A130824 -id:A130824 - cp's OEIS frontend

A110185 Coefficients of x in the partial quotients of the continued fraction expansion exp(1/x) = [1, x - 1/2, 12x, 5x, 28x, 9x, 44x, 13x, ...]. The partial quotients all have the form a(n)*x except the constant term of 1 and the initial partial quotient which equals (x - 1/2).

0, 1, 12, 5, 28, 9, 44, 13, 60, 17, 76, 21, 92, 25, 108, 29, 124, 33, 140, 37, 156, 41, 172, 45, 188, 49, 204, 53, 220, 57, 236, 61, 252, 65, 268, 69, 284, 73, 300, 77, 316, 81, 332, 85, 348, 89, 364, 93, 380, 97, 396, 101, 412, 105, 428, 109, 444, 113, 460, 117, 476

Offset: 0

Paul D. Hanna, Jul 14 2005

Simple continued fraction expansion of 2*(e - 1)/(e + 1) = 2*tanh(1/2) = 1/(1 + 1/(12 + 1/(5 + 1/(28 + ...)))). - Peter Bala, Oct 01 2023

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
D. H. Lehmer, Continued fractions containing arithmetic progressions, Scripta Mathematica, 29 (1973): 17-24. [Annotated copy of offprint]
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Cf. continued fraction expansions: A004273 ( tanh(1) ), A204877 ( 3*tanh(1/3) ), A130824 ( tanh(1/2) ).

a(n)=polcoeff(x*(1+12*x+3*x^2+4*x^3)/(1-x^2)^2+x*O(x^n),n)

G.f.: x*((1+3*x^2) + 4*x*(3+x^2))/(1-x^2)^2 = sum_{n>=0} a(n)*x^n.
From Carl R. White, Feb 11 2010: (Start)
a(n) = sign(n) * (2*n+1) * (3*cos(Pi*n)+5)/2.
a(2n+1) = a(2n-1) + 4, a(2n+2) = a(2n) + 16, with a(0)=0, a(1)=1, a(2)=12. (End)
a(n) = (5+3*(-1)^n)*(2*n-1)/2, with a(0)=0. Sum_{i=0..n} a(i) = A085787(A042948(n)). - Bruno Berselli, Jan 20 2012

A133653 A007318^(-1) * A003261.

Original entry on oeis.org

1, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42, 46, 50, 54, 58, 62, 66, 70, 74, 78, 82, 86, 90, 94, 98, 102, 106, 110, 114, 118, 122, 126, 130, 134, 138, 142, 146, 150, 154

Offset: 1

Gary W. Adamson, Sep 19 2007

It appears this sequence gives the positive integers m such that the sum of the first m Fibonacci numbers divides their product. For example, if n=2 and m=a(2)=6, we have the sum 1+1+2+3+5+8=20 which clearly divides the corresponding product 480. See A175553 for the analogous sequence when using the triangular numbers. Sum_{k=1..n} Fibonacci(k) divides Product_{k=1..n} Fibonacci(k). - John W. Layman, Jul 10 2010

			a(4) = 14 = (1, 3, 3, 1) dot (1, 5, -1, 1) = (1, 15, -3, 1).

Cf. A003261, A007318, A175553.

Essentially the same as A130824, A113127, A111284, A073760, A016825.

Inverse binomial transform of A003261: (1, 7, 23, 63, 159, 383, ...).
Binomial transform of [1, 5, -1, 1, -1, 1, ...].
"1" followed by 2 * [3, 5, 7, 9, 11, ...].
O.g.f.: x*(1+4x-x^2)/(1-x)^2. a(n) = 4n-2, n > 1. - R. J. Mathar, Jun 08 2008
1/(1+1/(6+1/(10+1/(14+1/(...(continued fraction)))))) = (e-1)/2 with e = 2.718281...- Philippe Deléham, Mar 09 2013

More terms from R. J. Mathar, Jun 08 2008

A047441 Numbers that are congruent to {0, 2, 5, 6} mod 8.

Original entry on oeis.org

0, 2, 5, 6, 8, 10, 13, 14, 16, 18, 21, 22, 24, 26, 29, 30, 32, 34, 37, 38, 40, 42, 45, 46, 48, 50, 53, 54, 56, 58, 61, 62, 64, 66, 69, 70, 72, 74, 77, 78, 80, 82, 85, 86, 88, 90, 93, 94, 96, 98, 101, 102, 104, 106, 109, 110, 112, 114, 117, 118, 120, 122, 125

Offset: 1

N. J. A. Sloane

Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Cf. A047615, A130824.

[n : n in [0..150] | n mod 8 in [0, 2, 5, 6]]; // Wesley Ivan Hurt, May 26 2016

A047441:=n->(8*n-7-I^(2*n)-I^(1-n)+I^(1+n))/4: seq(A047441(n), n=1..100); # Wesley Ivan Hurt, May 26 2016

Table[(8n-7-I^(2n)-I^(1-n)+I^(1+n))/4, {n, 80}] (* Wesley Ivan Hurt, May 26 2016 *)
LinearRecurrence[{1,0,0,1,-1},{0,2,5,6,8},100] (* Harvey P. Dale, Dec 25 2023 *)

G.f.: x^2*(2+3*x+x^2+2*x^3) / ( (1+x)*(x^2+1)*(x-1)^2 ). - R. J. Mathar, Dec 07 2011
From Wesley Ivan Hurt, May 26 2016: (Start)
a(n) = a(n-1) + a(n-4) - a(n-5) for n>5.
a(n) = (8*n-7-i^(2*n)-i^(1-n)+i^(1+n))/4 where i=sqrt(-1).
a(2k) = A130824(k) k>0, a(2k-1) = A047615(k). (End)
E.g.f.: (4 - sin(x) + (4*x - 3)*sinh(x) + 4*(x - 1)*cosh(x))/2. - Ilya Gutkovskiy, May 27 2016
a(n) = (8*n-7-cos(n*Pi)-2*sin(n*Pi/2))/4. - Wesley Ivan Hurt, Oct 05 2017
Sum_{n>=2} (-1)^n/a(n) = (sqrt(2)-1)*Pi/16 + (4-sqrt(2))*log(2)/16 + sqrt(2)*log(2+sqrt(2))/8. - Amiram Eldar, Dec 21 2021

cp's OEIS Frontend

A110185 Coefficients of x in the partial quotients of the continued fraction expansion exp(1/x) = [1, x - 1/2, 12x, 5x, 28x, 9x, 44x, 13x, ...]. The partial quotients all have the form a(n)*x except the constant term of 1 and the initial partial quotient which equals (x - 1/2).

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A133653 A007318^(-1) * A003261.

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A047441 Numbers that are congruent to {0, 2, 5, 6} mod 8.

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cp's OEIS Frontend

A110185 Coefficients of x in the partial quotients of the continued fraction expansion exp(1/x) = [1, x - 1/2, 12*x, 5*x, 28*x, 9*x, 44*x, 13*x, ...]. The partial quotients all have the form a(n)*x except the constant term of 1 and the initial partial quotient which equals (x - 1/2).

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Author

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Programs

PARI

Formula

A133653 A007318^(-1) * A003261.

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Author

Keywords

Comments

Examples

Crossrefs

Formula

Extensions

A047441 Numbers that are congruent to {0, 2, 5, 6} mod 8.

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Maple

Mathematica

Formula

A110185 Coefficients of x in the partial quotients of the continued fraction expansion exp(1/x) = [1, x - 1/2, 12x, 5x, 28x, 9x, 44x, 13x, ...]. The partial quotients all have the form a(n)*x except the constant term of 1 and the initial partial quotient which equals (x - 1/2).