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A261803
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a(n) is the smallest number satisfying a(n)^2+1 = p(n)*q(n), p(n) < q(n) both prime, such that q(n+1)/p(n+1) < q(n)/p(n) with the initial condition q(1)/p(1) < 3/2.
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1
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50, 334, 516, 670, 844, 1164, 1250, 1800, 2450, 9800, 14450, 20000, 24200, 101250, 105800, 135200, 162450, 168200, 204800, 304200, 336200, 451250, 480200, 490050, 530450, 696200, 924800, 966050, 1008200, 1125000, 1155200, 1428050, 1805000, 2332800, 2420000
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OFFSET
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1,1
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COMMENTS
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The sequence is probably infinite. [This follows from Schinzel's hypothesis H, for example. - Charles R Greathouse IV, Aug 31 2021]
A majority of numbers in the sequence are of the form 2*q^2 with q = 5, 25, 30, 35, 70, 85, 100, 110, 225, 230, 260, 285, 290, 320, 390, 410, ... So, it seems that {a(n)} = {334, 516, 670, 844, 1164} union {2*A109306(n)^2} where A109306 are the numbers k such that k^2 + (k-1)^2 and k^2 + (k+1)^2 are both primes.
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LINKS
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EXAMPLE
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a(1) = 50 because 50^2+1 = 41*61 => 61/41 = 1.4878... < 1.5
a(2) = 334 because 334^2+1 = 281*397 => 397/281 = 1.4128... < 1.4878...
a(3) = 516 because 516^2+1 = 449*593 => 593/449 = 1.3207... < 1.4128...
a(4) = 670 because 670^2+1 = 593*757 => 757/593 = 1.2765... < 1.3207...
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MAPLE
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with(numtheory):nn:=100:d:=1.5:
for n from 1 to nn do:
x:=factorset(n^2+1):n0:=bigomega(n^2+1):
if n0=2
then
q:=evalf(x[2]/x[1]):
if q<d then
d:=q:printf(`%d, `, n):
else fi:
fi :
od:
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MATHEMATICA
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(* Assumption: n>7 ==> a(n)=0 mod 50 *) a[n_] := a[n] = For[k = Which[n==1, 0, n <= 7, a[n-1]+1, True, a[n-1] + 50], True, k = Which[n <= 7, k+1, k == a[n-1]+1, k+49, True, k+50], f = FactorInteger[k^2+1]; If[Length[f] == 2, If[f[[All, 2]] == {1, 1}, {p1, q1} = f[[All, 1]]; If[q1/p1 < If[n == 1, 3/2, q[n-1]/p[n-1]], p[n] = p1; q[n] = q1; Return[k]]]]]; Table[Print["a(", n, ") = ", a[n], " p = ", p[n], " q = ", q[n], " q/p = ", N[q[n]/p[n], 10], " q-p = ", q[n]-p[n]]; a[n], {n, 1, 100}] (* Jean-François Alcover, Sep 28 2015 *)
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CROSSREFS
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KEYWORD
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nonn,nice
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AUTHOR
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STATUS
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approved
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