A182134 -id:A182134 - cp's OEIS frontend

A246782 Numbers k such that A182134(k)=2, i.e., there exist only two primes p with prime(k) < p < prime(k)^(1+1/k).

5, 6, 7, 9, 10, 11, 14, 15, 22, 23, 28, 29, 30, 45, 46, 61, 66, 216, 217, 367, 3793, 1319945, 1576499, 8040877, 17567976, 44405858, 445538764, 1478061204, 3643075047, 17440041685, 190836014732, 714573709895, 714573709896

Offset: 1

Farideh Firoozbakht, Oct 12 2014

Firoozbakht's conjecture says that for every n, there exists at least one prime p such that prime(n) < p < prime(n)^(1+1/n).
Let A(m) = {n | A182134(n) = m} where A182134(n) = #{p | p is prime and prime(n) < p < prime(n)^(1+1/n)}. This sequence gives the terms of A(2) and the sequence A246781 gives the terms of A(3).
The only known indices n for which A182134(n) = 1 are {1, 2, 3, 4, 8}.  It is conjectured that this is the complete set A(1).
Conjecture: For all m, where m is greater than one, A(m) is an infinite set.
a1 = 49749629143524, a2 = 1475067052906944 and a3 = 1475067052906945 are three large terms of the sequence. It is interesting that a3 - a2 = 1.
Conjecture: The sequence is infinite.
Next term is greater than 25000000.
a(34) > 10^12. - Robert Price, Nov 01 2014
The conjecture that A(1)={1, 2, 3, 4, 8} holds through 10^12. - Robert Price, Nov 01 2014

			5 is in the sequence since there exists only two primes p, prime(5) < p < prime(5)^(1+1/5). Note that prime(5) = 11, 11^(1+1/5) ~ 17.77 and 11 < 13 < 17 < 17.77.

A. Kourbatov, Verification of the Firoozbakht conjecture for primes up to four quintillion, arXiv:1503.01744 [math.NT], 2015
Wikipedia, Firoozbakht's conjecture

Cf. A000040, A002386, A005669, A182134, A246781, A249566.

a246782 n = a246782_list !! (n-1)
a246782_list = filter ((== 2) . a182134) [1..]
-- Reinhard Zumkeller, Nov 17 2014

np[n_]:=(a = Prime[n]; b = a^(1 + 1/n); Length[Select[Range[a+1,b], PrimeQ]]); Do[If[np[n] == 2,Print[n]], {n, 25000000}]

for(n=1,oo,2==primepi(prime(n)^(1+1/n))-n&&print1(n", ")) \\ M. F. Hasler, Nov 03 2014

a(26)-a(27) from Robert Price, Oct 24 2014

a(28)-a(33) from Robert Price, Nov 01 2014

A246785 a(n) is the least m>0 such that A182134(n - m) = m, or zero if there is no such m.

Original entry on oeis.org

1, 1, 1, 1, 0, 2, 2, 1, 0, 2, 2, 2, 0, 3, 2, 2, 0, 3, 0, 3, 0, 3, 2, 2, 0, 0, 4, 0, 2, 2, 2, 0, 3, 0, 4, 3, 0, 4, 4, 4, 3, 0, 4, 0, 4, 2, 2, 0, 0, 4, 0, 5, 4, 4, 3, 0, 0, 5, 0, 5, 3, 2, 0, 0, 4, 4, 2, 0, 0, 0, 5

Offset: 2

Farideh Firoozbakht, Oct 24 2014

nonn

Recall that A182134(k) is the number of primes p with prime(k) < p < prime(k)^(1+1/k). The record values up to n = 56000 are the positive integers up to 21 except 13 which first occurs after 14; A246790 gives indices of these record values.

Robert Price, Table of n, a(n) for n = 2..10000

Cf. A000040, A182134, A246790, A246793.

a246785 n = if null ms then 0 else head ms
            where ms = [m | m <- [1 .. n-1], a182134 (n - m) == m]
-- Reinhard Zumkeller, Nov 17 2014

h[n_]:= If[n==0, 0, (i=Prime[n]+1; j=Prime[n]^(1+1/n); Length[Select[Range[i,j], PrimeQ]])];a1[n_]:= (For[m=1, m<=n-1 && h[n-m]!= m, m++]; m); a[k_]:= If[c=a1[k]; c==k, 0, c]; Table[a[k],{k,2,90}]

A246781 Numbers n such that A182134(n) = 3, i.e., there exist only three primes p with prime(n) < p < prime(n)^(1 + 1/n).

Original entry on oeis.org

12, 13, 16, 18, 20, 21, 27, 31, 34, 39, 44, 53, 59, 60, 65, 96, 97, 98, 99, 136, 154, 202, 214, 215, 220, 221, 280, 324, 325, 326, 365, 366, 736, 780, 2146, 2225, 3792, 5946, 5947, 5948, 6902, 6903, 18524, 22078, 23510, 23511, 23512, 31542, 31544, 33606

Offset: 1

Farideh Firoozbakht, Oct 12 2014

nonn

Firoozbakht's conjecture states that for every n, there exists at least one prime p with prime(n) < p < prime(n)^(1+1/n).
The only known indices n for which A182134(n) = 1 are {1, 2, 3, 4, 8}.
See A246782 for indices n such that A182134(n) = 2.
This sequence lists numbers n such that A182134(n) = 3.

			12 is in the sequence since there exists only three primes p where, prime(12) < p < prime(12)^(1 + 1/12). Note that prime(12) = 37, 37^(1 + 1/12) ~ 49.99 and 37 < 41 < 43 < 47 < 49.99.

Robert Price, Table of n, a(n) for n = 1..170
Carlos Rivera, Conjecture 30. The Firoozbakht Conjecture PrimePuzzles.net.
A. Kourbatov, Verification of the Firoozbakht conjecture for primes up to four quintillion, arXiv:1503.01744 [math.NT], 2015
Wikipedia, Firoozbakht's conjecture

Cf. A000040, A182134, A246782.

Cf. A249566.

a246781 n = a246781_list !! (n-1)
a246781_list = filter ((== 3) . a182134) [1..]
-- Reinhard Zumkeller, Nov 17 2014

N:= 10^5: # to get all terms where prime(n)^(1+1/n) < N
Primes:= select(isprime,[2,seq(2*i+1,i=1..floor((N+1)/2))]):
filter:= proc(n) local t; t:= Primes[n]^(n+1); Primes[n+3]^n <= t and Primes[n+4]^n > t end proc:
select(filter, [$1..nops(Primes)-4]); # Robert Israel, Mar 23 2015

np[n_] := (a = Prime[n]; b = a^(1 + 1/n); Length[Select[Range[a + 1, b], PrimeQ]]); Select[Range[10000], np[#] == 3 &]

a(43)-a(50) from Robert Price, Oct 24 2014

A246793 a(n) is the largest m such that A182134(n - k) = k for A246785(n) <= k <= m, or zero if there is no such m.

Original entry on oeis.org

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

Offset: 1

Farideh Firoozbakht, Oct 24 2014

nonn

Recall that A182134(k) is the number of primes p with prime(k) < p < prime(k)^(1+1/k). Obviously a(n) = 0 if and only if A246785(n) = 0.

			A182134(217 - k) = k for k = 3, 4, ..., 9 since A246785(217) = 3 and a(217) = 9.

Robert Price, Table of n, a(n) for n = 1..10000

Cf. A000040, A182134, A246785, A246790.

np[n_]:= If[n==0, 0, (i=Prime[n]+1; j=Prime[n]^(1+1/n); Length[Select[Range[i,j], PrimeQ]])]; a1[n_]:= (For[m=1, m<=n-1&& np[n-m] != m, m++];m);a2[k_]:= If[c=a1[k]; c==k,0,c]; a[n_]:= If[a2[n]==0, 0, For[r=a2[n], np[n-r]==r, r++]; r-1]; Table[a[k], {k,2,90}]

A249566 Numbers n such that A182134(n) = 4, i.e., there exist exactly four primes p with prime(n) < p < prime(n)^(1+1/n).

Original entry on oeis.org

17, 19, 24, 26, 32, 33, 35, 36, 37, 38, 40, 42, 43, 47, 50, 51, 52, 58, 62, 63, 64, 76, 77, 78, 79, 90, 91, 93, 95, 121, 123, 124, 125, 126, 134, 135, 137, 150, 153, 185, 186, 187, 188, 189, 201, 203, 213, 218, 219, 238, 239, 259, 263, 278, 279, 289, 293

Offset: 1

Robert Price, Nov 01 2014

nonn

See A246782 for a more complete description of this sequence.
a(1136) > 10^12.
It is interesting that three consecutive integers n = 20004097201301075, n + 1 and n + 2 are in the sequence. Conjecture: The sequence is infinite. - Farideh Firoozbakht, Nov 01 2014

Robert Price, Table of n, a(n) for n = 1..1135
A. Kourbatov, Verification of the Firoozbakht conjecture for primes up to four quintillion, arXiv:1503.01744 [math.NT], 2015
Wikipedia, Firoozbakht's conjecture

Cf. A000040, A002386, A005669, A182134, A246781, A246782.

a249566 n = a249566_list !! (n-1)
a249566_list = filter ((== 4) . a182134) [1..]
-- Reinhard Zumkeller, Nov 17 2014

np[n_]:=(a = Prime[n]; b = a^(1 + 1/n); Length[Select[Range[a+1,b], PrimeQ]]); Do[If[np[n] == 4,Print[n]], {n, 293}]
np[n_]:=(a = Prime[n]; b = a^(1 + 1/n); Length[Select[Range[a+1,b], PrimeQ]]); Select[Range[293], np[#]==4&] (* Farideh Firoozbakht, Nov 01 2014 *)

for(n=1,9e9,primepi(prime(n)^(1+1/n))-n==4&&print1(n",")) \\ M. F. Hasler, Nov 03 2014

A246787 Indices of records in A182134.

Original entry on oeis.org

1, 5, 12, 17, 25, 55, 83, 169, 206, 384, 953, 1615, 2192, 2197, 3023, 10709, 10935, 29508, 62735, 94332, 196966, 314940, 608777, 1258688, 1767259, 2448973, 7939362, 9373134, 16854966, 16854967, 32881913, 41084049, 83715318, 90288054, 151449026, 315082003, 327952702, 384935466, 720004431

Offset: 1

Farideh Firoozbakht and Jahangeer Kholdi, Oct 10 2014

nonn

Robert Price, Table of n, a(n) for n = 1..53

Cf. A000040, A182134, A246786, A246810.

a(n) = A246810(A246786(n)).

a(31)-a(39) from Robert Price, Oct 24 2014

A246786 Record values of A182134.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17, 18, 19, 21, 23, 25, 26, 27, 28, 29, 30, 32, 33, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 47, 48, 50, 51, 52, 53, 55, 56, 57, 58, 60, 61, 62, 63, 64

Offset: 1

Farideh Firoozbakht and Jahangeer Kholdi, Oct 10 2014

It seems that a(n+1) <= a(n)+2 for all n.
Also: indices n for which A246810(k) > A246810(n) for all k > n. The missing numbers are {9, 16, 20, 22, 24, 31, 34, ...}, I conjecture that these are exactly the indices for which A246810(n+1) < A246810(n). - M. F. Hasler, Oct 16 2014
Searched A182134(n) through n=10^12. - Robert Price, Nov 01 2014

Cf. A000040, A182134, A246787, A246810.

a(31)-a(39) from Robert Price, Oct 24 2014

a(40)-a(53) from Robert Price, Nov 01 2014

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

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

A246782 Numbers k such that A182134(k)=2, i.e., there exist only two primes p with prime(k) < p < prime(k)^(1+1/k).

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A246785 a(n) is the least m>0 such that A182134(n - m) = m, or zero if there is no such m.

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A246781 Numbers n such that A182134(n) = 3, i.e., there exist only three primes p with prime(n) < p < prime(n)^(1 + 1/n).

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A246793 a(n) is the largest m such that A182134(n - k) = k for A246785(n) <= k <= m, or zero if there is no such m.

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A249566 Numbers n such that A182134(n) = 4, i.e., there exist exactly four primes p with prime(n) < p < prime(n)^(1+1/n).

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A246787 Indices of records in A182134.

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A246786 Record values of A182134.

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

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