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A214635 Period of A213437 mod n. 3
1, 1, 1, 1, 3, 1, 1, 1, 3, 3, 1, 1, 4, 1, 3, 1, 6, 3, 1, 3, 1, 1, 1, 1, 3, 4, 3, 1, 1, 3, 1, 1, 1, 6, 3, 3, 1, 1, 4, 3, 7, 1, 1, 1, 3, 1, 4, 1, 6, 3, 6, 4, 1, 3, 3, 1, 1, 1, 3, 3, 10, 1, 3, 1, 12, 1, 1, 6, 1, 3, 11, 3, 6, 1, 3, 1, 1, 4, 4, 3, 9, 7, 5, 1, 6, 1, 1, 1, 14, 3, 4, 1, 1, 4, 3, 1, 3, 6, 3 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,5

LINKS

Table of n, a(n) for n=1..99.

FORMULA

Empirically:

A214635(2^n) = 1,  A214635(5^n) = A214635(10^n) = 3, for all n>0.

A214635(3^n) = A214635(6^n) = (1, 3, 3, 9, 27, 81, ...) = 3^(n-2) for n>2.

A214635(15^n) = (3,3,3,9,27,81,...) = A214635(3^n) for n>1.

A214635(7^n) = (1,6,42,294,...) = 6*7^(n-2) for n>1.

A214635(11^n) = (1,20,220,2420,...) = 20*11^(n-2) for n>1. - M. F. Hasler, Jul 24 2012

PROG

(PARI) A214635(n, N=99)={my(a=[Mod(1, n)]); for(n=1, N-1, a=concat(a, a[n]+(a[n]+1)*prod(k=1, n-1, a[k]))); for(p=1, N\3, forstep(m=N, p+1, -1, a[m]==a[m-p]&next; 3*m>N&next(2); return(p)); return(p))} /* the 2nd optional parameter must be taken large enough, at least 3 times the period length and starting position. The script returns zero if the period is not found (probably due to these constraints). */

CROSSREFS

Cf. A213437, A214636.

Sequence in context: A295920 A176187 A180683 * A166030 A191523 A132890

Adjacent sequences:  A214632 A214633 A214634 * A214636 A214637 A214638

KEYWORD

nonn

AUTHOR

David Applegate, M. F. Hasler and N. J. A. Sloane, Jul 23 2012

STATUS

approved

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Last modified January 26 08:35 EST 2022. Contains 350578 sequences. (Running on oeis4.)