A181098 -id:A181098 - cp's OEIS frontend

A038822 Number of primes between 100n and 100n+99.

25, 21, 16, 16, 17, 14, 16, 14, 15, 14, 16, 12, 15, 11, 17, 12, 15, 12, 12, 13, 14, 10, 15, 15, 10, 11, 15, 14, 12, 11, 12, 10, 11, 15, 11, 14, 13, 12, 11, 11, 15, 9, 16, 9, 11, 12, 12, 12, 8, 15, 12, 11, 10, 10, 13, 13, 12, 10, 16, 7, 12, 11, 13, 15, 8, 11, 10, 12, 12, 13, 9, 10

Offset: 0

Jeff Burch

nonn

The number k first occurs in century A186311(k).

			a(3) = 16 because there are 16 primes between 300 and 399 (namely, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 367, 373, 379, 383, 389, 397).
a(4) = 17 because there are 17 primes between 400 and 499 (401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491, 499).

George P. Loweke, The Lore of Prime Numbers. New York: Vantage Press (1982): 91.

T. D. Noe, Table of n, a(n) for n = 0..9999

Cf. A028505.
Cf. A181098 (centuries without primes).
Cf. A186393-A186408 (centuries having 1 to 16 primes), A186509 (17 primes).
Cf. A094892, A098592.

with(numtheory); A038822 := n->pi(100*n+99)-pi(100*n); seq(A038822(k), k=0..100); # Wesley Ivan Hurt, Oct 03 2013

Table[PrimePi[100n + 99] - PrimePi[100n], {n, 0, 71}]
Differences[PrimePi[100 Range[0,100]]] (* Harvey P. Dale, Feb 18 2021 *)

a(n)=sum(i=100*n,100*n+99,isprime(i)) \\ Charles R Greathouse IV, Apr 28 2015

a(n)= my(r=0, p=100*n, q=p+99); while((p=nextprime(p+1))<=q, r+=isprime(p)); r; \\ Ruud H.G. van Tol, Nov 17 2024

a(n) = pi(100n+99) - pi(100n). - Wesley Ivan Hurt, Oct 03 2013

Edited, corrected and extended by Robert G. Wilson v, Jan 29 2003

A186311 Least k such that the interval 100k to 100k+99 has exactly n primes.

Original entry on oeis.org

16718, 1559, 3020, 588, 314, 188, 186, 59, 48, 41, 21, 13, 11, 19, 5, 8, 2, 4, 1228537713709, 14688670051164208, 203860951641372730864, 1

Offset: 0

T. D. Noe, Feb 22 2011

It is known that a(25)=0. Terms for n = 22 and 23 are unknown. Glaisher tabulates the number of centuries having 0, 1, 2, ... primes for numbers up to 9000000. Glaisher's 1883 book is still in print!
a(24) does not exist because the only century having 24 primes is 0 to 99 -- the same century having 25 primes. From A020497, we see that a range of 101 numbers is required to find 24 primes. Dickson's conjecture implies that a(n) exists for n=18..23. - Charles R Greathouse IV, Feb 24 2011
To see that Dickson's conjecture is applicable to the preceding statement, the appropriate general sequence to consult is A364678, which affirms that 23 primes are permissible between adjacent multiples of 100, as opposed to in an arbitrary interval of 99 integers. - Peter Munn, Sep 04 2023
a(n) for n = 18..23 is greater than 10^10. Ribenboim discusses Dickson's conjecture in two books. - T. D. Noe, Feb 24 2011
a(19) <= 1108851311300675700427. - Donovan Johnson, Feb 28 2011
a(20) <= 394338677302163715754576644. - Tim Johannes Ohrtmann, Aug 27 2015

James Glaisher, Factor Table for the Sixth Million, Taylor and Francis, London, 1883.
Paulo Ribenboim, The New Book of Prime Number Records, Springer-Verlag NY, 1995, p. 372.
Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY, 2004, p. 250.

Chris Caldwell, Prime Glossary: Dickson's Conjecture
L. E. Dickson, A new extension of Dirichlet's theorem on prime numbers, Messenger of Math., 33 (1904), 155-161.

Cf. A038822 (number of primes between 100n and 100n+99).
Cf. A181098 (centuries without primes).
Cf. A186393-A186408 (centuries having 1 to 16 primes).
Cf. A186509 (centuries having 17 primes).
Cf. A361723 (centuries having 18 primes).
Cf. A020497, A261571, A364678.

t = Differences[PrimePi[100*Range[0, 20000]]]; Flatten[Table[Position[t, n, 1, 1], {n, 0, 17}] - 1]

a(n)=for(k=0,9e99,if(sum(i=100*k+1,100*k+99,ispseudoprime(i))==n, return(k))) \\ Charles R Greathouse IV, Feb 24 2011

a(18) from Donovan Johnson, Feb 28 2011
a(19) from Brian Kehrig, Apr 08 2023
a(20)-a(21) from Brian Kehrig, May 28 2024

A186393 Numbers k such that there is 1 prime between 100k and 100k + 99.

Original entry on oeis.org

1559, 2683, 4133, 10048, 11400, 12727, 12800, 13572, 14223, 14443, 14514, 14680, 14913, 15536, 15619, 16538, 16557, 17334, 19043, 20452, 20465, 20522, 21162, 21663, 22440, 22832, 23055, 23144, 23214, 23460, 24833, 25139, 25278, 25980, 26207, 26257, 26702, 26747, 27536, 27878, 28448, 28671, 29180, 29873, 30212, 30232

Offset: 1

Tim Johannes Ohrtmann, Feb 20 2011

nonn

There are 40 possible prime patterns for centuries having 1 prime. - Tim Johannes Ohrtmann, Aug 27 2015

			1559 is in this sequence because there is 1 prime between 155900 and 155999 (155921).

T. D. Noe, Table of n, a(n) for n = 1..1000

Cf. A038822 (number of primes between 100n and 100n+99), A186311 (first occurrences).

Cf. A181098 (no primes), A186394-A186408 (2 to 16 primes), A186509 (17 primes), A361723 (18 primes).

Select[Range[31000],PrimePi[100 #+99]-PrimePi[100 #]==1&] (* Harvey P. Dale, Dec 16 2011 *)

isA186393(n)=my(k=nextprime(100*n)); n=100*(n+1); kn \\ Charles R Greathouse IV, Feb 21 2011

a(4)-a(46) from Charles R Greathouse IV, Feb 21 2011

A186408 Numbers k such that there are 16 primes between 100k and 100k + 99.

Original entry on oeis.org

2, 3, 6, 10, 42, 58, 194, 230, 12463, 8392963, 24662691, 37400476, 163061323, 205481131, 278399797, 313114319, 481863166, 494959102, 656914015, 776749247, 960655996, 980373049, 1097546872, 1156724143, 2013136112, 2245034146, 3416649829, 3606810631, 4141180699, 5928231877, 6569717174, 6594050440, 7240502155, 7492029097, 8995086259

Offset: 1

Tim Johannes Ohrtmann, Feb 20 2011

nonn

There are 6699888 possible prime patterns for centuries having 16 primes. - Tim Johannes Ohrtmann, Aug 27 2015

			2 is in this sequence because there are 16 primes between 200 and 299 (211, 223, 227, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 283 and 293).

Brian Kehrig, Table of n, a(n) for n = 1..1000

Cf. A038822 (number of primes between 100n and 100n+99), A186311 (first occurrences).

Cf. A181098 (no primes), A186393-A186407 (1 to 15 primes), A186509 (17 primes), A361723 (18 primes).

for(n=1, 1e6, if(sum(k=100*n,100*(n+1), ispseudoprime(k))==16, print1(n", "))); \\ Charles R Greathouse IV, Feb 21 2011

N=100; s=0; forprime(p=2, 4e9, if(p>N, if(s==16, print1((N\100)-1,", ")); s=1; N=100*(p\100+1),s++)) \\ Charles R Greathouse IV, Feb 21 2011

a(9)-a(12) from Charles R Greathouse IV, Feb 21 2011

a(13)-a(35) from T. D. Noe, Feb 23 2011

A186509 Numbers k such that there are 17 primes between 100k and 100k + 99.

Original entry on oeis.org

4, 14, 7837, 27049, 144997771, 651186838, 12779564974, 22369949923, 149621468452, 225012717952, 240728320642, 586832463472, 766964610742, 939742446571, 949543082647, 1908189311558, 2693729868901, 2701032171244, 3465208973035, 3489960850720, 3910908182851

Offset: 1

T. D. Noe, Feb 22 2011

nonn

There are 2829786 possible prime patterns for centuries having 17 primes. - Tim Johannes Ohrtmann, Aug 27 2015

			4 is in this sequence because there are 17 primes between 400 and 499 (401, 409, 419, 421, 431, 433, 439, 443, 449, 457, 461, 463, 467, 479, 487, 491 and 499).

Brian Kehrig, Table of n, a(n) for n = 1..365

Cf. A038822 (number of primes between 100n and 100n+99), A186311 (first occurrences).

Cf. A181098 (no primes), A186393-A186408 (1 to 16 primes), A361723 (18 primes).

a(7)-a(15) from Donovan Johnson, Feb 28 2011

Terms a(16) and beyond from Brian Kehrig, Mar 21 2023

A361723 Numbers k such that there are 18 primes between 100k and 100k + 99.

Original entry on oeis.org

1228537713709, 23352869714018, 28703237474266, 144785865481702, 161394923966449, 168975708209638, 174748809066898, 207552241231357, 278215179205531, 312303328909720, 592248982143877, 812939886634531, 939100782752014, 983930290209021, 1111161494544274

Offset: 1

Brian Kehrig, Mar 21 2023

nonn

There are A261571(18) = 948729 possible patterns for centuries having 18 primes.

			1228537713709 is in the sequence because there are 18 primes between 122853771370900 and 122853771370999: 122853771370900 + x, where x is one of (1, 3, 7, 19, 21, 27, 31, 33, 37, 49, 51, 61, 69, 73, 87, 91, 97, or 99).

Brian Kehrig, Table of n, a(n) for n = 1..39 (terms up to 10^16)
Note (Apr 24 2024): An older version of the b-file missed a(33) and a(38). The present b-file is correct.

Cf. A038822 (number of primes between 100n and 100n+99), A186311 (first occurrences).
Cf. A181098 (no primes), A186393-A186408 (1 to 16 primes), A186509 (17 primes).
Cf. A261571 (number of patterns for centuries with n primes).

isok(k) = sum(i=0, 99, isprime(100*k + i)) == 18; \\ Michel Marcus, Mar 23 2023

A190639 Centuries whose prime pattern repeats in the next century.

Original entry on oeis.org

473267, 726760, 1773439, 1808828, 1919128, 2131583, 2165420, 2339971, 2390652, 2518488, 2802591, 2844914, 2982584, 2996184, 3183263, 3193175, 3250986, 3418185, 3428241, 3633472, 3909324, 3953449, 4280455, 4303819, 4373399, 4658285, 4728653, 4978360, 5165402, 5254365

Offset: 1

M. F. Hasler, May 15 2011

Alternate definition: Numbers x such that for all N in [100x,100x+99], N is prime iff N+100 is prime.
Contains in particular the first of two consecutive prime-free centuries, i.e., N such that there is no prime in [100 N,100 (N+2)], cf. A181098.
x belongs to this sequence if and only if the primality character of (100 * x) + k is the same as (100 * (x+1)) + k for all k = 0..99. - V. Raman, Dec 09 2012

			The first value refers to two consecutive prime-free centuries (cf. A181098); the second value is such that 100*a(2)+17 and 100*a(2)+117 are the only primes between 100*a(2) and 100*(a(2)+2). See the link for more examples.

Donovan Johnson, Table of n, a(n) for n = 1..1000
J. K. Andersen, in reply to R. Wood, Re: First repetition of prime pattern within "centuries", Yahoo group "primenumbers", May 15, 2011.
Jens Kruse Andersen, Phil Carmody, Maximilian Hasler, First repetition of prime pattern within "centuries", digest of 11 messages in primenumbers Yahoo group, May 15, 2011.

Cf. A181098.

Cf. A219996 (upper century).

a(n) ~ n. In particular there are x - 200x/log x + O(x/log^2 x) members of this sequence below x. - Charles R Greathouse IV, Dec 09 2012

a(n) = A219996(n) - 1. - V. Raman, Dec 09 2012

a(1)-a(5) computed by J. K. Andersen, May 15 2011

a(6)-a(30) from Donovan Johnson, May 15 2011

A164987 First pair of primes (p1, p2) that begin centuries of primes having the same prime configuration, ordered by increasing p2. Each configuration is allowed only once.

Original entry on oeis.org

390503, 480803, 351121, 566821, 78901, 578701, 323623, 606223, 326701, 645901, 619471, 745471, 655717, 842617, 437321, 855821, 854713, 876913, 811337, 915437, 561409, 920509, 515401, 956401, 452401, 1023601, 805633, 1049333, 247141, 1092541, 1037903, 1127603

Offset: 1

Ki Punches, Sep 03 2009 through Dec 06 2009

Rearranged the pairs of numbers so that the sequence of values of p2 increases. The first pair is for the primes 390500 + {3, 27, 39, 53, 81} and 480803 + {3, 27, 39, 53, 81}. There is a large, but finite number of terms. How many terms are there? - T. D. Noe, Feb 10 2013
The sequence lists the small prime twin centuries. As exploration goes into higher primes many are found to be triples or even higher. Example: 1072009 is a twin with 5179509, a triple with 7183109, quadruple with 8284709, quintuple with 8462609, and sextuple with 9739309, and there could be infinitely more. - Ki Punches, Dec 17 2009
The first two centuries without any primes start with 1671800 and 2637800. These are not included in the sequence since they do not have a first prime. However, if they were to be included they would be the 136th pair. - Andrew Howroyd, Feb 25 2018

			The primes in 480800..480899 are 480803, 480827, 480839, 480853, 480881 ending with 03, 27, 39, 53, 81. The primes in 390500..390599 end with the same digits, and no earlier pair has this quality. Hence a(1) = 390503 and a(2) = 480803.

T. D. Noe, Table of n, a(n) for n = 1..10000

Cf. A038822 (number of primes between 100n and 100n+99).

Cf. A181098, A186393-A186408 (centuries having 0 to 16 primes).

pSig[n_] := Prime[Range[PrimePi[100 n] + 1, PrimePi[100 (n + 1)]]] - 100 n; t = {}; c = {}; found = {}; Do[s = pSig[n]; If[Length[s] > 0 && ! MemberQ[found, s] && MemberQ[c, s], d = Mod[s[[1]], 100]; AppendTo[found, s]; AppendTo[t, {Position[c, s][[1, 1]]*100 + d, n*100 + d}]]; AppendTo[c, s], {n, 11000}]; Flatten[t] (* T. D. Noe, Feb 10 2013 *)

sig(c)={my(s=0); for(v=0,49,if(isprime(100*c+2*v+1),s+=2^v)); s}
pairs(n)={my(L=List(),M=Map(),c=0); while(#L<2*n, c++; my(s=sig(c),f=0); if(mapisdefined(M,s,&f), if(f&&s,my(d=2*valuation(s,2)+1); listput(L,100*f+d); listput(L,100*c+d); mapput(M,s,0)), mapput(M,s,c))); Vec(L)}
pairs(20) \\ Andrew Howroyd, Feb 25 2018

Terms rearranged by T. D. Noe, Feb 10 2013

A219996 Centuries whose prime pattern is the same as prime pattern in the previous century.

Original entry on oeis.org

473268, 726761, 1773440, 1808829, 1919129, 2131584, 2165421, 2339972, 2390653, 2518489, 2802592, 2844915, 2982585, 2996185, 3183264, 3193176, 3250987, 3418186, 3428242, 3633473, 3909325, 3953450, 4280456, 4303820, 4373400, 4658286, 4728654, 4978361, 5165403, 5254366

Offset: 1

V. Raman, Dec 08 2012

x belongs to this sequence if and only if the primality character of (100 * (x-1)) + k is the same as (100 * x) + k for all k = 0..99.

Cf. A181098.

Cf. A190639 (lower century).

a(n) ~ n. In particular there are x - 200x/log x + O(x/log^2 x) members of this sequence below x. - Charles R Greathouse IV, Dec 09 2012

a(n) = A190639(n) + 1.

A186394 Numbers k such that there are 2 primes between 100k and 100k + 99.

Original entry on oeis.org

3020, 3709, 4484, 4617, 4806, 4921, 5072, 5423, 5616, 6041, 6194, 6231, 6452, 6485, 6683, 6828, 7101, 7365, 7454, 7532, 7839, 8096, 8157, 8728, 8738, 9221, 9486, 9635, 9796, 10152, 10506, 10720, 10852, 11261, 11621, 11736, 11953, 11992

Offset: 1

Tim Johannes Ohrtmann, Feb 20 2011

nonn

There are 780 possible prime patterns for centuries having 2 primes. - Tim Johannes Ohrtmann, Aug 27 2015

			3020 is in this sequence because there are 2 primes between 302000 and 302099 (302009 and 302053).

T. D. Noe, Table of n, a(n) for n = 1..1000

Cf. A038822 (number of primes between 100n and 100n+99), A186311 (first occurrences).

Cf. A181098 (no primes), A186393-A186408 (1 to 16 primes), A186509 (17 primes), A361723 (18 primes).

for(n=1, 1e6, if(sum(k=100*n, 100*(n+1), ispseudoprime(k))==2, print1(n", "))); \\ Charles R Greathouse IV, Feb 21 2011

N=100; s=0; forprime(p=2, 4e9, if(p>N, if(s==2, print1((N\100)-1, ", ")); s=1; N=100*(p\100+1), s++)) \\ Charles R Greathouse IV, Feb 21 2011

def is_A186394(n):
    np0 = next_prime(next_prime(100*n))
    np1 = next_prime(np0)
    return np0 <= 100*n+99 and np1 > 100*n+99  # D. S. McNeil, Feb 21 2011

cp's OEIS Frontend

A038822 Number of primes between 100n and 100n+99.

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A186311 Least k such that the interval 100k to 100k+99 has exactly n primes.

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A186393 Numbers k such that there is 1 prime between 100*k and 100*k + 99.

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A186408 Numbers k such that there are 16 primes between 100*k and 100*k + 99.

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A186509 Numbers k such that there are 17 primes between 100*k and 100*k + 99.

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A361723 Numbers k such that there are 18 primes between 100*k and 100*k + 99.

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A190639 Centuries whose prime pattern repeats in the next century.

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A186393 Numbers k such that there is 1 prime between 100k and 100k + 99.

A186408 Numbers k such that there are 16 primes between 100k and 100k + 99.

A186509 Numbers k such that there are 17 primes between 100k and 100k + 99.

A361723 Numbers k such that there are 18 primes between 100k and 100k + 99.

A186394 Numbers k such that there are 2 primes between 100k and 100k + 99.