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%I #20 Nov 16 2015 08:10:31
%S 13,33,37,53,61,73,85,89,93,109,113,133,141,145,153,157,173,181,193,
%T 201,205,213,229,233,245,253,257,273,277,293,297,301,313,317,325,333,
%U 349,353,369,373,393,397,401,405,413,421,425,433,445,453,469,473,481
%N Numbers which are the values of the quadratic polynomial 13+20*t+24*k+32*k*t at nonnegative integers.
%C For all these numbers a(n) we have the following Erdos-Straus decomposition: 4/p=4/(13+24*k+20*t+32*k*t) = 1/(6*k+8*k*t+4+6*t) + 1/((13+24*k+20*t+32*k*t)*(5+8*k)*(3*k+4*k*t+2+3*t)) + 1/(2*(5+8*k)*(3*k+4*k*t+2+3*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+4k.
%C In 1948 Erdos 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 M. Mizony and I. Gueye : Towards the proof of Erdős-Straus conjecture, B SO MA S S, Volume 1, Issue 2,p 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 W. Weisstein, <a href="http://mathworld.wolfram.com/TwinPythagoreanTriple.html">MathWorld: 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=12 the a(12)=133 solutions are {k = 0, t = 6},{k = 5, t = 0}.
%p G:=(n,p)->4/p = [2*(2*n+1)/(n*p+p+1), 4/p/(n*p+p+1), 2/(n*p+p+1)]:
%p cousin:=proc(p)
%p local n;
%p for n from 0 to 300 do
%p if n*p+p+1 mod 4*(2*n+1)=0 then return([p,n,G(n,p)]);fi:
%p od:
%p end:
%p L:=NULL:for m to 400 do L:=L,cousin(4*m+1): od:{L}[1..4];map(u->op(1,u),{L});
%Y Cf. A213340 (the quadratic polynomial 5+8t+12k+16kt).
%Y Cf. A001844 (centered square numbers: 2n(n+1)+1).
%Y Cf. A005408 (x values), A046092 (y values).
%Y Cf. 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 17 2012
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