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A342042
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When a digit d is even, the next digit is > d.
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8
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0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 30, 13, 14, 50, 15, 16, 70, 17, 18, 90, 19, 23, 24, 51, 25, 26, 71, 27, 28, 91, 29, 31, 32, 33, 34, 52, 35, 36, 72, 37, 38, 92, 39, 45, 46, 73, 47, 48, 93, 49, 53, 54, 55, 56, 74, 57, 58, 94, 59, 67, 68, 95, 69, 75, 76, 77, 78, 96, 79, 89, 97, 98, 99, 101
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graph;
refs;
listen;
history;
text;
internal format)
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OFFSET
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1,3
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COMMENTS
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The sequence is always extended with the smallest positive integer not yet present that doesn't lead to a contradiction.
Proof of the conjecture: By the definition of the sequence, it cannot contain a number in A347298. Thus, one must show that the converse is also true. That is, if a number is not in A347298, then it is in this sequence (1). Let P be the property of not being in A347298, so that a number n has the property P if every even digit of n (except possibly the last) is followed by a larger digit.
Let m be the smallest number not in B := {a(1), a(2), ..., a(n)} with the property P. If one can show that for every n, there is some n' > n such that a(n') = m, then (1) follows. If a(n') ends with an odd number for any n' > n, then m must appear in the sequence. This is because if a(n') ends with an odd digit, then a(n'+1) is the smallest number not yet appearing in the sequence with the property P.
Suppose that a(n') ends with an even number for every n' > n (2). Let d be an even digit such that S := {k > n | a(k) ends with d} is infinite (such a d must exist if (2) holds). Let k be in S. If a(k+1) ends with an even digit (as we assume) and a(k+1)+1 does not yet appear in the sequence, then a(k'+1) = a(k+1)+1 for k' being the smallest number larger than k in S. As B is finite but S is infinite, B cannot contain {a(k+1)+1 | k in S}. Thus, for some k in S, a(k+1)+1 is a term appearing after a(n). a(k+1)+1 ends with an odd digit, contradicting (2). Q.E.D. (End)
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LINKS
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PROG
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(PARI) See Links section.
(Python)
def cond(s, minfirst):
return all(s[i+1] > s[i] for i in range(len(s)-1) if s[i] in "02468")
def aupton(terms):
alst, seen = [0], {0}
while len(alst) < terms:
d = alst[-1]%10
an = minfirst = (1 - d%2)*(d+1)
stran = str(an)
while an in seen or not cond(stran, minfirst):
an += 1
stran = str(an)
if int(stran[0]) < minfirst:
an = minfirst*10**(len(stran)-1)
alst.append(an); seen.add(an)
return alst
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CROSSREFS
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KEYWORD
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base,nonn,nice
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AUTHOR
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STATUS
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approved
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