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For each n = 2, 3, 4, ..., a(n) is the most significant (nonzero) digit of the decimal expansion of P(n) := Sum_{p prime} 1/p^n, the prime zeta function at argument n.

The present sequence starts at n = 2, since the underlying series diverges for any integer less than 2.

It is conjectured that a(n) = A111395(n) for all n >= 10 (see "Is the leading digit of the decimal expansion of the prime zeta function at n equal to the first digit of 5^n, for all integers n >= 10?" in Links).

Write lg = log_10, let {x} denote the fractional part of x. Note that {k*lg(5)} = 1 - {k*lg(2)} and that {k*lg(8)} = {k*lg(2)} + 0, 1, or 2, so we have {k > 0 : 5^k and 8^k both start with a} = {k: {k*lg(2)} is in I_a}, where I_a = (1-lg(a+1), 1-lg(a)) intersect (((lg(a))/3, (lg(a+1))/3) U ((lg(a)+1)/3, (lg(a+1)+1)/3) U ((lg(a)+2)/3, (lg(a+1)+2)/3)). Note that I_1 = (1-lg(2), (lg(2)+2)/3), I_3 = ((lg(3)+1)/3, 1-lg(3)), I_5 = (lg(5)/3, lg(6)/3) and that I_a is empty otherwise. As a result, k > 0 is a term if and only if {k*lg(2)} is in (lg(5)/3, lg(6)/3) U ((lg(3)+1)/3, 1-lg(3)) U (1-lg(2), (lg(2)+2)/3). We see that when 5^k and 8^k both start with 1, 3, or 5, 2^k starts with 5, 3, or 1 respectively. - Jianing Song, Dec 26 2022

The first digit of 2^k is A008952(k) = floor(10^{k*log_10(2)}), the first digit of 5^k is A111395(k) = floor(10^{k*log_10(5)}), and the first digit of 8^k is A008952(3*k) = floor(10^{k*log_10(8)}), where "{x}" denotes the fractional part of x.

All numbers 2^k, 5^k, 8^k for k = a(n) > 0 start only with 3.

From Jianing Song, Dec 26 2022: (Start)

Write lg = log_10. Note that {k*lg(5)} = 1 - {k*lg(2)} and that {k*lg(8)} = {k*lg(2)} + 0, 1, or 2, so we have {k > 0 : 2^k, 5^k, 8^k all start with a} = {k: {k*lg(2)} is in I_a}, where I_a = (lg(a), lg(a+1)) intersect (1-lg(a+1), 1-lg(a)) intersect (((lg(a))/3, (lg(a+1))/3) U ((lg(a)+1)/3, (lg(a+1)+1)/3) U ((lg(a)+2)/3, (lg(a+1)+2)/3)). Note that I_3 = ((lg(3)+1)/3, 1-lg(3)) and I_a is empty otherwise (since (lg(a), lg(a+1)) intersect (1-lg(a+1), 1-lg(a)) is empty). As a result, k > 0 is a term if and only if (lg(3)+1)/3 < {k*lg(2)} < 1-lg(3).

Except for 0, also numbers k in A358196 such that 2^k starts with 3 (see the comment in A358196).

Claim: for n > 1, a(n+1) - a(n) = 10, 73, or 83. Proof: write a = (lg(3)+1)/3, b = 1-lg(3).

Step 1. If k > 0 is a term, then:

(a) {k*lg(2)} is in (a, b-(10*lg(2)-3)) => {(k+10)*lg(2)} is in (a+(10*lg(2)-3), b) => k+10 is a term;

(b) {k*lg(2)} is in (a+(25-83*lg(2)), b) => {(k+83)*lg(2)} is in (a, b-(25-83*lg(2))) => k+83 is a term.

Note that (a, b-(10*lg(2)-3)) U (a+(25-83*lg(2)), b) = (a, b), so at least one of k+10 and k+83 is a term.

Step 2. If k > 0 and k+m are both terms, 0 < m <= 83, then {k*lg(2)} and {(k+m)*lg(2)} are both in (a, b), so m*lg(2) is the range (N-(b-a), N+(b-a)) for some integer N, which implies that m = 10, 20, 73, or 83.

Note that if {k*lg(2)} < 1 - 2*(10*lg(2)-3), then {(k+10)*lg(2)} = {k*lg(2)} + (10*lg(2)-3), {(k+20)*lg(2)} = {k*lg(2)} + 2*(10*lg(2)-3), so {k*lg(2)} < {(k+10)*lg(2)} < {(k+20)*lg(2)}, which means that if k and k+20 are both terms, so is k+10. This shows that the next term after k is either 10, 73, or 83 larger.

We can show similarly that, for k > 0 being a term of this sequence:

(a) if k+73 is a term, then k+83 is a term;

(b) if k+83 is a term, then k+93 is a term;

(c) if k+166 is a term, then k+73 is a term;

(d) if k+10, k+73 are not terms (i.e., the next term is k+83), then k+176, k+196 are terms.

As a result, if we write out the sequence of the first differences, 73 is always followed by two 10's, and 83 is followed by one or two 10's; 83, 10, 10 is always followed by 73, 10, 10, and 83, 10 is always followed by 83, 10, 10. (End)

a(n) is also the first digit of n*5^n = A036291(n).

The sequence starts at n = 2 since the zeta function sum diverges for any integer n < 2.

Conjecture: a(n) = A111395(n) for all n >= 10 (for a weaker conjecture see the comments in the related sequence A385431 and the Mathematics Stack Exchange discussion).