|
%I #62 Jan 01 2024 02:04:40
%S 0,8,152,2736,49104,881144,15811496,283725792,5091252768,91358824040,
%T 1639367579960,29417257615248,527871269494512,9472265593285976,
%U 169972909409653064,3050040103780469184,54730748958638792256,982103441151717791432
%N a(n) = (Fibonacci(6*n+3) - 2)/4.
%C 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
%C It appears this sequence gives all nonnegative m such that 5*m^2 + 5*m + 1 is a square. - _Gerald McGarvey_, Apr 03 2005
%C sqrt(5*a(n)^2+5*a(n)+1) = A049629(n). - _Gerald McGarvey_, Apr 19 2005
%C 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
%C From _Russell Jay Hendel_, Apr 25 2015: (Start)
%C 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.
%C 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.
%C 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).
%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).
%C 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).
%C 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)
%H G. C. Greubel, <a href="/A053606/b053606.txt">Table of n, a(n) for n = 0..790</a>
%H F. Ellermann, <a href="/A001792/a001792.txt">Illustration of binomial transforms</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (19,-19,1).
%F a(n) = 8*A049664(n).
%F a(n+1) = 9*a(n) + 2*sqrt(5*(2*a(n)+1)^2-1) + 4. - _Richard Choulet_, Aug 30 2007
%F G.f.: 8*x/((1-x)*(1-18*x+x^2)). - _Richard Choulet_, Oct 09 2007
%F a(n) = 18*a(n-1) - a(n-2) + 8, n > 1. - _Gary Detlefs_, Dec 07 2010
%F a(n) = Sum_{k=0..n} A134492(k). - _Gary Detlefs_, Dec 07 2010
%F a(n) = (Fibonacci(6*n+6) - Fibonacci(6*n) - 8)/16. - _Gary Detlefs_, Dec 08 2010
%p A053606 := proc(n) add(combinat[fibonacci](6*k),k=0..n) ;end proc:
%p seq(A053606(n),n=0..30) ;
%t Table[(Fibonacci[6n+3] -2)/4, {n,0,30}] (* _Vladimir Joseph Stephan Orlovsky_, Jul 01 2011 *)
%o (Magma) [(Fibonacci(6*n+3)-2)/4: n in [0..30]]; // _Vincenzo Librandi_, Apr 20 2011
%o (PARI) a(n)=fibonacci(6*n+3)\4 \\ _Charles R Greathouse IV_, Jul 02 2013
%o (Sage) [(fibonacci(6*n+3)-2)/4 for n in (0..30)] # _G. C. Greubel_, May 16 2019
%o (GAP) List([0..30], n-> (Fibonacci(6*n+3)-2)/4) # _G. C. Greubel_, May 16 2019
%Y Cf. A049629.
%Y Related to sum of Fibonacci(kn) over n.. A000071, A027941, A099919, A058038, A138134.
%K nonn,easy
%O 0,2
%A _N. J. A. Sloane_, _James A. Sellers_, Jan 20 2000
| 