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A061438
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Number of steps for trajectory of n to reach 1 under the map that sends x -> x/13 if x mod 13 = 0, x -> 14x+13-(x mod 13) if x is not 0 mod 13 (for a 2nd time when n starts at 1).
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0
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25, 23, 21, 19, 17, 15, 13, 11, 9, 7, 5, 3, 1, 36, 42, 34, 40, 32, 38, 30, 36, 28, 34, 26, 32, 24, 30, 34, 113, 28, 32, 111, 26, 30, 109, 24, 28, 107, 22, 26, 105, 36, 28, 24, 103, 34, 26, 22, 101, 32, 24, 20, 99, 30, 22, 24, 166, 97, 28, 20, 22, 164, 95, 26, 18, 20, 162, 93
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,1
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COMMENTS
| This sequence is generated by the program below for m=13,p=14. Other values of m and p also converge but not necessarily to 1. For m =2 and p=1 we have the count of steps for the x+1 problem. m=prime and p=m+1 usually converge to 1 but break down for certain values of n. E.g. 17 locks at n=34, 23 at n=49, 29 at n=91. I verified m=13 for n up to 100000. 100000 requires 100 steps to reach 1.
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LINKS
| Cino Hilliard, The x+1 conjecture
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EXAMPLE
| x = 12: step 1: x = 12*14+13-12 = 169, step 2: x = 169/13 = 13, step 3: x = 13/13 = 1. Count = 3.
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PROG
| (PARI) countxp2(n, m, p) = { c=1; x=1; x=x*p+m-1; while(x>1, r = x%m; if(r==0, x=x/m, x=x*p+m-r); c++; ); print1(c" "); for(j=2, n, x=j; c=0; while(x>1, r = x%m; if(r==0, x=x/m, x=x*p+m-r); c++; \ print1(x" "); ); print1(c" ") ) }
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CROSSREFS
| Sequence in context: A104790 A171806 A038822 * A022981 A023467 A158501
Adjacent sequences: A061435 A061436 A061437 * A061439 A061440 A061441
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KEYWORD
| easy,nonn
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AUTHOR
| Cino Hilliard (hillcino368(AT)gmail.com), Mar 29 2003
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