A001467 -id:A001467 - cp's OEIS frontend

A073422 e = 1/a(0)+1/a(1)+1/a(2)+1/a(3)+... with each term a(n) being a positive or negative integer chosen so as to minimize the absolute difference between e and the partial sum.

1, 1, 2, 5, 55, 9999, 3620211522, -26596490130011501642, 6462025287494698350477135653962718736877, -532695733923048954868620962844302990205269539900643893905567041090276924142488084

Offset: 0

Henry Bottomley, Jul 30 2002

sign

			a(4)=55 since e-(1/1)-(1/1)-(1/2)-(1/5)=0.0182818... is closer to 1/55=0.0181818... than to 1/54=0.0185185...

Cf. A001467, A006525, A014015.

A156736 Signed greedy Egyptian fraction for Pi/2.

Original entry on oeis.org

1, 2, 14, -1582, -7497258, 303297921775458, -2646995089135122277190614296178, 82888930564911423983289917045230098319343306166666586941750246

Offset: 0

Jaume Oliver Lafont, Feb 14 2009

sign

The second and fourth convergents of Pi (22/7 and 355/113) appear when truncating the series to three and four terms.

			1+1/2+1/14=11/7=(1/2)(22/7)
1+1/2+1/14-1/1582=355/226=(1/2)(355/113)

Wikipedia, Greedy algorithm for Egyptian fractions

Cf. A001466, A156269, A156618.

Cf. A156750. [From Jaume Oliver Lafont, Mar 03 2009]

x=Pi/2; for (k=0,7, d=round(1/x); x=x-1/d; print1(d,", "))

Sum(n>=0,1/a(n))=Pi/2.

a(n) = 2*A001467(n+1). - R. J. Mathar, Apr 02 2011

cp's OEIS Frontend

A073422 e = 1/a(0)+1/a(1)+1/a(2)+1/a(3)+... with each term a(n) being a positive or negative integer chosen so as to minimize the absolute difference between e and the partial sum.

Views

Author

Keywords

Examples

Crossrefs

A156736 Signed greedy Egyptian fraction for Pi/2.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula