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A306526
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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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4
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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
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OFFSET
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2,1
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COMMENTS
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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.
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LINKS
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FORMULA
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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)).
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.
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EXAMPLE
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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, ...
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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