A246778 -id:A246778 - cp's OEIS frontend

A246780 Strictly increasing terms of the sequence A246778: a(1)= A246778(1) and for n>0 a(n+1) is next term greater than a(n) after that a(n) appears in A246778 for the first time.

2, 3, 4, 6, 8, 9, 11, 12, 13, 14, 15, 16, 17, 18, 19, 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, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70

Offset: 1

Farideh Firoozbakht, Sep 29 2014

nonn

I conjecture that, a(n)=n+5 for all n, n>15.

A000040, A246776, A246777, A246778, A246779.

A246789 a(n) = A246778(A005669(n)).

Original entry on oeis.org

2, 2, 4, 9, 18, 19, 34, 39, 42, 44, 74, 82, 87, 96, 129, 149, 150, 157, 184, 184, 194, 219, 259, 265, 293, 326, 343, 343, 370, 374, 418, 422, 441, 463, 468, 509, 539, 542, 548, 573, 627, 645, 659, 670, 671, 671, 687, 693, 708, 718, 750, 753, 771, 787, 845, 884, 904, 952, 999, 1040, 1055, 1169, 1193, 1193, 1428, 1446, 1475, 1547, 1552, 1579, 1590, 1601, 1604, 1657, 1704

Offset: 1

Farideh Firoozbakht, Oct 08 2014

Conjecture: prime(A005669(n))^(1+1/A005669(n)) - prime(A005669(n)) is a strictly increasing function of n.
The truth of the conjecture would imply that "this sequence is an increasing sequence" which is another conjecture not equivalent to the first conjecture.
Note that if n is in the set {1, 19, 27, 45, 63} then a(n) = a(n+1) but there is no n, where n is less than 75 and a(n+1) < a(n).

Cf. A000040, A002386, A005669, A246776, A246777, A246778.

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

Original entry on oeis.org

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

A246776 a(n) = floor(prime(n)^(1+1/n)) - prime(n+1).

Original entry on oeis.org

1, 0, 1, 0, 4, 2, 6, 4, 3, 9, 5, 8, 11, 9, 7, 8, 13, 9, 12, 14, 10, 13, 11, 10, 15, 17, 15, 17, 15, 5, 17, 15, 20, 11, 20, 16, 16, 19, 17, 17, 22, 13, 22, 20, 22, 12, 13, 22, 24, 22, 20, 24, 16, 21, 21, 21, 25, 21, 23, 25, 17, 14, 25, 27, 24, 14, 23, 20, 28, 26

Offset: 1

Farideh Firoozbakht, Sep 26 2014

nonn

The Firoozbakht Conjecture, "prime(n)^(1/n) is a strictly decreasing function of n" is true if and only if a(n) is nonnegative for all n, n>1.
A246777 is a hard subsequence of this sequence.
18 is not in the sequence. It seems that, 18 is the only nonnegative integer which is not in the sequence.

Paulo Ribenboim, The little book Of bigger primes, second edition, Springer, 2004, p. 185.

Alois P. Heinz, Table of n, a(n) for n = 1..10000 (first 4230 terms from Hal M. Switkay)
Carlos Rivera, Conjecture 30
Alexei Kourbatov, Verification of the Firoozbakht conjecture for primes up to four quintillion, arXiv:1503.01744 [math.NT], 2015-2023.
Alexei Kourbatov, Upper Bounds for Prime Gaps Related to Firoozbakht’s Conjecture, arXiv:1506.03042 [math.NT], 2015-2019.
Alexei Kourbatov, Upper bounds for prime gaps related to Firoozbakht's conjecture, J. Int. Seq. 18 (2015) 15.11.2
Wikipedia, Prime gap.
Wikipedia, Firoozbakht Conjecture.

Cf. A000040, A001223, A005669, A246777, A246778.

Cf. A249669.

a246776 n = a249669 n - a000040 (n + 1)
-- Reinhard Zumkeller, Nov 16 2014

Table[Floor[Prime[n]^(1+1/n)]-Prime[n+1],{n,70}]

a(n) = A249669(n) - A000040(n+1). - Reinhard Zumkeller, Nov 16 2014

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).

A246777 a(n) = A246776(A005669(n)): using the indices of maximal primes in A002386 in order to verify the Firoozbakht conjecture for 0 <= floor(prime(n)^(1+1/n)) - prime(n+1).

Original entry on oeis.org

1, 0, 0, 3, 10, 5, 16, 19, 20, 10, 38, 38, 35, 24, 43, 53, 38, 43, 66, 52, 46, 65, 79, 55, 73, 104, 109, 95, 120, 92, 130, 130, 121, 127, 114, 127, 155, 148, 92, 109, 159, 171, 173, 180, 171, 157, 171, 161, 174, 178, 168, 165, 169, 135, 171, 168, 138, 174, 195, 234, 149, 253, 269, 61, 244, 248, 255, 323, 304, 307, 262, 245, 234, 215, 228

Offset: 1

Farideh Firoozbakht, Sep 30 2014

nonn

a(1) > 0 and a(n) >= 0 for n < 76; this implies "if p=p(k) is in the sequence A002386 and p <= 1425172824437699411 then p(k+1)^(1/(k+1)) < p(k)^(1/k)."

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
Wikipedia, Firoozbakht's Conjecture

Cf. A000040, A005250, A005669, A002386, A246776, A246778, A246779, A246789.

f[n_] := Block[{d, i, j, m = 0}, Reap@ For[i = 1, i <= n, i++, d = Prime[i + 1] - Prime@ i; If[d > m, m = d; Sow@ i, False]] // Flatten // Rest] (* A005669 *); g[n_] := Floor[Prime[n]^(1 + 1/n)] - Prime[n + 1] (* A246776 *); g@ f@ 100000; (* Michael De Vlieger, Mar 24 2015, with code from A246776 by Farideh Firoozbakht *)

a(n) = A246776(A005669(n)).

A246779 Strictly increasing terms of the sequence A246776: a(1)= A246776(1) and for n>0 a(n+1) is the next term greater than a(n) after that a(n) appears in A246776 for the first time.

Original entry on oeis.org

0, 1, 4, 6, 9, 11, 13, 14, 15, 17, 20, 22, 24, 25, 27, 28, 30, 32, 33, 34, 37, 38, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85

Offset: 1

Farideh Firoozbakht, Sep 30 2014

nonn

I conjecture that, a(n)=n+17 for all n, n>22.

Cf. A000040, A246776, A246777, A246778, A246780.

A262061 Least prime(i) such that prime(i)^(1+1/i) - prime(i) > n.

Original entry on oeis.org

2, 3, 5, 7, 11, 11, 17, 17, 23, 29, 29, 37, 41, 53, 59, 67, 79, 89, 97, 127, 127, 137, 163, 179, 211, 223, 251, 293, 307, 337, 373, 419, 479, 521, 541, 587, 691, 727, 797, 853, 929, 1009, 1151, 1201, 1277, 1399, 1523, 1693, 1777, 1931, 2053, 2203, 2333, 2521, 2647, 2953, 3119, 3299, 3527, 3847, 4127

Offset: 1

Farideh Firoozbakht and Robert G. Wilson v, Sep 11 2015

nonn

Where A246778(i) first exceeds n, stated by p_i.
Similar to A245396.
Number of terms < 10^n: 4, 19, 41, 75, 120, 176, 242, 319, 407, 506, ..., .
Concerning Firoozbakht's Conjecture (1982): (prime(n+1))^(1/(n+1)) < prime(n)^(1/n), for all n = 1 or prime(n+1) < prime(n)^(1+1/n), which can be rewritten as: (log(prime(n+1))/log(prime(n)))^n < (1+1/n)^n. This suggests a weaker conjecture: (log(prime(n+1))/log(prime(n)))^n < e.
Prime index of a(n): 1, 1, 3, 4, 5, 5, 7, 7, 9, 10, 10, 12, 13, 16, 17, 19, 22, 24, 25, 31, 31, ..., .
All terms are unique for n > 21. Indices not unique: 1 & 2, 5 & 6, 7 & 8, 10 & 11 and 20 & 21.
The distribution of initial digits, 1...9, for a(n), n<508: 140, 91, 60, 50, 44, 36, 32, 27 and 26.

			a(20) = 127 since for all primes less than the 31st prime, 127, p_k^(32/31) - p_k are less than 20.
a(100) = 38113,
a(200) = 2400407,
a(300) = 57189007,
a(400) = 828882731,
a(500) = 8748565643,
a(1000) = 91215796479037,
a(1064) = 246842748060263, limit of Mathematica by direct computation, i.e., the first Mathematica line.

Paulo Ribenboim, The little book Of bigger primes, second edition, Springer, 2004, p. 185.

Farideh Firoozbakht and Robert G. Wilson v, Table of n, a(n) for n = 1..507
Alexei Kourbatov, Upper Bounds for Prime Gaps Related to Firoozbakht's Conjecture, arXiv:1506.03042 [math.NT], 2015.
Alexei Kourbatov, Verification of the Firoozbakht conjecture for primes up to four quintillion, arXiv:1503.01744 [math.NT], 2015

Cf. A144104, A182134, A182514, A245396, A246777, A246778, A246781, A246810, A248855, A249566.

f[n_] := Block[{p = 2, k = 1}, While[n > p^(1 + 1/k) - p, p = NextPrime@ p; k++]; p]; Array[f, 60] (* or  quicker *)
(* or quicker *) p = 2; i = 1; lst = {}; Do[ While[ p^(1 + 1/i) < n + p, p = NextPrime@ p; i++]; AppendTo[lst, p]; Print[{n, p}], {n, 100}]; lst

a(n) = {i = 0; forprime(p=2,, i++; if (p^(1+1/i) - p > n, return (p)););} \\ Michel Marcus, Oct 04 2015

Log(y) ~= g + x^(1/2) where g = Euler's Gamma.

a(2) corrected in b-file by Andrew Howroyd, Feb 22 2018

A263023 Largest integer k such that prime(n+1) < prime(n)^(1+1/k).

Original entry on oeis.org

1, 2, 4, 4, 14, 9, 25, 15, 13, 50, 19, 35, 77, 42, 32, 37, 122, 43, 72, 153, 54, 88, 63, 52, 113, 235, 121, 252, 130, 40, 156, 108, 339, 71, 375, 128, 134, 210, 144, 151, 466, 96, 504, 256, 523, 90, 96, 304, 618, 313, 214, 657, 134, 233, 240, 247, 755, 255

Offset: 1

Alexei Kourbatov, Oct 08 2015

nonn

Firoozbakht's conjecture: prime(n+1) < prime(n)^(1+1/n).
Firoozbakht's conjecture restated for this sequence: a(n) >= n.
I further conjecture that n = 1,2,4 are the only values of n with a(n) = n.
Record values of a(n) occur when prime(n) and prime(n+1) are twin primes.
Upper bound for all n: a(n) < (1/2)*(prime(n)+2)*log(prime(n)).

			prime(1)=2; a(1)=1 because k=1 is the largest k for which 3 < 2^(1+1/k).
prime(2)=3; a(2)=2 because k=2 is the largest k for which 5 < 3^(1+1/k).
prime(10)=29; a(10)=50 because k=50 is the largest k for which 31 < 29^(1+1/k).

Paulo Ribenboim, The little book of bigger primes, 2nd edition, Springer, 2004, p. 185.

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.
Carlos Rivera, Conjecture 30

Cf. A000040, A002386, A182514, A182519, A245396, A246776, A246778, A249669.

[Floor(Log(NthPrime(n))/(Log(NthPrime(n+1))-Log(NthPrime(n)))): n in [1..60]]; // Vincenzo Librandi, Oct 08 2015

Table[Floor[Log@ Prime@ n /(Log@ Prime[n + 1] - Log@ Prime@ n)], {n, 58}] (* Michael De Vlieger, Oct 08 2015 *)

a(n) = floor(log(prime(n))/(log(prime(n+1)) - log(prime(n)))) \\ Michel Marcus, Oct 10 2015

a(n) = floor(log(prime(n))/(log(prime(n+1)) - log(prime(n)))).

More terms from Vincenzo Librandi, Oct 08 2015

cp's OEIS Frontend

A246780 Strictly increasing terms of the sequence A246778: a(1)= A246778(1) and for n>0 a(n+1) is next term greater than a(n) after that a(n) appears in A246778 for the first time.

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A246789 a(n) = A246778(A005669(n)).

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

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A246776 a(n) = floor(prime(n)^(1+1/n)) - prime(n+1).

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

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A246777 a(n) = A246776(A005669(n)): using the indices of maximal primes in A002386 in order to verify the Firoozbakht conjecture for 0 <= floor(prime(n)^(1+1/n)) - prime(n+1).

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A246779 Strictly increasing terms of the sequence A246776: a(1)= A246776(1) and for n>0 a(n+1) is the next term greater than a(n) after that a(n) appears in A246776 for the first time.

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A262061 Least prime(i) such that prime(i)^(1+1/i) - prime(i) > n.

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PARI