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A242074
Numbers n such that n^2 - 1 is the product of four distinct Fibonacci numbers greater than 1.
1
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
OFFSET
1,1
COMMENTS
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, ...
EXAMPLE
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.
MAPLE
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):
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, Aug 14 2014
STATUS
approved