The OEIS is supported by the many generous donors to the OEIS Foundation. Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 A110259 Numerators in the coefficients that form the even-indexed partial quotients of the continued fraction representation of the inverse tangent of 1/x. 7
 3, 28, 704, 768, 311296, 1507328, 3145728, 130023424, 7516192768, 12884901888, 2954937499648, 12919261626368, 52776558133248, 774056185954304, 66428094503714816, 31525197391593472, 308982963234634989568 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Lim_{n->infinity} a(n)/A110260(n) = lim_{n->infinity} A110255(2*n)/A110256(2*n) = Pi. The continued fraction expansion arctan(z) = z/(1 + z^2/(3 + 4*z^2/(5 + 9*z^2/(7 + ...)))) is due to Lambert - see Roegel, Section 1.2.1. - Peter Bala, Dec 02 2021 LINKS Paul D. Hanna, Table of n, a(n) for n = 1..200 Denis Roegel, Lambert’s proof of the irrationality of Pi: Context and translation, [Research Report] LORIA, 2020, hal-02984214. FORMULA a(n) = A110255(2*n). EXAMPLE arctan(1/x) = 1/x - 1/(3*x^3) + 1/(5*x^5) - 1/(7*x^7) +- ... arctan(1/x) = [0; x, 3*x, (5/4)*x, (28/9)*x, (81/64)*x, (704/225)*x, (325/256)*x, (768/245)*x, (20825/16384)*x, (311296/99225)*x, (83349/65536)*x, (1507328/480249)*x, (1334025/1048576)*x, (3145728/1002001)*x, ...] arctan(1/x) = 1/(x + 1/(3*x + 1/(5/4*x + 1/(28/9*x + 1/(81/64*x + ...))))). The coefficients of x in the even-indexed partial quotients converge to Pi: {3, 28/9, 704/225, 768/245, 311296/99225, ...}. The coefficients of x in the odd-indexed partial quotients converge to 4/Pi: {1, 5/4, 81/64, 325/256, 20825/16384, ...}. From Peter Bala, Dec 02 2021: (Start) Making use of the expansion 2*arcsin(sqrt(x)/2)^2 = Sum_{n >= 1} x^n/ (n^2*binomial(2*n,n)) we calculate 3 + Pi = Sum_{n >= 1} (2^n)*n/binomial(2*n,n); 28 + 9*Pi = Sum_{n >= 3} (2^n)*n*(n-1)*(n-2)/binomial(2*n,n); 704 + 225*Pi = Sum_{n >= 5} (2^n)*n*(n-1)*...*(n-4)/binomial(2*n,n); 45*(768 + 245*Pi) = Sum_{n >= 7} (2^n)*n*(n-1)*...*(n-6)/binomial(2*n,n); 9*(311296 + 99225*Pi) = Sum_{n >= 9} (2^n)*n*(n-1)*...*(n-8)/ binomial(2*n,n). It appears that Sum_{n >= 2*k+1} (2^n)*n*(n-1)*...*(n-2*k)/binomial(2*n,n) = N(2*k) + D(2*k)*Pi, where the ratios N(2*k)/D(2*k) are equal to the even-indexed partial quotients of Lambert's continued fraction representation of the inverse tangent of 1/x. (End) PROG (PARI) {a(n)=numerator(subst((contfrac( sum(k=0, 2*n+2, (-1)^k/x^(2*k+1)/(2*k+1)), 2*n+2))[2*n+1], x, 1))} CROSSREFS Cf. A110260 (denominators), A110255/A110256 (continued fraction), A110257/A110258. Sequence in context: A331196 A181588 A084880 * A276745 A354664 A015474 Adjacent sequences:  A110256 A110257 A110258 * A110260 A110261 A110262 KEYWORD nonn,frac AUTHOR Paul D. Hanna, Jul 18 2005 STATUS approved

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recents
The OEIS Community | Maintained by The OEIS Foundation Inc.

Last modified September 29 07:07 EDT 2022. Contains 357082 sequences. (Running on oeis4.)