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A242074
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Numbers n such that n^2 - 1 is the product of four distinct Fibonacci numbers greater than 1.
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1
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25, 41, 64, 103, 131, 169, 271, 274, 281, 441, 713, 901, 1156, 1871, 3025, 4894, 7921, 12817, 20736, 21319, 33551, 54289, 87842, 142129, 229969, 372100, 602071, 974169, 1576238, 2550409, 4126649, 6677056, 10803703, 17480761, 28284466, 45765225, 74049689
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OFFSET
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1,1
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COMMENTS
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The sequence contains the squares of the Fibonacci numbers (A007598(n) for n >=5).
Proof:
Let F(m) be the m-th Fibonacci number. If n = F(m)^2, n^2 - 1 = F(m)^4-1.
For m > 1, F(m)^4 - 1 = F(m-2)*F(m-1)*F(m+1)*F(m+2) with the property F(m-2) + F(m-1) + F(m+1) + F(m+2) = F(m) + F(m+3) = 2*F(m+2). (See A244855.)
F(m)^2 - 1 = F(m-1)*F(m+1) if m odd, and F(m)^2 - 1 = F(m-2)*F(m+2)if m even;
F(m)^2 + 1 = F(m-2)*F(m+2) if m odd, and F(m)^2 + 1 = F(m-1)*F(m+1) if m even, hence the product (F(m)^2 - 1)*(F(m)^2 + 1) = F(m-2)*F(m-1)*F(m+1)*F(m+2).
The primes of the sequence are 41, 103, 131, 271, 281, 1871, 21319, ...
The composites (nonsquares) of the sequence are 274, 713, 901, 4894, 12817, 33551, 87842, ...
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LINKS
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EXAMPLE
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25^2 - 1 = 2*3*8*13 = F(5 - 2)*F(5 - 1)*F(5 + 1)*F(5 + 2) where F(5) = 5;
41^2 - 1 = 2*5*8*21;
64^2 - 1 = 3*5*13*21 = F(6 - 2)*F(6 - 1)*F(6 + 1)*F(6 + 2) where F(6) = 8;
103^2 - 1 = 3*8*13*34;
131^2 - 1 = 3*8*13*55;
169^2 - 1 = 5*8*21*34 = F(7 - 2)*F(7 - 1)*F(7 + 1)*F(7 + 2) where F(7) = 13;
271^2 - 1 = 3*5*34*144;
274^2 - 1 = 5*13*21*55;
281^2 - 1 = 2*5*8*987;
441^2 - 1 = 8*13*34*55 = F(8 - 2)*F(8 - 1)*F(8 + 1)*F(8 + 2) where F(8) = 21.
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MAPLE
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with(combinat, fibonacci):with(numtheory):nn:=100:lst:={}:T:=array(1..nn):
for n from 1 to nn do:
T[n]:=fibonacci(n):
od:
for p from 1 to nn-1 do:
for q from p+1 to nn-1 do:
for r from q+1 to nn-1 do:
for s from r+1 to nn-1 do:
f:=T[p]*T[q]*T[r]*T[s]+1:x:=sqrt(f):
if x=floor(x)and T[p]<>1
then
lst:=lst union {x}:
else
fi:
od:
od:
od:
od:
print(lst):
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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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