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A091569 a(1) = 1; for n > 1, a(n) is the smallest positive integer not already used such that a(n)*a(n-1) + 1 is a perfect square.

%I A091569 #15 Sep 23 2023 03:16:05
%S A091569 1,3,5,7,9,11,13,15,8,6,4,2,12,10,36,34,32,30,28,26,24,22,20,18,16,14,
%T A091569 52,50,48,35,33,31,29,27,25,23,21,19,17,64,62,60,40,38,148,146,144,37,
%U A091569 39,41,43,45,47,49,51,53,55,57,59,61,63,65,67,69,71,73,75,77,79,81,83
%N A091569 a(1) = 1; for n > 1, a(n) is the smallest positive integer not already used such that a(n)*a(n-1) + 1 is a perfect square.
%C A091569 Does this sequence contain every positive integer? We could get an equally interesting sequence by choosing a(1) to be any other positive integer.
%C A091569 A sequence with the same condition but without the requirement for a(n) to be distinct would end up repeating (1,3) or (2,4), depending on the initial term. - _Ivan Neretin_, May 26 2015
%H A091569 Ivan Neretin, <a href="/A091569/b091569.txt">Table of n, a(n) for n = 1..10000</a>
%e A091569 10 is followed by 36 because 10*36+1 = 19^2 and 8 and 12 were already used.
%p A091569 N:= 10^4: Used:= Vector(N,datatype=integer[4]):
%p A091569 a[1]:= 1: blocked:= false: Used[1]:= 1:
%p A091569 for n from 2 to 100 while not(blocked) do
%p A091569     ndone:= false;
%p A091569     if n = 2 then T:= [0]
%p A091569       else T:= select(t -> t^2 mod a[n-1] = 1, [$0..a[n-1]-1])
%p A091569       fi;
%p A091569     for s from 0 while not (ndone) do
%p A091569        for t in T while not (ndone) do
%p A091569          x:= s * a[n-1] + t;
%p A091569          if x <= 1 then next fi;
%p A091569          y:= (x^2-1)/a[n-1];
%p A091569          if y > N then blocked:= true; ndone:= true
%p A091569          elif Used[y] = 0 then
%p A091569             a[n]:= y;
%p A091569             Used[y]:= 1;
%p A091569             ndone:= true;
%p A091569             print(n,y);
%p A091569          fi
%p A091569        od
%p A091569     od
%p A091569 od:
%p A091569 seq(a[n],n=1..100); # _Robert Israel_, May 26 2015
%t A091569 a = {1}; Do[a = Join[a, Select[Complement[Range[(Max[a] + 1)*n], a], IntegerQ[Sqrt[#*a[[-1]] + 1]] &, 1]], {n, 2, 71}]; a (* _Ivan Neretin_, May 26 2015 *)
%o A091569 (MATLAB) A = zeros(1, 100); A(1) = 1; used = zeros(1, 1000); used(1) = 1; for i = 2:100; found = 0; k = 0; while found == 0; k = k + 1; if used(k) == 0; s = sqrt(k*A(i - 1) + 1); if s == floor(s); A(i) = k; used(k) = 1; found = 1; end; end; end; end; A
%Y A091569 Cf. A083203.
%K A091569 easy,nonn
%O A091569 1,2
%A A091569 _David Wasserman_, Mar 04 2004