A324155 -id:A324155 - cp's OEIS frontend

A324165 The number of primes <= A324155(n).

2, 94, 88572, 1431655764, 405311584472655, 375279801995072058162, 2392926627528494733661481601, 44505401644584236815975682821886536, 9818959014098676479127822164411318257546629, 1111111111111111111111111111111111111111111111111110

Offset: 2

Hieronymus Fischer, Mar 05 2019

nonn

Also the number of zerofree numbers <= A324155(n).
Expressed in base n - 1 and starting with n = 3, the sequence is 1011110, 11111111110, 1111111111111110, 411111111111111111110, 211111111111111111111111110, 211111111111111111111111111111110, 211111111111111111111111111111111111110, 1111111111111111111111111111111111111111111110, 1111111111111111111111111111111111111111111111111110, ....
Ostensibly, the reason for that is the calculation formula (see Formula section) for the number of zerofree numbers <= x^m + y, with y < (x^(m+1)-1)/(x-1) - x^m. But the deeper reason is the definition of sequence A324155. Each term A324155(n) marks a point of intersection between the curve numOfZerofreeNum_n(x) [the number of base-n zerofree numbers <= x] and the curve pi(x) [the number of prime numbers <= x]. Since numOfZerofreeNum_n(x) doesn't change for relatively large intervals at x = k*n^m (approx. a portion of > 1/(k*n)), but grows similar to pi(x) for regions outside, it is likely, that the point of intersection lies between x = k*n^m and x = n^m*(k + 1/n + 1/n^2 + 1/n^3 + ... + 1/n^m). The chance is maximal for k = 1, since the density of primes becomes smaller for greater x. Nevertheless, k > could also happen as we can see for n = 6, 7, 8 and 9.

			a(2) = 2, since there are 2 primes <= A324155(2) = 4.
a(3) = 94, since there are 94 primes <= A324155(3) = 498.

Cf. A011540, A052382, A324154, A324155, A324160, A324161, A324164.

a(n) = pi(A324155(n)).
a(n) = numOfZerofreeNum_n(A324155(n)), where numOfZerofreeNum_n(x) is the number of base-n zerofree numbers <= x (cf. A324161).
a(n) = k*(n-1)^m + ((n-1)^m - 1)/(n-2) - 1, where m = floor(log_n(A324155(n))), k = floor(A324155(n)/n^m), and provided A324155(n) - k*n^m < (n^(m+1)-1)/(n-1) - n^m.
With d := log(n-1)/log(n):
a(n) <= ((n - 1)*(A324155(n) + 1)^d - 1)/(n - 2) - 1.
a(n) >= (((n - 1)*A324155(n) + n)^d - 1)/(n - 2) - 1.
a(n) < A324155(n) / (log(A324155(n)) - 1.1), for n > 3.
a(n) > A324155(n) / (log(A324155(n)) - 1), for n > 3.

A324161 Number of zerofree nonnegative integers <= n.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 81, 82, 83

Offset: 0

Hieronymus Fischer, Feb 15 2019

This sequence represents the counting function of A052382 (for indices > 0). The offset is set to zero for compatibility with A324160.
For indices 1 < n < 10761677 we have a(n) > A324160(n), for all indices n > 20327615 we have a(n) < A324160(n), i.e., the number of zerofree numbers <= N is smaller than the number of zero containing numbers <= N for sufficiently large N.
There are exactly 20 indices for which a(n) = A324160(n).
The greatest number n = n_max such that a(n) >= pi(n) [the number of primes <= n] is in the range 1.0075615552421*10^45 < n_max < 1.0075622026833*10^45 (see A324155). Thus, for all indices n > n_max, we have a(n) < pi(n). For n = n_max the number of primes is pi(n) = 9818959014098676479127822164411318257546629.
The least number n = n_min such that a(n) <= pi(n) [the number of primes <= n] is in the range 1.0953002073198*10^44 < n_min < 1.0953009588121*10^44 (see A324154). Thus, for all indices n < n_min, we have a(n) > pi(n). For n = n_min the number of primes is pi(n) = 1090995446010964053236424684934590917505180.
Differs from A028904 first at a(100)=90 <> A028904(100)=81. - R. J. Mathar, Mar 03 2020
Differs from A081600 first at a(101)=90 <> A081600(101)=91. - R. J. Mathar, Mar 03 2020

			a(10) = 9, since there are 9 numbers <= 10 which contain no '0'-digit (1, 2, 3, 4, 5, 6, 7, 8 and 9).
a(100) = 90.
a(10^3) = 819.
a(10^4) = 7380.
a(10^5) = 66429.
a(10^6) = 597870.
a(10^7) = 5380839
a(10^8) = 48427560.
a(10^9) = 435848049.
a(10^10) = 3922632450.
a(10^20) = 13677373641439044900.
a(10^50) = 579799710823512747416018770986323931789870962250 = 5.79799...*10^47.
a(10^100) = 2.9881573748536...68682786424500*10^95.
a(10^1000) = 1.96635515818798306...435874245000*10^954.

Hieronymus Fischer, Table of n, a(n) for n = 0..10000

Cf. A011540, A052382, A324160, A324154, A324155.

A324161 := proc(n)
    option remember;
    if n = 0 then
        0;
    else
        convert(convert(n,base,10),set) ;
        if 0 in % then
            procname(n-1) ;
        else
            1+procname(n-1) ;
        end if;
    end if;
end proc: # R. J. Mathar, Mar 03 2020

a(n) = sum(k=1, n, vecmin(digits(k)) != 0); \\ Michel Marcus, Mar 20 2019

With m := floor(log_10(n)); k := Max_{j | j=1..m and (floor(n/10^j) mod 10)*j = 0} = digit position of the leftmost '0' in n counted from the right (starting with 0), k = 0 if there is no '0' digit; b(n,k) := floor(n/10^k)*10^k:
a(n) = n - 1 - Sum_{j=1..m} floor((b(n,k+1)-1)/10^j)*9^(j-1), if k = 0 (valid for n > 9),
a(n) = b(n,k) - 1 - Sum_{j=1..m} floor((b(n,k)-1)/10^j)*9^(j-1), if k > 0 (valid for n > 0),
a(n) = b(n,k) - 1 + ceiling(fract(n/10))*(1-ceiling(k/(m+1))) - Sum_{j=1..m} floor((b(n,k)-1)/10^j)*9^(j-1) (all k, valid for n > 0).
a(n) + A324160(n) = n + 1.
a(A052382(n)) = n.
A052382(a(n)) <= n, for n > 0.
A052382(a(n)) = n, iff n is a zerofree number.
a(10*n + k) >= 9*a(n) + k, k=0..9, equality holds, if n is a zerofree number (i.e., contains no '0'-digit).
a(10*A052382(n) + k) = 9*n + k, k=0..9, n > 0.
Values for special indices:
a(k*(10^n - 1)/9 - j) = k*(9^n - 1)/8 - j, n > 0, k = 1, 2, ... 9, j = 0, 1, 2, ... k.
a(k*10^n - j) = k*9^n + (9^n - 1)/8 - j, n >= 0, k = 1, 2, ... 10, j = 1, 2, ... 10.
a(k*10^n + j) = k*9^n + (9^n - 1)/8 - 1, n > 0, k = 1, 2, ... 10, j = 0, 1, 2, ... (10^(n+1)-1)/9 - 10^n - 1.
With: d := log_10(9) = 0.95424250943932...
Upper bound:
a(n) <= (9*(n+1)^d - 1)/8 - 1,
equality holds for n = 10^k - 1, k >= 0.
Lower bound:
a(n) >= ((9*n + 10)^d - 1)/8 - 1,
equality holds for n = (10^k - 1)/9 - 1, k > 0.
Asymptotic behavior:
a(n) <= (9/8)*n^d*(1 + O(1/n)) - 9/8.
a(n) >= (9^d/8)*n^d*(1 + O(1/n)) - 9/8.
a(n) = O(n^d) = O(n^0.954242509...).
Lower and upper limits:
lim inf a(n)/n^d = 9^d/8 = 1.0173931195971..., for n -> infinity.
lim sup a(n)/n^d = 9/8, for n -> infinity.
From Hieronymus Fischer, Apr 04 2019: (Start)
Formulas for general bases b > 2:
With m := floor(log_b(n)); k := Max_{j | j=1..m and (floor(n/b^j) mod b)*j = 0} = digit position of the leftmost '0' in n counted from the right (starting with 0), k = 0 if there is no '0' digit; b(n,k):= floor(n/b^k)*b^k:
a(n) = n - 1 - Sum_{j=1..m} floor((b(n,k+1)-1)/b^j)*(b-1)^(j-1), if k = 0, valid for  n > b-1;
a(n) = b(n,k) - 1 - Sum_{j=1..m} floor((b(n,k)-1)/b^j)*(b-1)^(j-1), if k > 0, valid for n > 0;
a(n) = b(n,k) - 1 + ceiling(fract(n/b))*(1-ceiling(k/(m+1))) - Sum_{j=1..m} floor((b(n,k)-1)/b^j)*(b-1)^(j-1), (all k, valid for n > 0).
Formula for base b = 2: a(n) = floor(log_2(n + 1)).
With d := d(b) := log(b - 1)/log(b).
Upper bound (b = 10 for this sequence):
a(n) <= ((b - 1)*(n + 1)^d - 1)/(b - 2) - 1,
equality holds for n = b^k - 1, k >= 0.
Lower bound (b = 10 for this sequence):
a(n) >= (((b - 1)*n + b)^d - 1)/(b - 2) - 1,
equality holds for n = (b^k - 1)/(b - 1) - 1, k > 0.
Asymptotic behavior (b = 10 for this sequence):
a(n) = O(n^d(b)), for b > 2,
a(n) = O(log(n)), for b = 2.
Lower and upper limits:
lim inf a(n)/n^d = (b - 1)^d/(b - 2), for n -> infinity, for b > 2.
lim sup a(n)/n^d = (b - 1)/(b - 2), for n -> infinity, for b > 2.
In case of b = 2:
lim a(n)/log(n) = 1/log(2), for n -> infinity.
(End)

A324154 Least number N such that the number of primes (<= N) >= the number of the base-n-zerofree numbers (<= N).

Original entry on oeis.org

2, 3, 344251, 33182655683

Offset: 2

Hieronymus Fischer, Feb 22 2019

Further terms:
4.1645275173242*10^15  < a(6)  < 4.164601237609*10^15,
1.0163657136*10^22     < a(7)  < 1.0163715977928*10^22,
8.4513797224747*10^28  < a(8)  < 8.4514006058085*10^28,
1.959502408617*10^36   < a(9)  < 1.9595048275153*10^36,
1.0953002073198*10^44  < a(10) < 1.0953009588121*10^44,
1.3480809599483*10^53  < a(11) < 1.3480814844466*10^53,
3.540916347013*10^61   < a(12) < 3.5409172310273*10^61,
2.080341784427*10^71   < a(13) < 2.0803421176765*10^71,
2.4843833393543*10^81  < a(14) < 2.4843836067277*10^81,
5.6615671922884*10^91  < a(15) < 5.6615676172791*10^91,
2.1556069128839*10^148 < a(20) < 2.1556069510899*10^148.
a(n) is always a prime number.
For n > 2, all terms are odd.
All terms a(n) are zero-containing numbers (in base n), a(n) - 1 is also a zero-containing number (in base n).
The numbers between p := max( k < a(n) | k is prime) and a(n) + 1 are zero-containing numbers, i.e., a(n) + 1 - j is a zero-containing number for 0 < j < a(n) + 1 - p.
From the equality A324164(5) = A324165(5) we can conclude that a(5) and A324155(5) + 1 are proximate primes. Same is true for a(11): a(11) and A324155(11) + 1 are proximate primes.
Conjecture: a(n) can be represented in the form a(n) = k*n^m + j, where j < (n^(m+1)-1)/(n-1) - n^m, and m > 1, 0 < k < n.

			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, ...

Cf. A011540, A052382, A324155, A324160, A324161, A324164, A324165.

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

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))).

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.

A306521 Least integer N > 2 such that the number of primes (<=N) <= the number of base-n-zero containing numbers (<=N).

Original entry on oeis.org

3, 3, 4, 28, 42, 104, 136, 329, 510, 856, 1449, 2212, 2782, 3434, 4188, 5042, 6001, 7082, 8276, 9604, 11062, 12666, 14405, 31651, 35694, 40061, 66427, 73966, 108764, 149756, 197516, 288280, 354924, 515538, 701002, 963687, 1318399, 1840377

Offset: 2

Hieronymus Fischer, Mar 29 2019

a(n) <= A306526(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) < A306526(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 further 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) = A306526(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), 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, ... (Hint: numbers <= 2 are out of scope for self-evident reasons).
a(3) = 3, since pi(3) = 2 <= 2 = numOfZeroNum_3(3), 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, ...
a(4) = 4, since pi(3) = 2 > 1 = numOfZeroNum_4(3), pi(4) = 2 <= 2 = numOfZeroNum_4(4), where numOfZeroNum_4(m) is the number of base-4-zero containing numbers <= m and pi(m) = number of primes <= m. The first base-4-zero containing numbers are 0 = 0_2, 4 = 10_4, 8 = 20_4, ...

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

Cf. A011540, A052382, A306195, A306442, A306526, A324154, A324155, A324160, A324161.

cz(m,n) = vecmin(digits(m, n))==0;
a(n) = {my(m=2, nbz=1+sum(k=1, 2, cz(k,n)), pmp=primepi(2)); for (m=3, oo, if (isprime(m), pmp++); if (cz(m,n), nbz++); if (pmp <= nbz, return (m)););} \\ Michel Marcus, Jun 10 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) = min(k > 2 | 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 minimum always exists (for n > 2). The case n = 2 is obvious. See A324160 regarding general formulas for numOfZeroNum_n(k).
Estimate of the n-th term:
a(n) > e*(1 + c1/c2*(1 + sqrt(1 + c2*c3/c1^2)))^(1/(1-d)), for n > 6,
where d := log(n-1)/log(n),
c0 := e^(1-d),
c1 := (n-1)^d/(n-2) - 1/e^(sqrt(n*log(n))) - d*c0,
c2 := (n-1)^d/(n-2) - 1/e^(sqrt(n*log(n))) + (1 - 1/sqrt(n*log(n)))*c0,
c3 := 2*(1-d)*c0.
Also, but less accurate, for n > 6,
a(n) > e*(1 + 1/(sqrt(n*log(n)) - 2))^(1/(1-d)).
a(n) > e*(1 + 1/(sqrt(n*log(n)) - 2))^((n-1/2)*log(n)).
a(n) <= A306526(n), see A306526 for further upper 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))+1/2) = 1, 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.

A324164 Number of primes <= A324154(n).

Original entry on oeis.org

1, 2, 29523, 1431655764, 119209289550780, 204698073815493849906, 1288498953284574087356182400, 23736214210444926301853697505006152, 1090995446010964053236424684934590917505180, 1111111111111111111111111111111111111111111111111110

Offset: 2

Hieronymus Fischer, Feb 22 2019

nonn

Also the number of zerofree numbers <= A324154(n).
Expressed in base n - 1 and starting with n = 3, the sequence is 10, 1111111110, 1111111111111110, 111111111111111111110, 111111111111111111111111110, 111111111111111111111111111111110, 111111111111111111111111111111111111110, 111111111111111111111111111111111111111111110, 1111111111111111111111111111111111111111111111111110, ....
Ostensibly, the reason for that is the calculation formula (see Formula section) for the number of zerofree numbers <= x^m + y, with y < (x^(m+1)-1)/(x-1) - x^m. But the deeper reason is the definition of sequence A324154. Each term A324154(n) marks a point of intersection between the curve numOfZerofreeNum_n(x) [the number of base-n zerofree numbers <= x] and the curve pi(x) [the number of prime numbers <= x]. Since numOfZerofreeNum_n(x) doesn't change for relatively large intervals at x = k*n^m (approx. a portion of > 1/(k*n)), but grows similar to pi(x) for regions outside, it is likely, that the point of intersection lies between x = k*n^m and x = n^m*(k + 1/n + 1/n^2 + 1/n^3 + ... + 1/n^m). The chance is maximal for k = 1, since the density of primes becomes smaller for greater x.

			a(2) = 1, since there is only one prime <= A324154(2) = 2.
a(3) = 2, since there are 2 primes <= A324154(3) = 3.

Cf. A011540, A052382, A324154, A324155, A324160, A324161, A324165.

a(n) = pi(A324154(n)).
a(n) = numOfZerofreeNum_n(A324154(n)), where numOfZerofreeNum_n(x) is the number of base-n zerofree numbers <= x (cf. A324161).
a(n) = k*(n-1)^m + ((n-1)^m - 1)/(n-2) - 1,
where m = floor(log_n(A324154(n))), k = floor(A324154(n)/n^m), and provided A324154(n) - k*n^m < (n^(m+1)-1)/(n-1) - n^m.
With d := log(n-1)/log(n):
a(n) <= ((n - 1)*(A324154(n) + 1)^d - 1)/(n - 2) - 1,
a(n) >= (((n - 1)*A324154(n) + n)^d - 1)/(n - 2) - 1.
a(n) < A324154(n) / (log(A324154(n)) - 1.1), for n > 3.
a(n) > A324154(n) / (log(A324154(n)) - 1), for n > 3.

A324156 Irregular triangle T(n,k) read by rows in which row n lists the numbers m such that the number of prime numbers <= m is equal to the number of base-n zerofree numbers <= m.

Original entry on oeis.org

2, 3, 4, 3, 113, 114, 115, 116, 117, 118, 119, 120, 199, 200, 201, 293, 294, 295, 296, 297, 298, 299, 300, 301, 302, 303, 304, 305, 306, 482, 483, 491, 492, 493, 494, 495, 496, 497, 498, 344251, 344252, 344253, 344254, 344255, 344256, 351902, 353501, 353502, 353503

Offset: 2

Hieronymus Fischer, Jul 16 2019

The offset is 2 since the least base (= row) for which the definition makes sense is n = 2.
The least term of row n is T(n,1) = A324154(n). The last term of row n is T(n,j) = A324155(n), where j = A324157(n) is the number of terms of the n-th row.
Terms of rows higher than 5 are unknown, but they are bounded by the above rule. For example, the first term of the 6th row is T(6,1) = A324154(n) = 4.1645*10^15, approximately. The last term of the 6th row is A324155(n) = 1.46705*10^16, approximately.

			T(2,1) = 2, since pi(2) = 1 = numOfZerofreeNum_2(2) where numOfZerofreeNum_n(k) = number of base-n zerofree numbers <= k.
T(2,2) = 3, since pi(3) = 2 = numOfZerofreeNum_2(3).
T(3,2) = 4, since pi(4) = 2 = numOfZerofreeNum_2(3).
T(3,1) = 3, since pi(3) = 2 = numOfZerofreeNum_3(3).
T(3,2) = 113, since pi(113) = 30 = numOfZerofreeNum_3(113).
Triangle T(n,k) begins:
2, 3, 4;
3, 113, 114, 115, 116, 117, 118, 119, 120, 199, 200, 201, 293, 294, 295, 296, 297, 298, 299, 300, 301, 302, 303, 304, 305, 306, 482, 483, 491, 492, 493, 494, 495, 496, 497, 498;
344251, 344252, 344253, 344254, 344255, 344256, 351902, 353501, 353502, 353503, 353504, 353505, 353506, 353507, 353508, 353509, 353510, 353511, 353512, 353513, 353514, 353515, 353516, 353517, 353518, 353519, 353520, 353521, 353522, 353523, 353524, 353525, 353526, 353631, 353632, 353633, 353634, 353635, 353636, 601379, 601380, 601381, 601382, 601383, 601384, 601385, 601386, 601387, 601388, 601389, 601390, 601391, 601392, 601393, 601394, 601395, 601396, 617903, 617904, 617905, 617906, 617907, 617908, 867281, 867282, 867283, 867284, 867285, 867286, 867287, 867288, 867289, 867290, 867291, 867292, 867293, 867294, 867295, 867296, 867297, 867298, 867299, 867300, 876414, 876431, 876432, 876437, 877213, 877214, 877215, 877216, 877217, 877218, 877219, 877220, 877221, 877222, 878014, 878021, 878022, 878037, 1139549, 1139550, 1139551, 1139552, 1139553, 1139554, 1139555, 1139556;
33182655683, 33182655684, 33182655685, 33182655686, 33182655687, 33182655688;

Cf. A324160, A324161, A324154, A324155, A324157.

cp's OEIS Frontend

A324165 The number of primes <= A324155(n).

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A324161 Number of zerofree nonnegative integers <= n.

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A324154 Least number N such that the number of primes (<= N) >= the number of the 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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A306521 Least integer N > 2 such that the number of primes (<=N) <= the number of base-n-zero containing 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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A324164 Number of primes <= A324154(n).

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A324156 Irregular triangle T(n,k) read by rows in which row n lists the numbers m such that the number of prime numbers <= m is equal to the number of base-n zerofree numbers <= m.

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A324157 Number of terms in the n-th row of the triangle A324156.

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