A073101 -id:A073101 - cp's OEIS frontend

A008846 Hypotenuses of primitive Pythagorean triangles.

5, 13, 17, 25, 29, 37, 41, 53, 61, 65, 73, 85, 89, 97, 101, 109, 113, 125, 137, 145, 149, 157, 169, 173, 181, 185, 193, 197, 205, 221, 229, 233, 241, 257, 265, 269, 277, 281, 289, 293, 305, 313, 317, 325, 337, 349, 353, 365, 373, 377, 389, 397, 401, 409, 421, 425, 433

Offset: 1

N. J. A. Sloane, Ralph Peterson (RALPHP(AT)LIBRARY.nrl.navy.mil)

Numbers of the form x^2 + y^2 where x is even, y is odd and gcd(x, y)=1. Essentially the same as A004613.
Numbers n for which there is no solution to 4/n = 2/x + 1/y for integers y > x > 0. Related to A073101. - T. D. Noe, Sep 30 2002
Discovered by Frénicle (on Pythagorean triangles): Méthode pour trouver ..., page 14 on 44. First text of Divers ouvrages ... Par Messieurs de l'Académie Royale des Sciences, in-folio, 6+518+1 pp., Paris, 1693. Also A020882 with only one of doubled terms (first: 65). - Paul Curtz, Sep 03 2008
All divisors of terms are of the form 4*k+1 (products of members of A002144). - Zak Seidov, Apr 13 2011
A024362(a(n)) > 0. - Reinhard Zumkeller, Dec 02 2012
Closed under multiplication. Primitive elements are in A002144. - Jean-Christophe Hervé, Nov 10 2013
Not only the square of these numbers is equal to the sum of two nonzero squares, but the numbers themselves also are; this sequence is then a subsequence of A004431. - Jean-Christophe Hervé, Nov 10 2013
Conjecture: numbers p for which sqrt(-1) exists in the p-adic numbering system. For example the 5-adic number ...2431212, when squared, gives ...4444444, which is -1, and 5 is in the sequence. - Thierry Banel, Aug 19 2022
The above conjecture was proven true by George Bergman. 3 known facts: (1) prime factors of a(n) are equal to 1 mod 4, (2) modulo such primes, sqrt(-1) exists, (3) if sqrt(m) exists mod r, r being odd, this extends to sqrt(m) in the r-adic ring. - Thierry Banel, Jul 04 2025

A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, pp. 10, 107.

Zak Seidov, Table of n, a(n) for n = 1..87881 (with a(n) up to 10^6).
Ron Knott, Pythagorean Triples and Online Calculators

Subsequence of A004431 and of A000404 and of A339952; primitive elements: A002144.
Cf. A004613, A020882, A073101.
Cf. A137409 (complement), disjoint union of A024409 and A120960.

a008846 n = a008846_list !! (n-1)
a008846_list = filter f [1..] where
   f n = all ((== 1) . (`mod` 4)) $ filter ((== 0) . (n `mod`)) [1..n]
-- Reinhard Zumkeller, Apr 27 2011

for x from 1 by 2 to 50 do for y from 2 by 2 to 50 do if gcd(x,y) = 1 then print(x^2+y^2); fi; od; od; [ then sort ].

Union[ Map[ Plus@@(#1^2)&, Select[ Flatten[ Array[ {2*#1, 2*#2-1}&, {10, 10} ], 1 ], GCD@@#1 == 1& ] ] ] (* Olivier Gérard, Aug 15 1997 *)
lst = {}; Do[ If[ GCD[m, n] == 1, a = 2 m*n; b = m^2 - n^2; c = m^2 + n^2; AppendTo[lst, c]], {m, 100}, {n, If[ OddQ@m, 2, 1], m - 1, 2}]; Take[ Union@ lst, 57] (* Robert G. Wilson v, May 02 2009 *)
Union[Sqrt[#[[1]]^2+#[[2]]^2]&/@Union[Sort/@({Times@@#,(Last[#]^2-First[#]^2)/2}&/@ (Select[Subsets[Range[1,33,2],{2}],GCD@@#==1&]))]] (* Harvey P. Dale, Aug 26 2012 *)

is(n)=Set(factor(n)[,1]%4)==[1] \\ Charles R Greathouse IV, Nov 06 2015

# for an array from the beginning
from math import gcd, isqrt
hypothenuses_upto = 433
A008846 = set()
for x in range(2, isqrt(hypothenuses_upto)+1):
    for y in range(min(x-1, (yy:=isqrt(hypothenuses_upto-x**2))-(yy%2 == x%2)) , 0, -2):
        if gcd(x,y) == 1: A008846.add(x**2 + y**2)
print(A008846:=sorted(A008846)) # Karl-Heinz Hofmann, Sep 30 2024

# for single k
from sympy import factorint
def A008846_isok(k): return not any([(pf-1) % 4 for pf in factorint(k)]) # Karl-Heinz Hofmann, Oct 01 2024

x^2 + y^2 where x is even, y is odd and gcd(x, y)=1. Essentially the same as A004613.

More terms from T. D. Noe, Sep 30 2002

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

A004194 Number of partitions of 1/n into 3 reciprocals of positive integers.

Original entry on oeis.org

3, 10, 21, 28, 36, 57, 42, 70, 79, 96, 62, 160, 59, 136, 196, 128, 73, 211, 80, 292, 245, 157, 93, 366, 156, 174, 230, 340, 106, 497, 90, 269, 322, 211, 453, 538, 85, 216, 378, 604, 121, 623, 104, 473, 648, 204, 135, 706, 227, 437, 387, 467, 125, 601, 561, 783, 385

Offset: 1

Scott Aaronson (philomath(AT)voicenet.com)

nonn

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

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

Justus Springer, Table of n, a(n) for n = 1..5000 (terms 1..100 from Robert G. Wilson)
K. S. Brown, Unit Fraction Partitions
Index entries for sequences related to Egyptian fractions

Cf. A227610, A226641, A226642, A192787, A226644, A226645, A226646.

a[n_] := Length@ Solve[ 1/n == 1/x + 1/y + 1/z && 1 <= x <= y <= z, {x, y, z}, Integers]; Array[a, 70] (* Allan C. Wechsler and Robert G. Wilson v, Aug 17 2013 *)

a(n)=my(t=1/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++))); s \\ Charles R Greathouse IV, Jun 12 2013

More terms from David W. Wilson, Aug 15 1996

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.

Original entry on oeis.org

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

A075246 y-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 x and z components are in A075245 and A075247.

Original entry on oeis.org

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

Offset: 3

T. D. Noe, Sep 10 2002

See A073101 for more details.

See A257840 for a variant that differs from a(89) on. - M. F. Hasler, Jul 03 2022

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

Cf. A073101 (number of solutions), A075245 (x values), A075247 (z values), A192787 (number of solutions with x <= y <= z), A257840 (variant: lex earliest solution, not largest z).

A075246:=proc() local t, n,a,b,t1,largey,largez; for n from 3 to 100 do t:=4/n; largez:=0; largey:=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),alargez then largez:=(1/(t1 - 1/b)); largey:=b; end if end if end do; end do; lprint(n, largey) 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++ ]]; yLst

apply( {A075246(n, c=1, a, 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, a]=[t-1/y, y])); a}, [3..99]) \\ M. F. Hasler, Jul 03 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

A075248 Number of solutions (x,y,z) to 5/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z.

Original entry on oeis.org

0, 1, 2, 1, 1, 3, 5, 9, 6, 3, 12, 5, 18, 15, 10, 5, 21, 11, 22, 18, 15, 8, 55, 30, 15, 20, 43, 20, 45, 5, 24, 35, 23, 36, 53, 10, 21, 52, 62, 6, 62, 12, 73, 69, 16, 11, 92, 38, 84, 34, 50, 11, 77, 56, 80, 45, 38, 34, 142, 6, 23, 96, 53, 53, 67, 15, 66, 70, 124, 12, 148, 21, 57

Offset: 2

T. D. Noe, Sep 10 2002

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 (A075249, A075250, A075251). See A073101 for the 4/n conjecture due to Erdős and Straus.

			a(4)=2 because there are two solutions: 5/4 = 1/1 + 1/5 + 1/20 and 5/4 = 1/1 + 1/6 + 1/12.

T. D. Noe, Table of n, a(n) for n = 2..1000
Ron Knott, Egyptian Fractions.
Eric Weisstein's World of Mathematics, Egyptian Fraction.

Cf. A073101, A075249, A075250, A075251.

m = 5; 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[5/n == 1/x + 1/y + 1/z && 0 < x < y < z, {x, y, z}, Integers]; Array[f, 74] (* Robert G. Wilson v, Jul 17 2013 *)

A227610 Number of ways 1/n can be expressed as the sum of three distinct unit fractions: 1/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z.

Original entry on oeis.org

1, 6, 15, 22, 30, 45, 36, 62, 69, 84, 56, 142, 53, 124, 178, 118, 67, 191, 74, 274, 227, 145, 87, 342, 146, 162, 216, 322, 100, 461, 84, 257, 304, 199, 435, 508, 79, 204, 360, 580, 115, 587, 98, 455, 618, 192, 129, 676, 217, 417, 369, 449, 119, 573, 543, 759, 367, 240, 166, 1236, 102, 261, 857, 428, 568, 717, 115, 537, 460, 1018, 155, 1126, 112, 276, 839

Offset: 1

Robert G. Wilson v, Jul 17 2013

nonn

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

			a(1)=1 because 1 = 1/2 + 1/3 + 1/6;
a(2)=6 because 1/2 = 1/3 + 1/7 + 1/42 = 1/3 + 1/8 + 1/24 = 1/3 + 1/9 + 1/18 = 1/3 + 1/10 + 1/15 = 1/4 + 1/5 + 1/20 = 1/4 + 1/6 + 1/12;
a(3)=15 because 1/3 = 1/x + 1/y + 1/z presented as {x,y,z}: {4,13,156}, {4,14,84}, {4,15,60}, {4,16,48}, {4,18,36}, {4,20,30}, {4,21,28}, {5,8,120}, {5,9,45}, {5,10,30}, {5,12,20}, {6,7,42}, {6,8,24}, {6,9,18}, {6,10,15}; etc.

Jud McCranie, Table of n, a(n) for n = 1..500
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
Eric Weisstein's World of Mathematics, Egyptian Fraction
Index entries for sequences related to Egyptian fractions

Cf. A002966, A073546.
Cf. A227611 (2/n), A075785 (3/n), A073101 (4/n), A075248 (5/n), A227612.
Cf. A347566, A347569.

f[n_] := Length@ Solve[1/n == 1/x + 1/y + 1/z && 0 < x < y < z, {x, y, z}, Integers]; Array[f, 70]

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]

A292581 Number of solutions to 4/n = 1/x + 1/y + 1/z in positive integers.

Original entry on oeis.org

0, 3, 12, 10, 12, 39, 36, 46, 30, 63, 48, 106, 24, 93, 216, 148, 24, 141, 60, 196, 162, 141, 120, 304, 60, 111, 162, 232, 42, 459, 108, 394, 174, 141, 372, 442, 54, 183, 420, 538, 42, 489, 78, 352, 540, 243, 198, 904, 102, 303, 294, 334, 78, 513

Offset: 1

Hugo Pfoertner, Sep 20 2017

nonn

Corrected version of A192786.
The Erdos-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.
After a(2) = 3, the values shown are all composite. [Jonathan Vos Post, Jul 17 2011]

			a(3)=12 because 4/3 can be expressed in 12 ways:
  4/3 =  1/1  + 1/4  + 1/12
  4/3 =  1/1  + 1/6  + 1/6
  4/3 =  1/1  + 1/12 + 1/4
  4/3 =  1/2  + 1/2  + 1/3
  4/3 =  1/2  + 1/3  + 1/2
  4/3 =  1/3  + 1/2  + 1/2
  4/3 =  1/4  + 1/1  + 1/12
  4/3 =  1/4  + 1/12 + 1/1
  4/3 =  1/6  + 1/1  + 1/6
  4/3 =  1/6  + 1/6  + 1/1
  4/3 =  1/12 + 1/1  + 1/4
  4/3 =  1/12 + 1/4  + 1/1
a(4) = 10 because 4/4 = 1 can be expressed in 10 ways:
  4/4=  1/2 + 1/3 + 1/6
  4/4=  1/2 + 1/4 + 1/4
  4/4=  1/2 + 1/6 + 1/3
  4/4=  1/3 + 1/2 + 1/6
  4/4=  1/3 + 1/3 + 1/3
  4/4=  1/3 + 1/6 + 1/2
  4/4=  1/4 + 1/2 + 1/4
  4/4=  1/4 + 1/4 + 1/2
  4/4=  1/6 + 1/2 + 1/3
  4/4=  1/6 + 1/3 + 1/2

Hugo Pfoertner, Table of n, a(n) for n = 1..1000
Christian Elsholtz and Terence Tao, Counting the number of solutions to the Erdos-Straus equation on unit fractions, arXiv:1107.1010 [math.NT], 2011-2015.
Allan Swett, The Erdos-Strauss Conjecture, 1999.
R. C. Vaughan, On a problem of Erdős, Straus and Schinzel, Mathematika 17 (1970), pp. 193-198.

For more references and links see A192787.

Cf. A073101, A192786, A292624, A337432.

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[n]]; Sow[an]]][[2, 1]] (* Jean-François Alcover, Dec 02 2018, adapted from PARI *)

\\ modified version of code by Charles R Greathouse IV in A192786
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) =
{
  local(t, t1, s, a, b, c);
  t = 4/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(a292581(n),", "))

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