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
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a249669 n = floor $ fromIntegral (a000040 n) ** (1 + recip (fromIntegral n))
-- Reinhard Zumkeller, Nov 16 2014
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[Floor(NthPrime(n)^(1+1/n)): n in [1..70]]; // Vincenzo Librandi, Nov 04 2014
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seq(floor(ithprime(n)^(1+1/n)), n=1..100); # Robert Israel, Nov 26 2015
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a(n)=prime(n)^(1+1/n)\1
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
. 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] .
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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]
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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
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
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
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.
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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
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a(n) = {i = 0; forprime(p=2,, i++; if (p^(1+1/i) - p > n, return (p)););} \\ Michel Marcus, Oct 04 2015
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
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.
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[Floor(Log(NthPrime(n))/(Log(NthPrime(n+1))-Log(NthPrime(n)))): n in [1..60]]; // Vincenzo Librandi, Oct 08 2015
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Table[Floor[Log@ Prime@ n /(Log@ Prime[n + 1] - Log@ Prime@ n)], {n, 58}] (* Michael De Vlieger, Oct 08 2015 *)
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a(n) = floor(log(prime(n))/(log(prime(n+1)) - log(prime(n)))) \\ Michel Marcus, Oct 10 2015
Showing 1-5 of 5 results.
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