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A181755
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a(1) = 1, a(2) = 5. For n >= 3, a(n) is found by concatenating 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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1, 5, 3, 51, 301, 51001, 30100001, 5100100000001, 301000010000000000001, 5100100000001000000000000000000001, 3010000100000000000010000000000000000000000000000000001
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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
... a(3) = 15/5 = 3
... a(4) = 153/3 = 51
... a(5) = 15351/51 = 301
... a(6) = 15351301/301 = 51001.
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LINKS
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FORMULA
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DEFINITION
a(1) = 1, a(2) = 5, and for n >= 3
(1)... a(n) = concatenate(a(1),a(2),...,a(n-1))/a(n-1).
RECURRENCE RELATION
For n >= 2
(2)... a(n+2) = 10^F(n)*a(n)+1,
where F(n) = A000045(n) are the Fibonacci numbers.
For n >= 2, a(n) has F(n-1) digits.
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MAPLE
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M:=11:
a:=array(1..M):s:=array(1..M):
a[1]:=1:a[2]:=5:
s[1]:=convert(a[1], string):
s[2]:=cat(s[1], convert(a[2], 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], string));
end do:
seq(a[n], n = 1..M);
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MATHEMATICA
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nxt[lst_]:=Module[{nt=FromDigits[Flatten[IntegerDigits/@lst]]/Last[ lst]}, Flatten[{lst, nt}]]; Nest[nxt[#]&, {1, 5}, 10] (* Harvey P. Dale, Aug 08 2014 *)
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CROSSREFS
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KEYWORD
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easy,nonn,base
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AUTHOR
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STATUS
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approved
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