OFFSET
1,1
COMMENTS
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)).
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.
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).
REFERENCES
I. Gueye and M. Mizony : Recent progress about Erdős-Straus conjecture, B SO MA S S, Volume 1, Issue 2, pp. 6-14.
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.
LINKS
P. Erdős, On a Diophantine equation, (Hungarian. Russian, English summaries), Mat. Lapok 1, 1950, pp. 192-210.
M. Mizony and M.-L. Gardes, Sur la conjecture d'Erdős et Straus, see pages 14-17.
Eric W. Weisstein, MathWorld: Twin Pythagorean Triple
K. Yamamoto, On the Diophantine Equation 4/n=1/x+1/y+1/z, Mem. Fac. Sci. Kyushu U. Ser. A 19, 37-47, 1965.
EXAMPLE
For n=12 the a(12)=133 solutions are {k = 0, t = 6},{k = 5, t = 0}.
MAPLE
G:=(n, p)->4/p = [2*(2*n+1)/(n*p+p+1), 4/p/(n*p+p+1), 2/(n*p+p+1)]:
cousin:=proc(p)
local n;
for n from 0 to 300 do
if n*p+p+1 mod 4*(2*n+1)=0 then return([p, n, G(n, p)]); fi:
od:
end:
L:=NULL:for m to 400 do L:=L, cousin(4*m+1): od:{L}[1..4]; map(u->op(1, u), {L});
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Mizony, Jun 17 2012
STATUS
approved