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A213686
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Numbers which are the values of the quadratic polynomial 13+20*t+24*k+32*k*t at nonnegative integers.
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0
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13, 33, 37, 53, 61, 73, 85, 89, 93, 109, 113, 133, 141, 145, 153, 157, 173, 181, 193, 201, 205, 213, 229, 233, 245, 253, 257, 273, 277, 293, 297, 301, 313, 317, 325, 333, 349, 353, 369, 373, 393, 397, 401, 405, 413, 421, 425, 433, 445, 453, 469, 473, 481
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OFFSET
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1,1
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COMMENTS
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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).
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REFERENCES
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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.
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LINKS
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EXAMPLE
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For n=12 the a(12)=133 solutions are {k = 0, t = 6},{k = 5, t = 0}.
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MAPLE
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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});
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CROSSREFS
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Cf. A213340 (the quadratic polynomial 5+8t+12k+16kt).
Cf. A001844 (centered square numbers: 2n(n+1)+1).
Cf. A073101 (number of solutions (x,y,z) to 4/n = 1/x + 1/y + 1/z satisfying 0 < x < y < z).
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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