A256985 Define a sequence {b(i), i >= 0} by b(i) = 1 for 0 <= i <= n, thereafter b(i+1) = (b(i)+b(i-n)) mod 10; consider the sequence of n-tuples [b(i-n+1), b(i-n+2),...,b(i)]; this repeats with period a(n).

60, 217, 520, 42, 196812, 2480437, 2232, 7128815, 1736327236, 124516392, 203450520, 40193528485, 14417724597564, 22856442972, 324145501174, 7946757, 193726348876699204, 206135768515040, 581179046630097612, 32289695739703771, 275114595439871720

Offset

1,1

Links

Hiroaki Yamanouchi, Table of n, a(n) for n = 1..105

Popular Computing (Calabasas, CA), Contest 9, Vol. 4 (No. 40, July 1976), scanned copy of page PC40-1.

Popular Computing (Calabasas, CA), Contest 9, Vol. 4 (No. 40, July 1976), scanned copy of page PC40-3.

Example

For n=1 we get the Fibonacci sequence (starting 1,1,2,3,...), A000045, which read mod 10 repeats with period 60 (see A003893), so a(1)=60.  The full period in this case is:
[1, 1, 2, 3, 5, 8, 3, 1, 4, 5, 9, 4, 3, 7, 0, 7, 7, 4, 1, 5, 6, 1, 7, 8, 5, 3, 8, 1, 9, 0, 9, 9, 8, 7, 5, 2, 7, 9, 6, 5, 1, 6, 7, 3, 0, 3, 3, 6, 9, 5, 4, 9, 3, 2, 5, 7, 2, 9, 1, 0].

Prog

(Python)
from itertools import accumulate # requires python 3.2 or higher
def A256985(n):
....ilist, k = [1]*(n+1), 1
....jlist = [d % 10 for d in accumulate(ilist)]
....jlist = [jlist[-1]]+ jlist[:-1]
....while ilist != jlist:
........k += 1
........jlist = [d % 10 for d in accumulate(jlist)]
........jlist = [jlist[-1]]+ jlist[:-1]
....return k # Chai Wah Wu, Apr 30 2015

Crossrefs

Cf. A000045, A003893.

Sequence in context: A126248 A224335 A068628 * A075287 A277990 A103741

Adjacent sequences: A256982 A256983 A256984 * A256986 A256987 A256988

Keyword

nonn

Author

N. J. A. Sloane, Apr 26 2015

Extensions

a(6)-a(11) from Chai Wah Wu, Apr 30 2015

a(12)-a(21) from Hiroaki Yamanouchi, May 04 2015

Status

approved