|
| |
|
|
A156736
|
|
Signed greedy Egyptian fraction for Pi/2
|
|
1
| | |
|
|
|
OFFSET
| 0,2
|
|
|
COMMENTS
| The second and fourth convergents of Pi (22/7 and 355/113) appear when truncating the series to three and four terms.
|
|
|
LINKS
| Wikipedia, Greedy algorithm for Egyptian fractions
|
|
|
FORMULA
| Sum(n>=0,1/a(n))=Pi/2.
a(n) = 2*A001467(n+1). - R. J. Mathar, Apr 02 2011
|
|
|
EXAMPLE
| 1+1/2+1/14=11/7=(1/2)(22/7)
1+1/2+1/14-1/1582=355/226=(1/2)(355/113)
|
|
|
PROG
| (PARI) x=Pi/2; for (k=0, 7, d=round(1/x); x=x-1/d; print1(d, ", "))
|
|
|
CROSSREFS
| Cf. A001466, A156269, A156618.
Cf. A156750. [From Jaume Oliver Lafont (joliverlafont(AT)gmail.com), Mar 03 2009]
Sequence in context: A032419 A190634 A130421 * A006266 A106484 A027739
Adjacent sequences: A156733 A156734 A156735 * A156737 A156738 A156739
|
|
|
KEYWORD
| sign
|
|
|
AUTHOR
| Jaume Oliver Lafont (joliverlafont(AT)gmail.com), Feb 14 2009
|
| |
|
|
