A053606 - OEIS

A053606

a(n) = (Fibonacci(6*n+3) - 2)/4.

0, 8, 152, 2736, 49104, 881144, 15811496, 283725792, 5091252768, 91358824040, 1639367579960, 29417257615248, 527871269494512, 9472265593285976, 169972909409653064, 3050040103780469184, 54730748958638792256, 982103441151717791432

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OFFSET

0,2

COMMENTS

Define a(1)=0, a(2)=8 with 5*(a(1)^2) + 5*a(1) + 1 = j(1)^2 = 1^2 and 5*(a(2)^2) + 5*a(2) + 1 = j(2)^2 = 19^2. Then a(n) = a(n-2) + 8*sqrt(5*(a(n-1)^2) + 5*a(n-1)+1). Another definition: a(n) such that 5*(a(n)^2) + 5*a(n) + 1 = j(n)^2. - Pierre CAMI, Mar 30 2005

It appears this sequence gives all nonnegative m such that 5*m^2 + 5*m + 1 is a square. - Gerald McGarvey, Apr 03 2005

sqrt(5*a(n)^2+5*a(n)+1) = A049629(n). - Gerald McGarvey, Apr 19 2005

a(n) is such that 5*a(n)^2 + 5*a(n) + 1 = j^2 = the square of A049629(n). Also A049629(n)/a(n) tends to sqrt(5) as n increases. - Pierre CAMI, Apr 21 2005

From Russell Jay Hendel, Apr 25 2015: (Start)

We prove the two McGarvey-CAMI conjectures mentioned at the beginning of the Comments section. Let, as usual, F(n)= A000045(n), the Fibonacci numbers. In the sequel we indicate equations with upper case letters ((A), (B), (C), (D)) for ease of reference.

Then we must prove (A), 5*((F(6n+3) -2)/4)^2 + 5 *((F(6n+3) -2)/4) +1 = ((F(6n+5)-F(6n+1)/4)^2. Let m=3n+1 so that 6n+1, 6n+3, and 6n+5 are 2m-1, 2m+1, and 2m+3 respectively. Define G(m)= F(6n+3)=F(2m+1) = A001519(m+1), the bisected Fibonacci numbers. We can now simplify equation (A) by i) multiplying the LHS and RHS by 16, ii) expanding squares, and iii) gathering like terms. This shows proof of (A) equivalent to proving (B), 5G(m)^2-4 = (G(m+1)-G(m-1))^2.

By Jarden's theorem (D. Jarden, Recurring sequences, 2nd ed. Jerusalem, Riveon Lematematika, (1966)), if {H(n)}_{n >=1} is any recursive sequence satisfying (C), H(n)=3H(n-1)-H(n-2), then {H(n)}^2_{n >=1} is also a recursive sequence satisfying (D), H(n)^2=8H(n-1)^2-8H(n-2)^2+H(n-3)^2. As noted in the Formula section of A001519, {G(m)}_{m >= 1} satisfies (C).

Proof of (B) is now straightforward. Since {G(m)}_{m >=1} satisfies (C), it follows that {G(m)^2}_{m >=1} satisfies (D), and therefore, {5G(m)^2-4}_{m >=1} also satisfies (D).

Similarly, since {G(m)}_{m >=1} satisfies (C), it follows that both {G(m+1)}_{m >=1}, {G(m-1)}_{m >=1} and their difference {G(m+1)-G(m-1)}_{m >=1} satisfy (C), and therefore {G(m+1)-G(m-1)}^2_{m >=1} satisfies (D).

But then the LHS and RHS of (B) are equal for m=1,2,3 and satisfy the same recursion, (D). Hence the LHS and RHS of (B) are equal for all m. This completes the proof. (End)

LINKS

G. C. Greubel, Table of n, a(n) for n = 0..790

F. Ellermann, Illustration of binomial transforms

Index entries for linear recurrences with constant coefficients, signature (19,-19,1).

FORMULA

a(n) = 8*A049664(n).

a(n+1) = 9*a(n) + 2*sqrt(5*(2*a(n)+1)^2-1) + 4. - Richard Choulet, Aug 30 2007

G.f.: 8*x/((1-x)*(1-18*x+x^2)). - Richard Choulet, Oct 09 2007

a(n) = 18*a(n-1) - a(n-2) + 8, n > 1. - Gary Detlefs, Dec 07 2010

a(n) = Sum_{k=0..n} A134492(k). - Gary Detlefs, Dec 07 2010

a(n) = (Fibonacci(6*n+6) - Fibonacci(6*n) - 8)/16. - Gary Detlefs, Dec 08 2010

MAPLE