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A053606 a(n) = (Fibonacci(6*n+3) - 2)/4.

%I A053606 #70 Jul 03 2025 10:54:27
%S A053606 0,8,152,2736,49104,881144,15811496,283725792,5091252768,91358824040,
%T A053606 1639367579960,29417257615248,527871269494512,9472265593285976,
%U A053606 169972909409653064,3050040103780469184,54730748958638792256,982103441151717791432
%N A053606 a(n) = (Fibonacci(6*n+3) - 2)/4.
%C A053606 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 A053606 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 A053606 sqrt(5*a(n)^2+5*a(n)+1) = A049629(n). - _Gerald McGarvey_, Apr 19 2005
%C A053606 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 A053606 From _Russell Jay Hendel_, Apr 25 2015: (Start)
%C A053606 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 A053606 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.
%C A053606 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 A053606 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 A053606 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 A053606 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 A053606 G. C. Greubel, <a href="/A053606/b053606.txt">Table of n, a(n) for n = 0..790</a>
%H A053606 F. Ellermann, <a href="/A001792/a001792.txt">Illustration of binomial transforms</a>
%H A053606 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (19,-19,1).
%F A053606 a(n) = 8*A049664(n).
%F A053606 a(n+1) = 9*a(n) + 2*sqrt(5*(2*a(n)+1)^2-1) + 4. - _Richard Choulet_, Aug 30 2007
%F A053606 G.f.: 8*x/((1-x)*(1-18*x+x^2)). - _Richard Choulet_, Oct 09 2007
%F A053606 a(n) = 18*a(n-1) - a(n-2) + 8, n > 1. - _Gary Detlefs_, Dec 07 2010
%F A053606 a(n) = Sum_{k=0..n} A134492(k). - _Gary Detlefs_, Dec 07 2010
%F A053606 a(n) = (Fibonacci(6*n+6) - Fibonacci(6*n) - 8)/16. - _Gary Detlefs_, Dec 08 2010
%p A053606 A053606 := proc(n) add(combinat[fibonacci](6*k),k=0..n) ;end proc:
%p A053606 seq(A053606(n),n=0..30) ;
%t A053606 Table[(Fibonacci[6n+3] -2)/4, {n,0,30}] (* _Vladimir Joseph Stephan Orlovsky_, Jul 01 2011 *)
%o A053606 (Magma) [(Fibonacci(6*n+3)-2)/4: n in [0..30]]; // _Vincenzo Librandi_, Apr 20 2011
%o A053606 (PARI) a(n)=fibonacci(6*n+3)\4 \\ _Charles R Greathouse IV_, Jul 02 2013
%o A053606 (Sage) [(fibonacci(6*n+3)-2)/4 for n in (0..30)] # _G. C. Greubel_, May 16 2019
%o A053606 (GAP) List([0..30], n-> (Fibonacci(6*n+3)-2)/4); # _G. C. Greubel_, May 16 2019
%Y A053606 Cf. A049629.
%Y A053606 Related to sum of Fibonacci(kn) over n.. A000071, A027941, A099919, A058038, A138134.
%K A053606 nonn,easy
%O A053606 0,2
%A A053606 _N. J. A. Sloane_, _James Sellers_, Jan 20 2000