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A277858
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The absolute difference between a(n) and a(n+1) is visible in a(n).
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4
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1, 2, 4, 8, 16, 10, 9, 18, 17, 24, 20, 22, 44, 40, 36, 30, 27, 25, 23, 21, 19, 28, 26, 32, 29, 31, 34, 37, 74, 67, 60, 54, 49, 45, 41, 42, 38, 35, 70, 63, 57, 50, 55, 110, 99, 90, 81, 73, 66, 72, 65, 59, 64, 58, 53, 48, 52, 47, 43, 39, 78, 71, 142, 100, 101, 91, 82, 80, 88, 96, 87, 79, 86, 92, 83, 75, 68, 62, 56, 51, 102
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OFFSET
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1,2
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COMMENTS
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"Is visible in" means "is a substring of". (For example, if a(n) = 123, the absolute difference may be any of { 1, 2, 3, 12, 23, 123 } but not 13.)
The sequence was not extended with the smallest unused integer so far that does not lead to a contradiction, as this method would produce the finite sequence: 1, 2, 4, 8, 16, 10, 9, 18, 17, 24, 20, 22, 44, 40, 36, 30, 27, 25, 23, 21, 19, 28, 26, 32, 29, 31, 34, 37, 74, 67, 60, 54, 49, 45, 41, 42, 38, 35, 70, 63, 57, 50, 55, 110, 99, 90, 81, 73, 66, 72, 65, 59, 64, 58, 53, 48, 52, 47, 43, 39, 78, 71, 142, 100, 101, 91, 82, 80, 88, 96, 87, 79, 86, 92, 83, 75, 68, 62, 56, 51 and 46. This 46 has no successor as 40, 42 and 92 are already in the sequence.
Instead, a "pseudo-backtracking" method was applied: extend the sequence with the smallest unused integer so far that does not lead to a contradiction iff the said integer has itself a successor not yet in the sequence. With this method the sequence becomes infinite and doesn't stop anymore with 46 [46 was replaced by 102 (which is 51, the previous term, times 2)].
It looks like the sequence is not a permutation of the integers > 0. Indeed, after computing 40000 terms, the set of the unused integers so far hasn't evolved since the first 20000 terms of the sequence were computed: 3, 5, 6, 7, 11, 12, 13, 14, 15, 33, 46, 61. Question: could this 12-integer set be reduced (to 11, 10 or fewer elements) starting the sequence with another a(1) than 1?
For the starting value a(1) = 3 the set of unused numbers becomes { 1, 2, 4, 5, 7, 8, 11, 71 } (8 elements), for a(1) = 5 it becomes { 1, 2, 3, 4, 6, 7, 8, 33, 55 } (9 elements). - M. F. Hasler, Nov 04 2016
It is easy to prove that this is not a permutation of the positive integers: indeed, for example the number 5 can never occur, since it has no possible successor (10 being already used). There are several variants of this sequence, some of which are permutations: The condition "is a substring of" could be relaxed to (strongest to weakest) "digits of |a(n+1)-a(n)| are a subsequence of the digits of a(n)" (e.g., a step of 13 would be allowed after 123), "... are a sub-multiset of ..." (a step of 21 would be allowed after 123), or "...are a subset of ..." (a step of 11 would be allowed after 123). The last variant (which is a permutation) is A276070. - M. F. Hasler, Nov 05 2016
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LINKS
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EXAMPLE
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The absolute difference between 1 and 2 is 1, which is visible in 1; the abs diff between 2 and 4 is 2 (visible in "2"); the abs diff between 4 and 8 is 4 (visible in "4"); the abs diff between 8 and 16 is 8 (visible in "8"); the abs diff between 16 and 10 is 6 (visible in the "6" of "16"); the abs diff between 10 and 9 is 1 (visible in the "1" of "10"), etc.
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PROG
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(PARI) u=[a=1]; (S(a, d=[a])=while(d[1]>9, for(k=10, d[1], d=concat(d, d[1]%k); k=k*10-1); d=concat(d[1..1]\10, d)); Set(select(t->t>0&&!setsearch(u, t), concat(apply(x->[a-x, a+x], Set(d)))))); for(n=1, 99, print1(a", "); u=setunion(u, [a]); for(k=1, #a=S(a), S(a[k])&&(a=a[k])&&break)) \\ M. F. Hasler, Nov 04 2016
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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