A001946 -id:A001946 - cp's OEIS frontend

A041226 Numerators of continued fraction convergents to sqrt(125).

11, 56, 67, 123, 682, 15127, 76317, 91444, 167761, 930249, 20633239, 104096444, 124729683, 228826127, 1268860318, 28143753123, 141987625933, 170131379056, 312119004989, 1730726404001, 38388099893011, 193671225869056, 232059325762067, 425730551631123

Offset: 0

N. J. A. Sloane

From Johannes W. Meijer, Jun 12 2010: (Start)
The a(n) terms of this sequence can be constructed with the terms of sequence A001946.
For the terms of the periodical sequence of the continued fraction for sqrt(125) see A010186. We observe that its period is five. (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,1364,0,0,0,0,1).

Cf. A041227, A041018, A041046, A041090, A041150, A041226, A041318, A041426, A041550.

Numerator[Convergents[Sqrt[125], 30]] (* Vincenzo Librandi, Oct 31 2013 *)

From Johannes W. Meijer, Jun 12 2010: (Start)
a(5n)   = A001946(3n+1),
a(5n+1) = (A001946(3n+2) - A001946(3n+1))/2,
a(5n+2) = (A001946(3n+2) + A001946(3n+1))/2,
a(5n+3) = A001946(3n+2),
a(5n+4) = A001946(3n+3)/2. (End)
G.f.: -(x^9 -11*x^8 +56*x^7 -67*x^6 +123*x^5 +682*x^4 +123*x^3 +67*x^2 +56*x +11) / ((x^2 +4*x -1)*(x^4 -7*x^3 +19*x^2 -3*x +1)*(x^4 +3*x^3 +19*x^2 +7*x +1)). - Colin Barker, Nov 08 2013

More terms from Colin Barker, Nov 08 2013

A065705 a(n) = Lucas(10*n).

Original entry on oeis.org

2, 123, 15127, 1860498, 228826127, 28143753123, 3461452808002, 425730551631123, 52361396397820127, 6440026026380244498, 792070839848372253127, 97418273275323406890123, 11981655542024930675232002, 1473646213395791149646646123, 181246502592140286475862241127

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Oct 25 2003

Lim_{n->infinity} a(n+1)/a(n) = (123 + sqrt(15125))/2 = 122.9918693812...
Lim_{n->infinity} a(n)/a(n+1) = (123 - sqrt(15125))/2 = 0.00813061875578...
From Peter Bala, Oct 14 2019: (Start)
Let F(x) = Product_{n >= 0} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let Phi = 1/2*(sqrt(5) - 1). This sequence gives the partial denominators in the simple continued fraction expansion of the number F(Phi^10) = 1.0081300769... = 1 + 1/(123 + 1/(15127 + 1/(1860498 + ...))).
Also F(-Phi^10) = 0.9918699143... has the continued fraction representation 1 - 1/(123 - 1/(15127 - 1/(1860498 - ...))) and the simple continued fraction expansion 1/(1 + 1/((123 - 2) + 1/(1 + 1/((15127 - 2) + 1/(1 + 1/((1860498 - 2) + 1/(1 + ...))))))).
F(Phi^10)*F(-Phi^10) = 0.9999338930... has the simple continued fraction expansion 1/(1 + 1/((123^2 - 4) + 1/(1 + 1/((15127^2 - 4) + 1/(1 + 1/((1860498^2 - 4) + 1/(1 + ...))))))).
1/2 + (1/2)*F(Phi^10)/F(-Phi^10) = 1.0081967213... has the simple continued fraction expansion 1 + 1/((123 - 2) + 1/(1 + 1/((1860498 - 2) + 1/(1 + 1/(28143753123 - 2) + 1/(1 + ...))))). (End)

			a(4) = 228826127 = 123*a(3) - a(2) = 123*1860498 - 15127=((123+sqrt(15125))/2)^4 + ( (123-sqrt(15125))/2)^4 =228826126.99999999562986 + 0.00000000437013 = 228826127.
a(4) = L(10 * 4) = L(40) = 228826127. - _Indranil Ghosh_, Feb 08 2017

J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 91.
R. P. Stanley. Enumerative combinatorics. Vol. 2, volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999.

Indranil Ghosh, Table of n, a(n) for n = 0..477
Tanya Khovanova, Recursive Sequences
A. V. Zarelua, On Matrix Analogs of Fermat's Little Theorem,Mathematical Notes, vol. 79, no. 6, 2006, pp. 783-796. Translated from Matematicheskie Zametki, vol. 79, no. 6, 2006, pp. 840-855.
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for linear recurrences with constant coefficients, signature (123,-1).

Cf. A000032: a(n) = A000032(10*n).

Cf. Lucas(k*n): A005248 (k = 2), A014448 (k = 3), A056854 (k = 4), A001946 (k = 5), A087215 (k = 6), A087281 (k = 7), A087265 (k = 8), A087287 (k = 9), A089772 (k = 11), A089775 (k = 12).

[Lucas(10*n): n in [0..90]]; // Vincenzo Librandi, Apr 14 2011

LucasL[10*Range[0, 20]] (* Paolo Xausa, Mar 04 2024 *)

a(n) = 123*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 123.
a(n) = ((123 + sqrt(15125))/2)^n + ((123 - sqrt(15125))/2)^n.
a(n)^2 = a(2*n) + 2.
G.f.: (2 - 123*x)/(1 - 123*x + x^2). -  Philippe Deléham, Nov 18 2008
From Peter Bala, Oct 14 2019: (Start)
a(n) = F(10*n+10)/F(10) - F(10*n-10)/F(10) = A049670(n+1) - A049670(n-1).
a(n) = trace(M^n), where M is the 2 X 2 matrix [0, 1; 1, 1]^10 = [34, 55; 55, 89].
Consequently the Gauss congruences hold: a(n*p^k) = a(n*p^(k-1)) ( mod p^k ) for all prime p and positive integers n and k. See Zarelua and also Stanley (Ch. 5, Ex. 5.2(a) and its solution).
121*Sum_{n >= 1} 1/(a(n) - 125/a(n)) = 1: (125 = Lucas(10) + 2 and 121 = Lucas(10) - 2)
125*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 121/a(n)) = 1.
x*exp(Sum_{n >= 1} a(n)*x^/n) = x + 123*x^2 + 15128*x^3 + ... is the o.g.f. for A049670. (End)
E.g.f.: exp((1/2)*(123 - 55*sqrt(5))*x)*(1 + exp(55*sqrt(5)*x)). - Stefano Spezia, Oct 18 2019
From Peter Bala, Apr 16 2025: (Start)
a(n) = Lucas(2*n)^5 - 5*Lucas(2*n)^3 + 5*Lucas(2*n) = 2*T(5, (1/2)*Lucas(2*n)), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind.
Sum_{n >= 1} 1/a(n) = (1/4) * (theta_3( (123 - sqrt(15125))/2 )^2 - 1) and
Sum_{n >= 1} (-1)^(n+1)/a(n) = (1/4) * (1 - theta_3( (sqrt(15125) - 123)/2 )^2),
where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). See Borwein and Borwein, Proposition 3.5 (i), p. 91. Cf. A153415 and A003499. (End)

A089775 Lucas numbers L(12n).

Original entry on oeis.org

2, 322, 103682, 33385282, 10749957122, 3461452808002, 1114577054219522, 358890350005878082, 115561578124838522882, 37210469265847998489922, 11981655542024930675232002, 3858055874062761829426214722, 1242282009792667284144565908482, 400010949097364802732720796316482

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Jan 09 2004

a(n+1)/a(n) converges to (322 + sqrt(103680))/2 = 321.996894379... a(0)/a(1) = 2/322; a(1)/a(2) = 322/103682; a(2)/a(3) = 103682/33385282; a(3)/a(4) = 33385282/10749957122; etc. Lim_{n -> inf} a(n)/a(n+1) = 0.00310562... = 2/(322 + sqrt(103680)) = (322 - sqrt(103680))/2.

			a(4) = 10749957122 = 322*a(3) - a(2) = 322*33385282 - 103682 = ((322 + sqrt(103680))/2)^4 + ((322 - sqrt(103680))/2)^4.

J. M. Borwein and P. B. Borwein, Pi and the AGM, Wiley, 1987, p. 91.

Nathaniel Johnston, Table of n, a(n) for n = 0..100
Tanya Khovanova, Recursive Sequences
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)
Index entries for linear recurrences with constant coefficients, signature (322, -1).

Cf. A000032, A060964.
a(n) = A000032(12n).
Row 9 * 2 of array A188644
Cf. Lucas(k*n): A005248 (k = 2), A014448 (k = 3), A056854 (k = 4), A001946 (k = 5), A087215 (k = 6), A087281 (k = 7), A087265 (k = 8), A087287 (k = 9), A065705 (k = 10), A089772 (k = 11).

[ Lucas(12*n) : n in [0..70]]; // Vincenzo Librandi, Apr 15 2011

Table[LucasL[12n], {n, 0, 13}] (* Indranil Ghosh, Mar 15 2017 *)

Vec((2 - 322*x)/(1 - 322*x + x^2) + O(x^14)) \\ Indranil Ghosh, Mar 15 2017

a(n) = 322*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 322
a(n) = ((322 + sqrt(103680))/2)^n + ((322 - sqrt(103680))/2)^n.
(a(n))^2 = a(2n) + 2.
G.f.: (2-322*x)/(1-322*x+x^2). - Philippe Deléham, Nov 02 2008
From Peter Bala, Apr 16 2025: (Start)
a(n) = Lucas(2*n)^6 - 6*Lucas(2*n)^4 + 9*Lucas(2*n)^2 - 2 = 2*T(6, (1/2)*Lucas(2*n)), where T(k, x) denotes the k-th Chebyshev polynomial of the first kind.
320*Sum_{n >= 1} 1/(a(n) - 324/a(n)) = 1: (324 = Lucas(12) + 2 and 320 = Lucas(12) - 2)
324*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 320/a(n)) = 1.
Sum_{n >= 1} 1/a(n) = (1/4) * (theta_3( (322 - sqrt(103680))/2 )^2 - 1) and
Sum_{n >= 1} (-1)^(n+1)/a(n) = (1/4) * (1 - theta_3( (sqrt(103680) - 322)/2 )^2),
where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). See Borwein and Borwein, Proposition 3.5 (i), p. 91. (End)

a(11) - a(13) from Vincenzo Librandi, Apr 15 2011

A124296 a(n) = 5F(n)^2 - 5F(n) + 1, where F(n) = Fibonacci(n).

Original entry on oeis.org

1, 1, 1, 11, 31, 101, 281, 781, 2101, 5611, 14851, 39161, 102961, 270281, 708761, 1857451, 4865911, 12744061, 33372361, 87382901, 228792301, 599019851, 1568309051, 4105974961, 10749725281, 28143378001, 73680695281, 192899171531

Offset: 0

Alexander Adamchuk, Oct 25 2006

11 = Lucas(5) divides a(3+10k), a(7+10k), and a(8+10k). The last digit of a(n) is 1, so a(n) mod 10 = 1. For odd n there exists the so-called Aurifeuillian factorization A001946(n) = Lucas(5n) = Lucas(n)*A(n)*B(n) = A000032(n)*A124296(n)*A124297(n), where A(n) = A124296(n) = 5*F(n)^2 - 5*F(n) + 1 and B(n) = A124297(n) = 5*F(n)^2 + 5*F(n) + 1, where F(n) = Fibonacci(n).

John Cerkan, Table of n, a(n) for n = 0..2375
Eric Weisstein's World of Mathematics, Aurifeuillean Factorization
Index entries for linear recurrences with constant coefficients, signature (4,-2,-6,4,2,-1).

Cf. A000032, A000045, A121171, A001946, A124297.

Bisections: A001604, A156094.

Table[5*Fibonacci[n]^2-5*Fibonacci[n]+1,{n,0,50}]
5#^2-5#+1&/@Fibonacci[Range[0,30]] (* Harvey P. Dale, Nov 29 2011 *)

a(n)=subst(5*t*(t-1)+1, t, fibonacci(n)) \\ Charles R Greathouse IV, Jan 03 2013

a(n) = 5*Fibonacci(n)^2 - 5*Fibonacci(n) + 1.

G.f.: -(x^5+9*x^4-15*x^3+x^2+3*x-1) / ((x-1)*(x+1)*(x^2-3*x+1)*(x^2+x-1)). [Colin Barker, Jan 03 2013]

A124297 a(n) = 5F(n)^2 + 5F(n) + 1, where F(n) = Fibonacci(n).

Original entry on oeis.org

1, 11, 11, 31, 61, 151, 361, 911, 2311, 5951, 15401, 40051, 104401, 272611, 712531, 1863551, 4875781, 12760031, 33398201, 87424711, 228859951, 599129311, 1568486161, 4106261531, 10750188961, 28144128251, 73681909211, 192901135711

Offset: 0

Alexander Adamchuk, Oct 25 2006

11 = Lucas(5) divides a(1+10k), a(2+10k), and a(9+10k). Last digit of a(n) is 1, or a(n) mod 10 = 1. For odd n there exists the so-called Aurifeuillian factorization A001946(n) = Lucas(5n) = Lucas(n)*A(n)*B(n) = A000032(n)*A124296(n)*A124297(n), where A(n) = A124296(n) = 5*F(n)^2 - 5*F(n) + 1 and B(n) = A124297(n) = 5*F(n)^2 + 5*F(n) + 1, where F(n) = Fibonacci(n).

John Cerkan, Table of n, a(n) for n = 0..2373
Eric Weisstein's World of Mathematics, Aurifeuillean Factorization
Index entries for linear recurrences with constant coefficients, signature (4,-2,-6,4,2,-1).

Cf. A000032, A000045, A121171, A001946, A124296.

Bisections: A001604, A156095.

Table[5*Fibonacci[n]^2+5*Fibonacci[n]+1,{n,0,50}]
LinearRecurrence[{4,-2,-6,4,2,-1},{1,11,11,31,61,151},30] (* Harvey P. Dale, Feb 23 2023 *)

a(n)=subst(5*t*(t+1)+1,t,fibonacci(n)) \\ Charles R Greathouse IV, Jan 03 2013

a(n) = 5*Fibonacci(n)^2 + 5*Fibonacci(n) + 1.

G.f.: -(11*x^5-21*x^4-15*x^3+31*x^2-7*x-1) / ((x-1)*(x+1)*(x^2-3*x+1)*(x^2+x-1)). [Colin Barker, Jan 03 2013]

cp's OEIS Frontend

A041226 Numerators of continued fraction convergents to sqrt(125).

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A065705 a(n) = Lucas(10*n).

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Magma

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A089775 Lucas numbers L(12n).

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A124296 a(n) = 5*F(n)^2 - 5*F(n) + 1, where F(n) = Fibonacci(n).

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A124297 a(n) = 5*F(n)^2 + 5*F(n) + 1, where F(n) = Fibonacci(n).

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A133320 Numbers k such that both A124296(k) = 5*F(k)^2 - 5*F(k) + 1 and A124297(k) = 5*F(k)^2 + 5*F(k) + 1 are prime, where F(k) = Fibonacci(k).

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A124296 a(n) = 5F(n)^2 - 5F(n) + 1, where F(n) = Fibonacci(n).

A124297 a(n) = 5F(n)^2 + 5F(n) + 1, where F(n) = Fibonacci(n).

A133320 Numbers k such that both A124296(k) = 5F(k)^2 - 5F(k) + 1 and A124297(k) = 5F(k)^2 + 5F(k) + 1 are prime, where F(k) = Fibonacci(k).