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A274696
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Variation on Fermat's Diophantine m-tuple: 1 + the LCM of any two distinct terms is a square.
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0
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0, 1, 3, 8, 15, 24, 120, 168, 840, 1680, 5040, 201600, 256032000
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OFFSET
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1,3
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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).
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MATHEMATICA
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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 *)
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PROG
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(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)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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