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A225538
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Let r(n) denote the reverse of n. For every n, consider the sequence n_1 = n + 1 + r(n+1), and for m >= 2, n_m = n_(m-1) + 1 + r(n_(m-1) + 1). a(n) is the least m for which n_m is a palindrome, or 0 if there is no such m.
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1
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1, 1, 1, 1, 2, 2, 2, 2, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 2, 2, 1, 1, 1, 1, 1, 2, 1, 2, 2, 4, 1, 1, 1, 1, 2, 1, 2, 2, 4, 7, 1, 1, 1, 2, 1, 2, 2, 4, 7, 10, 1, 1, 2, 1, 2, 2, 4, 7
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OFFSET
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0,5
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COMMENTS
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Conjecture: the least n's for which a(n) = 0 are 1895, 1985, 2894, 2984, 3893, and 3983. - Peter J. C. Moses, May 10 2013
See analogous numbers in A023108 for which the so-called Lychrel process "Reverse and Add!", apparently, never leads to a palindrome.
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LINKS
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EXAMPLE
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For n=8, 9 + 9 = 18, 19 + 91 = 110, 111 + 111 = 222 is a palindrome. Thus a(8)=3.
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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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