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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. 4
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 (list; graph; refs; listen; history; text; internal format)
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
FORMULA
It is tempting (in view of A054995) to conjecture that a(1)=1 and, for n>1, a(n) = (a(n-1)+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(n-1)+3) mod n +1;
The conjecture is true. After removing the 4th number, we are reduced to the n-1 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
Sequence in context: A188945 A281261 A102849 * A016538 A134226 A184050
KEYWORD
nonn,easy
AUTHOR
N. J. A. Sloane, Nov 13 2003
EXTENSIONS
More terms from David Wasserman, Aug 08 2005
STATUS
approved

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Last modified March 29 11:45 EDT 2024. Contains 371278 sequences. (Running on oeis4.)