cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

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

Views

Author

M. F. Hasler, Nov 03 2014

Keywords

Comments

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

Crossrefs

Programs

  • Haskell
    a249669 n = floor $ fromIntegral (a000040 n) ** (1 + recip (fromIntegral n))
    -- Reinhard Zumkeller, Nov 16 2014
  • Magma
    [Floor(NthPrime(n)^(1+1/n)): n in [1..70]]; // Vincenzo Librandi, Nov 04 2014
    
  • Maple
    seq(floor(ithprime(n)^(1+1/n)), n=1..100); # Robert Israel, Nov 26 2015
  • PARI
    a(n)=prime(n)^(1+1/n)\1
    

Formula

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

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

Views

Author

Reinhard Zumkeller, Nov 16 2014

Keywords

Comments

Length of n-th row = A182134(n);
T(n,1) = A000040(n+1); T(n,A182134(n)) = A245396(n).

Examples

			.   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] .
		

Crossrefs

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

Programs

  • Haskell
    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]
    
  • PARI
    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

Formula

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

A245722 Row products of table A244365.

Original entry on oeis.org

3, 5, 7, 11, 221, 323, 437, 23, 899, 1147, 1517, 82861, 107113, 2491, 3127, 241133, 21182921, 347261, 33984931, 478661, 583573, 7387, 8633, 107972737, 13710311357, 135745657, 1317919, 12317, 14351, 16637, 2494633, 428448457, 490995677, 3532343, 645328247
Offset: 1

Views

Author

Reinhard Zumkeller, Nov 18 2014

Keywords

Comments

a(n) = Product_{k=1..A182134(n)} A244365(n,k);
A020639(a(n)) = A000040(n+1);
A006530(a(n)) = A245396(n);
A001221(a(n)) = A001222(a(n)) = A182134(n); A008966(a(n)) = 1;
A001221(GCD(a(n),a(n+1))) = A182134(n) - 1.

Crossrefs

Programs

  • Haskell
    a245722 = product . a244365_row

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

Views

Author

Keywords

Comments

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.

Examples

			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.
		

References

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

Crossrefs

Programs

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

Formula

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

Extensions

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

Views

Author

Alexei Kourbatov, Oct 08 2015

Keywords

Comments

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)).

Examples

			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).
		

References

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

Crossrefs

Programs

  • Magma
    [Floor(Log(NthPrime(n))/(Log(NthPrime(n+1))-Log(NthPrime(n)))): n in [1..60]]; // Vincenzo Librandi, Oct 08 2015
    
  • Mathematica
    Table[Floor[Log@ Prime@ n /(Log@ Prime[n + 1] - Log@ Prime@ n)], {n, 58}] (* Michael De Vlieger, Oct 08 2015 *)
  • PARI
    a(n) = floor(log(prime(n))/(log(prime(n+1)) - log(prime(n)))) \\ Michel Marcus, Oct 10 2015

Formula

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

Extensions

More terms from Vincenzo Librandi, Oct 08 2015
Showing 1-5 of 5 results.