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The offset is 2 for compatibility with the definition of A324156.

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.

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(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.

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.

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.

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.

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.

Further terms:

14670238462896430 < a(6) < 14670469667698570;

1.88655928870547380*10^22 < a(7) < 1.8865698003644555*10^22;

1.5845871199286*10^29 < a(8) < 1.5845909238805*10^29;

3.7023896360635*10^36 < a(9) < 3.7023941021398*10^36;

1.0075615552422*10^45 < a(10) < 1.0075622026833*10^45;

1.3480809599483*10^53 < a(11) < 1.3480814844466*10^53;

3.9618565460983*10^62 < a(12) < 3.9618574860993*10^62;

7.8648507615953*10^71 < a(13) < 7.864851991241*10^71;

4.7945106758325*10^81 < a(14) < 4.7945111864185*10^81;

1.0953005932169*10^92 < a(15) < 1.0953006746693*10^92;

8.3149001821943*10^148 < a(20) < 8.3149003278317*10^148.

All terms are even, since a(n) + 1 is always an odd prime number.

The numbers a(n) + 1 and a(n) + 2 are zero containing numbers (in base n).

The numbers between a(n) and p := min( k > a(n) + 1 | k is prime) are zero containing numbers, i.e., a(n) + j is a zero containing number for 0 < j < p - a(n).

For numbers m > a(10) = 1.00756...*10^45, we have pi(m) > A324161(m) [= number of zerofree numbers <= m]; in general, the ratio A324161(m) to pi(m) is O(log(n)*n^d), where d := 1 - 1/(1 - log_10(9)) = -0.0457..., and thus tends to 0 for m --> infinity. Consequently, the share of primes <= m which have no '0'-digit become significantly smaller as m rises beyond that bound a(10). For m = 10^100, the share is not greater than 0.000688, for m = 10^1000, the share cannot exceed 4.52757*10^(-43). Conversely, the share of primes which contain a '0'-digit tends to 1 as m grows to infinity (cf. A011540).

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.

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.