cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A097048 a(n) = least denominator Y of the proper fractions X/Y which need n or more terms as an Egyptian fraction.

Original entry on oeis.org

2, 3, 5, 11, 17, 79, 733, 27539
Offset: 1

Views

Author

Ed Pegg Jr and Don Reble, Jul 21 2004

Keywords

Comments

These are the simplest proper fractions requiring n parts as an Egyptian fraction, where "simplest" means smallest denominator and the smallest numerator breaks ties: 1/2, 2/3, 4/5, 8/11, 16/17, 77/79, 732/733, ...
Checking just (p-1)/p for prime p finds no example requiring 9 parts for p <= 800399: see "results-single" in the github link. - Hugo van der Sanden, Feb 28 2015

Examples

			27538/27539 is the simplest rational that cannot be expressed as the sum of 7 or fewer distinct unit fractions. That is, no rational p/q requires 8 or more with 0 < p/q < 1, and either q < 27539 or (q = 27539 and p < 27538). - _Hugo van der Sanden_, Sep 14 2010

References

  • R. K. Guy, Unsolved Problems in Number Theory, D11

Links

Crossrefs

See A097049 for numerators.

Extensions

a(8) from Hugo van der Sanden, Sep 14 2010

A285261 a(2n-1) and a(2n) represent the smallest fraction a(2n-1)/a(2n) needing n unit fractions.

Original entry on oeis.org

1, 2, 2, 3, 4, 5, 8, 11, 16, 17, 77, 79, 732, 733, 27538, 27539
Offset: 1

Views

Author

Robert G. Wilson v, Apr 15 2017

Keywords

Comments

Huang Zhibin of China in April 2014 has verified that this fraction (27538/27539) needs 8 unit fractions.

Examples

			For n=1, 1/2 = 1/2;
for n=2, 2/3 = 1/3 + 1/3;
for n=3, 4/5 = 1/2 + 1/4 + 1/20 and 4/5 = 1/2 + 1/5 + 1/10;
for n=4, 8/11 = 1/2 + 1/6 + 1/22 + 1/66;
for n=5, 16/17 = 1/2 + 1/3 + 1/17 + 1/34 + 1/51;
for n=6, 77/79 = 1/2 + 1/3 + 1/8 + 1/79 + 1/474 + 1/632;
for n=7, 732/733 = 1/2 + 1/3 + 1/7 + 1/45 + 1/7330 + 1/20524 + 1/26388;
for n=8, 27538/27539 = 1/2 + 1/3 + 1/7 + 1/43 + 1/1933 + 1/14893663 + 1/1927145066572824 + 1/212829231672162931784; etc.

Links

Crossrefs

Cf. A097048, A097049.

Formula

a(2n-1) = A097049(n); a(2n) = A097048(n). - Jon E. Schoenfield, Jan 11 2020

A330808 Minimum number of unit fractions that must be added to 1/n to reach 1.

Original entry on oeis.org

0, 1, 2, 2, 3, 2, 3, 3, 3, 3, 4, 3, 4, 4, 3, 4, 5, 3, 4, 3, 4, 4, 5, 3, 4, 4, 4, 4, 5, 4, 5, 4, 4, 5, 4, 4, 5, 5, 5, 4, 5, 3, 4, 4, 4, 4, 5, 4, 4, 5, 4, 5, 5, 4, 5, 4, 5, 5, 5, 4, 5, 5, 4, 5, 5, 5, 5, 5, 5, 4, 5, 4, 5, 5, 5, 5, 5, 4, 5, 5, 5, 5, 5, 4, 5, 5, 5, 4, 5, 4, 4, 5, 5, 5, 5, 4, 5, 5, 4, 4, 5, 5, 6
Offset: 1

Views

Author

Jon E. Schoenfield, Jan 11 2020

Keywords

Comments

The unit fraction 1/n and the unit fractions to be added to it need not be distinct.
After a(1)=0, this sequence first differs from A097849 at n=42.
Record high values begin with a(1)=0, a(2)=1, a(3)=2, a(5)=3, a(11)=4, a(17)=5, a(103)=6, a(733)=7, a(27539)=8; of these, the greedy algorithm finds a decomposition of 1-1/n into a(n) unit fractions for all except the last:
1 - 1/1 = 0;
1 - 1/2 = 1/2;
1 - 1/3 = 2/3 = 1/2 + 1/6;
1 - 1/5 = 4/5 = 1/2 + 1/4 + 1/20;
1 - 1/11 = 10/11 = 1/2 + 1/3 + 1/14 + 1/231;
1 - 1/17 = 16/17 = 1/2 + 1/3 + 1/10 + 1/128 + 1/32640;
1 - 1/103 = 102/103 = 1/2 + 1/3 + 1/7 + 1/71 + 1/61430 + 1/4716994695;
1 - 1/733 = 732/733 = 1/2 + 1/3 + 1/7 + 1/45 + 1/4484 + 1/33397845 + 1/2305193137933140;
for 1 - 1/27539 = 27538/27539, the greedy algorithm gives 1/2 + 1/3 + 1/7 + 1/43 + 1/1933 + 1/14893663 + 1/1927127616646187 + 1/4212776934617443752169071350384 + 1/305910674290876542045680841765889946094783697598408841178664976, the sum of 9 unit fractions, but decompositions using only 8 unit fractions exist (e.g., 1/2 + 1/3 + 1/7 + 1/55 + 1/245 + 1/671 + 1/51423 + 1/758368982).

Examples

			For n=1, 1/n = 1/1 = 1, which is already at 1, so no additional unit fractions are needed, thus a(1)=0.
For n=2, 1/n = 1/2; adding the single unit fraction 1/2 gives 1/2 + 1/2 = 1, so a(2)=1.
There is no integer k such that 1/3 + 1/k = 1 (solving for k would give k = 3/2), so a(3) > 1. However, 1/3 + 1/2 + 1/6 = 1, so a(3)=2.
There is no integer k such that 1/5 + 1/k = 1, nor are there any two (not necessarily distinct) integers k1,k2 such that 1/5 + 1/k1 + 1/k2 = 1; however, 1/5 + 1/2 + 1/4 + 1/20 = 1, so a(5)=3.
There is no integer k such that 1/11 + 1/k = 1, no pair of integers k1,k2 such that 1/11 + 1/k1 + 1/k2 = 1, and no set of three integers k1,k2,k3 such that 1/11 + 1/k1 + 1/k2 + 1/k3 = 1, but 1/11 + 1/2 + 1/3 + 1/14 + 1/231 = 1, so a(11)=4.

Crossrefs

Cf. A097048, A097049, A097847, A192881, A285261.

Formula

a(n) = A097847(n, n-1).
Showing 1-3 of 3 results.

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