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A213687
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Numbers which are the values of the quadratic polynomial 3+4*k+7*t+8*k*t on nonnegative integers.
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0
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3, 7, 10, 11, 15, 17, 19, 22, 23, 24, 27, 31, 34, 35, 37, 38, 39, 43, 45, 46, 47, 51, 52, 55, 57, 58, 59, 63, 66, 67, 70, 71, 73, 75, 77, 79, 80, 82, 83, 87, 91, 94, 95, 97, 99, 101, 103, 106, 107, 108, 111, 112, 115, 117, 118, 119, 122, 123, 126, 127, 129
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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/(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).
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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)=31 solutions are {k = 0, t = 4},{k = 7, t = 0}.
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MAPLE
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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});
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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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