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 A291317 A variation of the 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, at k-th stage, move k places clockwise and delete the current number. 1
 1, 1, 1, 3, 4, 3, 7, 7, 6, 10, 7, 12, 3, 10, 11, 7, 11, 1, 12, 6, 21, 1, 7, 12, 25, 3, 25, 28, 16, 26, 25, 6, 32, 19, 15, 21, 28, 3, 12, 21, 24, 13, 21, 36, 17, 45, 41, 45, 8, 40, 11, 6, 25, 41, 23, 4, 43, 52, 51, 57, 28, 21, 11, 47, 26, 29, 57, 51, 48, 56, 12 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,4 COMMENTS In the classical Josephus problem (A006257), one moves one place clockwise at each stage, and in the A054995 version, one moves two places clockwise at each stage; here, on the other hand, the number of moves is progressive, and the resulting sequence seems random. No term belongs to A000096 (for the same reason that there are no even positive terms in A006257). See also A128982 for another variation of the Josephus problem. a(n) = 1 for n = 1, 2, 3, 18, 22, 171, 195, 234, 1262, 2136, ... a(n) = n for n = 1, 7, 10, 12, 21, 25, 28, 235, 822, ... More formally, for any function f over the natural numbers, let us define the function j_f with these rules: for any n > 0: - let L = (1, 2, ..., n) be the list of the first n natural numbers, - for k = 1 to n-1:    - for i = 1 to f(k): move the first element of L to the end,    - after these moves, discard the first element of L, - j_f(n) = the remaining element in L. In particular: - j_A000012 = A006257, - j_A007395 = A054995, - and j_A000027 = a (this sequence), - see also Links section for the scatterplots of j_f for certain classical or basic functions f. We have the following properties: - j_f(1) = 1, - if f(1) = 1 mod 2 then j_f(2) = 1 else j_f(2) = 2, - j_f(n) never equals k + Sum_{i=1..k} f(i), - iterating j_f(n), j_f(j_f(n)), ... eventually leads to a fixed point, - likely j_f = j_g iff f = g. LINKS Rémy Sigrist, Table of n, a(n) for n = 1..5000 EXAMPLE The different stages for n=6 are (where ^ indicates the counting reference position): - stage 1:  1^ 2  3  4  5  6 - stage 2:  1     3^ 4  5  6 - stage 3:  1     3  4     6^ - stage 4:  1     3        6^ - stage 5:        3^       6 - stage 6:        3^ Hence, a(6) = 3. PROG (PARI) a(n) = my (l = List(vector(n, i, i)), i = 0); for (k = 1, n-1, i += k; my (p = i \ #l); list pop(l, 1 + (i % #l)); i -= p); return (l[1]) CROSSREFS Cf. A000002, A000012, A000027, A000096, A000120, A006519, A007395, A006257, A020639, A054868, A054995, A128982. Sequence in context: A069915 A033634 A111970 * A141730 A127737 A299416 Adjacent sequences:  A291314 A291315 A291316 * A291318 A291319 A291320 KEYWORD nonn AUTHOR Rémy Sigrist, Aug 22 2017 STATUS approved

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Last modified May 25 01:17 EDT 2019. Contains 323534 sequences. (Running on oeis4.)