OFFSET
1,1
COMMENTS
Since the Fibonacci sequence mod 10^9 is periodic with period 1500000000, there is some positive M such that this sequence satisfies a(n+M) = a(n) + 1500000000. - Robert Israel, Jan 18 2015
REFERENCES
Clifford A. Pickover, "Wonders of Numbers".
LINKS
Norman Morton and Michael Satteson, Table of n, a(n) for n = 1..10000, (first 150 terms from Norman Morton)
EXAMPLE
The 541st Fibonacci number is:
51621 23292 73937 94428 28328 17223 02417 68441 62155 65352
08137 22196 49050 89439 99028 11978 84249 30258 98332 77779
69788 39725 641
which is pandigital 1-9 in its last 9 digits.
MAPLE
f:= proc(n) option remember; f(n-1)+f(n-2) mod 10^9 end proc:
f(0):= 0: f(1):= 1:
filter:= n -> convert(convert(f(n), base, 10), set)={$1..9};
select(filter, [$1..10^5]); # Robert Israel, Jan 18 2015
PROG
(J) NB. In J (www.jsoftware.com).
f=: 3 : '{."(1) 1e9&|@(+/\)@|.^:(<y.) 0 1'
I. (<'123456789')= /:~&.> ":&.> f n
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Roger Hui, Dec 22 2005
STATUS
approved