A069960 -id:A069960 - cp's OEIS frontend

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.

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

A069963 Define C(n) by the recursion C(0) = 6i where i^2 = -1, C(n+1) = 1/(1 + C(n)), then a(n) = 6(-1)^n/Im(C(n)) where Im(z) denotes the imaginary part of z.

Original entry on oeis.org

1, 37, 40, 153, 349, 964, 2473, 6525, 17032, 44641, 116821, 305892, 800785, 2096533, 5488744, 14369769, 37620493, 98491780, 257854777, 675072621, 1767363016, 4627016497, 12113686405, 31714042788, 83028441889, 217371282949, 569085406888, 1489884937785

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) = 6*i in C(n) = (F(n) + C(0)*F(n-1))/(F(n+1) + C(0)*F(n)) = (F(n)*(F(n+1) + 36*F(n-1)) + (-1)^n*6*i)/(F(n+1)^2 + 36*F(n)^2), where F(n) = Fibonacci(n), for which Im(C(n)) = 6*(-1)^n/(F(n+1)^2 + 36*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, A069962.

F:=Fibonacci; [F(n+1)^2 + 36*F(n)^2: n in [0..40]]; // G. C. Greubel, Aug 18 2022

a[n_]:= 36*Fibonacci[n]^2 +Fibonacci[n+1]^2; Table[a[n], {n,0,30}]

a(n)=36*fibonacci(n)^2+fibonacci(n+1)^2 \\ Charles R Greathouse IV, Jun 14 2013

a(n) = round((2^(-1-n)*(-35*(-1)^n*2^(2+n)-(3-sqrt(5))^n*(-75+sqrt(5))+(3+sqrt(5))^n*(75+sqrt(5))))/5) \\ Colin Barker, Sep 28 2016

Vec(-(x-1)*(36*x+1)/((x+1)*(x^2-3*x+1)) + O(x^30)) \\ Colin Barker, Sep 28 2016

f=fibonacci; [f(n+1)^2 +36*f(n)^2 for n in (0..40)] # G. C. Greubel, Aug 18 2022

a(n) = 36*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+36*x) / ((1+x)*(1-3*x+x^2)). (End)
a(n) = (2^(-1-n)*(-35*(-1)^n*2^(2+n) - (3-sqrt(5))^n*(-75+sqrt(5)) + (3+sqrt(5))^n*(75+sqrt(5))))/5. - Colin Barker, Sep 28 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

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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A069963 Define C(n) by the recursion C(0) = 6*i where i^2 = -1, C(n+1) = 1/(1 + C(n)), then a(n) = 6*(-1)^n/Im(C(n)) where Im(z) denotes the imaginary part of 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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Magma

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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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Author

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Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

PARI

SageMath

Formula

Extensions

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.

A069963 Define C(n) by the recursion C(0) = 6i where i^2 = -1, C(n+1) = 1/(1 + C(n)), then a(n) = 6(-1)^n/Im(C(n)) where Im(z) denotes the imaginary part of 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.