cp's OEIS Frontend

A307733 a(0) = a(1) = 1; a(n) = a(n-1) + a(n-2) + Sum_{k=0..n-1} a(k) * a(n-k-1).

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

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PARI

PARI

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cp's OEIS Frontend

A307733 a(0) = a(1) = 1; a(n) = a(n-1) + a(n-2) + Sum_{k=0..n-1} a(k) * a(n-k-1).

Views

Author

Keywords

Crossrefs

Programs

Mathematica

Formula

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)).

Views

Author

Keywords

Comments

Crossrefs

Programs

PARI

PARI

Formula

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