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%I #51 Apr 19 2017 09:01:25
%S 1,6,12,15,18,20,24,28,30,33,35,36,40,42,45,48,52,54,55,56,60,63,65,
%T 66,70,72,75,76,77,78,80,84,85,88,90,91,95,96,99,100,102,104,105,108,
%U 110,112,114,115,117,119,120,126,130,132,133,135,136,138,140,143,144,145
%N Numbers n such that there exists a solution to the equation 1 = 1/x_1 + ... + 1/x_k (for any k), 0 < x_1 < ... < x_k = n.
%C No prime or power of a prime is in this sequence. If n > 1 is in the sequence then all multiples of n are in the sequence. A multiple m*p of a prime p, with all prime factors of m < p, is in the sequence if p is a factor of the numerator of a sum 1/m + 1/x1 + ... + 1/xi, where x1,...,xi are distinct integers < m. See A093407 for the least m for each prime p. The Mathematica code uses backtracking to find one solution for a given n. If n is large or not in this sequence, the program will run for a long time. - _T. D. Noe_, Mar 30 2004
%C Conjecture (verified through n=2*10^5): For any n > 1, let P be the largest divisor of n that is either a prime (p) or prime power (p^e, where e > 1), and let m = n/P. Then n is in the sequence iff p is a factor of the numerator of a sum 1/m + 1/x_1 + ... + 1/x_i, where x_1,...,x_i are distinct integers < m. - _Jon E. Schoenfield_, Apr 06 2014
%D R. K. Guy, Unsolved Problems in Number Theory, 2nd Ed., New York, Springer-Verlag, 1994, Section D11.
%H Jon E. Schoenfield, <a href="/A092671/b092671.txt">Table of n, a(n) for n = 1..10000 (first 306 terms from Toshitaka Suzuki)</a>
%H Harry Ruderman and Paul Erdős, <a href="http://www.jstor.org/stable/2319578">Problem E2427: Bounds of Egyptian fraction partitions of unity</a>, Amer. Math. Monthly, Vol. 81, No. 7 (1974), 780-782.
%H Jon E. Schoenfield, <a href="/A092671/a092671_1.txt">Additional examples and notes</a>
%H Jon E. Schoenfield, <a href="/A092671/a092671_2.txt">Magma program</a>
%H <a href="/index/Ed#Egypt">Index entries for sequences related to Egyptian fractions</a>
%e From _Jon E. Schoenfield_, Apr 09 2017: (Start)
%e 6 is in the sequence because 1/2 + 1/3 + 1/6 = 1. (Note that the prime factorization of 6 is 2*3, and if we start with 1/6, adding 1/3 yields 1/2, which removes the factor 3 from the denominator; then adding 1/2 removes the 2.)
%e 23 cannot be in the sequence because it is a prime: for any positive integer j1 < 23, 1/j1 + 1/23 = (23 + j1)/(23*j1), which cannot be reduced; adding another 1/j2 to the sum (with j2 < 23) will give (23*(j1 + j2) + j1*j2)/(23*j1*j2), from which the factor of 23 in the denominator still cannot be removed by reduction (since 23 does not divide j1*j2, so 23 cannot divide the numerator); and adding any further reciprocals of integers < 23 similarly cannot remove the factor of 23 from the denominator.
%e 25 cannot be in the sequence because it is a prime power: for any positive integer j1 < 25, 1/j1 + 1/25 = (25 + j1)/(25*j1), which cannot be reduced unless 5 divides j1, but even then the denominator would remain divisible by 25 (and, as above, this would continue to be the case after the addition any number of reciprocals of other integers < 25).
%e For additional examples, including some ideas for heuristics for obtaining solutions for numbers n that are in the sequence, see the Links. (End)
%e For an inelegantly written Magma program that computes the first 1000 terms in about 0.3 seconds on the Magma Calculator, see the Links. - _Jon E. Schoenfield_, Apr 19 2017
%t n=55; try3[lev_, s_] := Module[{nmim, nmax, si, i}, AppendTo[soln, 0]; If[lev==1, nmin=2, nmin=1+soln[[ -2]]]; nmax=n-1; Do[If[i<n/2 || !PrimeQ[i], si=s+1/i; If[si==1, soln[[ -1]]=i; Print[soln]; Abort[]]; If[si<1, soln[[ -1]]=i; try3[lev+1, si]]], {i, nmin, nmax}]; soln=Drop[soln, -1]]; soln={n}; try3[1, 1/n] (* _T. D. Noe_, Mar 30 2004 *)
%Y Cf. A092669, A092672.
%Y Cf. A093407 (least m such that m*prime(n) is in this sequence).
%Y Cf. A128253 (primitive elements).
%K nonn
%O 1,2
%A _Max Alekseyev_, Mar 02 2004
%E More terms from _T. D. Noe_, Mar 30 2004
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