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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.
4
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, 52, 50, 48, 35, 33, 31, 29, 27, 25, 23, 21, 19, 17, 64, 62, 60, 40, 38, 148, 146, 144, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83
OFFSET
1,2
COMMENTS
Does this sequence contain every positive integer? We could get an equally interesting sequence by choosing a(1) to be any other positive integer.
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
LINKS
EXAMPLE
10 is followed by 36 because 10*36+1 = 19^2 and 8 and 12 were already used.
MAPLE
N:= 10^4: Used:= Vector(N, datatype=integer[4]):
a[1]:= 1: blocked:= false: Used[1]:= 1:
for n from 2 to 100 while not(blocked) do
ndone:= false;
if n = 2 then T:= [0]
else T:= select(t -> t^2 mod a[n-1] = 1, [$0..a[n-1]-1])
fi;
for s from 0 while not (ndone) do
for t in T while not (ndone) do
x:= s * a[n-1] + t;
if x <= 1 then next fi;
y:= (x^2-1)/a[n-1];
if y > N then blocked:= true; ndone:= true
elif Used[y] = 0 then
a[n]:= y;
Used[y]:= 1;
ndone:= true;
print(n, y);
fi
od
od
od:
seq(a[n], n=1..100); # Robert Israel, May 26 2015
MATHEMATICA
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 *)
PROG
(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
CROSSREFS
Cf. A083203.
Sequence in context: A187472 A187411 A189401 * A206545 A293703 A120890
KEYWORD
easy,nonn
AUTHOR
David Wasserman, Mar 04 2004
STATUS
approved