cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A174501 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A003499(n)) ), where A003499(n) = (3+sqrt(8))^n + (3-sqrt(8))^n.

This page as a plain text file.
%I A174501 #24 Jul 14 2021 17:07:48
%S A174501 1,4,1,32,1,196,1,1152,1,6724,1,39200,1,228484,1,1331712,1,7761796,1,
%T A174501 45239072,1,263672644,1,1536796800,1,8957108164,1,52205852192,1,
%U A174501 304278004996,1,1773462177792,1,10336495061764,1,60245508192800,1,351136554095044,1
%N A174501 Continued fraction expansion for exp( Sum_{n>=1} 1/(n*A003499(n)) ), where A003499(n) = (3+sqrt(8))^n + (3-sqrt(8))^n.
%H A174501 Colin Barker, <a href="/A174501/b174501.txt">Table of n, a(n) for n = 1..1000</a>
%H A174501 Peter Bala, <a href="/A174500/a174500_2.pdf">Some simple continued fraction expansions for an infinite product, Part 1</a>
%H A174501 <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,7,0,-7,0,1).
%F A174501 a(2n-1) = 1, a(2n) = A003499(n) - 2, for n>=1 [conjecture].
%F A174501 The above conjectures are correct. See the Bala link for details. - _Peter Bala_, Jan 08 2013
%F A174501 a(n) = 7*a(n-2)-7*a(n-4)+a(n-6). G.f.: -x*(x^4+4*x^3-6*x^2+4*x+1) / ((x-1)*(x+1)*(x^2-2*x-1)*(x^2+2*x-1)). - _Colin Barker_, Jan 20 2013
%F A174501 a(n) = (((-1-sqrt(2))^n+(1-sqrt(2))^n+(sqrt(2)-1)^n+(1+sqrt(2))^n-4))/2 for n even. - _Colin Barker_, May 11 2016
%e A174501 Let L = Sum_{n>=1} 1/(n*A003499(n)) or, more explicitly,
%e A174501 L = 1/6 + 1/(2*34) + 1/(3*198) + 1/(4*1154) + 1/(5*6726) +...
%e A174501 so that L = 0.1833074113563494600094468694966574405706183998044...
%e A174501 then exp(L) = 1.2011836088120841844713993433258934531421726294252...
%e A174501 equals the continued fraction given by this sequence:
%e A174501 exp(L) = [1;4,1,32,1,196,1,1152,1,6724,1,39200,1,...]; i.e.,
%e A174501 exp(L) = 1 + 1/(4 + 1/(1 + 1/(32 + 1/(1 + 1/(196 + 1/(1 +...)))))).
%e A174501 Compare these partial quotients to A003499(n), n=1,2,3,...:
%e A174501 [6,34,198,1154,6726,39202,228486,1331714,7761798,45239074,...].
%t A174501 LinearRecurrence[{0,7,0,-7,0,1},{1,4,1,32,1,196},50] (* _Harvey P. Dale_, Jul 14 2021 *)
%o A174501 (PARI) {a(n)=local(L=sum(m=1,2*n+1000,1./(m*round((3+sqrt(8))^m+(3-sqrt(8))^m))));contfrac(exp(L))[n]}
%o A174501 (PARI) Vec(-x*(x^4+4*x^3-6*x^2+4*x+1)/((x-1)*(x+1)*(x^2-2*x-1)*(x^2+2*x-1)) + O(x^50)) \\ _Colin Barker_, May 11 2016
%Y A174501 Cf. A003499, A174500, A174502.
%K A174501 cofr,nonn,easy
%O A174501 1,2
%A A174501 _Paul D. Hanna_, Mar 20 2010