A324154 Least number N such that the number of primes (<= N) >= the number of the base-n-zerofree numbers (<= N).
2, 3, 344251, 33182655683
Offset: 2
Examples
a(2) = 2, since pi(1) = 0 < 1 = numOfZerofreeNum_2(1), pi(2) = 1 >= 1 = numOfZerofreeNum_2(2), where numOfZerofreeNum_2(m) is the number of base-2-zerofree numbers <= m and pi(m) = number of primes <= m. The first base-2-zerofree numbers are 1 = 1_2, 3 = 11_2, 7 = 111_2, ... a(3) = 3, since pi(1) = 0 < 1 = numOfZerofreeNum_3(1), pi(2) = 1 < 2 = numOfZerofreeNum_3(2), pi(3) = 2 >= 2 = numOfZerofreeNum_3(3), where numOfZerofreeNum_3(m) is the number of base-3-zerofree numbers <= m and pi(m) = number of primes <= m. The first base-3-zerofree numbers are 1 = 1_3, 2 = 2_3, 4 = 11_3, 5 = 12_3, 7 = 21_3, ...
Programs
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PARI
a(n) = {my(k = 1, nbp = 0, nzf = 1); while(nbp < nzf, k++; if (isprime(k), nbp++); if (vecmin(digits(k, n)), nzf++);); k;} \\ Michel Marcus, Mar 20 2019
Formula
a(n) = min(k | pi(k) >= numOfZerofreeNum_n(k)), where numOfZerofreeNum_n(k) is the number of base-n-zerofree numbers <= k ((see A324161 for general formulas regarding numOfZerofreeNum_n(k))).
a(n) <= A324155(n) - 1.
Estimation for the n-th term (n > 2):
a(n) < e*(p*log(p*log((e/(e-1))*p*log(p))))^(1/(1-d)),
a(n) > e^1.1*(q*log(q*log(q*log(q))))^(1/(1-d)),
where p := (n-1)/((n-2)*(1-d))*e^(-(1-d)), q := (n-1)^d/((n-2)*(1-d))*e^(-1.1*(1-d)), d := d(n) := log(n-1)/log(n).
Also, but more imprecise:
a(n) > e^1.1*(q*log(q))^(1/(1-d)),
a(n) > (n/(n-1))*((n-1)*log(n)*log(n*log(n)))^((n-1/2))*log(n)).
Asymptotic behavior:
a(n) = O(n*((e/(e-1))*n*log(n)*log(n*log(n)))^(n*log(n))).
a(n) = O(n*(((e+1)/(e-1))*n*log(n)^2)^(n*log(n))).
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