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

1,2

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)

Links

Table of n, a(n) for n=1..14.

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

Example

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

Crossrefs

Subsequence of A179239.

Cf. A010888, A031347, A031286, A031346.

Cf. A064702, A239427.

Sequence in context: A138348 A278175 A261973 * A103878 A068154 A134874

Adjacent sequences: A357128 A357129 A357130 * A357132 A357133 A357134

Keyword

nonn,base

Author

Mohammed Yaseen, Sep 14 2022

Extensions

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

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

Status

approved