A192881 -id:A192881 - cp's OEIS frontend

A157248 'Greedy' sequence formed by summing unit fractions until the sum is 1, and repeating using up the 'left over' fractions.

1, 2, 3, 6, 4, 5, 7, 8, 9, 10, 15, 230, 57960, 11, 12, 13, 14, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 1544, 8242614, 92401258430373, 36895712779713620978746324067

Offset: 1

Jeremy Gardiner, Feb 25 2009

nonn

Subsequence of starting elements of each pass may be related to A002387 1,2,4,11,31,83,... - David W. Wilson

			1/2+1/3+1/6=1

H. Ibstedt, Computer Analysis of Number Sequences, American Research Press, 1998; Chapter VI.2 Integers represented as sums of terms of the harmonic series.

Jeremy Gardiner, Table of n, a(n) for n=1..94
K. S. Brown's Mathpages, The Greedy Algorithm for Unit Fractions

Cf. A192881.

{r=1;u=[];l=1;for(n=1,99,while(setsearch(u,l),l++);m=ceil(1/r);while(setsearch(u,m),m++);print1(m",");r-=1/m;r||r=1;u=setunion(u,Set(m)))} \\ M. F. Hasler

A294651 Least possible value for the highest denominator in the decomposition of unity as a sum of different unitary fractions the greatest of which is 1/n.

Original entry on oeis.org

1, 6, 15, 20, 24, 28, 33, 40, 48, 52, 65, 65, 75, 76, 85, 88, 91, 100, 105, 115, 115, 119, 132, 140, 144, 145, 155, 161, 162, 171, 217, 174, 182, 190, 195, 196, 296, 200, 207, 220, 246, 224, 301, 231, 238, 253, 329, 275, 280, 287, 288, 296, 371, 300, 304, 305

Offset: 1

Javier Múgica, Nov 06 2017

The decompositions need not be unique. E.g., for a(7) either 1/12 or 1/20 + 1/30 may be used in the decomposition indifferently.

For prime numbers p and any fixed epsilon < 1, a(p) > epsilon*p*log(p) for all sufficiently large p.

			1 = 1/3 + 1/4 + 1/6 + 1/10 + 1/12 + 1/15, and there is no such decomposition starting at 1/3 and having a greatest denominator smaller than 15, so a(3)=15.

Jon E. Schoenfield, All unitary decompositions (listed in lexicographic order) for n = 1..30. (Decompositions up to n = 18 originally found by J. Múgica.)
Javier Múgica, Values of a(n)/n.

Cf. A192881, which looks at decompositions with the least possible number of terms. Those from this sequence achieve those bounds up to a(7), with exception of a(3). However, n=7 is likely the last value of n for which this holds.

Cf. A272083.

a(18)-a(24) from Jon E. Schoenfield, Dec 22 2019

a(25)-a(56) from Jon E. Schoenfield, Jan 01 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

Jon E. Schoenfield, Jan 11 2020

nonn

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).

			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.

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

a(n) = A097847(n, n-1).

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A157248 'Greedy' sequence formed by summing unit fractions until the sum is 1, and repeating using up the 'left over' fractions.

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A294651 Least possible value for the highest denominator in the decomposition of unity as a sum of different unitary fractions the greatest of which is 1/n.

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

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