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A181865
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a(1) = 1, a(2) = 2. For n >= 3, a(n) is found by concatenating the cubes of the first n-1 terms of the sequence and then dividing the resulting number by a(n-1).
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9
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OFFSET
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1,2
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COMMENTS
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The calculations for the first few values of the sequence are
... 2^3 = 8 so a(3) = 18/2 = 9
... 9^3 = 729 so a(4) = 18729/9 = 2081
... 2081^3 = 9011897441 so a(5)=187299011897441/2081 = 90004330561.
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LINKS
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FORMULA
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DEFINITION
a(1) = 1, a(2) = 2, and for n >= 3
(1)... a(n) = concatenate(a(1)^3,a(2)^3,...,a(n-1)^3)/a(n-1).
RECURRENCE RELATION
For n >= 2
(2)... a(n+2) = a(n+1)^2 + 10^F(n,3)*a(n),
where F(n,3) is the Fibonacci polynomial F(n,x) evaluated at x = 3.
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MAPLE
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M:=7:
a:=array(1..M):s:=array(1..M):
a[1]:=1:a[2]:=2:
s[1]:=convert(a[1]^3, string):
s[2]:=cat(s[1], convert(a[2]^3, string)):
for n from 3 to M do
a[n] := parse(s[n-1])/a[n-1];
s[n]:= cat(s[n-1], convert(a[n]^3, string));
end do:
seq(a[n], n = 1..M);
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CROSSREFS
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Cf. A006190 (F(n,3)), A181754, A181755, A181756, A181864, A181866, A181867, A181868, A181869, A181870
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KEYWORD
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nonn,easy,base
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AUTHOR
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STATUS
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approved
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