A014742 -id:A014742 - cp's OEIS frontend

A069921 Define C(n) by the recursion C(0) = 1 + I where I^2 = -1, C(n+1) = 1/(1+C(n)); then a(n) = (-1)^n/Im(C(n)) where Im(z) is the imaginary part of the complex number z.

1, 5, 10, 29, 73, 194, 505, 1325, 3466, 9077, 23761, 62210, 162865, 426389, 1116298, 2922509, 7651225, 20031170, 52442281, 137295677, 359444746, 941038565, 2463670945, 6449974274, 16886251873, 44208781349, 115740092170, 303011495165, 793294393321

Offset: 0

Benoit Cloitre, May 05 2002

C(n) = (F(n) + F(n-1)*C(0))/(F(n+1) + F(n)*C(0)) = (3*F(n)*F(n+1) + (-1)^n*(1+I))/(3*F(n)*F(n+2) + (-1)^n).
a(n) = F(n+2)^2 + F(n)^2 is the square of the short sides L(n) of the parallelogram appearing in the dissection fallacy of the square F(n+3) X F(n+3), where F(n) is the Fibonacci number A000045(n). For n >= 0, floor(L(n)/h(n)) = A014742(n) (see the proof there), where h(n) is the perpendicular distance between the long sides LL(n) = L(n+1). a(n) is also the first difference of A014742(n). See the link with an illustration. - Kival Ngaokrajang, Jun 27 2014, edited by Wolfdieter Lang, Jul 16 2014
Re(C(n)) = A014742(n)/a(n), n >= 0. For Im(C(n)) see the name. - Wolfdieter Lang, Jul 16 2014

Kival Ngaokrajang, Illustration of initial terms
Eric Weisstein's World of Mathematics, Dissection Fallacy
Wikipedia, Missing square puzzle
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A000032, A000045, A014742, A059929, A069959-A069963, A224415.

[3*Fibonacci(n)*Fibonacci(n+2)+(-1)^n: n in [0..40]]; // Vincenzo Librandi, Sep 24 2015

a[n_] := 3Fibonacci[n]Fibonacci[n+2]+(-1)^n
(*A000045*) F[n_] := (((1 + Sqrt[5])/2)^n - ((1 - Sqrt[5])/2)^n)/Sqrt[5]; (*A000032*) L[n_] := ((1 + Sqrt[5])/2)^n + ((1 - Sqrt[5])/2)^n; Table[FullSimplify[ExpandAll[(F[n]^2 + L[n]^2)/2]], {n, 0, 50}] (* Roger L. Bagula, Nov 17 2008 *)
LinearRecurrence[{2, 2, -1}, {1, 5, 10}, 70] (* Vladimir Joseph Stephan Orlovsky, Feb 08 2012 *)

a(n)=([0,1,0; 0,0,1; -1,2,2]^n*[1;5;10])[1,1] \\ Charles R Greathouse IV, Sep 23 2015

a(n) = 3*F(n)*F(n+2) + (-1)^n = 3*A059929(n) +(-1)^n, where F(n) = A000045(n) is the n-th Fibonacci number.
a(n) = ceiling(3/5*(g/2)^(n+1))-(1+(-1)^n)/2, with g = 3 + sqrt(5).
a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3). - Vladeta Jovovic, May 06 2002
G.f.: (1+3*x-2*x^2)/((1+x)*(1-3*x+x^2)). - Vladeta Jovovic, May 06 2002
a(n) = F(n)^2 + F(n+2)^2. - Ron Knott, Aug 02 2004
a(n-1) = (A000045(n)^2 + A000032(n)^2)/2. - Roger L. Bagula, Nov 17 2008
a(n) = 2*F(n)*F(n+2) + F(n+1)^2 = F(n+1)*F(n+3) + F(n)^2 +(-1)^(n-1). - J. M. Bergot, Sep 15 2012
Equals the logarithmic derivative of A224415. - Paul D. Hanna, Apr 05 2013
2*a(n) = Fibonacci(n+1)^2 + Lucas(n+1)^2. - Bruno Berselli, Sep 26 2017

Edited by Dean Hickerson, May 08 2002

A264080 a(n) = 6F(n)F(n+1) + (-1)^n, where F = A000045.

Original entry on oeis.org

1, 5, 13, 35, 91, 239, 625, 1637, 4285, 11219, 29371, 76895, 201313, 527045, 1379821, 3612419, 9457435, 24759887, 64822225, 169706789, 444298141, 1163187635, 3045264763, 7972606655, 20872555201, 54645058949, 143062621645, 374542805987, 980565796315

Offset: 0

Bruno Berselli, Nov 03 2015

a(n) is prime for n = 1, 2, 5, 7, 14, 15, 29, 40, 49, 57, 70, 87, 105, 127, 175, 279, 362, 647, 727, ...

Bruno Berselli, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A000032, A000045; A001654, A002878.

Cf. similar sequences of the type k*F(n)*F(n+1)+(-1)^n: A226205 (k=1); A236428 (k=2); A014742 (k=3); A061647 (k=4); A002878 (k=5).

[6*Fibonacci(n)*Fibonacci(n+1)+(-1)^n: n in [0..30]];

a:= n-> (<<0|1|0>, <0|0|1>, <-1|2|2>>^n. <<1,5,13>>)[1, 1]:
seq(a(n), n=0..30);  # Alois P. Heinz, Sep 28 2016

Table[6 Fibonacci[n] Fibonacci[n + 1] + (-1)^n, {n, 0, 30}]
LinearRecurrence[{2,2,-1},{1,5,13},30] (* Harvey P. Dale, Jul 12 2019 *)

makelist(6*fib(n)*fib(n+1)+(-1)^n, n, 0, 30);

for(n=0, 30, print1(6*fibonacci(n)*fibonacci(n+1)+(-1)^n", "));

a(n) = round((2^(-n)*(-(-2)^n-3*(3-sqrt(5))^n*(-1+sqrt(5))+3*(1+sqrt(5))*(3+sqrt(5))^n))/5) \\ Colin Barker, Sep 28 2016

Vec((1+3*x+x^2)/((1+x)*(1-3*x+x^2)) + O(x^30)) \\ Colin Barker, Sep 28 2016

[6*fibonacci(n)*fibonacci(n+1)+(-1)^n for n in (0..30)]

G.f.: (1+3*x+x^2) / ((1+x)*(1-3*x+x^2)). - Corrected by Colin Barker, Sep 28 2016
a(n) = -a(-n-1) = 2*a(n-1) + 2*a(n-2) - a(n-3) for all n in Z.
a(n) = L(2*n+1) + F(n)*F(n+1) = A002878(n) + A001654(n). See similar identity for A061647.
a(n) = A001654(n+1) + 3*A001654(n) + A001654(n-1).
a(n) - a(n-1) = 2*A099016(n) with a(-1)=-1.
a(n) + a(n-1) = 2*A097134(n) for n>0.
Sum_{i>=0} 1/a(i) = 1.3232560865206157372628688449331...
a(n) = (2^(-n)*(-(-2)^n-3*(3-sqrt(5))^n*(-1+sqrt(5))+3*(1+sqrt(5))*(3+sqrt(5))^n))/5. - Colin Barker, Sep 28 2016
E.g.f.: (1/5)*exp(-x)*(-1 + 6*exp(5*x/2)*(cosh((sqrt(5)*x)/2) + sqrt(5)*sinh((sqrt(5)*x)/2))). - Stefano Spezia, Dec 09 2019

A245271 a(n) = floor(sqrt(F(n+2)^2 + F(n)^2)), where F(n) = A000045(n).

Original entry on oeis.org

1, 2, 3, 5, 8, 13, 22, 36, 58, 95, 154, 249, 403, 652, 1056, 1709, 2766, 4475, 7241, 11717, 18959, 30676, 49635, 80311, 129947, 210258, 340205, 550464, 890670, 1441135, 2331806, 3772941, 6104748, 9877690, 15982438, 25860128, 41842566, 67702694, 109545261, 177247955

Offset: 0

Kival Ngaokrajang, Jul 15 2014

a(n) is the length of the short side (rounded down) of the parallelogram appearing in the dissection fallacy using the square F(n+3) X F(n+3) (see the links and references). Let the actual length of the short side be L(n) and the one of the long side LL(n), then L(n) = LL(n-1). See the Ngaokrajang link for an illustration. Also floor(LL(n)*L(n)) = A014742(n), n >= 1 (proof by Wolfdieter Lang given there).

Note that F(n+2)^2 + F(n)^2 = 3*F(n+1)^2 - 2*(-1)^n = A069921(n). It appears that for n > 1, a(n) = floor(sqrt(3)*F(n+1)). - Robert Israel, Jul 16 2014

T. Koshy, Fibonacci and Lucas Numbers with Applications, John Wiley & Sons, 2001, ch. 6, pp. 100-108.

Kival Ngaokrajang, Illustration of initial terms.
Eric Weisstein's World of Mathematics, Dissection Fallacy.
Wikipedia, Missing square puzzle.

Cf. A000045, A000196, A014742, A069921.

A245271 := n -> floor(sqrt(3*combinat:-fibonacci(n+1)^2 - 2*(-1)^n)):
seq(A245271(n), n=0..100); # Robert Israel, Jul 16 2014

Table[Floor[Sqrt[Fibonacci[n + 2]^2 + Fibonacci[n]^2]], {n, 0, 50}] (* Wesley Ivan Hurt, Jul 17 2014 *)

a(n) = sqrtint(fibonacci(n+2)^2 + fibonacci(n)^2)
for (n=0,50,print1(a(n),", "))

a(n) = floor(sqrt(F(n+2)^2 + F(n)^2)), n >= 0, with F(n) = A000045(n), and F(n+2)^2 + F(n)^2 = A069921(n).

a(n) = A000196(A069921(n)). - Jason Yuen, Nov 10 2024

A069921 added to Crossrefs and to the Robert Israel comment by Wolfdieter Lang, Jul 17 2014

cp's OEIS Frontend

A069921 Define C(n) by the recursion C(0) = 1 + I where I^2 = -1, C(n+1) = 1/(1+C(n)); then a(n) = (-1)^n/Im(C(n)) where Im(z) is the imaginary part of the complex number z.

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A264080 a(n) = 6*F(n)*F(n+1) + (-1)^n, where F = A000045.

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A245271 a(n) = floor(sqrt(F(n+2)^2 + F(n)^2)), where F(n) = A000045(n).

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A264080 a(n) = 6F(n)F(n+1) + (-1)^n, where F = A000045.