A306526 a(n) = greatest integer N such that (number of primes <= N) >= (number of numbers <= N that contain a zero in base n).

3, 9, 31, 50, 107, 147, 257, 406, 701, 1091, 1731, 2213, 2782, 3434, 4188, 5042, 6001, 7082, 8276, 18543, 21383, 24521, 27932, 46917, 52924, 59437, 88034, 122055, 162060, 208619, 262334, 359458, 471733, 600588, 839889, 1114547, 1481920, 2076185

Offset: 2

Author

Hieronymus Fischer, Mar 29 2019

Keywords

• nonn
• base

Comments

a(n) >= A306521(n), equality holds for n = 2, 14, 15, 16, 17, 18, 19, 20, 40, 41, 42, 44, 45, 46, 47, 48, 49, 50, 51, 52 (but a(n) > A306521(n) for all other indices n up to 82). For sufficiently large n, equality holds true for those bases n which satisfy 1/2 <= fract(sqrt(n/log(n)) + O(sqrt(log(n)/n))) < 3/4. This is true for infinitely many indices, at least for all bases n = ceiling(x), where x is a solution of x/log(x) = k-th triangular number + 1/4, k > 1. For k = 2..10 the corresponding bases are n = 19, 48, 92, 152, 230, 326, 440, 574, 727. Let e(n) be the number of bases m <= n for which a(m) = A306521(m), then lim_{n->infinity} e(n)/n >= 1/4. Conjecture: lim_{n->infinity} e(n)/n = 1/4.

Examples

			a(2) = 3, since pi(3) = 2 >= 2 = numOfZeroNum_2(3), and pi(k) < numOfZeroNum_2(k) for all k > 3, where numOfZeroNum_2(m) is the number of base-2-zero-containing-numbers <= m and pi(m) = number of primes <= m. The first base-2-zero-containing-numbers are 0 = 0_2, 2 = 10_2, 4 = 100_2, ...
a(3) = 9, since pi(9) = 4 >= 4 = numOfZeroNum_3(9), and pi(k) < numOfZeroNum_3(k) for all k > 9, where numOfZeroNum_3(m) is the number of base-3-zero-containing-numbers <= m and pi(m) = number of primes <= m. The first base-3-zero-containing-numbers are 0 = 0_2, 3 = 10_3, 6 = 20_3, 9 = 100_3, 10 = 101_3, 11 = 102_3, 12 = 120_3, ...
		

Links

• Hieronymus Fischer, Table of n, a(n) for n = 2..100

Crossrefs

Cf. A011540, A052382, A306195, A306442, A306521, A324154, A324155, A324160, A324161.

Programs

PARI

lbz(n, b) = my(d = log(b - 1)/log(b)); n + 2 - ((b-1)*(n+1)^d - 1)/(b-2);
ubp(n) = n/(log(n) - 4);
f(b) = if (b==2, 10, ceil(solve(x=100, 10^100, lbz(x, b) - ubp(x))));
cz(m, n) = vecmin(digits(m, n))==0;
getpos(vdiff) = {forstep (k=#vdiff, 1, -1, if (vdiff[k]  == 0, return (k)););}
a(n) = {my(ub = f(n), vdiff = vector(ub), nbz = 1, pmp = 0); for (m=1, ub, if (cz(m, n), nbz++); if (isprime(m), pmp++); vdiff[m] = nbz - pmp;); getpos(vdiff);} \\ Michel Marcus, Jun 14 2019

Formula

With numOfZeroNum_n(k) [= the number of base-n-zero containing numbers <= k] and pi(k) [= the number of primes <= k] and d := log(n-1)/log(n):

a(n) = max(k | pi(k) >= numOfZeroNum_n(k)). Because of d = d(n) < 1, numOfZeroNum_n(k) = k*(1 + O(k^(d-1)), pi(k) = k/log(k)*(1+o(1)), and pi(3) = 2 >= 2 = numOfZeroNum_n(3) this maximum always exists (for n > 2). The case n = 2 is obvious. See A324160 regarding general formulas for numOfZeroNum_n(k).

Estimation for the n-th term (n > 2):

a(n) < e^alpha*(1 + c1/c2*(1 + sqrt(1 + c2*c3/c1^2)))^(1/(1-d)),

where d := log(n-1)/log(n), alpha := 1.1,

c0 := e^(alpha*(1-d)),

c1 := (n-1)/(n-2) - d*c0,

c2 := (n-1)/(n-2) + (1 - 1/sqrt(n*log(n)))*c0,

c3 := 2*(1-d)*c0.

Also, but less accurate, n > 2,

a(n) < e^alpha*(1 + (1 + sqrt(1 + 4*(n-2)^2/(n*log(n))))/(1 + (n-2)*(2-1/sqrt(n*log(n)))))^((n-1/2)*log(n)).

a(n) >= A306521(n), see A306521 for further lower bound estimations.

Asymptotic behavior:

a(n) = O(sqrt(n)*e^sqrt(n*log(n))).

lim sup a(n)/e^(sqrt(n*log(n))+(log(n)+1)/2) = 1, for n --> infinity.

lim inf a(n)/e^(sqrt(n*log(n))+log(log(n))/2+1) = 1, for n --> infinity.