A006243 -id:A006243 - cp's OEIS frontend

A112845 Recurrence a(n) = a(n-1)^3 - 3*a(n-1) with a(0) = 6.

6, 198, 7761798, 467613464999866416198, 102249460387306384473056172738577521087843948916391508591105798

Offset: 0

Eric W. Weisstein, Sep 21 2005

Identical to A006243 apart from the initial term. For some general remarks on this recurrence see A001999. - Peter Bala, Nov 13 2012

G. C. Greubel, Table of n, a(n) for n = 0..6
E. B. Escott, Rapid method for extracting a square root, Amer. Math. Monthly, 44 (1937), 644-646.
N. J. Fine, Infinite products for k-th roots, Amer. Math. Monthly Vol. 84, No. 8, Oct. 1977, 629-630.
Eric Weisstein's World of Mathematics, Pierce Expansion

Cf. A006275, A006276.
Cf. A006243. - R. J. Mathar, Aug 15 2008
Cf. A001999, A219160, A219161.

RecurrenceTable[{a[n] == a[n - 1]^3 - 3*a[n - 1], a[0] == 6}, a, {n,
  0, 5}] (* G. C. Greubel, Dec 30 2016 *)
NestList[#^3-3#&,6,5] (* Harvey P. Dale, Jul 23 2025 *)

a(n) = -2*cos(3^n*arccos(-3)).
From Peter Bala, Nov 13 2012: (Start)
a(n) = (3 + 2*sqrt(2))^(3^n) + (3 - 2*sqrt(2))^(3^n).
Product {n = 0..inf} (1 + 2/(a(n) - 1)) = sqrt(2).
(End)

A006242 Extracting a square root.

Original entry on oeis.org

10, 970, 912670090, 760223786832147978143718730, 439363892017598816969702791108195858981800447259539613873486126455827777484460810

Offset: 1

N. J. A. Sloane

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..7
E. B. Escott, Rapid method for extracting a square root, Amer. Math. Monthly, 44 (1937), 644-646.

Cf. A006243.

[n eq 1 select 10 else Self(n-1)^3-3*Self(n-1): n in [1..5]]; // Vincenzo Librandi, Feb 09 2017

RecurrenceTable[{a[1]==10, a[n]==a[n-1]^3 - 3 a[n-1]}, a, {n, 8}] (* Vincenzo Librandi, Feb 09 2017 *)

a(1) = 10, a(n) = a(n-1)^3 - 3*a(n-1) [From Escott]. - Sean A. Irvine, Feb 08 2017
a(n) = (5 + 2*sqrt(6))^(3^(n-1)) + (5 - 2*sqrt(6))^(3^(n-1)). - Bruno Berselli, Feb 10 2017
a(n) = 2*T(3^(n-1),5), where T(n,x) deotes the n-th Chebyshev polynomial of the first kind. - Peter Bala, Mar 29 2022

New offset and a(5) from Sean A. Irvine, Feb 08 2017

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

A112845 Recurrence a(n) = a(n-1)^3 - 3*a(n-1) with a(0) = 6.

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A006242 Extracting a square root.

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

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