A069921 -id:A069921 - cp's OEIS frontend

A224415 G.f.: exp( Sum_{n>=1} 5L(n)x^n/n ) where L(n) = Fibonacci(n-1)^2 + Fibonacci(n+1)^2 = A069921(n-1).

1, 5, 25, 100, 380, 1348, 4610, 15250, 49250, 155860, 485228, 1489780, 4520475, 13577775, 40423155, 119413496, 350336200, 1021523000, 2962214500, 8547193700, 24551057380, 70231278200, 200150437000, 568435763000, 1609247086325, 4542394525369, 12786764813645

Offset: 0

Paul D. Hanna, Apr 05 2013

nonn

Given g.f. A(x), note that A(x)^(1/5) does not yield an integer series.

Compare to: exp( Sum_{n>=1} Lucas(n)*x^n/n ) = 1/(1-x-x^2) where Lucas(n) = Fibonacci(n-1) + Fibonacci(n+1).

			G.f.: 1 + 5*x + 25*x^2 + 100*x^3 + 380*x^4 + 1348*x^5 + 4610*x^6 +...
where
log(A(x))/5 = x + 5*x^2/2 + 10*x^3/3 + 29*x^4/4 + 73*x^5/5 + 194*x^6/6 + 505*x^7/7 + 1325*x^8/8 +...+ A069921(n-1)*x^n/n +...

Index entries for linear recurrences with constant coefficients, signature (5,0,-25,5,48,5,-25,0,5,-1).

Cf. A069921.

{L(n)=fibonacci(n-1)^2+fibonacci(n+1)^2}
{a(n)=polcoeff(exp(sum(m=1, n, 5*L(m)*x^m/m)+x*O(x^n)), n)}
for(n=0, 30, print1((a(n)), ", "))

{a(n)=polcoeff(1/((1+x)^4*(1-3*x+x^2)^3+x*O(x^n)),n)}
for(n=0, 30, print1((a(n)), ", "))

G.f.: 1 / ( (1+x)^4 * (1 - 3*x + x^2)^3 ).

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

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

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

A014742 Expansion of (1+x^2)/(1 - 2x - 2x^2 + x^3).

Original entry on oeis.org

1, 2, 7, 17, 46, 119, 313, 818, 2143, 5609, 14686, 38447, 100657, 263522, 689911, 1806209, 4728718, 12379943, 32411113, 84853394, 222149071, 581593817, 1522632382, 3986303327, 10436277601, 27322529474, 71531310823, 187271402993, 490282898158, 1283577291479

Offset: 0

N. J. A. Sloane

Let M = a triangle with (1,1,1,3,3,5,5,7,7,...) as the left border and (0,1,2,3,4,5,...) as all other columns. A014742 = lim_{n->infinity} M^n, the left-shifted vector considered as a sequence. - Gary W. Adamson, Jul 26 2010
For n >= 1, a(n) is the ratio of L/h (rounded down), where L = length of short sides of parallelogram appearing in dissection fallacy of square F(n+2) X F(n+2), F(n) is Fibonacci number, and h = perpendicular distance between the long sides LL. The first differences of A069921 give L^2. See illustration. - Kival Ngaokrajang, Jun 27 2014
From Wolfdieter Lang, Jul 15 2014: (Start)
The preceding comment is a conjecture using a(n) = floor(LL(n)*L(n)) with LL(n) = sqrt(F(n+2)^2 + F(n)^2) and L(n) = LL(n-1), n >= 1 (its author agreed with this in an email). See also, e.g., the Koshy reference for the dissection fallacy, sect. 6, 100 - 108.
The proof of the conjecture uses first the identity (LL(n)*LL(n-1))^2 - a(n)^2 = + 1 with a(n) = F(n-1)*F(n) + F(n+1)*F(n+2) (see the formula section for a(n)). This identity is due to the factorization of the left-hand side which is A(n)^2 with A(n) = F(n)*F(n+1) - F(n-1)*F(n+2). But A(n) = (-1)^(n+1) is a special instance of a well known Fibonacci identity (Koshy, p. 88, Nr. 19 for h=-1, k=2, F(-1) = 1). Now one has (LL(n)*LL(n-1))^2 = 1 + a(n)^2, that is LL(n)*LL(n-1) = sqrt(1 + a(n)^2). Because a(n) < sqrt(1 + a(n)^2) <  a(n) + 1 (just square both inequalities using a(n) > 0) one has now proved that floor(LL(n)*LL(n-1)) = a(n), n >= 1. (End)
a(n) = numerator(Re(C(n))), with the complex sequence C(n) defined in the name of A069921. - Wolfdieter Lang, Jul 16 2014

			a(2) = F(1)*F(2) + F(3)*F(4) = 1*1 + 2*3 = 7. - _James R. Buddenhagen_, Jan 06 2009

T. Koshy, Fibonacci and Lucas Numbers with Applications, John Wiley & Sons, 2001.

Colin Barker, Table of n, a(n) for n = 0..1000
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. A069921. - Kival Ngaokrajang, Jun 27 2014
Cf. similar sequences of the type k*F(n)*F(n+1) + (-1)^n listed in A264080.
Cf. A000032, A000045.

seq(combinat[fibonacci](n-1)*combinat[fibonacci](n)+combinat[fibonacci](n+1)*combinat[fibonacci](n+2), n=0..50); # will give first 50 terms - James R. Buddenhagen, Jan 06 2009

CoefficientList[Series[(1 + x^2)/(1 - 2*x - 2*x^2 + x^3), {x, 0, 30}], x] (* Wesley Ivan Hurt, Jun 27 2014 *)
LinearRecurrence[{2, 2, -1},{1, 2, 7},30] (* Ray Chandler, Sep 23 2015 *)

Vec((1+x^2)/(1-2*x-2*x^2+x^3)+O(x^99)) \\ Charles R Greathouse IV, Sep 26 2012

a(n) = round((2^(-1-n)*((-1)^n*2^(2+n)-3*(3-sqrt(5))^n*(-1+sqrt(5))+3*(1+sqrt(5))*(3+sqrt(5))^n))/5) \\ Colin Barker, Sep 29 2016

a(n) = F(n-1)*F(n) + F(n+1)*F(n+2), where F = A000045. - James R. Buddenhagen, Jan 06 2009
From Wolfdieter Lang, Jul 15 2014: (Start)
G.f.: (1+x^2)/(1 - 2*x - 2*x^2 + x^3) = (1+x^2)/((1+x)*(1 - 3*x + x^2)) = (2/(1+x) + 3*(1+x)/(1 - 3*x + x^2))/5 (see the name).
a(n) = (2*(-1)^n + 3*(F(2*n) + F(2*(n+1))))/5,
a(n) = (2*(-1)^n + L(2*n-1) + L(2*n+3))/5 with L(n) = A000032(n) and L(-1) = -1. (End)
a(n) = 3*F(n)*F(n+1) + (-1)^n. - Bruno Berselli, Oct 30 2015
a(n) = (2^(-1-n)*((-1)^n*2^(2+n) - 3*(3-sqrt(5))^n*(-1+sqrt(5)) + 3*(1+sqrt(5))*(3+sqrt(5))^n))/5. - Colin Barker, Sep 29 2016
a(n) = Fibonacci(n + 2)^2 - 2*Fibonacci(n)^2. - Detlef Meya, Jun 29 2024
E.g.f.: exp(-x)*(2 + 3*exp(5*x/2)*(cosh(sqrt(5)*x/2) + sqrt(5)*sinh(sqrt(5)*x/2)))/5. - Stefano Spezia, Jun 30 2024

Buddenhagen's Jan 06 2009 entries adjusted for offset 0 by Wolfdieter Lang, Jul 15 2014

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

A380696 a(n) = A007598(floor(n/2) - (-1)^n).

Original entry on oeis.org

1, 1, 0, 1, 1, 4, 1, 9, 4, 25, 9, 64, 25, 169, 64, 441, 169, 1156, 441, 3025, 1156, 7921, 3025, 20736, 7921, 54289, 20736, 142129, 54289, 372100, 142129, 974169, 372100, 2550409, 974169, 6677056, 2550409, 17480761, 6677056, 45765225, 17480761, 119814916

Offset: 0

Benjamin G. Brunsden, Jan 30 2025

The Fibonacci spiral is produced by creating a quarter circle of radius 1, then adding successive quarter circles such that the radius of the new quarter circle is the sum of the radii of the previous two quarter circles, and that the circumference of the new quarter circle continues where the previous quarter circle ended. When the center of the first quarter circle is at 0,0 the circumference turns clockwise from -1,0, and terms after n=1 are given signs - + + - repeating, these are the x coordinates where the circumferences meet. The y coordinates are the golden rectangle numbers (A001654) with the same pattern of alternation (x,a,b,x), and the same pattern of signs shifted backward one.

Paolo Xausa, Table of n, a(n) for n = 0..1000
Benjamin G. Brunsden, Sequence shown as coordinates on the Fibonacci spiral
Index entries for linear recurrences with constant coefficients, signature (0,2,0,2,0,-1).

Cf. A000045, A007598, A053602, A069921, A123231, A272912, A001654.

A380696[n_] := Fibonacci[Floor[n/2] - (-1)^n]^2; Array[A380696, 50, 0] (* or *)
LinearRecurrence[{0, 2, 0, 2, 0, -1}, {1, 1, 0, 1, 1, 4}, 50] (* Paolo Xausa, Mar 27 2025 *)

from sympy import fibonacci
def A380696(n): return fibonacci(n+1>>1 if n&1 else (n>>1)-1)**2 # Chai Wah Wu, Mar 26 2025

a(n) = Fibonacci(floor(n/2)-(-1)^n)^2.
a(n) = A053602(n-2)^2 for n >= 2.
a(n) = A272912(n)^2 for n >= 3.
G.f. ( 1+x-x^3-x^4-2*x^2 ) / ( (1+x^2)*(x^2-x-1)*(x^2+x-1) ).
a(2*n) + a(2*n+1) = A069921(n-1) for n>=1.

cp's OEIS Frontend

A224415 G.f.: exp( Sum_{n>=1} 5*L(n)*x^n/n ) where L(n) = Fibonacci(n-1)^2 + Fibonacci(n+1)^2 = A069921(n-1).

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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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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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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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A224415 G.f.: exp( Sum_{n>=1} 5L(n)x^n/n ) where L(n) = Fibonacci(n-1)^2 + Fibonacci(n+1)^2 = A069921(n-1).

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.

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.

A014742 Expansion of (1+x^2)/(1 - 2x - 2x^2 + x^3).