A357131 Numbers m such that A010888(m) = A031347(m) = A031286(m) = A031346(m); only the least of the anagrams are considered.

0, 137, 11126, 111134, 111278, 1111223, 11111447, 111112247, 1111122227, 111111111137, 11111111111126, 111111111111134, 1111111111111223, 111111111111111111111111111111111111111111111111111111111111111111111111111111111111278

Offset: 1

Author

Mohammed Yaseen, Sep 14 2022

Keywords

• nonn
• base

Comments

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

a(14) > 10^50. - Michael S. Branicky, Sep 16 2022

From Michael S. Branicky, Sep 17 2022: (Start)

(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.

a(2)-a(14) and all terms of the forms above have 2 = A010888(m) = A031347(m) = A031286(m) = A031346(m).

(R_{t})277 where t+2+7+7 = 4 followed by 55555 9's is a term with 4 = A010888(m) = A031347(m) = A031286(m) = A031346(m).

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.

However, there are no terms with A010888(m) = A031347(m) = A031286(m) = A031346(m) = 1 or 3. (End)

Examples

			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
		

Links

• Eric Weisstein's World of Mathematics, Additive Persistence
• Eric Weisstein's World of Mathematics, Digital Root
• Eric Weisstein's World of Mathematics, Multiplicative Digital Root
• Eric Weisstein's World of Mathematics, Multiplicative Persistence

Crossrefs

Subsequence of A179239.

Cf. A010888, A031347, A031286, A031346.

Cf. A064702, A239427.

Extensions

a(8)-a(13) from Pontus von Brömssen, Sep 14 2022

a(14) confirmed by Michael S. Branicky, Sep 17 2022