A195909 First numerator and then denominator in a fraction expansion of log(2) - Pi/8.

1, 2, -1, 3, 1, 12, 1, 30, -1, 35, 1, 56, 1, 90, -1, 99, 1, 132, 1, 182, -1, 195, 1, 240, 1, 306, -1, 323, 1, 380, 1, 462, -1, 483, 1, 552, 1, 650, -1, 675, 1, 756, 1, 870, -1, 899, 1, 992, 1, 1122, -1, 1155, 1

Offset

1,2

References

Mohammad K. Azarian, Problem 1218, Pi Mu Epsilon Journal, Vol. 13, No. 2, Spring 2010, p. 116. Solution published in Vol. 13, No. 3, Fall 2010, pp. 183-185.

Granino A. Korn and Theresa M. Korn, Mathematical Handbook for Scientists and Engineers, McGraw-Hill Book Company, New York (1968).

Links

Table of n, a(n) for n=1..53.

Formula

log(2) - Pi/8 = Sum_{n>=1} (-1)^(n+1)*(1/n) + (-1/2)*Sum_{n>=0} (-1)^n*(1/(2*n+1)).

Empirical g.f.: x*(1+2*x-2*x^2+x^3+2*x^4+9*x^5-2*x^6+14*x^7+2*x^8+3*x^9-2*x^10+3*x^11+x^12) / ((1-x)^3*(1+x)^3*(1-x+x^2)^2*(1+x+x^2)^2). - Colin Barker, Dec 17 2015

Example

1/2 - 1/3 + 1/12 + 1/30 - 1/35 + 1/56 + 1/90 - 1/99 + 1/132 + 1/182 - 1/195 + 1/240 + ... = [(1 - 1/2) + (1/3 - 1/4) + (1/5 - 1/6) + (1/7 - 1/8) + (1/9 - 1/10) + (1/11 - 1/12) + ... ] - (1/2)*[(1 - 1/3) + (1/5 - 1/7) + (1/9 - 1/11) + (1/13 - 1/15) + ... ] = log(2) - Pi/8.

Crossrefs

Cf. A195913, A195697, A195947, A164833, A118324, A098289, A075549, A016655, A019675, A161685, A144981, A168056, A004772.

Sequence in context: A098570 A122048 A046208 * A195697 A137764 A344752

Adjacent sequences: A195906 A195907 A195908 * A195910 A195911 A195912

Keyword

frac,sign

Author

Mohammad K. Azarian, Sep 26 2011

Status

approved