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0, 137, 11126, 111134, 111278, 1111223, 11111447, 111112247, 1111122227, 111111111137, 11111111111126, 111111111111134, 1111111111111223, 111111111111111111111111111111111111111111111111111111111111111111111111111111111111278
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OFFSET
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1,2
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COMMENTS
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The infinite sequence {(R_1)37, (R_91)37, (R_991)37, (R_9991)37, (R_99991)37, ...} is a subsequence, where R_k is the repunit of length k. Hence this sequence is infinite.
a(14) <= (R_84)278.
Some additional terms < 10^100: (R_84)278, (R_86)447, (R_86)2247, (R_86)22227, (R_91)37, (R_93)26, (R_94)34, (R_93)278, (R_94)223, (R_95)447, (R_95)2247, (R_95)22227.
If a term k > 0 then k cannot contain a digit 0 as if it does A031347(k) = 0 while A031346(k) = 1, contradicting equality. - David A. Corneth, Sep 15 2022
(R_{10^k})37 and (R_{2*10^k - 10})37 also form infinite subsequences for k >= 0.
Indeed, terms of the form (R_k)e form infinite subsequences for each e in 26, 34, 37, 223, 278, 447, 2247, 22227 for k such that A007953(k + A007953(e)) = 2.
Likewise, there exists a term of the form (R_{t})5579 with 5 = A010888(m) = A031347(m) = A031286(m) = A031346(m), where t+26 is part of the additive persistence chain ending ..., 5999999, 59, 14, 5. Likewise for 888899 and 6, and so on.
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LINKS
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EXAMPLE
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137 is in the sequence as A010888(137) = 137 mod 9 = 2, A031347(137) = 2 via 1*3*7 = 21 -> 2*1 = 2 < 10, A031286(137) = 2 via 1+3+7 = 11 -> 1+1 = 2 so two steps, A031346(137) = 2 via 1*3*7 = 21 -> 2*1 = 2 so two steps. As all these outcomes are 2, 137 is a term. - David A. Corneth, Sep 15 2022
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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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EXTENSIONS
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STATUS
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approved
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