|
%I #28 Mar 03 2023 22:34:01
%S 5,13,17,21,29,37,41,45,53,61,65,69,77,85,89,93,97,101,109,113,117,
%T 125,133,137,141,149,153,157,161,165,173,181,185,189,197,205,209,213,
%U 221,229,233,237,241,245,253,257
%N Numbers which are the values of the quadratic polynomial 5+8t+12k+16kt on nonnegative integers.
%C For all these numbers a(n) we have the following Erdős-Straus decomposition: 4/p = 4/(5+8*t+12*k+16*k*t) = 1/(2*(2*k+1)*(2+3*t+3*k+4*k*t)) + 1/(2+3*t+3*k+4*k*t) + 1/(2*(5+8*t+12*k+16*k*t)*(2*k+1)*(2+3*t+3*k+4*k*t)).
%C Moreover this sequence is related to irreducible twin Pythagorean triples: the associated Pythagorean triple is [2n(n+1), 2n+1,2n(n+1)+1], where n=2+4t+6k+8kt.
%C In 1948 Erdős and Straus conjectured that for any positive integer n >= 2 the equation 4/n = 1/x + 1/y +1/z has a solution with positive integers x, y and z (without the additional requirement 0 < x < y < z).
%C For the solution (x,y,z) having the largest z value, see (A075245, A075246, A075247).
%D I. Gueye and M. Mizony, Recent progress about Erdős-Straus conjecture, B SO MA S S, Volume 1, Issue 2, pp. 6-14.
%D I. Gueye and M. Mizony, Towards the proof of Erdős-Straus conjecture, B SO MA S S, Volume 1, Issue 2, pp. 141-150.
%H P. Erdős, <a href="https://www.renyi.hu/~p_erdos/1950-02.pdf">On a Diophantine equation</a>, (Hungarian. Russian, English summaries), Mat. Lapok 1, 1950, pp. 192-210.
%H M. Mizony and M.-L. Gardes, <a href="http://math.univ-lyon1.fr/~mizony/SurErdos_Straus2.pdf"> Sur la conjecture d'Erdős et Straus</a>, see pages 14-17.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/TwinPythagoreanTriple.html">Twin Pythagorean Triple</a>.
%H K. Yamamoto, <a href="http://dx.doi.org/10.2206/kyushumfs.19.37">On the Diophantine Equation 4/n=1/x+1/y+1/z</a>, Mem. Fac. Sci. Kyushu U. Ser. A 19, 37-47, 1965.
%e For n=5 the a(5)=29 solutions are {k=0, t=3}, {k=2, t=0}.
%p G:=(p,d)->4/p = [4*d/(p+d)/(p+1), 4/(p+d), 4*d/(p+d)/(p+1)/p]:
%p cousin:=proc(p)
%p local d;
%p for d from 3 by 4 to 100 do
%p if ((p+1)/2) mod d=0 and (p+d)*(p+1) mod d=0 then
%p return([p,G(p,d)]) fi;od;
%p end:
%p for k to 20 do cousin(4*k+1) od;
%Y Cf. A001844 (centered square numbers: 2n(n+1)+1).
%Y Cf. A005408 (x values), A046092 (y values).
%Y Cf. A195770 (positive integers a for which there is a 1-Pythagorean triple (a,b,c) satisfying a<=b).
%Y A073101 number of solutions (x,y,z) to 4/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z.
%K nonn
%O 1,1
%A _Michel Mizony_, Jun 09 2012
|
