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Quotient Fibonacci(5*n)/(5*Fibonacci(n)), where Fibonacci(n) = A000045(n).
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%I #35 Jan 14 2024 11:49:37

%S 1,11,61,451,3001,20801,141961,974611,6675901,45768251,313671601,

%T 2150012161,14736206161,101003973851,692290189501,4745031073651,

%U 32522917584361,222915417520961,1527884938291801,10472279325329251,71778069881360701,491974211042344811,3372041404278257761

%N Quotient Fibonacci(5*n)/(5*Fibonacci(n)), where Fibonacci(n) = A000045(n).

%C The sequences {Fibonacci(k*n)/(Fibonacci(k)*Fibonacci(n)): n >= 1} are integral in the three cases k = 1 (A000012), k = 2 (A000032) and k = 5 (the present sequence). See Young, Section 4. - _Peter Bala_, Jan 09 2023

%H Chatchawan Panraksa and Aram Tangboonduangjit, <a href="https://www.fq.math.ca/Papers1/55-1/PanraksaTang09282016.pdf">On Some Arithmetic Properties of a Sequence Related to the Quotient of Fibonacci Numbers</a>, Fibonacci Quart. 55 (2017), no. 1, 21-28.

%H Paul Thomas Young, <a href="https://www.fq.math.ca/Scanned/32-1/young.pdf">p-adic congruences for generalized Fibonacci sequences</a>, The Fibonacci Quarterly, Vol. 32, No. 1, 1994.

%F a(n) = 5*Fib(n)^2*(Fib(n)^2 + (-1)^n) + 1 = 5*A007598(n)*A059929(n+1) + 1.

%F a(n) = A103326(n) / 5.

%F G.f.: -x*(x^4-4*x^3-9*x^2+6*x+1) / ((x-1)*(x^2-7*x+1)*(x^2+3*x+1)). - _Colin Barker_, Jul 16 2013

%F The expansion of exp(Sum_{n >= 1} a(n)*x^n/n) = 1 + x + 6*x^2 + 26*x^3 + 151*x^4 + 851*x^5 + 5101*x^6 + ... has integral coefficients and is equal to G(x)^(1/5), where G(x) is the o.g.f. of A001656. See Young, Theorem 3. - _Peter Bala_, Jan 09 2023

%o (PARI) a(n)=fibonacci(5*n)/(5*fibonacci(n)); \\ _Joerg Arndt_, Jul 16 2013

%Y Cf. A000032, A000045, A001656, A103326.

%K nonn,easy

%O 1,2

%A _Lekraj Beedassy_, Nov 17 2003