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

Offset: 0

Author

N. J. A. Sloane, James Sellers, Jan 20 2000

Keywords

• nonn
• easy

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(6*n+3)-2)/4)^2 + 5*((F(6*n+3)-2)/4) + 1 = ((F(6*n+5)-F(6*n+1))/4)^2. Let m = 3*n+1 so that 6*n+1, 6*n+3, and 6*n+5 are 2*m-1, 2*m+1, and 2*m+3 respectively. Define G(m) = F(6*n+3) = F(2*m+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), 5*G(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).

Crossrefs

Cf. A049629.

Related to sum of Fibonacci(kn) over n.. A000071, A027941, A099919, A058038, A138134.

Programs

GAP

List([0..30], n-> (Fibonacci(6*n+3)-2)/4); # G. C. Greubel, May 16 2019

Magma

[(Fibonacci(6*n+3)-2)/4: n in [0..30]]; // Vincenzo Librandi, Apr 20 2011

Maple

A053606 := proc(n) add(combinat[fibonacci](6*k),k=0..n) ;end proc:
seq(A053606(n),n=0..30) ;

Mathematica

Table[(Fibonacci[6n+3] -2)/4, {n,0,30}] (* Vladimir Joseph Stephan Orlovsky, Jul 01 2011 *)

PARI

a(n)=fibonacci(6*n+3)\4 \\ Charles R Greathouse IV, Jul 02 2013

Sage

[(fibonacci(6*n+3)-2)/4 for n in (0..30)] # G. C. Greubel, May 16 2019

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