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A305324 Number of one-sided 'divisible' polyominoes of size 2^(n-1), where a 'divisible' polyomino is either a monomino (square) or a polyomino which can be separated into two identical 'divisible' polyominoes.

Original entry on oeis.org

1, 1, 6, 90, 3356, 232283, 27964488

Offset: 1

Josh Marza, May 30 2018

a(n) is nonzero for any n >= 1. Proof is trivial by induction.

a(n) <= A000988(2^(n-1)) as any polyomino counted here is also counted in A000988.

			For n = 3 (polyominoes of size 4), the 'divisible' polyominoes are the I, O, J, L, S and Z tetrominoes. The T tetromino is not 'divisible'.

Cf. A000988.

Definition changed and more terms added by John Mason, Sep 20 2022

A293010 a(n) is the smallest x > 2 to satisfy pi(x-1)/(x-1)^n < pi(x)/x^n, where pi(x) is the prime counting function (A000720).

Original entry on oeis.org

3, 11, 29, 127, 347, 1087, 3079, 8419, 23531, 64553, 175211, 480881, 1304519, 3523901, 9558533, 25874767, 70115473, 189961193, 514272463, 1394193607, 3779851091, 10246935679, 27788566133, 75370121191, 204475052401, 554805820477, 1505578023841, 4086199302077

Offset: 1

Josh Marza, Sep 27 2017

nonn

The integer 2 satisfies the inequality for all values of n (as pi(1) = 0), so it is omitted. With n=0 the sequence is clearly satisfied by all primes.
Conjecture: a(n) exists for all n, that is, for all n, there exists at least one integer which satisfies the inequality.
Occurs when examining convergence of alternating sum to infinity of (-1^x)* pi(x)/(x^n).
If a(n) exists it is prime. Proof: If a(n) is composite then pi(x - 1) = pi(x), so pi(x-1)/((x-1)^n) > pi(x)/(x^n), a contradiction. - David A. Corneth, Oct 02 2017
From Chai Wah Wu, Apr 24 2018: (Start)
Conjecture above is true.
Theorem: a(n) exists for all n and satisfies prime(floor(e^W(e^n))) < a(n) < prime(ceiling(e^W(e^(n+1)))), where W is Lambert W function.
Proof: for a fixed n, let x = a(n) if it exists. Since x is prime, pi(x-1) = pi(x)-1 and thus the condition is m-1/(x-1)^n < m/x^n, where m = pi(x). This simplifies to 1-1/m < (1-1/x)^n. A result of Dusart in 1999 shows that x > m(log(m log(m))-1) when m > 1. This implies that (1-1/x)^n > (1-1/(m(log(m log(m))-1)))^n >= 1-n/(m(log(m log(m))-1)) where the last inequality is due to Bernoulli's inequality.
Thus (1-1/x)^n > 1-1/m if log(m log(m))-1 >= n which is satisfied if m >= e^W(e^(n+1)).
The lower bound on a(n) follows analogously from the 1941 upper bound on x due to Rosser: x < m log(m log(m)) when m > 5.
(End)

			For n=3, the first integer which satisfies pi(x-1)/((x-1)^3) < pi(x)/(x^3) is 29 = a(3).

Cf. A000040, A000720.

For[j = 1, j < 11, j++, For[i = 2, i < 1000000  i++, If[(PrimePi[i]/(i^j)) - (PrimePi[i-1]/((i-1)^j)) > 0, Print[i] Break[]]]]

a(n) = my(x=3); while(primepi(x-1)/(x-1)^n >= primepi(x)/x^n, x++); x; \\ Michel Marcus, Oct 02 2017

upto(u)=my(t = 1, n = 1, logt = 0, logtm1, logp, logpm1, res = List()); forprime(p = 3, u, t++; logtm1 = logt; logt = log(t); logp = log(p); logpm1 = log(p - 1); if(logtm1 + n * logp < logt + n*logpm1, listput(res, p); n++)); res \\ David A. Corneth, Oct 02 2017

a(20)-a(28) from Chai Wah Wu, Apr 24 2018

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User: Josh Marza

A305324 Number of one-sided 'divisible' polyominoes of size 2^(n-1), where a 'divisible' polyomino is either a monomino (square) or a polyomino which can be separated into two identical 'divisible' polyominoes.

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