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A213687 Numbers which are the values of the quadratic polynomial 3+4*k+7*t+8*k*t on nonnegative integers. 0

%I #16 Nov 16 2015 04:24:01

%S 3,7,10,11,15,17,19,22,23,24,27,31,34,35,37,38,39,43,45,46,47,51,52,

%T 55,57,58,59,63,66,67,70,71,73,75,77,79,80,82,83,87,91,94,95,97,99,

%U 101,103,106,107,108,111,112,115,117,118,119,122,123,126,127,129

%N Numbers which are the values of the quadratic polynomial 3+4*k+7*t+8*k*t on nonnegative integers.

%C For all these numbers a(n) we have the following Erdos-Straus decomposition: 4/p= 4/(3+4*k+7*t+8*k*t) = 1/(2*(3+4*k+7*t+8*k*t)*(1+k)) + 1/((1+k)*(2*t+1)) + 1/(2*(1+k)*(2*t+1)*(3+4*k+7*t+8*k*t)));

%C Moreover this sequence is related to irreducible twin Pythagorean triples: the associated Pythagorean triple is [2t(t+1),2t+1, 2t(t+1)+1].

%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)=31 solutions are {k = 0, t = 4},{k = 7, t = 0}.

%p H:=(k, t) -> 4/(3+4*k+7*t+8*k*t) = [1/2*1/((3+4*k+7*t+8*k*t)*(1+k)), 1/((1+k)*(2*t+1)), 1/2*1/((1+k)*(2*t+1)*(3+4*k+7*t+8*k*t))]:

%p cousin:=proc(p)

%p local n,k;

%p for n from 0 to (p-3)/7 do

%p if (p-3-7*n) mod (4+8*n)=0 then k:=(p-3-7*n)/(4+8*n):

%p return([p,n,H(k,n)]) fi;od;

%p end:

%p L:=NULL:for p from 2 to 500 do L:=L,cousin(p): od:{L}[1..10];map(u->op(1,u),{L});map(u->op(2,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 18 2012

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