A006267 - OEIS

%I M3699 #59 Nov 19 2022 20:59:00

%S 1,4,76,439204,84722519070079276,

%T 608130213374088941214747405817720942127490792974404

%N Continued cotangent for the golden ratio.

%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

%H Harry J. Smith, <a href="/A006267/b006267.txt">Table of n, a(n) for n = 0..7</a>

%H Mohammad K. Azarian, <a href="https://doi.org/10.35834/1998/1003176">Problem 123</a>, Missouri Journal of Mathematical Sciences, Vol. 10, No. 3 (Fall 1998), p. 176; <a href="https://doi.org/10.35834/2000/1201050">Solution</a>, ibid., Vol. 12, No. 1 (Winter 2000), pp. 61-62.

%H Jeffrey Shallit, <a href="http://archive.org/details/jresv80Bn2p285">Predictable regular continued cotangent expansions</a>, J. Res. Nat. Bur. Standards Sect. B, Vol. 80B, No. 2 (1976), pp. 285-290.

%H Zalman Usiskin, <a href="https://fq.math.ca/Scanned/11-3/elementary11-3.pdf">Problem B-266</a>, Elementary Problems and Solutions, The Fibonacci Quarterly, Vol. 11, No. 3 (1973), p. 334; <a href="https://www.fq.math.ca/Scanned/12-3/elementary12-3.pdf">Lucas Numbers for Powers of 3</a>, Solution to Problem B-266 by David Zeitlin, ibid., Vol. 12, No. 3 (1974), p. 315-316.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/LehmerCotangentExpansion.html">Lehmer Cotangent Expansion</a>.

%F (1+sqrt(5))/2 = cot(Sum_{n>=0} (-1)^n*acot(a(n))); let b(0) = (1+sqrt(5))/2, b(n) = (b(n-1)*floor(b(n-1))+1)/(b(n-1)-floor(b(n-1)) then a(n) = floor(b(n)). - _Benoit Cloitre_, Apr 10 2003

%F a(n) = A000204(3^n). - _Benoit Cloitre_, Sep 18 2005

%F a(n) = Round(c^(3^n)) where c = GoldenRatio = 1.6180339887498948482... = (sqrt(5)+1)/2 (A001622). - _Artur Jasinski_, Sep 22 2008

%F Recurrence a(n+1) = a(n)^3 + 3*a(n), a(0) = 4. - _Artur Jasinski_, Sep 24 2008

%F a(n+1) = Product_{k = 0..n} A002813(k). Thus a(n) divides a(n+1). - _Peter Bala_, Nov 22 2012

%F Sum_{n>=0} a(n)^2/A045529(n+1) = 1. - _Amiram Eldar_, Jan 12 2022

%F a(n) = Product_{k=0..n-1} (Lucas(2*3^k) + 1) (Usiskin, 1973). - _Amiram Eldar_, Jan 29 2022

%F From _Peter Bala_, Nov 15 2022: (Start)

%F a(n) = Lucas(3^n) for n >= 1.

%F a(n) == 1 (mod 3) for n >= 1.

%F a(n+1) == a(n) (mod 3^(n+1)) for n >= 1 (a particular case of the Gauss congruences for the Lucas numbers).

%F The smallest positive residue of a(n) mod 3^n = A268924(n).

%F In the ring of 3-adic integers the limit_{n -> oo} a(n) exists and is equal to A271223. Cf. A006266. (End)

%p a := proc(n) option remember; if n = 1 then 4 else a(n-1)^3 + 3*a(n-1) end if; end: seq(a(n), n = 1..5); # _Peter Bala_, Nov 15 2022

%t c = N[GoldenRatio, 1000]; Table[Round[c^(3^n)], {n, 1, 8}] (* _Artur Jasinski_, Sep 22 2008 *)

%t a = {}; x = 4; Do[AppendTo[a, x]; x = x^3 + 3 x, {n, 1, 10}]; a (* _Artur Jasinski_, Sep 24 2008 *)

%o (PARI) bn=vector(100); b(n)=if(n<0,0,bn[n]); bn[1]=(1+sqrt(5))/2; for(n=2,10,bn[n]=(b(n-1)*floor(b(n-1))+1)/(b(n-1)-floor(b(n-1)))) a(n)=floor(b(n+1))

%o (PARI) { default(realprecision, 10000); bn=vector(8); bn[1]=(1+sqrt(5))/2; for(n=2, 8, bn[n]=(bn[n-1]*floor(bn[n-1]) + 1)/(bn[n-1] - floor(bn[n-1]))); for (n=1, 8, write("b006267.txt", n-1, " ", floor(bn[n]))); } \\ _Harry J. Smith_, May 04 2009

%Y Cf. A000032, A000204, A001622, A001999, A002666, A002667, A002668, A006266, A002813, A045529, A271223, A268924.

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_

%E The next term is too large to include.