A245722 -id:A245722 - cp's OEIS frontend

A182134 Number of primes p such that prime(n) < p < prime(n)^(1 + 1/n).

1, 1, 1, 1, 2, 2, 2, 1, 2, 2, 2, 3, 3, 2, 2, 3, 4, 3, 4, 3, 3, 2, 2, 4, 5, 4, 3, 2, 2, 2, 3, 4, 4, 3, 4, 4, 4, 4, 3, 4, 5, 4, 4, 3, 2, 2, 4, 5, 5, 4, 4, 4, 3, 5, 6, 5, 5, 4, 3, 3, 2, 4, 4, 4, 3, 2, 5, 5, 5, 5, 5, 5, 5, 5, 5, 4, 4, 4, 4, 5, 6, 5, 7, 6, 6, 5, 5, 5, 5, 4, 4, 5, 4, 5, 4, 3, 3, 3, 3, 5, 5, 5, 5, 6, 5

Offset: 1

Thomas Ordowski, Apr 20 2012

nonn

Firoozbakht's conjecture: prime(n+1)^(1/(n+1)) < prime(n)^(1/n), for all n >= 1.
According to Firoozbakht's conjecture, all terms of this sequence are positive. - Jahangeer Kholdi, Jul 30 2014
Conjecture: a(n)=1 only for n = 1, 2, 3, 4, and 8. - Farideh Firoozbakht, Oct 18 2014
See A246782 and A246781 for indices such that a(n)=2 resp. a(n)=3. - M. F. Hasler, Oct 19 2014
Length of n-th row in A244365; a(n) = A001221(A245722(n)). - Reinhard Zumkeller, Nov 18 2014
a(n) = 2 for n = 5, 6, 7, 9, 10, 11, 14, 15, 22, 23, 28, 29, 30, 45, 46, 61, 66, 216, 217, 367, 3793, 1319945, ... = A246782. - Robert G. Wilson v, Feb 20 2015
a(n) = 3 for n = 12, 13, 16, 18, 20, 21, 27, 31, 34, 39, 44, 53, 59, 60, 65, 96, 97, 98, 99, 136, 154, 202, ... = A246781. - Robert G. Wilson v, Feb 20 2015
First occurrence of k: 1, 5, 12, 17, 25, 55, 83, 169, 207, 206, 384, 953, ... = A246810. - Robert G. Wilson v, Feb 20 2015
Conjecture: lim sup n->oo a(n) = oo. - John W. Nicholson, Feb 28 2015
a(n) is unbounded (that is, the above conjecture is true). In particular, there is a constant c > 1 such that a(n) > c log n infinitely often (by Maier's theorem). - Thomas Ordowski and Charles R Greathouse IV, Apr 09 2015

			a(25) = 5, because p(25) = 97 and there are 5 primes p such that 97 < p < 97^(1 + 1/25) = 121.9299290...: 101, 103, 107, 109, 113.

T. D. Noe, Table of n, a(n) for n = 1..10000
Alexei Kourbatov, Verification of the Firoozbakht conjecture for primes up to four quintillion, arXiv:1503.01744 [math.NT], 2015.
Alexei Kourbatov, Upper Bounds for Prime Gaps Related to Firoozbakht’s Conjecture, arXiv:1506.03042 [math.NT], 2015.
Alexei Kourbatov, Upper bounds for prime gaps related to Firoozbakht's conjecture, J. Int. Seq. 18 (2015) 15.11.2.
Wikipedia, Firoozbakht's conjecture

Cf. A010051, A000040, A249669, A244365, A246810.

a182134 = length . a244365_row  -- Reinhard Zumkeller, Nov 16 2014

a:= n-> numtheory[pi](ceil(ithprime(n)^(1+1/n))-1)-n:
seq(a(n), n=1..100);  # Alois P. Heinz, Apr 21 2012

Table[i = Prime[n] + 1; j = Floor[Prime[n]^(1 + 1/n)]; Length[Select[Range[i, j], PrimeQ]], {n, 100}] (* T. D. Noe, Apr 21 2012 *)
f[n_] := PrimePi[ Prime[n]^(1 + 1/n)] - n; Array[f, 105] (* Robert G. Wilson v, Feb 20 2015 *)

A182134(n)=primepi(prime(n)^(1+1/n))-n \\ M. F. Hasler, Nov 03 2014

from sympy import primepi, prime
def a(n): return primepi(prime(n)**(1 + 1/n)) - n # Indranil Ghosh, Apr 23 2017

a(n) = Sum_{m=A000040(n+1)..A249669(n)} A010051(m). - Reinhard Zumkeller, Nov 16 2014

a(n) = primepi(prime(n)^(1+1/n)) - n (see PARI program). - John W. Nicholson, Feb 11 2015

More terms from Alois P. Heinz, Apr 21 2012

A245396 Largest prime not exceeding prime(n)^(1 + 1/n).

Original entry on oeis.org

3, 5, 7, 11, 17, 19, 23, 23, 31, 37, 41, 47, 53, 53, 59, 67, 73, 73, 83, 83, 89, 89, 97, 107, 113, 113, 113, 113, 127, 131, 139, 151, 157, 157, 167, 173, 179, 181, 181, 193, 199, 199, 211, 211, 211, 223, 233, 241, 251, 251, 257, 263, 263, 277, 283, 283, 293, 293, 293, 307, 307, 317, 331, 337

Offset: 1

M. F. Hasler, Nov 03 2014

nonn

Firoozbakht's conjecture, prime(n+1) < prime(n)^(1 + 1/n), is equivalent to a(n) > prime(n). See also A182134.
Here prime(n) = A000040(n). The conjecture is also equivalent to a(n) - prime(n) >= A001223(n), the n-th gap between primes. See also A246778(n) = floor(prime(n)^(1 + 1/n)) - prime(n).
It is also conjectured that the equality a(n) - prime(n) = A001223(n) holds only for n in the set {1, 2, 3, 4, 8}, see A246782. a(n) is also largest prime less than prime(n)^(1 + 1/n), since prime(n)^(1 + 1/n) is never prime. - Farideh Firoozbakht, Nov 03 2014
a(n) = A007917(A249669(n)) = A244365(n,A182134(n)) = A006530(A245722(n)). - Reinhard Zumkeller, Nov 18 2014

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, J. Int. Seq. 18 (2015) 15.11.2
Nilotpal Kanti Sinha, On a new property of primes that leads to a generalization of Cramer's conjecture, arXiv:1010.1399 [math.NT]
Wikipedia, Firoozbakht's conjecture

Cf. A000040, A001223, A007917, A182134, A246778, A249669.

Cf. A244365.

a245396 n = a244365 n (a182134 n)  -- Reinhard Zumkeller, Nov 16 2014

seq(prevprime(ceil(ithprime(n)^(1+1/n))),n=1..100); # Robert Israel, Nov 03 2014

Table[NextPrime[Prime[n]^(1 + 1/n), -1], {n, 64}] (* Farideh Firoozbakht, Nov 03 2014 *)

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

a(n)=precprime(sqrtnint(prime(n)^(n+1),n)) \\ Charles R Greathouse IV, Oct 29 2018

A245396 = A007917 o A249669, i.e., a(n) = A007917(A249669(n)). Although one could say "less than" in the definition of this sequence, one cannot use A151799 in this formula because for n = 2 and n = 4, one has a(n) = A249669(n).

A244365 Table read by rows: row n contains all primes p such that prime(n) < p <= floor(prime(n)^(1+1/n)).

Original entry on oeis.org

3, 5, 7, 11, 13, 17, 17, 19, 19, 23, 23, 29, 31, 31, 37, 37, 41, 41, 43, 47, 43, 47, 53, 47, 53, 53, 59, 59, 61, 67, 61, 67, 71, 73, 67, 71, 73, 71, 73, 79, 83, 73, 79, 83, 79, 83, 89, 83, 89, 89, 97, 97, 101, 103, 107, 101, 103, 107, 109, 113, 103, 107, 109

Offset: 1

Reinhard Zumkeller, Nov 16 2014

Length of n-th row = A182134(n);

T(n,1) = A000040(n+1); T(n,A182134(n)) = A245396(n).

			.   n | A182134(n) | A249669(n) |  T(n,1) ... T(n,A182134(n))
. ----+------------+------------+----------------------------
.   1 |          1 |          4 |  [3]
.   2 |          1 |          5 |  [5]
.   3 |          1 |          8 |  [7]
.   4 |          1 |         11 |  [11]
.   5 |          2 |         17 |  [13, 17]
.   6 |          2 |         19 |  [17, 19]
.   7 |          2 |         25 |  [19, 23]
.   8 |          1 |         27 |  [23]
.   9 |          2 |         32 |  [29, 31]
.  10 |          2 |         40 |  [31, 37]
.  11 |          2 |         42 |  [37, 41]
.  12 |          3 |         49 |  [41, 43, 47]
.  13 |          3 |         54 |  [43, 47, 53]
.  14 |          2 |         56 |  [47, 53]
.  15 |          2 |         60 |  [53, 59]
.  16 |          3 |         67 |  [59, 61, 67]
.  17 |          4 |         74 |  [61, 67, 71, 73]
.  18 |          3 |         76 |  [67, 71, 73]
.  19 |          4 |         83 |  [71, 73, 79, 83]
.  20 |          3 |         87 |  [73, 79, 83]
.  21 |          3 |         89 |  [79, 83, 89]
.  22 |          2 |         96 |  [83, 89]
.  23 |          2 |        100 |  [89, 97]
.  24 |          4 |        107 |  [97, 101, 103, 107]
.  25 |          5 |        116 |  [101, 103, 107, 109, 113] .

Reinhard Zumkeller, Rows n = 1..1000 of triangle, flattened

Cf. A182134 (row lengths), A245722 (row products), A245396, A249669, A010051, A000040.

a244365 n k = a244365_tabf !! (n-1) !! (k-1)
a244365_row n = a244365_tabf !! (n-1)
a244365_tabf = zipWith farideh (map (+ 1) a000040_list) a249669_list
               where farideh u v = filter ((== 1) .  a010051') [u..v]

row(n) = my(list=List(), p=prime(n)); forprime(q=nextprime(p+1), p^(1+1/n), listput(list, q)); Vec(list); \\ Michel Marcus, Jan 24 2022

T(n,k) = A000040(n+k) for k = 1 .. A182134(n).

cp's OEIS Frontend

A182134 Number of primes p such that prime(n) < p < prime(n)^(1 + 1/n).

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A245396 Largest prime not exceeding prime(n)^(1 + 1/n).

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Haskell

Maple

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PARI

PARI

Formula

A244365 Table read by rows: row n contains all primes p such that prime(n) < p <= floor(prime(n)^(1+1/n)).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

PARI

Formula