

A306195


Least integer N > 1 such that the number of basenzero containing numbers [<= N] >= the number of basenzerofree numbers [<= N].


4



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

2,1


COMMENTS

For numbers 1 < k < a(n) the number of basenzero containing numbers <= k is always smaller than the number of basenzerofree 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 basenzero containing numbers <= k may be greater or smaller than the number of basenzerofree 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((n1/2)*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.


LINKS



FORMULA

With numOfZeroNum_n(k) [= the number of basenzero containing numbers <= k] and numOfZerofreeNum_n(k) [the number of basenzerofree numbers <= k] and d := log(n1)/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^(d1)) 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 nth term (n > 3): a(n) > (2*(n1)^d/(n2))^(1/(1d)), where d := log(n1)/log(n).
Also, but less accurate,
a(n) > 2^((n1/2)*log(n).
a(n) > 2^((n1/2)*log(n)*e^((11*log(n)+12)/(12*n).
Asymptotic behavior:
a(n) = O(n*2^((n1/2)*log(n))).
Lower and upper limits:
lim sup a(n)/(n*2^((n1/2)*log(n))) = 1, for n > infinity.
lim inf a(n)/(2^((n1/2)*log(n)) = 1, for n > infinity.


EXAMPLE

a(2) = 2, since numOfZeroNum_2(2) [= the number of base2zero containing numbers <= 2)] is greater than or equal to numOfZerofreeNum_2(2) [the number of base2zerofree 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.


CROSSREFS



KEYWORD

nonn,base


AUTHOR



STATUS

approved



