A069963 -id:A069963 - 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

A069959 Define C(n) by the recursion C(0) = 2i where i^2 = -1, C(n+1) = 1/(1 + C(n)), then a(n) = 2(-1)^n/Im(C(n)) where Im(z) denotes the imaginary part of the complex number z.

Original entry on oeis.org

1, 5, 8, 25, 61, 164, 425, 1117, 2920, 7649, 20021, 52420, 137233, 359285, 940616, 2462569, 6447085, 16878692, 44188985, 115688269, 302875816, 792939185, 2075941733, 5434886020, 14228716321, 37251262949, 97525072520, 255323954617, 668446791325, 1750016419364

Offset: 0

Benoit Cloitre, Apr 28 2002

If we define C(n) with C(0) = i in the recurrence C(n+1) = 1/(1 + C(n)) then Im(C(n)) = 1/Fibonacci(2*n+1).

Here, C(n) is defined with C(0) = 2*i in C(n) = (F(n) + C(0)*F(n-1))/(F(n+1) + C(0)*F(n)) = (F(n)*(F(n+1) + 4*F(n-1)) + (-1)^n*2*i)/(F(n+1)^2 + 4*F(n)^2), where F(n) = Fibonacci(n), for which Im(C(n)) = 2*(-1)^n/(F(n+1)^2 + 4*F(n)^2).

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A000045, A069921, A069960, A069961, A069962, A069963.

F:=Fibonacci; [F(n+1)^2 +4*F(n)^2: n in [0..40]]; // G. C. Greubel, Aug 17 2022

a[n_]:= 4Fibonacci[n]^2+Fibonacci[n+1]^2;
4#[[1]]^2+#[[2]]^2&/@Partition[Fibonacci[Range[0,30]],2,1] (* or *) LinearRecurrence[{2,2,-1},{1,5,8},30] (* Harvey P. Dale, Aug 25 2017 *)

a(n) = round((2^(-1-n)*(-3*(-1)^n*2^(2+n)-(3-sqrt(5))^n*(-11+sqrt(5))+(3+sqrt(5))^n*(11+sqrt(5))))/5) \\ Colin Barker, Sep 28 2016

Vec(-(x-1)*(4*x+1)/((x+1)*(x^2-3*x+1)) + O(x^40)) \\ Colin Barker, Sep 28 2016

f=fibonacci; [f(n+1)^2+4*f(n)^2 for n in (0..40)] # G. C. Greubel, Aug 17 2022

a(n) = 4*F(n)^2 + F(n+1)^2, where F(n) = A000045(n) is the n-th Fibonacci number.
From Colin Barker, Jun 14 2013: (Start)
a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
G.f.: (1-x)*(1+4*x) / ((1+x)*(1-3*x+x^2)). (End)
a(n) = (2^(-1-n)*(-3*(-1)^n*2^(2+n) - (3-sqrt(5))^n*(-11+sqrt(5)) + (3+sqrt(5))^n*(11+sqrt(5))))/5. - Colin Barker, Sep 28 2016

Edited by Dean Hickerson, May 08 2002

A069960 Define C(n) by the recursion C(0) = 3i where i^2 = -1, C(n+1) = 1/(1 + C(n)), then a(n) = 3(-1)^n/Im(C(n)) where Im(z) denotes the imaginary part of the complex number z.

Original entry on oeis.org

1, 10, 13, 45, 106, 289, 745, 1962, 5125, 13429, 35146, 92025, 240913, 630730, 1651261, 4323069, 11317930, 29630737, 77574265, 203092074, 531701941, 1392013765, 3644339338, 9541004265, 24978673441, 65395016074, 171206374765, 448224108237, 1173465949930

Offset: 0

Benoit Cloitre, Apr 28 2002

If we define C(n) with C(0) = i then Im(C(n)) = 1/F(2*n+1) where F(k) are the Fibonacci numbers.

Here, C(n) = (F(n) + C(0)*F(n-1))/(F(n+1) + C(0)*F(n)) = (F(n)*(F(n+1) + 9*F(n-1)) + 3*i*(-1)^n)/(F(n+1)^2 + 9*F(n)^2), where F(n) = Fibonacci(n), for which Im(C(n)) = 3*(-1)^n/(F(n+1)^2 + 9*F(n)^2).

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A000045, A069921, A069959, A069961, A069962, A069963.

F:=Fibonacci; [F(n+1)^2 +9*F(n)^2: n in [0..40]]; // G. C. Greubel, Aug 17 2022

a[n_] := 9Fibonacci[n]^2+Fibonacci[n+1]^2
9*First[#]+Last[#]&/@(Partition[Fibonacci[Range[0,30]],2,1]^2) (* Harvey P. Dale, Mar 06 2012 *)

a(n) = round((2^(-1-n)*(-(-1)^n*2^(5+n)-(3-sqrt(5))^n*(-21+sqrt(5))+(3+sqrt(5))^n*(21+sqrt(5))))/5) \\ Colin Barker, Sep 30 2016

Vec(-(x-1)*(9*x+1)/((x+1)*(x^2-3*x+1)) + O(x^30)) \\ Colin Barker, Sep 30 2016

f=fibonacci; [f(n+1)^2 +9*f(n)^2 for n in (0..40)] # G. C. Greubel, Aug 17 2022

a(n) = 9*F(n)^2 + F(n+1)^2, where F(n) = A000045(n) is the n-th Fibonacci number.
From Colin Barker, Jun 14 2013: (Start)
a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
G.f.: (1-x)*(1+9*x) / ((1+x)*(1-3*x+x^2)). (End)
a(n) = (2^(-1-n)*(-(-1)^n*2^(5+n) - (3-sqrt(5))^n*(-21+sqrt(5)) + (3+sqrt(5))^n*(21+sqrt(5))))/5. - Colin Barker, Sep 30 2016

Edited by Dean Hickerson, May 08 2002

A069961 Define C(n) by the recursion C(0) = 4i where i^2 = -1, C(n+1) = 1/(1 + C(n)), then a(n) = 4(-1)^n/Im(C(n)) where Im(z) denotes the imaginary part of z.

Original entry on oeis.org

1, 17, 20, 73, 169, 464, 1193, 3145, 8212, 21521, 56321, 147472, 386065, 1010753, 2646164, 6927769, 18137113, 47483600, 124313657, 325457401, 852058516, 2230718177, 5840095985, 15289569808, 40028613409, 104796270449, 274360197908, 718284323305, 1880492771977

Offset: 0

Benoit Cloitre, Apr 28 2002

If we define C(n) with C(0) = i then Im(C(n)) = 1/F(2*n+1) where F(k) are the Fibonacci numbers.

Here, C(n) = (F(n) + C(0)*F(n-1))/(F(n+1) + C(0)*F(n)) = (F(n)*(F(n+1) + 16*F(n-1)) + (-1)^n*4*i)/(F(n+1)^2 + 16*F(n)^2), where F(n) = Fibonacci(n), for which Im(C(n)) = 4*(-1)^n/(F(n+1)^2 + 16*F(n)^2).

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A000045, A069921, A069959, A069960, A069962, A069963.

F:=Fibonacci; [F(n+1)^2 + 16*F(n)^2: n in [0..40]]; // G. C. Greubel, Aug 17 2022

a[n_]:= 16Fibonacci[n]^2+Fibonacci[n+1]^2; Array[a,30,0]
16First[#]^2+Last[#]^2&/@Partition[Fibonacci[Range[0,30]],2,1] (* Harvey P. Dale, Nov 08 2011 *)

a(n) = round((2^(-1-n)*(-15*(-1)^n*2^(2+n)-(3-sqrt(5))^n*(-35+sqrt(5))+(3+sqrt(5))^n*(35+sqrt(5))))/5) \\ Colin Barker, Oct 01 2016

Vec(-(x-1)*(16*x+1)/((x+1)*(x^2-3*x+1)) + O(x^30)) \\ Colin Barker, Oct 01 2016

f=fibonacci; [f(n+1)^2 +16*f(n)^2 for n in (0..40)] # G. C. Greubel, Aug 17 2022

a(n) = 16*F(n)^2 + F(n+1)^2, where F(n) = A000045(n) is the n-th Fibonacci number.
From Colin Barker, Jun 14 2013: (Start)
a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
G.f.: (1-x)*(1+16*x) / ((1+x)*(1-3*x+x^2)). (End)
a(n) = (2^(-1-n)*(-15*(-1)^n*2^(2+n) - (3-sqrt(5))^n*(-35+sqrt(5)) + (3+sqrt(5))^n*(35+sqrt(5))))/5. - Colin Barker, Oct 01 2016

Edited by Dean Hickerson, May 08 2002

A069962 Define C(n) by the recursion C(0) = 5i where i^2 = -1, C(n+1) = 1/(1 + C(n)), then a(n) = 5(-1)^n/Im(C(n)) where Im(z) denotes the imaginary part of z.

Original entry on oeis.org

1, 26, 29, 109, 250, 689, 1769, 4666, 12181, 31925, 83546, 218761, 572689, 1499354, 3925325, 10276669, 26904634, 70437281, 184407161, 482784250, 1263945541, 3309052421, 8663211674, 22680582649, 59378536225, 155455026074, 406986541949, 1065504599821

Offset: 0

Benoit Cloitre, Apr 28 2002

If we define C(n) with C(0) = i then Im(C(n)) = 1/F(2*n+1) where F(k) are the Fibonacci numbers.

Here, C(n) is defined with C(0) = 5*i in C(n) = (F(n) + C(0)*F(n-1))/(F(n+1) + C(0)*F(n)) = (F(n)*(F(n+1) + 25*F(n-1)) + (-1)^n*2*i)/(F(n+1)^2 + 25*F(n)^2), where F(n) = Fibonacci(n), for which Im(C(n)) = 2*(-1)^n/(F(n+1)^2 + 25*F(n)^2).

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,2,-1).

Cf. A000045, A069921, A069959, A069960, A069961, A069963.

F:=Fibonacci; [F(n+1)^2 + 25*F(n)^2: n in [0..40]]; // G. C. Greubel, Aug 17 2022

a[n_]:= 25*Fibonacci[n]^2+Fibonacci[n+1]^2; Table[a[n], {n,0,30}]
LinearRecurrence[{2,2,-1},{1,26,29},30] (* Harvey P. Dale, Mar 18 2018 *)

a(n) = round((2^(-1-n)*(-3*(-1)^n*2^(5+n)-(3-sqrt(5))^n*(-53+sqrt(5))+(3+sqrt(5))^n*(53+sqrt(5))))/5) \\ Colin Barker, Oct 01 2016

Vec(-(x-1)*(25*x+1)/((x+1)*(x^2-3*x+1)) + O(x^30)) \\ Colin Barker, Oct 01 2016

f=fibonacci; [f(n+1)^2 +25*f(n)^2 for n in (0..40)] # G. C. Greubel, Aug 17 2022

a(n) = 25*F(n)^2 + F(n+1)^2, where F(n) = A000045(n).
From Colin Barker, Jun 14 2013: (Start)
a(n) = 2*a(n-1) + 2*a(n-2) - a(n-3).
G.f.: (1-x)*(1+25*x) / ((1+x)*(1-3*x+x^2)). (End)
a(n) = (2^(-1-n)*(-3*(-1)^n*2^(5+n) - (3-sqrt(5))^n*(-53+sqrt(5)) + (3+sqrt(5))^n*(53+sqrt(5))))/5. - Colin Barker, Oct 01 2016

Edited by Dean Hickerson, May 08 2002

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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A069959 Define C(n) by the recursion C(0) = 2*i where i^2 = -1, C(n+1) = 1/(1 + C(n)), then a(n) = 2*(-1)^n/Im(C(n)) where Im(z) denotes the imaginary part of the complex number z.

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A069960 Define C(n) by the recursion C(0) = 3*i where i^2 = -1, C(n+1) = 1/(1 + C(n)), then a(n) = 3*(-1)^n/Im(C(n)) where Im(z) denotes the imaginary part of the complex number z.

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A069961 Define C(n) by the recursion C(0) = 4*i where i^2 = -1, C(n+1) = 1/(1 + C(n)), then a(n) = 4*(-1)^n/Im(C(n)) where Im(z) denotes the imaginary part of z.

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A069962 Define C(n) by the recursion C(0) = 5*i where i^2 = -1, C(n+1) = 1/(1 + C(n)), then a(n) = 5*(-1)^n/Im(C(n)) where Im(z) denotes the imaginary part of z.

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A069959 Define C(n) by the recursion C(0) = 2i where i^2 = -1, C(n+1) = 1/(1 + C(n)), then a(n) = 2(-1)^n/Im(C(n)) where Im(z) denotes the imaginary part of the complex number z.

A069960 Define C(n) by the recursion C(0) = 3i where i^2 = -1, C(n+1) = 1/(1 + C(n)), then a(n) = 3(-1)^n/Im(C(n)) where Im(z) denotes the imaginary part of the complex number z.

A069961 Define C(n) by the recursion C(0) = 4i where i^2 = -1, C(n+1) = 1/(1 + C(n)), then a(n) = 4(-1)^n/Im(C(n)) where Im(z) denotes the imaginary part of z.

A069962 Define C(n) by the recursion C(0) = 5i where i^2 = -1, C(n+1) = 1/(1 + C(n)), then a(n) = 5(-1)^n/Im(C(n)) where Im(z) denotes the imaginary part of z.