

A088333


A version of Josephus problem: a(n) is the surviving integer under the following elimination process. Arrange 1,2,3,...,n in a circle, increasing clockwise. Starting with i=1, delete the integer 3 places clockwise from i. Repeat, counting 3 places from the next undeleted integer, until only one integer remains.


3



1, 1, 2, 2, 1, 5, 2, 6, 1, 5, 9, 1, 5, 9, 13, 1, 5, 9, 13, 17, 21, 3, 7, 11, 15, 19, 23, 27, 2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, 3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, 55, 59, 63, 67, 2, 6, 10, 14, 18, 22, 26, 30, 34, 38, 42
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OFFSET

1,3


COMMENTS

If one counts only one place (resp. two places) at each stage to determine the element to be deleted, we get A006257 (resp. A054995).


REFERENCES

See A054995 for references and links.


LINKS

Table of n, a(n) for n=1..79.


FORMULA

It is tempting (in view of A054995) to conjecture that a(1)=1 and, for n>1, a(n) = (a(n1)+4) mod n. The conjecture is false; counterexample: a(21)=21; a(20)=17; (a(20)+4)mod 21=0; corrected formula: a(n)=(a(n1)+3) mod n +1;
The conjecture is true. After removing the 4th number, we are reduced to the n1 case, but starting with 5 instead of 1.  David Wasserman, Aug 08 2005
a(n) = A032434(n,4) if n>=4.  R. J. Mathar, May 04 2007


CROSSREFS

Cf. A006257, A054995, A032434, A005427, A005428, A006257, A007495, A000960, A056530.
Sequence in context: A188945 A281261 A102849 * A016538 A134226 A184050
Adjacent sequences: A088330 A088331 A088332 * A088334 A088335 A088336


KEYWORD

nonn,easy


AUTHOR

N. J. A. Sloane, Nov 13 2003


EXTENSIONS

More terms from David Wasserman, Aug 08 2005


STATUS

approved



