A246776 -id:A246776 - cp's OEIS frontend

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).

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.

A246778 a(n) = floor(prime(n)^(1+1/n)) - prime(n).

Original entry on oeis.org

2, 2, 3, 4, 6, 6, 8, 8, 9, 11, 11, 12, 13, 13, 13, 14, 15, 15, 16, 16, 16, 17, 17, 18, 19, 19, 19, 19, 19, 19, 21, 21, 22, 21, 22, 22, 22, 23, 23, 23, 24, 23, 24, 24, 24, 24, 25, 26, 26, 26, 26, 26, 26, 27, 27, 27, 27, 27, 27, 27, 27, 28, 29, 29, 28, 28, 29, 30, 30, 30

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) - A001223(n) is nonnegative for all n. The conjecture is true for all primes p where p < 4.0*10^18. (See A. Kourbatov link.)
0, 1, 5, 7, 10 & 20 are not in the sequence. It seems that these six integers are all the nonnegative integers which are not in the sequence.
From Alexei Kourbatov, Nov 27 2015: (Start)
Theorem: if prime(n+1) - prime(n) < prime(n)^(3/4), then every integer > 20 is in this sequence.
Proof: Let f(n) = prime(n)^(1+1/n) - prime(n). Then a(n) = floor(f(n)).
Define F(x) = log^2(x) - log(x) - 1. Using the upper and lower bounds for f(n) established in Theorem 5 of J. Integer Sequences Article 15.11.2; arXiv:1506.03042 we have F(prime(n))-3.83/(log prime(n)) < f(n) < F(prime(n)) for n>10^6; so f(n) is unbounded and asymptotically equal to F(prime(n)).
Therefore, for every n>10^6, jumps in f(n) are less than F'(x)*x^(3/4)+3.83/(log x) at x=prime(n), which is less than 1 as x >= prime(10^6)=15485863. Thus jumps in a(n) cannot be more than 1 when n>10^6. Separately, we verify by direct computation that a(n) takes every value from 21 to 256 when 30 < n <= 10^6. This completes the proof.
(End)

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

Robert G. Wilson v, 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.
A. Kourbatov, Upper bounds for prime gaps related to Firoozbakht's conjecture, Journal of Integer Sequences, 18 (2015), Article 15.11.2.
Carlos Rivera, Conjecture 30
Wikipedia, Firoozbakht's conjecture.
Wikipedia, Prime gap.

Cf. A000040, A001223, A246776, A246777, A246779, A246780, A249669.

[Floor(NthPrime(n)^(1+1/n)) - NthPrime(n): n in [1..70]]; // Vincenzo Librandi, Mar 24 2015

N:= 10^4: # to get entries corresponding to all primes <= N
Primes:= select(isprime, [2,seq(2*i+1,i=1..floor((N-1)/2))]):
seq(floor(Primes[n]^(1+1/n) - Primes[n]), n=1..nops(Primes)); # Robert Israel, Mar 23 2015

f[n_] := Block[{p = Prime@ n}, Floor[p^(1 + 1/n)] - p]; Array[f, 75]

first(m)=vector(m,i,floor(prime(i)^(1+1/i)) - prime(i)) \\ Anders Hellström, Sep 06 2015

a(n) = A249669(n) - A000040(n). - M. F. Hasler, Nov 03 2014

a(n) = (log(prime(n)))^2 - log(prime(n)) + O(1), see arXiv:1506.03042. - Alexei Kourbatov, Sep 06 2015

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.

Original entry on oeis.org

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.

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

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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A246778 a(n) = floor(prime(n)^(1+1/n)) - prime(n).

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

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