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%I #16 Mar 14 2020 17:14:14
%S 0,1,3,8,15,24,120,168,840,1680,5040,201600,256032000
%N Variation on Fermat's Diophantine m-tuple: 1 + the LCM of any two distinct terms is a square.
%C a(1) = 0; for n>1, a(n) = smallest integer > a(n-1) such that lcm(a(n),a(i))+1 is square for all 1 <= i <= n-1.
%e After a(1)=0, a(2)=1, a(3)=3, we want m, the smallest number > 3 such that lcm(0,m)+1, lcm(2,m)+1 and lcm(3,m)+1 are squares: this is m = 8 = a(4).
%t a = {0}; Do[AppendTo[a, SelectFirst[Range[Max@ a + 1, 3*10^5], Function[k, Times @@ Boole@ Map[IntegerQ@ Sqrt[LCM[a[[#]], k] + 1] &, Range[n - 1]] == 1]]], {n, 2, 12}]; a (* _Michael De Vlieger_, Jul 05 2016, Version 10 *)
%o (Sage)
%o seq = [0]
%o prev_element = 0
%o max_n = 13
%o for n in range(2, max_n+1):
%o next_element = prev_element + 1
%o while True:
%o all_match = True
%o for element in seq:
%o x = lcm( element, next_element ) + 1
%o if not is_square(x):
%o all_match = False
%o break
%o if all_match:
%o seq.append( next_element )
%o print(seq)
%o break
%o next_element += 1
%o prev_element = next_element
%o print(seq)
%Y Cf. A030063.
%K nonn
%O 1,3
%A _Robert C. Lyons_, Jul 05 2016
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