A348625 -id:A348625 - cp's OEIS frontend

A002966 Egyptian fractions: number of solutions of 1 = 1/x_1 + ... + 1/x_n where 0 < x_1 <= ... <= x_n.

1, 1, 3, 14, 147, 3462, 294314, 159330691

Offset: 1

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
From R. J. Mathar, May 06 2010: (Start)
This is the leading edge of the triangle A156869. This is also the row n=1 of an array T(n,m) which gives the number of ways to write 1/n as a sum over m (not necessarily distinct) unit fractions:
   1, 1,  3,  14,   147,   3462, 294314, ...
   1, 2, 10, 108,  2892, 270332, ...
   1, 2, 21, 339, 17253, ...
   1, 3, 28, 694, 51323, ...
   ...
T(.,2) = A018892. T(.,3) = A004194. T(.,4) = A020327, T(.,5) = A020328. T(2,6) is computed by D. S. McNeil, who conjectures that the 2nd row is A003167. (End)
If on the other hand, all x_k must be unique, see A006585. - Robert G. Wilson v, Jul 17 2013

			For n=3 the 3 solutions are {2,3,6}, {2,4,4}, {3,3,3}.
For n=4 the solutions are: {2,3,7,42}, {2,3,8,24}, {2,3,9,18}, {2,3,10,15}, {2,3,12,12}, {2,4,5,20}, {2,4,6,12}, {2,4,8,8}, {2,5,5,10}, {2,6,6,6}, {3,3,4,12}, {3,3,6,6}, {3,4,4,6}, {4,4,4,4}. [Neven Juric, May 14 2008]

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D. Singmaster, The number of representations of one as a sum of unit fractions, unpublished manuscript, 1972.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).

Matthew Brendan Crawford, On the Number of Representations of One as the Sum of Unit Fractions, Master's Thesis, Virginia Polytechnic Institute and State University (2019).
Yuya Dan, Representation of one as the sum of unit fractions, International Mathematical Forum 6:1 (2011), pp. 25-30.
Jacques Le Normand, C++ code for a(8) [Broken link]
Jacques Le Normand, C++ code for a(8) [Cached copy]
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.
David Singmaster, The number of representations of one as a sum of unit fractions, Unpublished M.S., 1972
R. G. Wilson, v, Fax to N. J. A. Sloane, Sep 9, 1994, with copy of Scientific American column by Ian Stewart
Index entries for sequences related to Egyptian fractions

Cf. A002967, A006585, A000058, A348625.

a(n,rem=1,mn=1)=if(n==1,return(numerator(rem)==1)); sum(k=max(1\rem+1,mn), n\rem, a(n-1,rem-1/k,k)) \\ Charles R Greathouse IV, Jan 04 2015

a(n) <= binomial(A007018(n), n-1). - Charles R Greathouse IV, Jul 29 2024

a(7) from Jud McCranie, Nov 15 1999. Confirmed by Marc Paulhus.

a(8) from John Dethridge (jcd(AT)ms.unimelb.edu.au) and Jacques Le Normand (jacqueslen(AT)sympatico.ca), Jan 06 2004

A348626 Greedy Egyptian fraction representation of 1 with square denominators.

Original entry on oeis.org

2, 2, 2, 3, 3, 7, 12, 49, 340, 6153, 362275, 234314697, 4303312007019, 8064823505928103487, 21034270897045389505182033301, 13184627067084215135862894820778146400791573, 36011454158212923548860166370685543204871921069986403871775848271, 6820216143160044256325325882329406136711110111012515344838677137010956148075846307036940303634819

Offset: 1

Max Alekseyev, Oct 25 2021

nonn

Greedy representation 1 = 1/a(1)^2 + 1/a(2)^2 + ... constructed in a similar way to Sylvester's sequence (A000058). Let s(n) = Sum_{i=1..n} 1/a(i)^2, and let g(n) = 1 - s(n). Each a(n) is taken be the smallest positive integer satisfying s(n) < 1. [Revised by N. J. A. Sloane, Apr 20 2025]
Comment from David desJardins, Apr 20 2025 (Start)
We know that s(n) < 1 and s(n)-1/a(n)^2+1/(a(n)-1)^2 > 1. Then, arguing heuristically, the gap g(n) = 1-s(n) \approx 1/a(n+1)^2. This implies
   0 < g(n) < 1/(a(n)-1)^2 - 1/a(n)^2 \approx 2/a(n)^3.
So a(n)^3/a(n+1)^2 should be roughly uniform on [0,2].
Let L(n) = ln(a(n)). Then 3*L(n) - 2*L(n+1) \approx ln(2) - e(n+1), where e(i) has an exponential distribution.
So L(n+1) \approx (3/2)*L(n) + (1/2)*(e(n+1)-ln(2)).
This gives us the conjecture that L(n) = C * (3/2)^n * (1+o(1)), as n -> oo.
The plot of L(n)*(2/3)^n (see link) shows that the conjecture is plausible, with C \approx 0.15113.
(End)

David desJardins, Table of n, a(n) for n = 1..24
David desJardins, Plot of log(a(n))*(2/3)^n vs n
David desJardins, Comma-separated list of first 40 terms

Cf. A000058, A348625.

s=1; for(n=1,20, t=sqrtint(floor(1/s))+1; s-=1/t^2; print1(t,", "));

from math import isqrt
from fractions import Fraction
def A348626List():
    s = Fraction(1, 1)
    while True:
        t = isqrt(1 // s) + 1
        yield t
        s -= Fraction(1, t * t)
a = A348626List()
print([next(a) for  in range(18)])  # _Peter Luschny, Oct 26 2021

A369470 a(n) = number of integer solutions to 1 <= x1 <= x2 <= ... <= xn to 1/x1 + ... + 1/xn = (1 - 1/x1) * ... * (1 - 1/xn).

Original entry on oeis.org

1, 1, 2, 35, 455, 13624, 1176579

Offset: 1

Max Alekseyev, Jan 23 2024

For any n, a(n) >= A369469(n) >= 1 (see A369607).

Max Muller et al., The diophantine equation \sum_{n=1}^{N} \frac{1}{x_{n}} = \prod_{k=1}^{N} \left(1-\frac{1}{x_{k}} \right), MathOverflow, 2024.

Cf. A002966, A007850, A054377, A085098, A118086, A158649, A164014, A342267, A348625, A369469, A369607.

A378270 Number of partitions of 1 into {1/1^2, 1/2^2, 1/3^2, ..., 1/n^2}.

Original entry on oeis.org

1, 2, 3, 7, 8, 58, 59, 259, 664, 3427, 3428, 73351, 73352, 298785, 7060868, 43070304, 43070305, 901194373, 901194374, 32808600352, 1204438945226, 2459587779124, 2459587779125, 96010353352980

Offset: 1

Ilya Gutkovskiy, Nov 21 2024

From David A. Corneth, Nov 24 2024: (Start)
Primes n/2 < p <= n occur in exactly one solution namely (p^2) * (1/p^2).
Proof: If the numerator k of k/p^2 is less than p^2 then p divides the denominator of the sum of the Egyptian fractions as p divides no other number <= n. But the goal is have sum 1 i.e. denominator 1 so p cannot be a divisor of the denominator. Consequently this can be reduced to "Number of partitions of 1 into {1/1^2, 1/2^2, ..., 1/(n/2)^2, ..., 1/n^2}" plus number of primes n/2 < p <= n. The denominators for the first part can be cleared turning this into a partitioning problem of positive integers. (End)

			a(4) = 7 because we have 16 * (1/16) = 12 * (1/16) + 1/4 = 8 * (1/16) + 2 * (1/4) = 4 * (1/16) + 3 * (1/4) = 9 * (1/9) = 4 * (1/4) = 1.
From _David A. Corneth_, Nov 24 2024: (Start)
To find a(12) we can rewrite the problem as "Number of partitions of 1 into {1/1^2, 1/2^2, 1/3^2, 1/4^2, 1/5^2, 1/6^2, 1/8^2, 1/9^2, 1/10^2, 1/12^2} + |{7, 11}|". The lcm of (1^2, 2^2, 3^2, 4^2, 5^2, 6^2, 8^2, 9^2, 10^2, 12^2) is 129600. So this comes a partition problem of (number of partitions of 129600 into parts 129600, 32400, 14400, 8100, 5184, 3600, 2025, 1600, 1296, 900) + |{7, 11}|. (End)

Index entries for sequences related to Egyptian fractions

Cf. A000290, A020473, A038034, A348625, A378271.

a(p) = a(p-1) + 1 for prime p. - David A. Corneth, Nov 22 2024

a(12)-a(21) from David A. Corneth, Nov 22 2024

a(22)-a(24) from Jinyuan Wang, Dec 11 2024

A369607 Greedy solution a(1) < a(2) < ... to 1/a(1) + 1/a(2) + ... = (1 - 1/a(1)) * (1 - 1/a(2)) * ....

Original entry on oeis.org

3, 6, 29, 803, 643727, 414383582243, 171713753231982206218247, 29485613049014079571725771288849499850026859243, 869401376876189366008603664962520703088459987798626788985159595026678611496977754082506135887

Offset: 1

Max Alekseyev, Jan 27 2024

nonn

For any n, (x1, x2, ..., xn) = (a(1), a(2), ..., a(n-1), a(n)-1) forms a solution to 1/x1 + ... + 1/xn = (1 - 1/x1) * ... * (1 - 1/xn), proving that A369470(n) >= A369469(n) >= 1.

Max Muller et al., The diophantine equation Sum_{n=1..N} 1/x_n = Product_{k=1..N} (1-1/x_k), MathOverflow, 2024.

Cf. A000058, A348625, A348626, A369469, A369470.

a(n+2) = a(n+1)^2 + (a(n) - 2)*a(n+1) - a(n)^3 + 2*a(n)^2 - 2*a(n) + 2.

A379528 Number of compositions (ordered partitions) of 1 into {1/1^2, 1/2^2, 1/3^2, ..., 1/n^2}.

Original entry on oeis.org

1, 2, 3, 97, 98, 40917543, 40917544, 2901109178066823, 81221415992592163051371926, 373220766236315864054296758124337507430, 373220766236315864054296758124337507431

Offset: 1

Ilya Gutkovskiy, Dec 24 2024

Index entries for sequences related to Egyptian fractions

Cf. A000290, A020473, A038034, A348625, A378270, A378842.

a(p) = a(p-1) + 1 for p prime. - Chai Wah Wu, Dec 27 2024

a(6)-a(11) from Alois P. Heinz, Dec 26 2024

cp's OEIS Frontend

A002966 Egyptian fractions: number of solutions of 1 = 1/x_1 + ... + 1/x_n where 0 < x_1 <= ... <= x_n.

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A348626 Greedy Egyptian fraction representation of 1 with square denominators.

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A369470 a(n) = number of integer solutions to 1 <= x1 <= x2 <= ... <= xn to 1/x1 + ... + 1/xn = (1 - 1/x1) * ... * (1 - 1/xn).

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A378270 Number of partitions of 1 into {1/1^2, 1/2^2, 1/3^2, ..., 1/n^2}.

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A369607 Greedy solution a(1) < a(2) < ... to 1/a(1) + 1/a(2) + ... = (1 - 1/a(1)) * (1 - 1/a(2)) * ....

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A379528 Number of compositions (ordered partitions) of 1 into {1/1^2, 1/2^2, 1/3^2, ..., 1/n^2}.

Views

Author

Keywords

Links

Crossrefs

Formula

Extensions