A073101 -id:A073101 - cp's OEIS frontend

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.

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

A257840 y-value of the lexicographically first integer solution (x,y,z) of 4/n = 1/x + 1/y + 1/z with 0 < x < y < z, or 0 if there is no such solution. Corresponding x and z values are in A257839 and A257841.

Original entry on oeis.org

0, 0, 4, 3, 4, 7, 15, 7, 10, 16, 34, 13, 18, 29, 61, 21, 30, 46, 96, 31, 43, 67, 139, 43, 60, 92, 190, 57, 78, 121, 249, 73, 100, 154, 316, 91, 124, 191, 391, 111, 154, 232, 474, 133, 181, 277, 565, 157, 99, 326, 664, 183, 248, 379, 771, 211, 286, 436, 886, 241, 326, 497, 1009, 273, 370, 562, 1140, 307, 415, 631, 1279, 343, 210, 704, 1426, 381, 514, 781, 1581, 421

Offset: 1

M. F. Hasler, May 16 2015

nonn

See A073101 for more details.

This differs from A075246 starting with a(89)=690 vs A075246(89)=306, corresponding to the representations 4/89 = 1/23 + 1/690 + 1/61410 = 1/24 + 1/306 + 1/108936.

M. F. Hasler, Table of n, a(n) for n = 1..1000

Cf. A073101, A075245, A075246, A075247.

apply( {A257840(n, t)=for(x=n\4+1, 3*n\4, for(y=max(1\t=4/n-1/x, x)+1, ceil(2/t)-1, numerator(t-1/y)==1 && return(y)))}, [1..99]) \\ improved by M. F. Hasler, Jul 03 2022

A292624 Number of solutions to 4/p = 1/x + 1/y + 1/z in positive integers, where p is the n-th prime.

Original entry on oeis.org

3, 12, 12, 36, 48, 24, 24, 60, 120, 42, 108, 54, 42, 78, 198, 78, 156, 66, 96, 234, 42, 216, 156, 60, 48, 96, 156, 144, 90, 78, 192, 186, 102, 210, 108, 180, 144, 138, 384, 156, 276, 102, 396, 36, 138, 246, 174, 342, 216, 120, 114, 630, 48, 300

Offset: 1

Hugo Pfoertner, Sep 20 2017

nonn

Corrected version of A192788.

			a(3) = 12 because 4/(3rd prime) = 4/5 can be expressed in the following 12 ways:
  4/5 =  1/2  + 1/4  + 1/20
  4/5 =  1/2  + 1/5  + 1/10
  4/5 =  1/2  + 1/10 + 1/5
  4/5 =  1/2  + 1/20 + 1/4
  4/5 =  1/4  + 1/2  + 1/20
  4/5 =  1/4  + 1/20 + 1/2
  4/5 =  1/5  + 1/2  + 1/10
  4/5 =  1/5  + 1/10 + 1/2
  4/5 =  1/10 + 1/2  + 1/5
  4/5 =  1/10 + 1/5  + 1/2
  4/5 =  1/20 + 1/2  + 1/4
  4/5 =  1/20 + 1/4  + 1/2

For references and links see A192787.

Hugo Pfoertner, Table of n, a(n) for n = 1..1000
Christian Elsholtz, Terence Tao, Counting the number of solutions to the Erdos-Straus equation on unit fractions, arXiv:1107.1010 [math.NT], 2011-2015.
Christian Elsholtz, Terence Tao, Counting the number of solutions to the Erdos-Straus equation on unit fractions, Journal of the Australian Mathematical Society, 94(1), 50-105, 2013. doi:10.1017/S1446788712000468.

Cf. A073101, A192787, A192789, A292581.

checkmult[a_, b_, c_] := If[Denominator[c] == 1, If[a == b && a == c && b == c, Return[1], If[a != b && a != c && b != c, Return[6], Return[3]]], Return[0]];
a292581[n_] := Module[{t, t1, s, a, b, c, q = Quotient}, t = 4/n; s = 0; For[a = q[1, t] + 1, a <= q[3, t], a++, t1 = t - 1/a; For[b = Max[q[1, t1] + 1, a], b <= q[2, t1], b++, c = 1/(t1 - 1/b); s += checkmult[a, b, c]]]; Return[s]];
Reap[For[n = 1, n <= 54, n++, Print[n, " ", an = a292581[Prime[n]]]; Sow[an]]][[2, 1]] (* Jean-François Alcover, Dec 02 2018, adapted from PARI *)

checkmult (a,b,c) =
{
  if(denominator(c)==1,
     if(a==b && a==c && b==c,
        return(1),
        if(a!=b && a!=c && b!=c,
           return(6),
           return(3)
          )
       ),
     return(0)
     )
}
a292624(n) =
{
  local(t, t1, s, a, b, c);
  t = 4/prime(n);
  s = 0;
  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);
          s+=checkmult(a,b,c);
         )
      );
  return(s);
}
for (n=1,54,print1(a292624(n),", "))

a(n) = A292581(A000040(n)).

A349083 The number of three-term Egyptian fractions of rational numbers x/y, 0 < x/y < 1, ordered as below. The sequence is the number of (p,q,r) such that x/y = 1/p + 1/q + 1/r where p, q, and r are integers with p < q < r.

Original entry on oeis.org

6, 15, 5, 22, 6, 3, 30, 9, 7, 2, 45, 15, 6, 5, 1, 36, 14, 6, 5, 3, 1, 62, 22, 16, 6, 5, 3, 2, 69, 21, 15, 4, 9, 5, 2, 1, 84, 30, 15, 9, 6, 7, 2, 2, 1, 56, 22, 13, 7, 3, 5, 2, 0, 0, 0, 142, 45, 22, 15, 12, 6, 9, 5, 3, 1, 2, 53, 17, 8, 4, 5, 1, 6, 3, 1, 1, 1, 0, 124, 36, 27, 14, 18, 6, 6, 5, 2, 3, 1, 1, 0

Offset: 1

Jud McCranie, Nov 09 2021

The sequence are the terms in a triangle, where the rows correspond to the denominator of the rational number (starting with row 2, column 1) and the columns correspond to the numerators:
            x = 1   2   3   4   5   Rationals x/y:
Row 1: (y=2)    6                   1/2
Row 2: (y=3)   15,  5               1/3, 2/3
Row 3: (y=4)   22,  6,  3           1/4, 2/4, 3/4
Row 4: (y=5)   30,  9,  7,  2       1/5, 2/5, 3/5, 4/5
Row 5: (y=6)   45, 15,  6,  5,  1   1/6, 2/6, 3/6, 4/6, 5/6
Alternatively, order the rational numbers, x/y, 0 < x/y < 1, in this order: 1/2, 1/3, 2/3, 1/4, 2/4, 3/4, 1/5, 2/5, ... The numerators of the n-th rational number are A002260(n) and the denominators are A003057(n).

			The sixth rational number is 3/4;
  3/4 = 1/2 + 1/5 + 1/20
      = 1/2 + 1/6 + 1/12
      = 1/3 + 1/4 + 1/5,
so a(6)=3.

Jud McCranie, Table of n, a(n) for n = 1..990

Cf. A002260, A003057, A349082.

Columns x=1..5: A227610, A227611, A075785, A073101, A075248.

Efrac3(x,y)=sum(p=if(y%x,y\x,y\x+1),3*y\x, my(N=x/y-1/p); sum(q=max(if(numerator(N)==1,1\N+1,1\N),p+1),2\N, my(M=N-1/q,r=1/M); type(r)=="t_INT" && qCharles R Greathouse IV, Nov 09 2021

A075785 Number of ways to express 3/n as Egyptian fractions in just three terms.

Original entry on oeis.org

0, 1, 1, 3, 7, 6, 6, 16, 15, 15, 13, 22, 8, 27, 30, 26, 21, 45, 8, 59, 36, 29, 20, 62, 32, 45, 69, 67, 27, 84, 14, 59, 56, 44, 129, 142, 15, 45, 53, 130, 31, 124, 21, 131, 178, 40, 29, 118, 38, 88, 67, 102, 43, 191, 102, 180, 74, 57, 43, 274, 21, 75, 227, 86, 144, 145, 23, 121, 87

Offset: 1

Robert G. Wilson v, Aug 18 2002

nonn

Christian Elsholtz, Sums Of k Unit Fractions
David Eppstein, Algorithms for Egyptian Fractions

Cf. A073101, A347566, A347569.

Needs["MyOwn`Egypt`"]; Table[ Length[ EgyptianFraction[3/n, Method -> Lexicographic, MaxTerms -> 3, MinTerms -> 3, Duplicates -> Disallow, OutputFormat -> Plain]], {n, 5, 70}]
f[n_] := Length@ Solve[3/n == 1/x + 1/y + 1/z && 0 < x < y < z, {x, y, z}, Integers]; Array[f, 69] (* Robert G. Wilson v, Jul 17 2013 *)

A257839 Smallest possible x such that 4/n = 1/x + 1/y + 1/z with 0 < x < y < z all integers, or 0 if there is no such solution. Corresponding y and z values are in A257840 and A257841.

Original entry on oeis.org

0, 0, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 8, 8, 8, 8, 9, 9, 9, 9, 10, 10, 10, 10, 11, 11, 11, 11, 12, 12, 12, 12, 13, 14, 13, 13, 14, 14, 14, 14, 15, 15, 15, 15, 16, 16, 16, 16, 17, 17, 17, 17, 18, 18, 18, 18, 19, 20, 19, 19, 20, 20, 20, 20, 21

Offset: 1

M. F. Hasler, May 16 2015

nonn

Otherwise said, x-value of the lexicographically first solution (x,y,z) to the given equation.
See A073101 for more details about these sequences related to the Erdős-Straus conjecture.
This differs from A075245 starting with a(89)=23 vs A075245(89)=24, respective solutions being 1/23 + 1/690 + 1/61410 = 1/24 + 1/306 + 1/108936 = 4/89.

M. F. Hasler, Table of n, a(n) for n = 1..1000

Cf. A073101, A075245, A075246, A075247.

Cf. A257840, A257841.

apply( {A257839(n, t)=for(x=n\4+1, 3*n\4, for(y=max(1\t=4/n-1/x, x)+1, ceil(2/t)-1, numerator(t-1/y)==1 && return(x)))}, [1..99]) \\ improved by M. F. Hasler, Jul 03 2022

Conjecture: a(n) = floor(n/4) + d with d = 1 for all n > 2 except some n = 24k + 1 (k = 2, 3, 7, 8, 10, 13, 15, 17, 18, 23, 25, 28, 30, 32, 33, 37, 40, 43, ...) where d = 2. - M. F. Hasler, Jul 03 2022

A257841 z-value of the lexicographically first solution (x,y,z) of 4/n = 1/x + 1/y + 1/z with 0 < x < y < z all integers, or 0 if there is no such solution. Corresponding x and y values are in A257839 and A257840.

Original entry on oeis.org

0, 0, 12, 6, 20, 42, 210, 42, 90, 240, 1122, 156, 468, 812, 3660, 420, 510, 2070, 9120, 930, 1806, 4422, 19182, 1806, 2100, 8372, 35910, 3192, 9048, 14520, 61752, 5256, 9900, 23562, 99540, 8190, 22940, 36290, 152490, 12210, 6314, 53592, 224202, 17556, 32580, 76452, 318660, 24492, 9702, 105950, 440232, 33306, 92008, 143262, 593670, 44310, 81510, 189660, 784110, 57840

Offset: 1

M. F. Hasler, May 16 2015

nonn

See A073101 for more details.

This differs from A075247 starting with a(89) = 61410 vs. A075247(89) = 108936, corresponding to the representations 4/89 = 1/23 + 1/690 + 1/61410 = 1/24 + 1/306 + 1/108936.

M. F. Hasler, Table of n, a(n) for n = 1..1000

Cf. A073101, A075245, A075246, A075247.

apply( {A257841(n, t)=for(x=n\4+1, 3*n\4, for(y=max(1\t=4/n-1/x, x)+1, ceil(2/t)-1, numerator(t-1/y)==1 && return(y/(t*y-1))))}, [1..99]) \\ improved by M. F. Hasler, Jul 03 2022

cp's OEIS Frontend

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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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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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.

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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.

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A257840 y-value of the lexicographically first integer solution (x,y,z) of 4/n = 1/x + 1/y + 1/z with 0 < x < y < z, or 0 if there is no such solution. Corresponding x and z values are in A257839 and A257841.

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PARI

A292624 Number of solutions to 4/p = 1/x + 1/y + 1/z in positive integers, where p is the n-th prime.

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A349083 The number of three-term Egyptian fractions of rational numbers x/y, 0 < x/y < 1, ordered as below. The sequence is the number of (p,q,r) such that x/y = 1/p + 1/q + 1/r where p, q, and r are integers with p < q < r.

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PARI

A075785 Number of ways to express 3/n as Egyptian fractions in just three terms.

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A257839 Smallest possible x such that 4/n = 1/x + 1/y + 1/z with 0 < x < y < z all integers, or 0 if there is no such solution. Corresponding y and z values are in A257840 and A257841.

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