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A261803 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.

%I A261803 #28 Sep 04 2021 04:36:49
%S A261803 50,334,516,670,844,1164,1250,1800,2450,9800,14450,20000,24200,101250,
%T A261803 105800,135200,162450,168200,204800,304200,336200,451250,480200,
%U A261803 490050,530450,696200,924800,966050,1008200,1125000,1155200,1428050,1805000,2332800,2420000
%N A261803 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.
%C A261803 The sequence is probably infinite. [This follows from Schinzel's hypothesis H, for example. - _Charles R Greathouse IV_, Aug 31 2021]
%C A261803 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.
%H A261803 Jean-François Alcover, <a href="/A261803/b261803.txt">Table of n, a(n) for n = 1..500</a>
%e A261803 a(1) = 50 because 50^2+1 = 41*61 => 61/41 = 1.4878... < 1.5
%e A261803 a(2) = 334 because 334^2+1 = 281*397 => 397/281 = 1.4128... < 1.4878...
%e A261803 a(3) = 516 because 516^2+1 = 449*593 => 593/449 = 1.3207... < 1.4128...
%e A261803 a(4) = 670 because 670^2+1 = 593*757 => 757/593 = 1.2765... < 1.3207...
%p A261803 with(numtheory):nn:=100:d:=1.5:
%p A261803 for n from 1 to nn do:
%p A261803   x:=factorset(n^2+1):n0:=bigomega(n^2+1):
%p A261803    if n0=2
%p A261803    then
%p A261803    q:=evalf(x[2]/x[1]):
%p A261803    if q<d then
%p A261803    d:=q:printf(`%d, `,n):
%p A261803    else fi:
%p A261803    fi :
%p A261803   od:
%t A261803 (* 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 *)
%Y A261803 Cf. A002496, A109306, A206328.
%Y A261803 Subsequence of A085722.
%K A261803 nonn,nice
%O A261803 1,1
%A A261803 _Michel Lagneau_, Sep 01 2015