A073101 -id:A073101 - cp's OEIS frontend

A075259 Number of solutions (x,y,z) to 3/(2n+1) = 1/x + 1/y + 1/z satisfying 0 < x < y < z and odd x, y, z.

0, 1, 2, 1, 1, 5, 2, 1, 3, 5, 1, 12, 8, 3, 3, 5, 14, 8, 6, 4, 7, 20, 1, 9, 6, 3, 22, 11, 3, 11, 31, 24, 5, 10, 3, 11, 16, 20, 6, 23, 2, 35, 7, 3, 35, 15, 25, 16, 47, 8, 12, 54, 3, 9, 8, 4, 42, 41, 22, 11, 8, 25, 8, 15, 5, 61, 92, 3, 7, 16, 28, 47, 37, 7, 10, 40, 23, 13, 11, 29, 11, 75, 3

Offset: 1

T. D. Noe, Sep 10 2002

N. J. A. Sloane and R. H. Hardin conjecture a(n) > 0 for n > 1. All of the solutions can be printed by removing the comment symbols from the Mathematica program. For the solution (x,y,z) having the largest z value, see (A075260, A075261, A075262). See A073101 for the 4/n conjecture due to Erdős and Straus.

The conjecture was proved by Thomas Hagedorn and Gary Mulkey. - T. D. Noe, Jan 03 2005

			a(3)=2 because there are two solutions: 3/7 = 1/3+1/11+1/231 and 3/7 = 1/3+1/15+1/35.

R. K. Guy, Unsolved Problems in Theory of Numbers, Springer-Verlag, Third Edition, 2004, D11.

T. D. Noe, Table of n, a(n) for n=1..499
Thomas R. Hagedorn, A proof of a conjecture on Egyptian fractions, Amer. Math Monthly, 107 (2000), 62-63.
Stan Wagon, Problem of the Week 848: An Odd Egyptian Puzzle
Eric Weisstein's World of Mathematics, Egyptian Fraction

Cf. A073101, A075260, A075261, A075262.

m = 3; For[lst = {}; n = 3, n <= 200, n = n + 2, cnt = 0; xr = n/m; If[IntegerQ[xr], xMin = xr + 1, xMin = Ceiling[xr]]; If[IntegerQ[3xr], xMax = 3xr - 1, xMax = Floor[3xr]]; For[x = xMin, x <= xMax, x++, yr = 1/(m/n - 1/x); If[IntegerQ[yr], yMin = yr + 1, yMin = Ceiling[yr]]; If[IntegerQ[2yr], yMax = 2yr + 1, yMax = Ceiling[2yr]]; For[y = yMin, y <= yMax, y++, zr = 1/(m/n - 1/x - 1/y); If[y > x && zr > y && IntegerQ[zr], z = zr; If[OddQ[x y z], cnt++;(*Print[n, " ", x, " ", y, " ", z]*)]]]]; AppendTo[lst, cnt]]; lst

More terms from T. D. Noe, Oct 15 2002

A227611 Number of ways 2/n can be expressed as the sum of three distinct unit fractions: 2/n = 1/x + 1/y + 1/z with 0 < x < y < z.

Original entry on oeis.org

0, 1, 5, 6, 9, 15, 14, 22, 21, 30, 22, 45, 17, 36, 72, 62, 22, 69, 29, 84, 77, 56, 39, 142, 48, 53, 82, 124, 30, 178, 34, 118, 94, 67, 176, 191, 29, 74, 151, 274, 37, 227, 37, 145, 220, 87, 57, 342, 80, 146, 138, 162, 39, 216, 214, 322, 134, 100, 73, 461, 31, 84, 316, 257, 197, 304, 47, 199, 166, 435, 69, 508, 34, 79, 317

Offset: 1

Robert G. Wilson v, Jul 17 2013

nonn

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

Cf. A002966, A073546.

Cf. A227610 (1/n), A075785 (3/n), A073101 (4/n), A075248 (5/n), A227612.

f[n_] := Length@ Solve[2/n == 1/x + 1/y + 1/z && 0 < x < y < z, {x, y, z}, Integers]; Array[f, 75]

A227612 Table read by antidiagonals: Number of ways m/n can be expressed as the sum of three distinct unit fractions, i.e., m/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z and read by antidiagonals.

Original entry on oeis.org

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

Offset: 1

Robert G. Wilson v, Jul 17 2013

The main diagonal is 1, 1, 1, 1, 1, 1, 1, ..., ; i.e., 1 = 1/2 + 1/3 + 1/6.

			  m\n| 1  2   3   4   5   6   7   8   9  10  11   12  13   14   15
  ---+------------------------------------------------------------
   1 | 1  6  15  22  30  45  36  62  69  84  56  142  53  124  178  A227610
   2 | 0  1   5   6   9  15  14  22  21  30  22   45  17   36   72  A227611
   3 | 0  1   1   3   7   6   6  16  15  15  13   22   8   27   30  A075785
   4 | 0  0   1   1   2   5   5   6   4   9   7   15   4   14   33  A073101
   5 | 0  0   1   2   1   1   3   5   9   6   3   12   5   18   15  A075248
   6 | 0  0   0   1   1   1   1   3   5   7   5    6   1    6    9  n/a
   7 | 0  0   0   1   1   4   1   2   2   2   2    9   6    6    7  n/a
   8 | 0  0   0   0   1   1   1   1   1   2   0    5   3    5   15  n/a
   9 | 0  0   0   0   0   1   1   3   1   1   0    3   1    2    7  n/a
  10 | 0  0   0   0   0   1   0   2   2   1   0    1   1    3    5  n/a
.
Antidiagonals are {1}, {0, 6}, {0, 1, 15}, {0, 1, 5, 22}, {0, 0, 1, 6, 30}, {0, 0, 1, 3, 9, 45}, ...

G. C. Greubel, Table of n, a(n) for the first 50 rows, flattened
Christian Elsholtz, Sums Of k Unit Fractions, Trans. Amer. Math. Soc. 353 (2001), 3209-3227.
David Eppstein, Algorithms for Egyptian Fractions
David Eppstein, Ten Algorithms for Egyptian Fractions, Wolfram Library Archive.
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. A002966, A073546, A227610 (1/n), A227611 (2/n), A075785 (3/n), A073101 (4/n), A075248 (5/n).

f[m_, n_] := Length@ Solve[m/n == 1/x + 1/y + 1/z && 0 < x < y < z, {x, y, z}, Integers]; Table[ f[n, m - n + 1], {m, 12}, {n, m, 1, -1}]

A337432 a(n) is the least value of z such that 4/n = 1/x + 1/y + 1/z with 0 < x <= y <= z has at least one solution.

Original entry on oeis.org

2, 3, 3, 10, 6, 14, 6, 9, 10, 33, 9, 52, 14, 12, 12, 102, 18, 57, 15, 21, 22, 138, 18, 50, 26, 27, 21, 232, 24, 248, 24, 33, 34, 30, 27, 370, 38, 39, 30, 164, 35, 258, 33, 36, 46, 329, 36, 98, 50, 51, 39, 742, 54, 44, 42, 57, 58, 885, 45, 549, 62, 56, 48, 60, 66, 603

Offset: 2

Hugo Pfoertner, Oct 13 2020

See A073101 and A192787 for the history of the problem, references, and links.

			a(6)=6 because it is the least denominator z in the A192787(6)=8 solutions
  [x, y, z]: [2, 7, 42], [2, 8, 24], [2, 9, 18], [2, 10, 15], [2, 12, 12],
  [3, 4, 12], [3, 6, 6], [4, 4, 6];
a(13)=52 because the minimum of z in the A192787(13)=4 solutions is 52:
  [4, 18, 468], [4, 20, 130], [4, 26, 52], [5, 10, 130].

Hugo Pfoertner, Table of n, a(n) for n = 2..10000 (first 576 terms from Robert Israel)

Cf. A073101, A192787, A292581.

f:= proc(n) local z,x,y;
  for z from floor(n/4)+1 do
    for x from floor(n*z/(4*z-n))+1 to z do
      y:= n*x*z/(4*x*z-n*x-n*z);
      if y::posint and y >= x and y <= z then return z fi
  od od
end proc:
map(f, [$2..100]); # Robert Israel, Oct 14 2020

a[n_] := For[z = Floor[n/4] + 1, True, z++, For[x = Floor[n(z/(4z - n))] + 1, x <= z, x++, y = n x z/(4 x z - n x - n z); If[IntegerQ[y] && x <= y <= z, Print[z]; Return [z]]]];
a /@ Range[2, 100] (* Jean-François Alcover, Oct 23 2020, after Robert Israel *)

a337432(n)={my(target=4/n,a,b,c,m=oo);for(a=1\target+1,3\target,my(t=target-1/a);for(b=max(1\t+1,a),2\t,c=1/(t-1/b);if(denominator(c)==1,m=min(m,max(a,max(b,c))))));m};
for(k=2,67,print1(a337432(k),", "))

A257843 Numbers n for which the lexicographically first integer solution (x,y,z) of 4/n = 1/x + 1/y + 1/z with 0 < x < y < z, is different from the solution having the largest z-value.

Original entry on oeis.org

89, 113, 161, 233, 281, 329, 353, 401, 409, 449, 473, 521, 593, 641, 689, 713, 761, 769, 809, 929, 953, 1049, 1073, 1121, 1129, 1169, 1193, 1241, 1249, 1313, 1321, 1361, 1369, 1409, 1433, 1481, 1513, 1529, 1553, 1561, 1601, 1609, 1649, 1673, 1721, 1769

Offset: 1

M. F. Hasler, May 16 2015

Related to the Erdős-Straus conjecture, see A073101 for more details.

This lists indices for which (A075245, A075246, A075247) differ from (A257839, A257840, A257841).

			For n=89, 4/89 = 1/23 + 1/690 + 1/61410 = 1/24 + 1/306 + 1/108936 are the representations with the largest resp. smallest unit fraction.

Manfred Scheucher, Table of n, a(n) for n = 1..62

Cf. A073101, A075245, A075246, A075247, A257839, A257840, A257841.

is(n,s=0)=for(c=n\4+1,n*3\4,for(b=c+1,ceil(2/(t=4/n-1/c))-1,numerator(t-1/b)==1||next;!s&&(s=t-1/b)&&next(2);t-1/b

More terms from Manfred Scheucher, May 24 2015

A154962 The terms of this sequence are integer values of consecutive denominators (with signs) from the fractional expansion (using only fractions with numerators to be positive 1's) of the BBP polynomial ( 4/(8k+1) - 2/(8k+4) - 1/(8k+5) - 1/(8k+6) ) for all k (starting from 0 to infinity); for k>=1 the Erdos-Straus conjecture is applied to the first fraction - so it is always replaced by exactly three fractions.

Original entry on oeis.org

1, 1, 1, 1, -2, -5, -6, 3, 10, 90, -5, -13, -14, 5, 30, 510, -10, -21, -22, 7, 60, 2100, -14, -29, -30

Offset: 0

Alexander R. Povolotsky, Jan 18 2009, corrected Jan 20 2009

sign

This sequence is different from A154925, where the first fraction for k>=1 is expanded with Egyptians fractions, using R.Knott's converter calculator #1 (http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fractions/egyptian.html#calc1)

Cf. A073101, A075245, A075246, A075247, A154925

A213340 Numbers which are the values of the quadratic polynomial 5+8t+12k+16kt on nonnegative integers.

Original entry on oeis.org

5, 13, 17, 21, 29, 37, 41, 45, 53, 61, 65, 69, 77, 85, 89, 93, 97, 101, 109, 113, 117, 125, 133, 137, 141, 149, 153, 157, 161, 165, 173, 181, 185, 189, 197, 205, 209, 213, 221, 229, 233, 237, 241, 245, 253, 257

Offset: 1

Michel Mizony, Jun 09 2012

nonn

For all these numbers a(n) we have the following Erdős-Straus decomposition: 4/p = 4/(5+8*t+12*k+16*k*t) = 1/(2*(2*k+1)*(2+3*t+3*k+4*k*t)) + 1/(2+3*t+3*k+4*k*t) + 1/(2*(5+8*t+12*k+16*k*t)*(2*k+1)*(2+3*t+3*k+4*k*t)).
Moreover this sequence is related to irreducible twin Pythagorean triples: the associated Pythagorean triple is [2n(n+1), 2n+1,2n(n+1)+1], where n=2+4t+6k+8kt.
In 1948 Erdős and Straus conjectured that for any positive integer n >= 2 the equation 4/n = 1/x + 1/y +1/z has a solution with positive integers x, y and z (without the additional requirement 0 < x < y < z).
For the solution (x,y,z) having the largest z value, see (A075245, A075246, A075247).

			For n=5 the a(5)=29 solutions are {k=0, t=3}, {k=2, t=0}.

I. Gueye and M. Mizony, Recent progress about Erdős-Straus conjecture, B SO MA S S, Volume 1, Issue 2, pp. 6-14.
I. Gueye and M. Mizony, Towards the proof of Erdős-Straus conjecture, B SO MA S S, Volume 1, Issue 2, pp. 141-150.

P. Erdős, On a Diophantine equation, (Hungarian. Russian, English summaries), Mat. Lapok 1, 1950, pp. 192-210.
M. Mizony and M.-L. Gardes, Sur la conjecture d'Erdős et Straus, see pages 14-17.
Eric Weisstein's World of Mathematics, Twin Pythagorean Triple.
K. Yamamoto, On the Diophantine Equation 4/n=1/x+1/y+1/z, Mem. Fac. Sci. Kyushu U. Ser. A 19, 37-47, 1965.

Cf. A001844 (centered square numbers: 2n(n+1)+1).
Cf. A005408 (x values), A046092 (y values).
Cf. A195770 (positive integers a for which there is a 1-Pythagorean triple (a,b,c) satisfying a<=b).
A073101 number of solutions (x,y,z) to 4/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z.

G:=(p,d)->4/p = [4*d/(p+d)/(p+1), 4/(p+d), 4*d/(p+d)/(p+1)/p]:
cousin:=proc(p)
local d;
for d from 3 by 4 to 100 do
if ((p+1)/2) mod d=0 and (p+d)*(p+1) mod d=0 then
return([p,G(p,d)]) fi;od;
end:
for k to 20 do cousin(4*k+1) od;

A213686 Numbers which are the values of the quadratic polynomial 13+20t+24k+32kt at nonnegative integers.

Original entry on oeis.org

13, 33, 37, 53, 61, 73, 85, 89, 93, 109, 113, 133, 141, 145, 153, 157, 173, 181, 193, 201, 205, 213, 229, 233, 245, 253, 257, 273, 277, 293, 297, 301, 313, 317, 325, 333, 349, 353, 369, 373, 393, 397, 401, 405, 413, 421, 425, 433, 445, 453, 469, 473, 481

Offset: 1

Michel Mizony, Jun 17 2012

nonn

For all these numbers a(n) we have the following Erdos-Straus decomposition:  4/p=4/(13+24*k+20*t+32*k*t) = 1/(6*k+8*k*t+4+6*t) + 1/((13+24*k+20*t+32*k*t)*(5+8*k)*(3*k+4*k*t+2+3*t)) + 1/(2*(5+8*k)*(3*k+4*k*t+2+3*t)).
Moreover this sequence is related to irreducible twin Pythagorean triples: the associated Pythagorean triple is [2n(n+1),2n+1,2n(n+1)+1], where n=2+4k.
In 1948 Erdos and Straus conjectured that for any positive integer n >= 2 the equation 4/n = 1/x + 1/y +1/z has a solution with positive integers x, y and z (without the additional requirement 0 < x < y < z).
For the solution (x,y,z) having the largest z value, see (A075245, A075246, A075247).

			For n=12 the a(12)=133  solutions are {k = 0, t = 6},{k = 5, t = 0}.

I. Gueye and M. Mizony : Recent progress about Erdős-Straus conjecture, B SO MA S S, Volume 1, Issue 2, pp. 6-14.
M. Mizony and I. Gueye : Towards the proof of Erdős-Straus conjecture, B SO MA S S, Volume 1, Issue 2,p pp 141-150.

P. Erdős, On a Diophantine equation, (Hungarian. Russian, English summaries), Mat. Lapok 1, 1950, pp. 192-210.
M. Mizony and M.-L. Gardes, Sur la conjecture d'Erdős et Straus, see pages 14-17.
Eric W. Weisstein, MathWorld: Twin Pythagorean Triple
K. Yamamoto, On the Diophantine Equation 4/n=1/x+1/y+1/z, Mem. Fac. Sci. Kyushu U. Ser. A 19, 37-47, 1965.

Cf. A213340 (the quadratic polynomial 5+8t+12k+16kt).
Cf. A001844 (centered square numbers: 2n(n+1)+1).
Cf. A005408 (x values), A046092 (y values).
Cf. A073101 (number of solutions (x,y,z) to 4/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z).

G:=(n,p)->4/p = [2*(2*n+1)/(n*p+p+1), 4/p/(n*p+p+1), 2/(n*p+p+1)]:
cousin:=proc(p)
local n;
for n from 0 to 300 do
if n*p+p+1 mod 4*(2*n+1)=0 then return([p,n,G(n,p)]);fi:
od:
end:
L:=NULL:for m to 400 do L:=L,cousin(4*m+1): od:{L}[1..4];map(u->op(1,u),{L});

A213687 Numbers which are the values of the quadratic polynomial 3+4k+7t+8kt on nonnegative integers.

Original entry on oeis.org

3, 7, 10, 11, 15, 17, 19, 22, 23, 24, 27, 31, 34, 35, 37, 38, 39, 43, 45, 46, 47, 51, 52, 55, 57, 58, 59, 63, 66, 67, 70, 71, 73, 75, 77, 79, 80, 82, 83, 87, 91, 94, 95, 97, 99, 101, 103, 106, 107, 108, 111, 112, 115, 117, 118, 119, 122, 123, 126, 127, 129

Offset: 1

Michel Mizony, Jun 18 2012

nonn

For all these numbers a(n) we have the following Erdos-Straus decomposition: 4/p = 4/(3+4*k+7*t+8*k*t) = 1/(2*(3+4*k+7*t+8*k*t)*(1+k)) + 1/((1+k)*(2*t+1)) + 1/(2*(1+k)*(2*t+1)*(3+4*k+7*t+8*k*t));
Moreover this sequence is related to irreducible twin Pythagorean triples: the associated Pythagorean triple is [2t(t+1),2t+1, 2t(t+1)+1].
In 1948 Erdos and Straus conjectured that for any positive integer n >= 2 the equation 4/n = 1/x + 1/y +1/z has a solution with positive integers x, y and z (without the additional requirement 0 < x < y < z).
For the solution (x,y,z) having the largest z value, see (A075245, A075246, A075247).

			31 is a term because the solutions to 3+4*k+7*t+8*k*t = 31 are {k = 0, t = 4}, {k = 7, t = 0}.

I. Gueye and M. Mizony, Recent progress about Erdős-Straus conjecture, B SO MA S S, Volume 1, Issue 2, pp. 6-14.
M. Mizony and I. Gueye, Towards the proof of Erdős-Straus conjecture, B SO MA S S, Volume 1, Issue 2, pp. 141-150.

P. Erdős, On a Diophantine equation, (Hungarian. Russian, English summaries), Mat. Lapok 1, 1950, pp. 192-210.
M. Mizony and M.-L. Gardes, Sur la conjecture d'Erdős et Straus, see pages 14-17.
Eric Weisstein's World of Mathematics, Twin Pythagorean Triple
K. Yamamoto, On the Diophantine Equation 4/n=1/x+1/y+1/z, Mem. Fac. Sci. Kyushu U. Ser. A 19, 37-47, 1965.

Cf. A213340 (the quadratic polynomial 5+8t+12k+16kt).
Cf. A001844 (centered square numbers: 2n(n+1)+1).
Cf. A005408 (x values), A046092 (y values).
Cf. A073101 (number of solutions (x,y,z) to 4/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z).

H:=(k, t) -> 4/(3+4*k+7*t+8*k*t) = [1/2*1/((3+4*k+7*t+8*k*t)*(1+k)), 1/((1+k)*(2*t+1)), 1/2*1/((1+k)*(2*t+1)*(3+4*k+7*t+8*k*t))]:
cousin:=proc(p)
local n,k;
for n from 0 to (p-3)/7 do
if (p-3-7*n) mod (4+8*n)=0  then k:=(p-3-7*n)/(4+8*n):
return([p,n,H(k,n)]) fi;od;
end:
L:=NULL:for p from 2 to 500 do L:=L,cousin(p): od:{L}[1..10];map(u->op(1,u),{L});map(u->op(2,u),{L});

A287116 Nonsquare integers that cannot be represented in the form 4M-d, where (a*b)|M and d|(a+b) for some positive integers a,b.

Original entry on oeis.org

288, 336, 4545

Offset: 1

Max Alekseyev, May 19 2017

If there are no more terms, the Erdos-Straus conjecture would follow.
This sequence together with the squares (A000290) and E(4) form a partition of the nonnegative integers. That is, this sequence gives nonsquare non-elements of E(4) (see Dubickas and Novikas, 2012).
There are no other terms below 2*10^9.

A. Dubickas, A. Novikas, On integers expressible by some special linear form, Acta Math. Univ. Comenianae LXXXI:2 (2012), 203-209.
Wikipedia, Erdos-Straus conjecture.

Cf. A073101, A075245, A075246, A075247, A075248, A292581, A192787, A292624, A192789.

cp's OEIS Frontend

A075259 Number of solutions (x,y,z) to 3/(2n+1) = 1/x + 1/y + 1/z satisfying 0 < x < y < z and odd x, y, z.

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A227611 Number of ways 2/n can be expressed as the sum of three distinct unit fractions: 2/n = 1/x + 1/y + 1/z with 0 < x < y < z.

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A227612 Table read by antidiagonals: Number of ways m/n can be expressed as the sum of three distinct unit fractions, i.e., m/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z and read by antidiagonals.

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A337432 a(n) is the least value of z such that 4/n = 1/x + 1/y + 1/z with 0 < x <= y <= z has at least one solution.

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A257843 Numbers n for which the lexicographically first integer solution (x,y,z) of 4/n = 1/x + 1/y + 1/z with 0 < x < y < z, is different from the solution having the largest z-value.

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A213340 Numbers which are the values of the quadratic polynomial 5+8t+12k+16kt on nonnegative integers.

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A213686 Numbers which are the values of the quadratic polynomial 13+20*t+24*k+32*k*t at nonnegative integers.

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A213686 Numbers which are the values of the quadratic polynomial 13+20t+24k+32kt at nonnegative integers.

A213687 Numbers which are the values of the quadratic polynomial 3+4k+7t+8kt on nonnegative integers.