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

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

Offset: 2

Author

Hieronymus Fischer, Mar 05 2019

Keywords

• nonn

Comments

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.

Examples

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

Crossrefs

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

Formula

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.