A306521 -id:A306521 - cp's OEIS frontend

A306195 Least integer N > 1 such that the number of base-n-zero containing numbers [<= N] >= the number of base-n-zerofree numbers [<= N].

2, 3, 77, 679, 2809, 18659, 274511, 1123471, 10761677, 222222219, 1329025059, 11257702583, 298693399003, 8722140365427, 18535191127229, 600479950316063, 21047228319925113, 44095690303774235, 1686791892208310919

Offset: 2

Hieronymus Fischer, Mar 26 2019

For numbers 1 < k < a(n) the number of base-n-zero containing numbers <= k is always smaller than the number of base-n-zerofree numbers <= k. The boundary a(n) is rapidly growing as the base n rises (see formula section).
For numbers k >= a(n) the number of base-n-zero containing numbers <= k may be greater or smaller than the number of base-n-zerofree numbers <= k, also both numbers may be equal. Example 1: for base n = 2 we have numOfZeroNum_2(2) > numOfZerofreeNum_2(2), numOfZeroNum_2(3) = numOfZerofreeNum_2(3), but numOfZeroNum_2(k) > numOfZerofreeNum_2(k) for k > 3. Example 2: for base n = 3 we have numOfZeroNum_3(k) = numOfZerofreeNum_3(k), for k = 1, 3, 11, 13, 15, 19, 23, 25, 27, but numOfZeroNum_2(k) < numOfZerofreeNum_2(k) for k = 2, 4..10, 14, 16, 17, 18, 26, and numOfZeroNum_2(k) > numOfZerofreeNum_2(k) for k = 12, 20, 21, 22, 24 and for k > 27.
The number of indices k = k(n) for which numOfZeroNum_n(k) = numOfZerofreeNum_n(k) forms the sequence 2, 9, 9, 1, 27, 20, 1, 68, 20, 1, 103, 40, ... (starting with n = 2).
All terms a(n) are zero containing numbers (in base n).
All terms are odd for n > 2. Proof: The definition implies numOfZeroNum_n(a(n)) = numOfZerofreeNum_n(a(n)), for n > 2. In general, we have numOfZeroNum_n(k) + numOfZerofreeNum_n(k) = k + 1. It follows that a(n) = 2*numOfZeroNum_n(a(n)) - 1.
a(n) <= A306442(n), equality holds for n = 5, 8, 11, 14, 15, 17, 18, 21, 24, 27, 28, 30, 31, 34, 37, 40, 41, 43, 44, 47, 50, 51, 53, 54, 56, 57, 60, 63, 64, 66, 67, 69, 70, 73, 76, 77, 79, 80, 82, 83, 86, 89, 90, 92, 93, 96, 99, ...  For significantly large n, equality holds true for those bases which satisfy fract((n-1/2)*log(2) + O(1/n)) < 1/2  + O(1/n). This is true for infinitely many indices n. Let e(n) be the number of bases m <= n for which a(m) = A306442(m), then lim_{n->infinity} e(n)/n = 1/2, i.e., for large n, on average, every second term of this sequence is also a term of A306442.

			a(2) = 2, since numOfZeroNum_2(2) [= the number of base-2-zero containing numbers <= 2] is greater than or equal to numOfZerofreeNum_2(2) [the number of base-2-zerofree numbers <= 2], i.e., numOfZeroNum_2(2) = 2 >= 1 = numOfZerofreeNum_2(2), and indices < 2 are out of focus by definition. Hint: the zero numbers <= 2 in base 2 are 0 = 0_2 and 2 = 10_2, the only zerofree numbers <= 2 in base 2 is 1 = 1_2.
a(3) = 3, since numOfZeroNum_3(3) = 2 <= 2 = numOfZerofreeNum_3(3) but numOfZeroNum_3(k) > numOfZerofreeNum_3(k) for k > 3. Hint: the zero numbers <= 3 in base 3 are 0_3 = 0, and 10_3 = 3, the zerofree numbers <= 3 in base 3 are 1_3 = 1 and 2_3 = 2.

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

Cf. A011540, A052382, A324160, A324161, A306442, A306521, A306526.

With numOfZeroNum_n(k) [= the number of base-n-zero containing numbers <= k] and numOfZerofreeNum_n(k) [the number of base-n-zerofree numbers <= k] and d := log(n-1)/log(n):
a(n) = min(k > 1 | numOfZeroNum_n(k) >= numOfZerofreeNum_n(k)).
Because of d = d(n) < 1, numOfZeroNum_n(k) =  k*(1 + O(k^(d-1)) and numOfZerofreeNum _n(k) =  O(k^d) this minimum always exists (for n > 2).
For all bases n >= 2: numOfZeroNum_n(1) = numOfZerofreeNum_n(1).
See A324160 and A324161 for general formulas regarding numOfZeroNum_n(k) and numOfZerofreeNum_n(k).
a(n) = min(k > 1 | numOfZeroNum_n(k) = (n + 1)/2).
a(n) = min(k > 1 | numOfZerofreeNum_n(k) = (n + 1)/2).
Estimate of the n-th term (n > 3): a(n) > (2*(n-1)^d/(n-2))^(1/(1-d)), where d := log(n-1)/log(n).
Also, but less accurate,
a(n) > 2^((n-1/2)*log(n).
a(n) > 2^((n-1/2)*log(n)*e^((11*log(n)+12)/(12*n).
a(n) <= A306442(n), for further upper bound estimations see A306442.
Asymptotic behavior:
a(n) = O(n*2^((n-1/2)*log(n))).
Lower and upper limits:
lim sup a(n)/(n*2^((n-1/2)*log(n))) = 1, for n --> infinity.
lim inf a(n)/(2^((n-1/2)*log(n)) = 1, for n --> infinity.

A306442 Greatest integer N such that the number of base-n-zero containing numbers [<= N] <= the number of base-n-zerofree numbers [<= N].

Original entry on oeis.org

3, 27, 131, 679, 7809, 34211, 274511, 4793487, 20327615, 222222219, 5187484917, 31896823991, 298693399003, 8722140365427, 70433726283479, 600479950316063, 21047228319925113, 252325338960485915, 3284805263774079161, 68985263157894736839

Offset: 2

Hieronymus Fischer, Mar 26 2019

For numbers k > a(n) the number of base-n-zero containing numbers <= k is always greater than the number of base-n-zerofree numbers <= k. This boundary is rapidly growing as the base n rises (see formula section).
The quotient numOfZerofreeNum_n(k)/numOfZeroNum_n(k) tends to 0 for k --> infinity and each fixed base n. Formally, numOfZerofreeNum_n(k)/numOfZeroNum_n(k) = O(k^c) with a constant c := c(n) = log(n-1)/(log(n) - 1 < 0, if n > 2. For n = 2 we have numOfZerofreeNum_2(k) = floor(log_2(k+1)), numOfZeroNum_2(k) = (k + 1 - floor(log_2(k+1)), thus numOfZerofreeNum_2(k)/ numOfZeroNum_2(k) = (k + 1)^(-1) * floor(log_2(k+1)) / (1 - floor(log_2(k+1))/(k+1)) = O(log(k)/k). Example: n = 3, numOfZerofreeNum_3(k)/numOfZeroNum_3(k) = O(k^(-0.369070...)); example: n = 10, numOfZerofreeNum _10(k)/numOfZeroNum_10(k) = O(k^(-0.045757490...)).
The first term a(2) = 3 = 11_2 is the only one which is a zerofree (i.e., a zeroless) number (in base 2), all the other terms a(n) are zero containing numbers (in base n). In any case, a(n) + 1 is always a zero containing number (in base n).
All terms are odd. Proof: The definition implies numOfZeroNum_n(a(n)) = numOfZerofreeNum_n(a(n)). In general, we have numOfZeroNum_n(k) + numOfZerofreeNum_n(k) = k + 1. It follows a(n) = 2*numOfZeroNum_n(a(n)) - 1.
a(n) >= A306195(n), equality holds for n = 5, 8, 11, 14, 15, 17, 18, 21, 24, 27, 28, 30, 31, 34, 37, 40, 41, 43, 44, 47, 50, 51, 53, 54, 56, 57, 60, 63, 64, 66, 67, 69, 70, 73, 76, 77, 79, 80, 82, 83, 86, 89, 90, 92, 93, 96, 99, .... For significantly large n, equality holds true for those bases which satisfy fract((n-1/2)*log(2) + O(1/n)) < 1/2 + O(1/n). This is true for infinitely many indices n. Let e(n) be the number of bases m <= n for which a(m) = A306195(m), then lim_{n->infinity} e(n)/n > 1/2, i.e., for large n, on average, at least every second term of this sequence is also a term of A306195.

			a(2) = 3, since numOfZeroNum_2(3) [= the number of zero numbers <= 3, in base 2] is less than or equal to numOfZerofreeNum_2(3) [the number of zerofree numbers <= 3, in base 2], i.e., numOfZeroNum_2(3) = 2 <= 2 = numOfZerofreeNum_2(3), but numOfZeroNum_2(k) > numOfZerofreeNum_2(k) for k > 3. Hint: the zero numbers <= 3 in base 2 are 0 = 0_2 and 2 = 10_2, the zerofree numbers <= 3 in base 2 are 1 = 1_2 and 3 = 11_2.
a(3) = 27, since numOfZeroNum_3(27) = 14 <= 14 = numOfZerofreeNum_3(27) but numOfZeroNum_3(k) > numOfZerofreeNum_3(k) for k > 27. Hint: the zero numbers <= 27 in base 3 are 0_3, 10_3, 20_3, 100_3, 101_3, 102_3, 110_3 120_3, 200_3, 201_3, 202_3, 210_3, 220_3 and 1000_3 = 27, the zerofree numbers <= 27 in base 3 are 1_3, 2_3, 11_3, 12_3, 21_3, 22_3, 111_3, 112_3, 121_3, 122_3, 211_3, 212_3, 221_3 and 222_3 = 26.

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

Cf. A011540, A052382, A306195, A306521, A306526, A324154, A324155, A324160, A324161.

With numOfZeroNum_n(k) [= the number of base-n-zero containing numbers <= k] and numOfZerofreeNum_n(k) [the number of base-n-zerofree numbers <= k] and d := log(n-1)/log(n):
a(n) = max(k | numOfZeroNum_n(k) <= numOfZerofreeNum_n(k)).
Because of d = d(n) < 1, numOfZeroNum_n(k) = k*(1 + O(k^(d-1)), numOfZerofreeNum _n(k) = O(k^d) and numOfZeroNum_n(1) = 1 = numOfZerofreeNum_n(1) this maximum always exists (for n > 2). This is also true for the case n = 2, since numOfZeroNum_2(k) = k*(1 + O(log(k)/k)) and numOfZerofreeNum_2(k) = O(log(k)).
a(n) = max(k > 1 | numOfZeroNum_n(k) = (n + 1)/2).
a(n) = max(k > 1 | numOfZerofreeNum _n(k) = (n + 1)/2).
See A324160 and A324161 for general formulas regarding numOfZeroNum_n(k) and numOfZerofreeNum_n(k).
Estimate of the n-th term (n > 2):
a(n) < (2*(n-1)/(n-2))^(1/(1-d)) - 1,
where d := log(n-1)/log(n).
Also, but less accurate,
a(n) < (2*(n-1)/(n-2))^((n-1/2)*log(n)), n > 2,
a(n) < n*2^(n*log(n)), n > 1.
a(n) >= A306195(n), for further lower bound estimations see A306195.
Asymptotic behavior:
a(n) = O(n*2^((n-1/2)*log(n))).
Lower and upper limits:
lim sup a(n)/(n*2^((n-1/2)*log(n))) = 1, for n --> infinity.
lim inf a(n)/(log(n)*2^((n-1/2)*log(n)) = e, for n --> infinity.

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

Original entry on oeis.org

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

Hieronymus Fischer, Mar 29 2019

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.

			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, ...

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

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

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

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.

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A306195 Least integer N > 1 such that the number of base-n-zero containing numbers [<= N] >= the number of base-n-zerofree numbers [<= N].

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A306442 Greatest integer N such that the number of base-n-zero containing numbers [<= N] <= the number of base-n-zerofree numbers [<= N].

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A306526 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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