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A338807 Numbers k such that the process starting at (k, 0) mapping (k, t) to (k+1, t+k) if t = 0 (mod k), and (k-1, t+k) otherwise, eventually reaches (1, T) for some T. 0
1, 2, 8, 9, 11, 14, 18, 20, 23, 32, 35, 38, 40, 47, 49, 50, 53, 56, 57, 58, 59, 62, 67, 71, 73, 74, 77, 89, 91, 92, 95, 98, 101, 104, 106, 114, 116, 128, 134, 135, 137, 140, 148, 149, 152, 155, 156, 158, 159, 162, 164, 169, 172, 173, 185, 188, 191, 194, 197 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

At each step, let the state of the process be (i,t), i_max the greatest i seen so far and i_min > 1 the least i seen so far. Consider the triples (i,j,t%j) for 2 <= j <= i_max, where i_max is the largest i seen so far. If all of those triples have been seen in previous steps, then the next step will not produce a new triple either, so the process will never reach i=1.

LINKS

Table of n, a(n) for n=1..59.

EXAMPLE

For k = 8, the process stops at T = 57: (8,0), (9,8), (8,17), (7,25), (6,32), (5,38), (4,43), (3,47), (2,50), (3,52), (2,55), (1,57).

For k = 4, the process never stops: (4,0), (5,4), (4,9), (3,13), (2,16), (3,18), (4,21), ...

PROG

(Python)

def isok(n):

    t = 0

    seen = set()

    maxn = n

    steps = 0

    while n>1:

        maxn = max(maxn, n)

        tuples = set((n, m, t%m) for m in range(2, maxn+1))

        if tuples <= seen:

            break

        seen = seen.union(tuples)

        t += n

        if t%n==0:

            n += 1

        else:

            n -= 1

    return n==1

CROSSREFS

Sequence in context: A345019 A116039 A030334 * A020676 A086678 A289146

Adjacent sequences:  A338804 A338805 A338806 * A338808 A338809 A338810

KEYWORD

easy,nonn

AUTHOR

Christian Perfect, Nov 10 2020

STATUS

approved

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Last modified October 4 21:22 EDT 2022. Contains 357240 sequences. (Running on oeis4.)