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%I #14 Oct 12 2017 08:21:59
%S 9,10,11,14,15,18,26,30,34,35,38,42,54,55,59,62,63,70,74,82,90,95,98,
%T 102,105,122,126,131,135,138,143,158,159,170,179,190,194,195,202,203,
%U 210,215,227,230,234,238,251,255,258,266,270,278,294,297,298,310,315
%N Integers representable as the product of the sum of three positive integers with the sum of their reciprocals: n=(x+y+z)*(1/x+1/y+1/z).
%C All terms of this sequence occur also in A085514. Bremner et al. have shown that the problem is equivalent to finding rational points of infinite order on the elliptic curve E_n : u^2 = v^3 + (n^2 - 6*n - 3)*v^2 + 16*n*v
%C The only values of n < 1000 with positive representations are shown in bold type in Table 1 in Section 8 of Bremner et al.'s paper (except for the singular value n=9 and the case n=10) - Herman Jamke (hermanjamke(AT)fastmail.fm), Jan 09 2008
%H A. Bremner, R. K. Guy and R. Nowakowski, <a href="https://doi.org/10.1090/S0025-5718-1993-1189516-5">Which integers are representable as the product of the sum of three integers with the sum of their reciprocals?</a>, Math. Comp. 61 (1993) 117-130.
%H A. MacLeod, <a href="http://web.archive.org/web/20090628095732/http://maths.paisley.ac.uk/allanm/ecrnt/knight/knintro.htm">The Knight's Problem</a>
%H A. MacLeod, <a href="http://web.archive.org/web/20090226001126/http://maths.paisley.ac.uk/allanm/ECRNT/Ecrnt.htm">Elliptic Curves</a>
%e a(2)=(1+1+2)*(1/1+1/1+1/2)=10.
%e a(3)=(1+2+3)*(1/1+1/2+1/3)=6*(11/6)=11.
%e a(4)=(2+3+10)*(1/2+1/3+1/10)=14.
%e a(12)=(561+6450+13889)*(1/561+1/6450+1/13889)=42.
%Y Cf. A085514 (also negative x, y, z admitted).
%K nonn
%O 1,1
%A _Hugo Pfoertner_, Jul 19 2003
%E Corrected and extended by _David J. Rusin_, Jul 30 2003
%E More terms from Herman Jamke (hermanjamke(AT)fastmail.fm), Jan 09 2008
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