A112371 Numbers n such that the last 9 decimal digits of the n-th Fibonacci number is pandigital 1-9.

541, 919, 1788, 6355, 16257, 17799, 20411, 24347, 28837, 36485, 40784, 43450, 45136, 45196, 51973, 54453, 54833, 57128, 57969, 63692, 67188, 67952, 69931, 74765, 76259, 78102, 78196, 78826, 81070, 81726, 87123, 87362, 91636, 91932

Offset: 1

Author

Roger Hui, Dec 22 2005

Keywords

• nonn
• base

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

Examples

			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.
		

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)

Crossrefs

Cf. A000045, A112516.

Programs

J

NB. In J (www.jsoftware.com).
f=: 3 : '{."(1) 1e9&|@(+/\)@|.^:( ":&.> f n

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