login
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
OFFSET
1,5
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
Sequence in context: A180683 A375360 A365331 * A166030 A351149 A191523
KEYWORD
nonn
AUTHOR
STATUS
approved