A004194 -id:A004194 - cp's OEIS frontend

A002966 Egyptian fractions: number of solutions of 1 = 1/x_1 + ... + 1/x_n where 0 < x_1 <= ... <= x_n.

1, 1, 3, 14, 147, 3462, 294314, 159330691

Offset: 1

All denominators in the expansion 1 = 1/x_1 + ... + 1/x_n are bounded by A000058(n-1), i.e., 0 < x_1 <= ... <= x_n < A000058(n-1). Furthermore, for a fixed n, x_i <= (n+1-i)*(A000058(i-1)-1). - Max Alekseyev, Oct 11 2012
From R. J. Mathar, May 06 2010: (Start)
This is the leading edge of the triangle A156869. This is also the row n=1 of an array T(n,m) which gives the number of ways to write 1/n as a sum over m (not necessarily distinct) unit fractions:
   1, 1,  3,  14,   147,   3462, 294314, ...
   1, 2, 10, 108,  2892, 270332, ...
   1, 2, 21, 339, 17253, ...
   1, 3, 28, 694, 51323, ...
   ...
T(.,2) = A018892. T(.,3) = A004194. T(.,4) = A020327, T(.,5) = A020328. T(2,6) is computed by D. S. McNeil, who conjectures that the 2nd row is A003167. (End)
If on the other hand, all x_k must be unique, see A006585. - Robert G. Wilson v, Jul 17 2013

			For n=3 the 3 solutions are {2,3,6}, {2,4,4}, {3,3,3}.
For n=4 the solutions are: {2,3,7,42}, {2,3,8,24}, {2,3,9,18}, {2,3,10,15}, {2,3,12,12}, {2,4,5,20}, {2,4,6,12}, {2,4,8,8}, {2,5,5,10}, {2,6,6,6}, {3,3,4,12}, {3,3,6,6}, {3,4,4,6}, {4,4,4,4}. [Neven Juric, May 14 2008]

R. K. Guy, Unsolved Problems in Number Theory, D11.
D. Singmaster, The number of representations of one as a sum of unit fractions, unpublished manuscript, 1972.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Matthew Brendan Crawford, On the Number of Representations of One as the Sum of Unit Fractions, Master's Thesis, Virginia Polytechnic Institute and State University (2019).
Yuya Dan, Representation of one as the sum of unit fractions, International Mathematical Forum 6:1 (2011), pp. 25-30.
Jacques Le Normand, C++ code for a(8) [Broken link]
Jacques Le Normand, C++ code for a(8) [Cached copy]
Joel Louwsma, On solutions of Sum_{i=1..n} 1/x_i = 1 in integers of the form 2^a*k^b, where k is a fixed odd positive integer, arXiv:2402.09515 [math.NT], 2024.
David Singmaster, The number of representations of one as a sum of unit fractions, Unpublished M.S., 1972
R. G. Wilson, v, Fax to N. J. A. Sloane, Sep 9, 1994, with copy of Scientific American column by Ian Stewart
Index entries for sequences related to Egyptian fractions

Cf. A002967, A006585, A000058, A348625.

a(n,rem=1,mn=1)=if(n==1,return(numerator(rem)==1)); sum(k=max(1\rem+1,mn), n\rem, a(n-1,rem-1/k,k)) \\ Charles R Greathouse IV, Jan 04 2015

a(n) <= binomial(A007018(n), n-1). - Charles R Greathouse IV, Jul 29 2024

a(7) from Jud McCranie, Nov 15 1999. Confirmed by Marc Paulhus.

a(8) from John Dethridge (jcd(AT)ms.unimelb.edu.au) and Jacques Le Normand (jacqueslen(AT)sympatico.ca), Jan 06 2004

A192787 Number of distinct solutions of 4/n = 1/a + 1/b + 1/c in positive integers satisfying 1 <= a <= b <= c.

Original entry on oeis.org

0, 1, 3, 3, 2, 8, 7, 10, 6, 12, 9, 21, 4, 17, 39, 28, 4, 26, 11, 36, 29, 25, 21, 57, 10, 20, 29, 42, 7, 81, 19, 70, 31, 25, 65, 79, 9, 32, 73, 96, 7, 86, 14, 62, 93, 42, 34, 160, 18, 53, 52, 59, 13, 89, 98, 136, 41, 33, 27, 196, 11, 37, 155, 128, 49, 103, 17, 73, 55, 185, 40, 211, 7, 32, 129, 80, 97, 160, 37, 292

Offset: 1

Charles R Greathouse IV, Jul 10 2011

nonn

The Erdős-Straus conjecture is that a(n) > 0 for n > 1. Swett verified the conjecture for n < 10^14.
Vaughan shows that the number of n < x with a(n) = 0 is at most x exp(-c * (log x)^(2/3)) for some c > 0.
See A073101 for the 4/n conjecture due to Erdős and Straus.

			a(1) = 0, since 4/1 = 4 cannot be expressed as the sum of three reciprocals.
a(2) = 1 because 4/2 = 1/1 + 1/2 + 1/2, and there are no other solutions.
a(3) = 3 since 4/3 = 1 + 1/4 + 1/12 = 1 + 1/6 + 1/6 = 1/2 + 1/2 + 1/3.
a(4) = 3 = A002966(3).

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
Christian Elsholtz, Terence Tao, Counting the number of solutions to the Erdős-Straus equation in unit fractions, arXiv:1107.1010 [math.NT], 2010-2015.
Maria Monks, Ameya Velingker, On the Erdős-Straus conjecture: Properties of solutions to its underlying diophantine equation. See also.
Alan Swett, The Erdos-Strauss Conjecture, 1999.
R. C. Vaughan, On a problem of Erdős, Straus and Schinzel, Mathematika 17 (1970), pp. 193-198.
Wikipedia, Erdős-Straus conjecture.
Konstantine Zelator, An ancient Egyptian problem: the diophantine equation 4/n=1/x+1/y+1/z, n>or=2, arXiv:0912.2458 [math.GM], 2009.

A292581 counts the solutions with multiplicity. A073101 counts solutions with a, b, and c distinct.
Cf. A292581, A292624, A192789, A075245, A075246, A075247, A075248.
Cf. A004194, A226641, A226642, A226644, A226645, A226646.
Cf. A337432 (solutions with minimal c).

A192787 := proc(n) local t,a,b,t1,count; t:= 4/n; count:= 0; for a from floor(1/t)+1 to floor(3/t) do t1:= t - 1/a; for b from max(a,floor(1/t1)+1) to floor(2/t1) do if type( 1/(t1 - 1/b),integer) then count:= count+1; end if end do end do; count; end proc; # Robert Israel, Feb 19 2013

f[n_] := Length@ Solve[ 4/n == 1/x + 1/y + 1/z && 1 <= x <= y <= z, {x, y, z}, Integers]; Array[f, 70] (* Allan C. Wechsler and Robert G. Wilson v, Jul 19 2013 *)

a(n, show=0)=my(t=4/n, t1, s, c); for(a=1\t+1, 3\t, t1=t-1/a; for(b=max(1\t1+1, a), 2\t1, c=1/(t1-1/b); if(denominator(c)==1&&c>=b, s++; show&&print("4/",n," = 1/",a," + 1/",b," + 1/",c)))); s \\ variant with print(...) added by Robert Munafo, Feb 19 2013, both combined through option "show" by M. F. Hasler, Jul 02 2022

Corrected at the suggestion of Allan C. Wechsler by Charles R Greathouse IV, Feb 19 2013

Examples and cross-references added by M. F. Hasler, Feb 19 2013

A063520 Sum divides product: number of solutions (r,s,t), r>=s>=t>0, to the equation rst = n(r+s+t).

Original entry on oeis.org

1, 3, 6, 5, 8, 8, 8, 14, 13, 9, 14, 17, 8, 18, 23, 18, 14, 17, 13, 33, 23, 10, 19, 36, 15, 22, 32, 22, 19, 26, 17, 39, 24, 18, 50, 45, 8, 22, 39, 38, 22, 27, 13, 50, 45, 16, 27, 52, 24, 39, 38, 27, 20, 50, 45, 72, 24, 12, 31, 58, 15, 28, 69, 45, 49, 39, 12, 52, 40, 33, 33, 66, 12, 33, 64

Offset: 1

Jud McCranie and Vladeta Jovovic, Aug 01 2001

nonn

Number of solutions (r,s) in positive integers to the equation rs = n(r+s) is tau(n^2), cf. A048691. Number of solutions (r,s), r>=s>0, to the equation rs = n(r+s) is (tau(n^2)+1)/2, cf. A018892.

Conjecturally, includes all positive integers except 2, 4, 7 and 11 - David W. Wilson

			There are 8 such solutions to rst = 5(r+s+t): (5, 4, 3), (7, 5, 2), (10, 4, 2), (11, 10, 1), (15, 8, 1), (20, 7, 1), (25, 3, 2), (35, 6, 1).

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000
M. J. Pelling, Problem 10745, Amer. Math. Monthly, vol. 106 (1999), p. 587.
M. J. Pelling and F. W. Roush, The Sum Divides the Product: Problem 10745, Amer. Math. Monthly, vol. 108, (no. 7, Aug. 2001), pp. 668-669. [Gives upper bound]

Cf. A018892, A004194, A063525.

(* Assuming s <= 2n and t <= n*(n+2) *) redu[n_] := Reap[ Do[ red = Reduce[0 < r <= s <= t && r*s*t == n*(r+s+t), r, Integers]; If[red =!= False, Sow[{r, s, t} /. ToRules[red] ] ], {s, 1, 2*n}, {t, s, n*(n+2)}] ][[2, 1]]; a[n_] := redu[n] // Length; a[1] = 1; Table[ Print[n, " ", an = a[n]]; an, {n, 1, 75}] (* Jean-François Alcover, Feb 22 2013 *)

a(n)=sum(t=1,sqrtint(3*n),sum(s=t,sqrtint(n^2+t)+n,my(N=n*(s+t), D=s*t-n);D&&denominator(N/D)==1&&N/D>=s)) \\ Charles R Greathouse IV, Feb 22 2013

More terms from David W. Wilson, Aug 01 2001

A241883 Number of ways 1/n can be expressed as the sum of four distinct unit fractions: 1/n = 1/w + 1/x + 1/y + 1/z satisfying 0 < w < x < y < z.

Original entry on oeis.org

6, 71, 272, 586, 978, 1591, 1865, 3115, 3772, 4964, 4225, 8433, 4987, 10667, 13659, 10845, 7513, 17360, 9569, 28554, 23309, 17220, 12326, 37554, 19984, 24091, 31056, 42343, 16095, 57001, 15076, 42655, 46885, 38416, 77887, 71959, 16692, 42054, 68894, 95914, 24566, 100023, 24224, 99437, 108756, 41907, 29711, 127069, 52811, 94745, 83433

Offset: 1

Robert G. Wilson v, Apr 30 2014

nonn

			1/1 = 1/2 + 1/3 + 1/7  + 1/42
    = 1/2 + 1/3 + 1/8  + 1/24
    = 1/2 + 1/3 + 1/9  + 1/18
    = 1/2 + 1/3 + 1/10 + 1/15
    = 1/2 + 1/4 + 1/5  + 1/20
    = 1/2 + 1/4 + 1/6  + 1/12
so a(1) = 6.

Jud McCranie, Table of n, a(n) for n = 1..644

Cf. A002966, A004194, A020327, A227610.

a[n_] := Length@ Solve[1/n == 1/w + 1/x + 1/y + 1/z && 0 < w < x < y < z, {w, x, y, z}, Integers]; Array[f, 21]

A226645 Number of ways to express 6/n as Egyptian fractions in just three terms; i.e., 6/n = 1/x + 1/y + 1/z satisfying 1<=x<=y<=z.

Original entry on oeis.org

0, 1, 1, 3, 3, 3, 1, 6, 8, 10, 7, 10, 1, 9, 12, 20, 8, 21, 2, 21, 17, 16, 11, 28, 7, 11, 26, 33, 9, 36, 3, 31, 25, 24, 60, 57, 2, 11, 20, 68, 10, 42, 6, 35, 81, 23, 15, 70, 10, 37, 25, 51, 14, 79, 33, 76, 32, 30, 20, 96, 2, 17, 86, 65, 48, 62, 9, 50, 42, 138, 35, 160, 2, 18, 53, 51, 52, 59, 8, 142, 89, 34, 23, 136, 37, 24, 33, 140, 23, 196, 30, 46, 37, 32, 75, 128, 5, 43, 103, 98

Offset: 1

Allan C. Wechsler and Robert G. Wilson v, Aug 17 2013

nonn

See A073101 for the 4/n conjecture due to Erdős and Straus.

Daniel Leary, Table of n, a(n) for n = 1..1000

Cf. A226640, A004194, A226642, A192787, A226644, A226646.

a[n_] := Length@ Solve[ 6/n == 1/x + 1/y + 1/z && 1 <= x <= y <= z, {x, y, z}, Integers]; Array[a, 70]

A015995 a(n) = (tau(n^3)+2)/3.

Original entry on oeis.org

1, 2, 2, 3, 2, 6, 2, 4, 3, 6, 2, 10, 2, 6, 6, 5, 2, 10, 2, 10, 6, 6, 2, 14, 3, 6, 4, 10, 2, 22, 2, 6, 6, 6, 6, 17, 2, 6, 6, 14, 2, 22, 2, 10, 10, 6, 2, 18, 3, 10, 6, 10, 2, 14, 6, 14, 6, 6, 2, 38, 2, 6, 10, 7, 6, 22, 2, 10, 6, 22, 2, 24, 2, 6, 10, 10, 6, 22, 2, 18, 5, 6, 2, 38, 6, 6

Offset: 1

Robert G. Wilson v

nonn

Antti Karttunen, Table of n, a(n) for n = 1..10000

Cf. A004194, A018892, A015996, A015999, A016001, A016002, A016003, A016005-A016009, A016012, A016017, A016018, A016020, A016025, A048785.

Table[(DivisorSigma[0,n^3]+2)/3,{n,90}] (* Harvey P. Dale, Apr 30 2018 *)

a(n)=(numdiv(n^3)+2)/3 \\ Charles R Greathouse IV, May 01 2013

a(n)=my(f=factor(n)[,2]);prod(i=1,#f,3*f[i]+1)\3+1 \\ Charles R Greathouse IV, May 01 2013

a(n) = (2+A048785(n))/3. - R. J. Mathar, May 07 2021

Definition corrected by Vladeta Jovovic, Sep 03 2005

A226641 Number of ways to express 2/n as Egyptian fractions in just three terms; i.e., 2/n = 1/x + 1/y + 1/z satisfying 1<=x<=y<=z.

Original entry on oeis.org

1, 3, 8, 10, 12, 21, 17, 28, 26, 36, 25, 57, 20, 42, 81, 70, 25, 79, 32, 96, 86, 62, 42, 160, 53, 59, 89, 136, 33, 196, 37, 128, 103, 73, 185, 211, 32, 80, 160, 292, 40, 245, 40, 157, 235, 93, 60, 366, 85, 156, 147, 174, 42, 230, 223, 340, 143, 106, 76, 497, 34, 90, 331, 269, 206, 322, 50, 211, 175, 453, 72, 538, 37, 85, 332, 216, 260, 378, 69, 604, 167, 121, 79, 623, 204, 104, 203, 473, 59, 648, 253, 204, 166, 135, 318, 706, 46, 227, 427, 437

Offset: 1

Allan C. Wechsler and Robert G. Wilson v, Aug 17 2013

nonn

Daniel Leary, Table of n, a(n) for n = 1..1000
Index entries for sequences related to Egyptian fractions

Cf. A227611, A004194, A226642, A192787, A226644, A226645, A226646.

See A073101 for the 4/n conjecture due to Erdős and Straus.

a[n_] := Length@ Solve[ 2/n == 1/x + 1/y + 1/z && 1 <= x <= y <= z, {x, y, z}, Integers]; Array[a, 70]

A226642 Number of ways to express 3/n as Egyptian fractions in just three terms; i.e., 3/n = 1/x + 1/y + 1/z satisfying 1<=x<=y<=z.

Original entry on oeis.org

1, 3, 3, 6, 10, 10, 9, 20, 21, 21, 16, 28, 11, 33, 36, 31, 24, 57, 11, 68, 42, 35, 23, 70, 37, 51, 79, 76, 30, 96, 17, 65, 62, 50, 138, 160, 18, 51, 59, 142, 34, 136, 24, 140, 196, 46, 32, 128, 43, 98, 73, 111, 46, 211, 111, 192, 80, 63, 46, 292, 24, 81, 245

Offset: 1

Allan C. Wechsler and Robert G. Wilson v, Aug 17 2013

nonn

See A073101 for the 4/n conjecture due to Erdős and Straus.

Cf. A075785, A004194, A226641, A192787, A226644, A226645, A226646.

f[n_] := Length@ Solve[ 3/n == 1/x + 1/y + 1/z && 1 <= x <= y <= z, {x, y, z}, Integers]; Array[f, 70]

A226644 Number of ways to express 5/n as Egyptian fractions in just three terms; i.e., 5/n = 1/x + 1/y + 1/z satisfying 1<=x<=y<=z.

Original entry on oeis.org

0, 1, 2, 4, 3, 4, 4, 7, 12, 10, 3, 17, 6, 21, 21, 12, 6, 26, 13, 28, 22, 18, 9, 61, 36, 18, 24, 48, 22, 57, 5, 27, 38, 26, 42, 60, 11, 24, 56, 70, 6, 71, 13, 79, 79, 19, 12, 99, 41, 96, 38, 55, 12, 84, 62, 86, 50, 41, 36, 160, 6, 26, 104, 57, 59, 76, 16, 71, 74, 136, 12, 158, 22, 60, 196, 52, 65, 103, 25, 128, 46, 30, 15, 224, 73, 32, 58, 141, 38, 211, 71, 67, 59, 41, 80, 151, 24, 97, 222, 292

Offset: 1

Allan C. Wechsler and Robert G. Wilson v, Aug 17 2013

nonn

See A073101 for the 4/n conjecture due to Erdős and Straus.

Cf. A075248, A004194, A226641, A226642, A192787, A226645, A226646.

f[n_] := Length@ Solve[ 5/n == 1/x + 1/y + 1/z && 1 <= x <= y <= z, {x, y, z}, Integers]; Array[f, 70]

A226646 Number of ways to express m/n as Egyptian fractions in just three terms, that is, m/n = 1/x + 1/y + 1/z satisfying 1 <= x <= y <= z and read by antidiagonals.

Original entry on oeis.org

3, 1, 10, 1, 3, 21, 0, 3, 8, 28, 0, 1, 3, 10, 36, 0, 1, 3, 6, 12, 57, 0, 1, 2, 3, 10, 21, 42, 0, 0, 1, 4, 2, 10, 17, 70, 0, 0, 1, 3, 3, 8, 9, 28, 79, 0, 0, 0, 1, 3, 4, 7, 20, 26, 96, 0, 0, 1, 1, 2, 3, 4, 10, 21, 36, 62, 0, 0, 0, 1, 1, 7, 1, 7, 6, 21, 25, 160, 0, 0, 0, 1, 0, 3, 3, 6, 12, 12, 16, 57, 59

Offset: 1

Allan C. Wechsler and Robert G. Wilson v, Aug 17 2013

See A073101 for the 4/n conjecture due to Erdös and Straus.
The first upper diagonal is 10, 8, 6, 2, 4, 1, 2, 1, 2, 0, 3, 0, 0, 1, 0, 0, 1, 0, 1, 0,... .
The main diagonal is: 3, 3, 3, 3, 3, 3, ... since 1 = 1/2 + 1/3 + 1/6 = 1/2 + 1/4 + 1/4 = 1/3 + 1/3 + 1/3. See A002966(3).
The first lower diagonal is 1, 3, 3, 4, 3, 7, 3, 5, 4, 6, 3, 10, 3, 6, 6, 6, 3, 9, 3, 9, ... .
The antidiagonal sum is 3, 11, 25, 39, 50, 79, 79, 104, 131, 157, 140, 229, 169, 220, 295, 282, ... .

			../n
m/ 1...2...3...4...5...6...7...8...9..10..11...12..13...14...15 =Allocation nbr.
.1 3..10..21..28..36..57..42..70..79..96..62..160..59..136..196 A004194
.2 1...3...8..10..12..21..17..28..26..36..25...57..20...42...81 A226641
.3 1...3...3...6..10..10...9..20..21..21..16...28..11...33...36 A226642
.4 0...1...3...3...2...8...7..10...6..12...9...21...4...17...39 A192787
.5 0...1...2...4...3...4...4...7..12..10...3...17...6...21...21 A226644
.6 0...1...1...3...3...3...1...6...8..10...7...10...1....9...12 A226645
.7 0...0...1...1...2...7...3...2...3...5...2...13...8...10....9 n/a
.8 0...0...0...1...1...3...3...3...1...2...0....8...3....7...19 n/a
.9 0...0...1...1...0...3...2...5...3...2...0....6...2....4...10 n/a
10 0...0...0...1...1...2...0...4...4...3...0....4...1....4....8 n/a
Triangle (by antidiagonals):
  {3},
  {1, 10},
  {1, 3, 21},
  {0, 3, 8, 28},
  {0, 1, 3, 10, 36},
  {0, 1, 3, 6, 12, 57},
  ...

Christian Elsholtz, Sums Of k Unit Fractions
David Eppstein, Algorithms for Egyptian Fractions
David Eppstein, Ten Algorithms for Egyptian Fractions
Ron Knott, Egyptian Fractions
Oakland University, The Erdős Number Project
Eric Weisstein's World of Mathematics, Egyptian Fraction
Index entries for sequences related to Egyptian fractions

Cf. A227612, A226640, A226641, A226642, A192787, A226644, A226645.

f[m_, n_] := Length@ Solve[m/n == 1/x + 1/y + 1/z && 1 <= x <= y <= z, {x, y, z}, Integers]; Table[f[n, m - n + 1], {m, 12}, {n, m, 1, -1}] // Flatten

cp's OEIS Frontend

A002966 Egyptian fractions: number of solutions of 1 = 1/x_1 + ... + 1/x_n where 0 < x_1 <= ... <= x_n.

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A192787 Number of distinct solutions of 4/n = 1/a + 1/b + 1/c in positive integers satisfying 1 <= a <= b <= c.

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A063520 Sum divides product: number of solutions (r,s,t), r>=s>=t>0, to the equation rst = n(r+s+t).

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A241883 Number of ways 1/n can be expressed as the sum of four distinct unit fractions: 1/n = 1/w + 1/x + 1/y + 1/z satisfying 0 < w < x < y < z.

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A226645 Number of ways to express 6/n as Egyptian fractions in just three terms; i.e., 6/n = 1/x + 1/y + 1/z satisfying 1<=x<=y<=z.

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Mathematica

A015995 a(n) = (tau(n^3)+2)/3.

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PARI

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Extensions

A226641 Number of ways to express 2/n as Egyptian fractions in just three terms; i.e., 2/n = 1/x + 1/y + 1/z satisfying 1<=x<=y<=z.

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Mathematica

A226642 Number of ways to express 3/n as Egyptian fractions in just three terms; i.e., 3/n = 1/x + 1/y + 1/z satisfying 1<=x<=y<=z.

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