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a(n) = greatest integer N such that (number of primes <= N) >= (number of numbers <= N that contain a zero in base n).
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%I #31 Jun 14 2019 14:24:27

%S 3,9,31,50,107,147,257,406,701,1091,1731,2213,2782,3434,4188,5042,

%T 6001,7082,8276,18543,21383,24521,27932,46917,52924,59437,88034,

%U 122055,162060,208619,262334,359458,471733,600588,839889,1114547,1481920,2076185

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

%C 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.

%H Hieronymus Fischer, <a href="/A306526/b306526.txt">Table of n, a(n) for n = 2..100</a>

%F 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):

%F 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).

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

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

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

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

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

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

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

%F Also, but less accurate, n > 2,

%F 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)).

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

%F Asymptotic behavior:

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

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

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

%e 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, ...

%e 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, ...

%o (PARI) lbz(n, b) = my(d = log(b - 1)/log(b)); n + 2 - ((b-1)*(n+1)^d - 1)/(b-2);

%o ubp(n) = n/(log(n) - 4);

%o f(b) = if (b==2, 10, ceil(solve(x=100, 10^100, lbz(x, b) - ubp(x))));

%o cz(m, n) = vecmin(digits(m, n))==0;

%o getpos(vdiff) = {forstep (k=#vdiff, 1, -1, if (vdiff[k] == 0, return (k)););}

%o 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

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

%K nonn,base

%O 2,1

%A _Hieronymus Fischer_, Mar 29 2019