A006242 -id:A006242 - cp's OEIS frontend

A087799 a(n) = 10*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 10.

2, 10, 98, 970, 9602, 95050, 940898, 9313930, 92198402, 912670090, 9034502498, 89432354890, 885289046402, 8763458109130, 86749292044898, 858729462339850, 8500545331353602, 84146723851196170, 832966693180608098, 8245520207954884810

Offset: 0

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Oct 11 2003

a(n+1)/a(n) converges to (5+sqrt(24)) = 9.8989794... a(0)/a(1)=2/10; a(1)/a(2)=10/98; a(2)/a(3)=98/970; a(3)/a(4)=970/9602; ... etc. Lim a(n)/a(n+1) as n approaches infinity = 0.10102051... = 1/(5+sqrt(24)) = (5-sqrt(24)).
Except for the first term, positive values of x (or y) satisfying x^2 - 10xy + y^2 + 96 = 0. - Colin Barker, Feb 25 2014
A triangle whose sides are a(n) - 1, a(n) and a(n) + 1 is nearly Fleenor-Heronian since its area is the product of an integer and the square root of 2. See A003500. - Charlie Marion, Dec 18 2020

			a(4) = 9602 = 10*a(3) - a(2) = 10*970 - 98 = (5+sqrt(24))^4 + (5-sqrt(24))^4.

T. D. Noe, Table of n, a(n) for n = 0..200
Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
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 (10,-1).

Cf. A005248, A006242, A036336, A086927, A135927, A174503.

I:=[2,10]; [n le 2 select I[n] else 10*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Nov 07 2018

a[0] = 2; a[1] = 10; a[n_] := 10a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 17}] (* Robert G. Wilson v, Jan 30 2004 *)
LinearRecurrence[{10,-1}, {2,10}, 30] (* G. C. Greubel, Nov 07 2018 *)

polsym(x^2 - 10*x + 1,20) \\ Charles R Greathouse IV, Jun 11 2011

{a(n) = 2 * real( (5 + 2 * quadgen(24))^n )}; /* Michael Somos, Feb 25 2014 */

[lucas_number2(n,10,1) for n in range(27)] # Zerinvary Lajos, Jun 25 2008

a(n) = (5+sqrt(24))^n + (5-sqrt(24))^n.
G.f.: (2-10*x)/(1-10*x+x^2). - Philippe Deléham, Nov 02 2008
From Peter Bala, Jan 06 2013: (Start)
Let F(x) = Product_{n = 0..inf} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let alpha = 5 - sqrt(24). This sequence gives the simple continued fraction expansion of 1 + F(alpha) = 2.09989 80642 72052 68138 ... = 2 + 1/(10 + 1/(98 + 1/(970 + ...))).
Also F(-alpha) = 0.89989 78538 78393 34715 ... has the continued fraction representation 1 - 1/(10 - 1/(98 - 1/(970 - ...))) and the simple continued fraction expansion 1/(1 + 1/((10-2) + 1/(1 + 1/((98-2) + 1/(1 + 1/((970-2) + 1/(1 + ...))))))).
F(alpha)*F(-alpha) has the simple continued fraction expansion 1/(1 + 1/((10^2-4) + 1/(1 + 1/((98^2-4) + 1/(1 + 1/((970^2-4) + 1/(1 + ...))))))). Cf. A174503 and A005248. (End)
a(-n) = a(n). - Michael Somos, Feb 25 2014
From Peter Bala, Oct 16 2019: (Start)
8*Sum_{n >= 1} 1/(a(n) - 12/a(n)) = 1.
12*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 8/a(n)) = 1.
Series acceleration formulas for sums of reciprocals:
Sum_{n >= 1} 1/a(n) = 1/8 - 12*Sum_{n >= 1} 1/(a(n)*(a(n)^2 - 12))  and
Sum_{n >= 1} (-1)^(n+1)/a(n) = 1/12 + 8*Sum_{n >= 1} (-1)^(n+1)/(a(n)*(a(n)^2 + 8)).
Sum_{n >= 1} 1/a(n) = ( (theta_3(5-sqrt(24)))^2 - 1 )/4 and
Sum_{n >= 1} (-1)^(n+1)/a(n) = ( 1 - (theta_3(sqrt(24)-5))^2 )/4, where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). Cf. A153415 and A003499. (End)
E.g.f.: 2*exp(5*x)*cosh(2*sqrt(6)*x). - Stefano Spezia, Oct 18 2019
From Peter Bala, Mar 29 2022: (Start)
a(n) = 2*T(n,5), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind.
a(2^n) = A135927(n+1) and a(3^n) = A006242(n+1), both for n >= 0. (End)

More terms from Colin Barker, Feb 25 2014

A006243 Extracting a square root.

Original entry on oeis.org

198, 7761798, 467613464999866416198, 102249460387306384473056172738577521087843948916391508591105798

Offset: 1

N. J. A. Sloane

nonn

Jeffrey Shallit, personal communication.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 1..6
E. B. Escott, Rapid method for extracting a square root, Amer. Math. Monthly, 44 (1937), 644-646.

Cf. A006242, A112845 (differs only for the initial 6).

[n eq 1 select 198 else Self(n-1)^3-3*Self(n-1): n in [1..5]]; // Vincenzo Librandi, Feb 09 2017

RecurrenceTable[{a[1]==198, a[n]==a[n-1]^3 - 3 a[n-1]}, a, {n, 8}] (* Vincenzo Librandi, Feb 09 2017 *)

a(1) = 198, a(n) = a(n-1)^3 - 3*a(n-1). - Sean A. Irvine, Feb 08 2017

a(n) = (99 + 70*sqrt(2))^(3^(n-1)) + (99 - 70*sqrt(2))^(3^(n-1)). - Bruno Berselli, Feb 10 2017

Offset changed by Sean A. Irvine, Feb 08 2017

A018844 Arises from generalized Lucas-Lehmer test for primality.

Original entry on oeis.org

4, 10, 52, 724, 970, 10084, 95050, 140452, 1956244, 9313930, 27246964, 379501252, 912670090, 5285770564, 73621286644, 89432354890, 1025412242452, 8763458109130, 14282150107684, 198924689265124

Offset: 1

Robert G. Wilson v

Apparently this was suggested by an article by R. M. Robinson.
Starting values for Lucas-Lehmer test that result in a zero term (mod Mersenne prime Mp) after P-1 steps. - Jason Follas (jfollas_mersenne(AT)hotmail.com), Aug 01 2004
m belongs to the sequence iff m-2 is twice a square and m+2 is either three or six times a square. - René Gy, Jan 10 2019
A006242 is a subsequence. - Davide Rotondo, Oct 21 2024

Jeppe Stig Nielsen, Table of n, a(n) for n = 1..1370
D. H. Lehmer, An Extended Theory of Lucas' Functions, Ann. Math. 31 (1930), 419-448. See p. 445.
E. L. Roettger and H. C. Williams, Some Remarks Concerning the Lucas-Lehmer Primality Test, Journal of Integer Sequences, Vol. 28 (2025), Article 25.2.5. See pp. 3, 13, 24.
Herb Savage et al., Re: Mersenne: starting values for LL-test.

Cf. A006242, A206257.

listUpTo(n)=a=List([4,52]);while(1,m=14*a[#a]-a[#a-1];m>n&&break();listput(a,m));b=List([10,970]);while(1,m=98*b[#b]-b[#b-1];m>n&&break();listput(b,m));setunion(Set(a),Set(b)) \\ Jeppe Stig Nielsen, Aug 03 2020

Union of sequences a_1=4, a_2=52, a_{n}=14*a_{n-1} - a_{n-2} and b_1=10, b_2=970, b_{n}=98*b_{n-1} - b_{n-2}.
a[1]=14 (mod Mp), a[2]=52 (mod Mp), a[n]=(14*a[n-1]-a[n-2]) (mod Mp). - Jason Follas (jfollas_mersenne(AT)hotmail.com), Aug 01 2004
Though originally noted as the union of two sequences, when the first sequence (14*a[n-1]-a[n-2]) is evaluated modulo a Mersenne prime, the terms of the second sequence (98*b[n-1]-b[n-2]) will occur naturally (just not in numerical order). - Jason Follas (jfollas_mersenne(AT)hotmail.com), Aug 01 2004
a(n) = sqrt(A206257(n) + 2). - Arkadiusz Wesolowski, Feb 08 2012

A282180 a(n+1) = a(n)*(a(n)^2 - 3) with a(0) = 8.

Original entry on oeis.org

8, 488, 116212808, 1569502402942700328379688, 3866214585126515728777536857817155683642224883875510905654220958052649608

Offset: 0

Vincenzo Librandi, Feb 10 2017

nonn

E. B. Escott, Rapid method for extracting a square root, Amer. Math. Monthly, 44 (1937), 644-646.

Cf. similar sequences with initial value k: A001999 (k=3), A219160 (k=4), A219161 (k=5), A112845 (k=6), A002000 (k=7), this sequence (k=8), A282181 (k=9), A006242 (k=10), A006243 (k=198).

[n eq 1 select 8 else Self(n-1)^3 - 3*Self(n-1): n in [1..6]];

RecurrenceTable[{a[0] == 8, a[n] == a[n-1]^3 - 3 a[n-1]}, a, {n, 8}]

a(n) = (4 + sqrt(15))^(3^n) + (4 - sqrt(15))^(3^n). - Bruno Berselli, Feb 10 2017

a(n) = -2*cos(3^n * arccos(-4)). - Daniel Suteu, Feb 10 2017

cp's OEIS Frontend

A087799 a(n) = 10*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 10.

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A006243 Extracting a square root.

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A018844 Arises from generalized Lucas-Lehmer test for primality.

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Programs

PARI

Formula

A282180 a(n+1) = a(n)*(a(n)^2 - 3) with a(0) = 8.

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Magma

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