A059921 -id:A059921 - cp's OEIS frontend

A249669 a(n) = floor(prime(n)^(1+1/n)).

4, 5, 8, 11, 17, 19, 25, 27, 32, 40, 42, 49, 54, 56, 60, 67, 74, 76, 83, 87, 89, 96, 100, 107, 116, 120, 122, 126, 128, 132, 148, 152, 159, 160, 171, 173, 179, 186, 190, 196, 203, 204, 215, 217, 221, 223, 236, 249, 253, 255, 259, 265, 267, 278, 284, 290, 296, 298, 304, 308, 310, 321

Offset: 1

M. F. Hasler, Nov 03 2014

nonn

Firoozbakht's conjecture (prime(n)^(1/n) is a decreasing function), is equivalent to say that prime(n+1) <= a(n). (One has equality for n=2 and n=4.) See also A182134 and A245396.
This is not A059921 o A000040, i.e., a(n) != A059921(prime(n)), since the base is prime(n) but the exponent is n.
A245396(n) = A007917(a(n)). - Reinhard Zumkeller, Nov 16 2014

Robert Israel, Table of n, a(n) for n = 1..10000
A. Kourbatov, Verification of the Firoozbakht conjecture for primes up to four quintillion, arXiv:1503.01744 [math.NT], 2015
A. Kourbatov, Upper bounds for prime gaps related to Firoozbakht's conjecture, arXiv:1506.03042 [math.NT], 2015 and J. Int. Seq. 18 (2015) 15.11.2
Wikipedia, Firoozbakht's conjecture

Cf. A245396, A007917, A244365, A246778, A263023.

a249669 n = floor $ fromIntegral (a000040 n) ** (1 + recip (fromIntegral n))
-- Reinhard Zumkeller, Nov 16 2014

[Floor(NthPrime(n)^(1+1/n)): n in [1..70]]; // Vincenzo Librandi, Nov 04 2014

seq(floor(ithprime(n)^(1+1/n)), n=1..100); # Robert Israel, Nov 26 2015

PARI
```
a(n)=prime(n)^(1+1/n)\1
```

a(n) = prime(n) + (log(prime(n)))^2 - log(prime(n)) + O(1), see arXiv:1506.03042, Theorem 5. - Alexei Kourbatov, Nov 26 2015

A269020 a(n) = ceiling(n^(1+1/n)).

Original entry on oeis.org

1, 3, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70

Offset: 1

Bob Selcoe, Feb 17 2016

			a(5)=7 because 5^(6/5) ~ 6.8986; a(6)=9 because 6^(7/6) ~ 8.088.

Cf. A059921 (floor(n^(1+1/n))), A269023 (complementary sequence).

[Ceiling(n^(1+1/n)): n in [1..70]]; // Vincenzo Librandi, Feb 18 2016

Table[Ceiling[n^(1 + 1 / n)], {n, 100}] (* Vincenzo Librandi, Feb 18 2016 *)

from sympy import integer_nthroot
def A269020(n):
    a, b = integer_nthroot(n**(n+1),n)
    return a+(b^1) # Chai Wah Wu, Sep 04 2024

a(n) = A059921(n) + 1, n>=2.

A269023 Complement of A269020: numbers not of the form ceiling(n^(1+1/n)).

Original entry on oeis.org

2, 4, 8, 19, 51, 141, 392, 1079, 2957, 8072, 21987, 59825, 162695, 442342, 1202521, 3268920, 8885999, 24154826, 65659826, 178482140

Offset: 1

Bob Selcoe, Feb 18 2016

The limiting ratio is e (see comment in A059921).

			The term 8 appears because A269020(5)=7 and A269020(6)=9.

Cf. A059921, A269020.

Complement[Range[1, 100000], Table[Ceiling[n^(1 + 1/n)], {n, 100000}]] (* Vaclav Kotesovec, Mar 12 2016 *)

a269020(n) = ceil(n^(1+1/n))
for(n=1, 1e20, if(a269020(n+1)-a269020(n) > 1, print1(a269020(n)+1, ", "))) \\ Felix Fröhlich, Mar 12 2016

from itertools import count
def A269023(n):
    def bisection(f,kmin=0,kmax=1):
        while f(kmax) > kmax: kmax <<= 1
        while kmax-kmin > 1:
            kmid = kmax+kmin>>1
            if f(kmid) <= kmid:
                kmax = kmid
            else:
                kmin = kmid
        return kmax
    def f(x):
        if x==1: return n+1
        z = x**x
        for y in count(x,-1):
            if y**(y+1) <= z:
                return n+y
            z //= x
    return bisection(f,n,n) # Chai Wah Wu, Sep 10 2024

a(18)-a(20) from Felix Fröhlich, Mar 12 2016

A095394 a(n) = Floor[n^((n)/(n+1))], integer part of n^x where x = n/(n+1) < 1.

Original entry on oeis.org

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

Offset: 1

Labos Elemer, Jun 17 2004

nonn

1, 3, 9, 24, 61, 160, 420, 1120, ... appear in the sequence twice. The ratio of such numbers converges to e. [Charles R Greathouse IV, Oct 26 2011]

Cf. A059921.

Table[Floor[N[x^((x+1)/(x)), 50]], {x, 1, 15}]

a(n)=floor(n^(n/(n+1.))) \\ Charles R Greathouse IV, Oct 26 2011

A095395 a(n) = floor(n^((n+1)/(n))) - floor(n^((n)/(n+1))).

Original entry on oeis.org

0, 1, 2, 2, 3, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9

Offset: 1

Labos Elemer, Jun 17 2004

nonn

Table[Floor[N[x^((x+1)/(x)), 50]]-Floor[N[x^((x)/(x+1)), 50]], {x, 1, 15}]

a(n) = A059921(n) - A095394(n).

cp's OEIS Frontend

A249669 a(n) = floor(prime(n)^(1+1/n)).

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A269020 a(n) = ceiling(n^(1+1/n)).

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A269023 Complement of A269020: numbers not of the form ceiling(n^(1+1/n)).

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A095394 a(n) = Floor[n^((n)/(n+1))], integer part of n^x where x = n/(n+1) < 1.

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A095395 a(n) = floor(n^((n+1)/(n))) - floor(n^((n)/(n+1))).

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