A075246 -id:A075246 - cp's OEIS frontend

A075245 x-value of the solution (x,y,z) to 4/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z and having the largest z-value. The y and z components are in A075246 and A075247.

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, 21, 21

Offset: 3

T. D. Noe, Sep 10 2002

See A073101 for more details.

a(n) = floor(n/4) + 1, at least up to n = 2000, except for some n = 8k+1 (k = 6, 9, 11, 14, 20, 21, 24, 29, 30, 35, 39, 41, 44, 45, 50, ...), where a(n) is one larger than a(n-1) and a(n+1). - M. F. Hasler, Jul 02 2022

			For n = 3, we have a(3) = 1 = x in 4/3 = 1/x + 1/y + 1/z with y = 4 and z = 12 which is the largest possible z: Indeed, x < y < z gives 4/3 < 3/x, so only x = 1 and 2 are possible, and then with y < z, 2/y > 4/3 - 1/x is impossible for x = 2 < y < 12/5 and for x = 1 < y < 6 only y = 4 gives a solution.

M. F. Hasler, Table of n, a(n) for n = 3..2000

Cf. A073101, A075246, A075247, A192787.

A075245:= proc () local t, n, a, b, t1, largex, largez; for n from 3 to 100 do t := 4/n; largez := 0; largex := 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 `and`(type(1/(t1-1/b), integer), a < b, b < 1/(t1-1/b)) then if largez < 1/(t1-1/b) then largez := 1/(t1-1/b); largex := a end if end if end do end do; lprint(n, largex) end do end proc; # [program derived from A192787] Patrick J. McNab, Aug 20 2014

For[xLst={}; yLst={}; zLst={}; n=3, n<=100, n++, cnt=0; xr=n/4; If[IntegerQ[xr], x=xr+1, x=Ceiling[xr]]; While[yr=1/(4/n-1/x); If[IntegerQ[yr], y=yr+1, y=Ceiling[yr]]; cnt==0&&y>x, While[zr=1/(4/n-1/x-1/y); cnt==0&&zr>y, If[IntegerQ[zr], z=zr; cnt++; AppendTo[xLst, x]; AppendTo[yLst, y]; AppendTo[zLst, z]]; y++ ]; x++ ]]; xLst

apply( {A075245(n,c=1,a)=for(x=n\4+1, 3*n\4, my(t=4/n-1/x); for(y=max(1\t,x)+1, ceil(2/t)-1, t-1/y >= c && break; numerator(t-1/y)==1 && [c,a]=[t-1/y,x])); a}, [3..99]) \\ M. F. Hasler, Jul 02 2022

Conjecture: a(n) = floor(n/4) + d, with d = 1 except for some n = 8k+1 (k = 6, 9, 11, 14, 20, 21, 24, 29, 30, 35, 39, ...) where d = 2 . - M. F. Hasler, Jul 02 2022

A075247 Largest possible z-value of an integer solution (x,y,z) to 4/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z. The x and y components are in A075245 and A075246.

Original entry on oeis.org

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

Offset: 3

T. D. Noe, Sep 10 2002

See A073101 for more details.

			For n = 6 we have a(n) = 42 the largest possible z in a solution of 4/n = 2/3 = 1/x + 1/y + 1/z with 0 < x < y < z in the integers. Indeed, from 1/x < 2/3 < 3/x we have 3/2 < x < 9/2. For x = 2 we get 2/y > 2/3 - 1/2 = 1/6 > 1/y <=> 6 < y < 12, and each of these y except y = 11 yields a solution, with z = 42, 24, 18, 15 and 12. There are no other possible z values: x = 3 gives 2/y > 1/3 <=> y < 6 and indeed y = 4 gives a solution with z = 12, no solution for y = 5; finally, x = 4 gives 2/y > 5/12 <=> y < 24/5, impossible with y > x.

M. F. Hasler, Table of n, a(n) for n = 3..2000

Cf. A075245 (x values), A075246 (y values), A073101 (number of solutions), A192787 (solutions with x <= y <= z).

A075247:= proc () local t, n, a, b, t1, largey, largez; for n from 3 to 100 do t := 4/n; largez := 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 `and`(type(1/(t1-1/b), integer), a < b, b < 1/(t1-1/b)) then if largez < 1/(t1-1/b) then largez := 1/(t1-1/b) end if end if end do end do; lprint(n, largez) end do end proc; # [program derived from A192787] Patrick J. McNab, Aug 20 2014

For[xLst={}; yLst={}; zLst={}; n=3, n<=100, n++, cnt=0; xr=n/4; If[IntegerQ[xr], x=xr+1, x=Ceiling[xr]]; While[yr=1/(4/n-1/x); If[IntegerQ[yr], y=yr+1, y=Ceiling[yr]]; cnt==0&&y>x, While[zr=1/(4/n-1/x-1/y); cnt==0&&zr>y, If[IntegerQ[zr], z=zr; cnt++; AppendTo[xLst, x]; AppendTo[yLst, y]; AppendTo[zLst, z]]; y++ ]; x++ ]]; zLst

apply( {A075247(n, c=1, t)=for(x=n\4+1, 3*n\4, for(y=max(1\t=4/n-1/x, x)+1, ceil(2/t)-1, t-1/y >= c && break; numerator(t-1/y)==1 && c=t-1/y)); 1/c}, [3..99]) \\ M. F. Hasler, Jul 02 2022

A073101 Number of integer solutions (x,y,z) to 4/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z.

Original entry on oeis.org

0, 0, 1, 1, 2, 5, 5, 6, 4, 9, 7, 15, 4, 14, 33, 22, 4, 21, 9, 30, 25, 22, 19, 45, 10, 17, 25, 36, 7, 72, 17, 62, 27, 22, 59, 69, 9, 29, 67, 84, 7, 77, 12, 56, 87, 39, 32, 142, 16, 48, 46, 53, 13, 82, 92, 124, 37, 30, 25, 178, 11, 34, 147, 118, 49, 94, 15, 67, 51, 176, 38, 191, 7

Offset: 1

Robert G. Wilson v, Aug 18 2002

nonn

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). 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 (A075245, A075246, A075247). See A075248 for Sierpiński's conjecture for 5/n.

See (A257839, A257840, A257841) for the lexicographically smallest solutions, and A257843 for the differences between these and those with largest z-value. - M. F. Hasler, May 16 2015

			a(5)=2 because there are two solutions: 4/5 = 1/2 + 1/4 + 1/20 and 4/5 = 1/2 + 1/5 + 1/10.

T. D. Noe, Table of n, a(n) for n = 1..1000, (corrected by _Peter Luschny_, Jan 19 2019)
Christian Elsholtz, Sums Of k Unit Fractions, Trans. Amer. Math. Soc. 353 (2001), 3209-3227.
David Eppstein, Algorithms for Egyptian Fractions
Paul Erdős, Az 1/z_1 + 1/z_2 + ... + 1/z_n = a/b egyenlet egész számú megoldásairól, (On a Diophantine equation), Mat. Lapok, 1:192-210, 1050. Math. Rev. 13:208b.
Ron Knott, Egyptian Fractions
Eric Weisstein's World of Mathematics, Egyptian Fraction
Wikipedia, Erdős-Straus conjecture

Cf. A075245, A075246, A075247, A075248.

Cf. A192787 (# distinct solutions with x <= y <= z).

import Data.Ratio ((%), numerator, denominator)
a073101 n = length [(x,y) |
   x <- [n `div` 4 + 1 .. 3 * n `div` 4],   let y' = recip $ 4%n - 1%x,
   y <- [floor y' + 1 .. floor (2*y') + 1], let z' = recip $ 4%n - 1%x - 1%y,
   denominator z' == 1 && numerator z' > y && y > x]
-- Reinhard Zumkeller, Jan 03 2011

A:= proc(n)
   local x,t, p,q,ds,zs,ys,js, tot,j;
tot:= 0;
for x from 1+floor(n/4) to ceil(3*n/4)-1 do
    t:= 4/n - 1/x;
    p:= numer(t);
    q:= denom(t);
    ds:= convert(select(d -> (d < q) and d + q mod p = 0,
          numtheory:-divisors(q^2)),list);
    ys:= map(d -> (d+q)/p, ds);
    zs:= map(d -> (q^2/d+q)/p, ds);
    js:= select(j -> ys[j] > x,[$1..nops(ds)]);
    tot:= tot + nops(js);
od;
tot;
end proc:
seq(A(n),n=2..100); # Robert Israel, Aug 22 2014

(* download Egypt.m from D. Eppstein's site and put it into MyOwn directory underneath Mathematica\AddOns\StandardPackages *) Needs["MyOwn`Egypt`"]; Table[ Length[ EgyptianFraction[4/n, Method -> Lexicographic, MaxTerms -> 3, MinTerms -> 3, Duplicates -> Disallow, OutputFormat -> Plain]], {n, 5, 80}]
m = 4; For[lst = {}; n = 2, n <= 100, n++, 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; cnt++; (*Print[n, " ", x, " ", y, " ", z]*)]]]; AppendTo[lst, cnt]]; lst
f[n_] := Length@ Solve[4/n == 1/x + 1/y + 1/z && 0 < x < y < z, {x, y, z}, Integers]; Array[f, 72, 2] (* Robert G. Wilson v, Jul 17 2013 *)

A073101(n)=sum(c=n\4+1,n*3\4,sum(b=c+1,ceil(2/(t=4/n-1/c))-1,numerator(t-1/b)==1)) \\ M. F. Hasler, May 15 2015

Edited by T. D. Noe, Sep 10 2002

Extended to offset 1 with a(1) = 0 by M. F. Hasler, May 16 2015

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

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

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

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;

cp's OEIS Frontend

A075245 x-value of the solution (x,y,z) to 4/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z and having the largest z-value. The y and z components are in A075246 and A075247.

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A075247 Largest possible z-value of an integer solution (x,y,z) to 4/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z. The x and y components are in A075245 and A075246.

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A073101 Number of integer solutions (x,y,z) to 4/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z.

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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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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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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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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.

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