A002966 -id:A002966 - cp's OEIS frontend

A357765 Smallest positive integer that can be represented as the sum of n of its (possibly equal) divisors in the maximum number of ways (=A002966(n)).

1, 2, 12, 2520, 48348686786400, 10543141534556403817127800577537146514577188497111149855093902265479066128013109211427715400552367011213513440000

Offset: 1

David A. Corneth and Max Alekseyev, Oct 12 2022

			a(3) = 12 since for n = 3 the tuples forming the solutions of 1 = 1/x_1 + 1/x_2 + 1/x_3 are (x_1, x_2, x_3) in {(2, 3, 6), (2, 4, 4), (3, 3, 3)}. All these terms combined have an lcm of 12. The ways to write 12 as a sum of 3 of its divisors are therefore 12 = 12/2 + 12/3 + 12/6 = 6 + 4 + 2. Similarily we have 12 = 6 + 3 + 3 and 12 = 4 + 4 + 4.

Max Alekseyev, Table of n, a(n) for n = 1..7

Apparently coincides with A181700 for n >= 4.

Cf. A002966, A006585.

a(n) = LCM of all denominators of Egyptian fractions enumerated by A002966(n).

A051908 Number of ways to express 1 as the sum of unit fractions such that the sum of the denominators is n.

Original entry on oeis.org

1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 1, 0, 3, 0, 1, 1, 1, 1, 2, 3, 2, 2, 1, 2, 2, 2, 4, 5, 5, 2, 4, 5, 5, 9, 4, 4, 6, 4, 4, 7, 8, 4, 10, 9, 9, 11, 8, 13, 13, 15, 16, 21, 18, 16, 22, 19, 18, 30, 24, 19, 26, 28, 26, 29, 35, 29, 44, 28, 47, 48

Offset: 1

Jud McCranie, Dec 16 1999

nonn

Also the number of partitions of n whose reciprocal sums to 1; "exact partitions". - Robert G. Wilson v, Sep 30 2009

			1 = 1/2 + 1/2, the sum of denominators is 4, and this is the only expression of 1 as unit fractions with denominator sum 4, so a(4)=1.
The a(22) = 3 partitions whose reciprocal sum is 1 are (12,4,3,3), (10,5,5,2), (8,8,4,2). - _Gus Wiseman_, Jul 16 2018

Derrick Niederman, "Number Freak, From 1 to 200 The Hidden Language of Numbers Revealed", a Perigee Book, Penguin Group, NY, 2009, pp. 82-83. [From Robert G. Wilson v, Sep 30 2009]

David A. Corneth, Table of n, a(n) for n = 1..200 (terms a(1)-a(86) from Jud McCranie, a(87)-a(88) from Robert G. Wilson v, a(89)-a(100) from Seiichi Manyama)
David A. Corneth, Tuples up to n = 170
Gus Wiseman, Sequences counting and ranking integer partitions by their reciprocal sums
Index entries for sequences related to Egyptian fractions

A028229 lists n such that a(n)=0.

Cf. A002966, A058360, A270599, A316854, A316855, A316888-A316904.

(* first do *) << "Combinatorica`"; (* then *) f[n_] := Block[{c = i = 0, k = PartitionsP@n, p = {n}}, While[i < k, If[1 == Plus @@ (1/p), c++ ]; i++; p = NextPartition@p]; c]; Array[f, 88] (* Robert G. Wilson v, Sep 30 2009 *)
Table[Length[Select[IntegerPartitions[n],Sum[1/m,{m,#}]==1&]],{n,30}] (* Gus Wiseman, Jul 16 2018 *)

def partition(n, min, max)
  return [[]] if n == 0
  [max, n].min.downto(min).flat_map{|i| partition(n - i, min, i).map{|rest| [i, *rest]}}
end
def A051908(n)
  ary = [1]
  (2..n).each{|m|
    cnt = 0
    partition(m, 2, m).each{|ary|
      cnt += 1 if ary.inject(0){|s, i| s + 1 / i.to_r} == 1
    }
    ary << cnt
  }
  ary
end
p A051908(100) # Seiichi Manyama, May 31 2016

a(n) > 0 for n > 23.

A316888 Heinz numbers of aperiodic integer partitions into relatively prime parts whose reciprocal sum is 1.

Original entry on oeis.org

2, 195, 3185, 6475, 10527, 16401, 20445, 20535, 21045, 25365, 46155, 164255, 171941, 218855, 228085, 267883, 312785, 333925, 333935, 335405, 343735, 355355, 414295, 442975, 474513, 527425, 549575, 607475, 633777, 691041, 711321, 722425, 753865, 804837, 822783

Offset: 1

Gus Wiseman, Jul 16 2018

nonn

The reciprocal sum of (y_1, ..., y_k) is 1/y_1 + ... + 1/y_k.
The Heinz number of an integer partition (y_1, ..., y_k) is prime(y_1) * ... * prime(y_k).
A partition is aperiodic if its multiplicities are relatively prime.
Does not contain 29888089, which belongs to A316890 and is the Heinz number of a periodic partition.

			The partition (6,4,4,3) with Heinz number 3185 is aperiodic, has relatively prime parts, and 1/6 + 1/4 + 1/4 + 1/3 = 1, so 3185 belongs to the sequence.
The sequence of partitions whose Heinz numbers belong to the sequence begins: (1), (6,3,2), (6,4,4,3), (12,4,3,3), (10,5,5,2), (20,5,4,2), (15,10,3,2), (12,12,3,2), (18,9,3,2), (24,8,3,2), (42,7,3,2).

Gus Wiseman, Sequences counting and ranking integer partitions by their reciprocal sums

Cf. A000837, A002966, A007916, A051908, A100953, A289509, A296150, A316855, A316856, A316857, A316888-A316904.

Select[Range[2,100000],And[GCD@@FactorInteger[#][[All,2]]==1,GCD@@PrimePi/@FactorInteger[#][[All,1]]==1,Sum[m[[2]]/PrimePi[m[[1]]],{m,FactorInteger[#]}]==1]&]

A316904 Heinz numbers of aperiodic integer partitions into relatively prime parts whose reciprocal sum is an integer.

Original entry on oeis.org

2, 18, 72, 162, 195, 250, 288, 294, 390, 500, 588, 648, 780, 1125, 1152, 1176, 1458, 1560, 1755, 2000, 2250, 2352, 2592, 2646, 3120, 3185, 3510, 4000, 4500, 4608, 4704, 4802, 5292, 6240, 6370, 6475, 7020, 8450, 9000, 9408, 10125, 10368, 10527, 10584, 12480

Offset: 1

Gus Wiseman, Jul 16 2018

nonn

The reciprocal sum of (y_1, ..., y_k) is 1/y_1 + ... + 1/y_k.
The Heinz number of an integer partition (y_1, ..., y_k) is prime(y_1) * ... * prime(y_k).
A partition is aperiodic if its multiplicities are relatively prime.

			The sequence of partitions whose Heinz numbers belong to this sequence begins: (1), (221), (22111), (22221), (632), (3331), (2211111), (4421), (6321), (33311), (44211), (2222111).

Gus Wiseman, Sequences counting and ranking integer partitions by their reciprocal sums

Cf. A000837, A002966, A051908, A058360, A100953, A296150, A316854, A316855, A316856, A316857, A316888-A316904.

Select[Range[2,20000],And[GCD@@FactorInteger[#][[All,2]]==1,GCD@@PrimePi/@FactorInteger[#][[All,1]]==1,IntegerQ[Sum[m[[2]]/PrimePi[m[[1]]],{m,FactorInteger[#]}]]]&]

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

A006585 Egyptian fractions: number of solutions to 1 = 1/x_1 + ... + 1/x_n in positive integers x_1 < ... < x_n.

Original entry on oeis.org

1, 0, 1, 6, 72, 2320, 245765, 151182379

Offset: 1

N. J. A. Sloane

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

If on the other hand, x_k need not be unique, see A002966. - Robert G. Wilson v, Jul 17 2013

			The 6 solutions for n=4 are 2,3,7,42; 2,3,8,24; 2,3,9,18; 2,3,10,15; 2,4,5,20; 2,4,6,12.

Marc LeBrun, personal communication.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

M. Le Brun, Email to N. J. A. Sloane, Jul 1991
S. V. Konyagin, Double exponential lower bound for the number of representations of unity by Egyptian fractions. Mathematical Notes, 95:1-2 (2014), 277-281.
T. D. Browning and C. Elsholtz, The number of representations of rationals as a sum of unit fractions, Illinois J. Math. 55:2 (2011), 685-696.
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.
Index entries for sequences related to Egyptian fractions

Cf. A000058, A002966, A002967, A280518.

a(n) = A280520(n,1).

a(1)-a(7) are confirmed by Jud McCranie, Dec 11 1999

a(8) from John Dethridge (jcd(AT)ms.unimelb.edu.au), Jan 08 2004

A355200 Numbers k that can be written as the sum of 3 divisors of k (not necessarily distinct).

Original entry on oeis.org

3, 4, 6, 8, 9, 12, 15, 16, 18, 20, 21, 24, 27, 28, 30, 32, 33, 36, 39, 40, 42, 44, 45, 48, 51, 52, 54, 56, 57, 60, 63, 64, 66, 68, 69, 72, 75, 76, 78, 80, 81, 84, 87, 88, 90, 92, 93, 96, 99, 100, 102, 104, 105, 108, 111, 112, 114, 116, 117, 120, 123, 124, 126, 128, 129, 132, 135

Offset: 1

Wesley Ivan Hurt, Jun 23 2022

From Bernard Schott, Aug 06 2023: (Start)
Equivalently: positive numbers that are divisible by 3 or by 4.
Proof (similar to the proof proposed by Robert Israel in A355641).
If k is divisible by 3, then k is in the sequence because k = k/3 + k/3 + k/3.
If k is divisible by 4, then k is in the sequence because k = k/2 + k/4 + k/4.
Moreover, if k is positive and divisible by 6 (A008588), then k = k/3 + k/3 + k/3, but k is also in the sequence because k = k/2 + k/3 + k/6.
Conversely, to show that every term of this sequence is divisible by 3 or by 4, we consider all positive integer solutions of the equation 1 = 1/a + 1/b + 1/c. Without loss of generality, we may assume a <= b <= c, then 3/a >= 1/a + 1/b + 1/c = 1. So a <= 3. Similarly, given a, we have 2/b >= 1/b + 1/c = 1 - 1/a, so b <= 2/(1 - 1/a).
-> if a = 1, then 1 = 1 + 1/b + 1/c; this equation has clearly no solution.
-> if a = 2, then 1/2 = 1/b + 1/c with b <= 2/(1 - 1/2) = 4; in this case, there are two solutions: (a,b,c) = (2,3,6) or (a,b,c) = (2,4,4).
-> if a = 3, then 2/3 = 1/b + 1/c with b <= 2/(1 - 1/3) = 3; in this case, there is one solution: (a,b,c) = (3,3,3).
It turns out that there are only 3 solutions with a <= b <= c. Each corresponds to a possible pattern k = k/a + k/b + k/c for writing k as the sum of 3 of its divisors, which works when k is divisible by 3 or by 4. (End)
From David A. Corneth, Aug 07 2023: (Start)
Proof that a(n + 6) = a(n) + 12.
As k is in the sequence, k = k/d1 + k/d2 + k/d3 where d1, d2 and d3 | k and they are not necessarily distinct. By discussion above from Bernard Schott, Aug 06 2023, (d1, d2, d3) are in {(2, 3, 6), (2, 4, 4), (3, 3, 3)}. The lcm of these tuples are 6, 4 and 3 respectively. So any number k in the sequence is divisible by 3, 4 or 6.
This tells us that if k is in the sequence then k + 12 is in the sequence since k + 12 is divisible by one of 3, 4 or 6 since lcm(3, 4, 6) = 12.
So we can write a(n + m) = a(n) + 12 for some m. Inspection gives m = 6 so a(n + 6) = a(n) + 12. (End)

			6 is in the sequence since it can be written as the sum of 3 of its (not necessarily distinct) divisors: 6 = 1+2+3 = 2+2+2 with 1|6, 2|6, and 3|6.

Ray Chandler, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (2,-2,2,-2,2,-1).

Cf. A002966, A008588, A348536.
Numbers k that can be written as the sum of j divisors of k (not necessarily distinct) for j=1..10: A000027 (j=1), A299174 (j=2), this sequence (j=3), A354591 (j=4), A355641 (j=5), A356609 (j=6), A356635 (j=7), A356657 (j=8), A356659 (j=9), A356660 (j=10).
Equals positive terms of A008585 union A008586.
Cf. A005843, A134667.

q[n_, k_] := AnyTrue[Tuples[Divisors[n], k], Total[#] == n &]; Select[Range[135], q[#, 3] &] (* Amiram Eldar, Aug 21 2022 *)
Table[2n + (Sin[Pi*n/3] + Sin[2*Pi*n/3])/Sqrt[3], {n, 100}] (* Wesley Ivan Hurt, Oct 30 2023 *)
CoefficientList[Series[(3 - 2*x + 4*x^2 - 2*x^3 + 3*x^4)/((x - 1)^2*(1 + x^2 + x^4)), {x, 0, 80}], x] (* Wesley Ivan Hurt, Jul 17 2025 *)

isok(k) = my(d=divisors(k)); forpart(p=k, if (setintersect(d, Set(p)) == Set(p), return(1)), , [3,3]); \\ Michel Marcus, Aug 21 2022

is(n) = my(v = [3,4,6]); sum(i = 1, 3, n%v[i] == 0) > 0 \\ David A. Corneth, Oct 08 2022

a(n + 6) = a(n) + 12. - David A. Corneth, Oct 08 2022
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*log(2)/3 - log(3)/4. - Amiram Eldar, Sep 10 2023
From Wesley Ivan Hurt, Oct 30 2023: (Start)
a(n) = 2*n + (sin(Pi*n/3) + sin(2*Pi*n/3))/sqrt(3).
a(n) = A005843(n) + A134667(n). (End)
From Wesley Ivan Hurt, Jul 17 2025: (Start)
G.f.: x*(3-2*x+4*x^2-2*x^3+3*x^4)/((x-1)^2*(1+x^2+x^4)).
a(n) = 2*a(n-1) - 2*a(n-2) + 2*a(n-3) - 2*a(n-4) + 2*a(n-5) - a(n-6). (End)

A002967 Egyptian fractions: number of solutions of 1 = 1/x_1 + ... + 1/x_n in positive integers.

Original entry on oeis.org

1, 1, 10, 215, 12231, 2025462, 1351857641, 6255560531733

Offset: 1

N. J. A. Sloane

Solutions differing only in the order of the x_i are counted as distinct.

All denominators in the expansion 1 = 1/x_1 + ... + 1/x_n are bounded by A000058(n-1) = A129871(n). - Max Alekseyev, Dec 30 2003

			For n=3 the 10 solutions are {2,3,6} (6 ways), {2,4,4} (3 ways), {3,3,3} (1 way).

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).

Zachary Harris and Joel Louwsma, On Arithmetical Structures on Complete Graphs, arXiv:1909.02022 [math.NT], 2019. See also Involve (2020) Vol. 13, No. 2, 345-355.
Claire Levaillant, On arithmetical structures on K9, arXiv:2311.03668 [math.NT], 2023.
Putnam Competition, 58th Putnam Mathematical Competition, 1997, Problem A-5
D. Singmaster, The number of representations of one as a sum of unit fractions, Unpublished M.S., 1972.
Carlos E. Valencia and R. R. Villagrán, Algorithmic aspects of arithmetical structures, arXiv:2101.05238 [math.NT], 2021.
Index entries for sequences related to Egyptian fractions

Cf. A002966, A006585.

Cf. A000058.

a(7) from Jud McCranie

a(8) from John Dethridge, Jan 11 2004

A325618 Numbers m such that there exists an integer partition of m whose reciprocal factorial sum is 1.

Original entry on oeis.org

1, 4, 11, 18, 24, 31, 37, 44, 45, 50, 52, 57, 58, 65, 66, 70, 71, 73, 76, 78, 79, 83, 86, 87, 89, 91, 92, 94, 96, 97, 99, 100, 102, 104, 107, 108, 109, 110, 112, 113, 114, 115, 117, 118, 119, 120, 121, 122, 123, 125, 126, 127, 128, 130, 131

Offset: 1

Gus Wiseman, May 13 2019

nonn

The reciprocal factorial sum of an integer partition (y_1,...,y_k) is 1/y_1! + ... + 1/y_k!.

Conjecture: 137 is the greatest integer not in this sequence. - Charlie Neder, May 14 2019

			The sequence of terms together with an integer partition of each whose reciprocal factorial sum is 1 begins:
   1: (1)
   4: (2,2)
  11: (3,3,3,2)
  18: (3,3,3,3,3,3)
  24: (4,4,4,4,3,3,2)
  31: (4,4,4,4,3,3,3,3,3)
  37: (4,4,4,4,4,4,4,4,3,2)
  44: (4,4,4,4,4,4,4,4,3,3,3,3)
  45: (5,5,5,5,5,4,4,4,3,3,2)
  50: (4,4,4,4,4,4,4,4,4,4,4,4,2)

Charlie Neder, Table of n, a(n) for n = 1..1000

Factorial numbers: A000142, A002982, A007489, A011371, A022559, A064986, A115627, A284605, A325616.

Reciprocal factorial sum: A002966, A051908, A058360, A316854, A316855, A325619, A325620, A325621, A325622, A325623, A325624.

a(11)-a(55) from Charlie Neder, May 14 2019

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]

cp's OEIS Frontend

A357765 Smallest positive integer that can be represented as the sum of n of its (possibly equal) divisors in the maximum number of ways (=A002966(n)).

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A051908 Number of ways to express 1 as the sum of unit fractions such that the sum of the denominators is n.

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A316888 Heinz numbers of aperiodic integer partitions into relatively prime parts whose reciprocal sum is 1.

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A316904 Heinz numbers of aperiodic integer partitions into relatively prime parts whose reciprocal sum is an integer.

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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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A006585 Egyptian fractions: number of solutions to 1 = 1/x_1 + ... + 1/x_n in positive integers x_1 < ... < x_n.

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A355200 Numbers k that can be written as the sum of 3 divisors of k (not necessarily distinct).

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A002967 Egyptian fractions: number of solutions of 1 = 1/x_1 + ... + 1/x_n in positive integers.