A274696 Variation on Fermat's Diophantine m-tuple: 1 + the LCM of any two distinct terms is a square.

0, 1, 3, 8, 15, 24, 120, 168, 840, 1680, 5040, 201600, 256032000

Offset: 1

Author

Robert C. Lyons, Jul 05 2016

Keywords

• nonn

Comments

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.

Examples

			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).
		

Crossrefs

Cf. A030063.

Programs

Mathematica

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 *)

Sage

seq = [0]
prev_element = 0
max_n = 13
for n in range(2, max_n+1):
    next_element = prev_element + 1
    while True:
        all_match = True
        for element in seq:
            x = lcm( element, next_element ) + 1
            if not is_square(x):
                all_match = False
                break
        if all_match:
            seq.append( next_element )
            print(seq)
            break
        next_element += 1
    prev_element = next_element
print(seq)