OFFSET
1,1
COMMENTS
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)));
Moreover this sequence is related to irreducible twin Pythagorean triples: the associated Pythagorean triple is [2t(t+1),2t+1, 2t(t+1)+1].
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)=31 solutions are {k = 0, t = 4},{k = 7, t = 0}.
MAPLE
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))]:
cousin:=proc(p)
local n, k;
for n from 0 to (p-3)/7 do
if (p-3-7*n) mod (4+8*n)=0 then k:=(p-3-7*n)/(4+8*n):
return([p, n, H(k, n)]) fi; od;
end:
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});
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Mizony, Jun 18 2012
STATUS
approved