Displaying 1-10 of 26 results found.
Number of primes between 100n and 100n+99.
+10
32
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
COMMENTS
The number k first occurs in century A186311(k).
REFERENCES
George P. Loweke, The Lore of Prime Numbers. New York: Vantage Press (1982): 91.
EXAMPLE
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).
MATHEMATICA
Table[PrimePi[100n + 99] - PrimePi[100n], {n, 0, 71}]
Differences[PrimePi[100 Range[0, 100]]] (* Harvey P. Dale, Feb 18 2021 *)
CROSSREFS
Cf. A181098 (centuries without primes).
Least k such that the interval 100k to 100k+99 has exactly n primes.
+10
24
16718, 1559, 3020, 588, 314, 188, 186, 59, 48, 41, 21, 13, 11, 19, 5, 8, 2, 4, 1228537713709, 14688670051164208, 203860951641372730864, 1
COMMENTS
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
REFERENCES
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.
MATHEMATICA
t = Differences[PrimePi[100*Range[0, 20000]]]; Flatten[Table[Position[t, n, 1, 1], {n, 0, 17}] - 1]
PROG
(PARI) 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
CROSSREFS
Cf. A038822 (number of primes between 100n and 100n+99). Cf. A181098 (centuries without primes). Cf. A186509 (centuries having 17 primes). Cf. A361723 (centuries having 18 primes).
Numbers k such that there is 1 prime between 100*k and 100*k + 99.
+10
23
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
EXAMPLE
1559 is in this sequence because there is 1 prime between 155900 and 155999 (155921).
MATHEMATICA
Select[Range[31000], PrimePi[100 #+99]-PrimePi[100 #]==1&] (* Harvey P. Dale, Dec 16 2011 *)
PROG
(PARI) isA186393(n)=my(k=nextprime(100*n)); n=100*(n+1); k<n&nextprime(k+1)>n \\ Charles R Greathouse IV, Feb 21 2011
CROSSREFS
Cf. A038822 (number of primes between 100n and 100n+99), A186311 (first occurrences).
Numbers k such that there are 16 primes between 100*k and 100*k + 99.
+10
23
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
COMMENTS
There are 6699888 possible prime patterns for centuries having 16 primes. - Tim Johannes Ohrtmann, Aug 27 2015
EXAMPLE
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).
PROG
(PARI) for(n=1, 1e6, if(sum(k=100*n, 100*(n+1), ispseudoprime(k))==16, print1(n", "))); \\ Charles R Greathouse IV, Feb 21 2011 (PARI) 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
CROSSREFS
Cf. A038822 (number of primes between 100n and 100n+99), A186311 (first occurrences).
Numbers k such that there are 17 primes between 100*k and 100*k + 99.
+10
22
4, 14, 7837, 27049, 144997771, 651186838, 12779564974, 22369949923, 149621468452, 225012717952, 240728320642, 586832463472, 766964610742, 939742446571, 949543082647, 1908189311558, 2693729868901, 2701032171244, 3465208973035, 3489960850720, 3910908182851
COMMENTS
There are 2829786 possible prime patterns for centuries having 17 primes. - Tim Johannes Ohrtmann, Aug 27 2015
EXAMPLE
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).
CROSSREFS
Cf. A038822 (number of primes between 100n and 100n+99), A186311 (first occurrences).
Numbers k such that there are 18 primes between 100*k and 100*k + 99.
+10
20
1228537713709, 23352869714018, 28703237474266, 144785865481702, 161394923966449, 168975708209638, 174748809066898, 207552241231357, 278215179205531, 312303328909720, 592248982143877, 812939886634531, 939100782752014, 983930290209021, 1111161494544274
COMMENTS
There are A261571(18) = 948729 possible patterns for centuries having 18 primes.
LINKS
Note (Apr 24 2024): An older version of the b-file missed a(33) and a(38). The present b-file is correct.
EXAMPLE
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).
PROG
(PARI) isok(k) = sum(i=0, 99, isprime(100*k + i)) == 18; \\ Michel Marcus, Mar 23 2023
CROSSREFS
Cf. A038822 (number of primes between 100n and 100n+99), A186311 (first occurrences). Cf. A261571 (number of patterns for centuries with n primes).
Centuries whose prime pattern repeats in the next century.
+10
11
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
COMMENTS
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
FORMULA
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
EXAMPLE
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.
EXTENSIONS
a(1)-a(5) computed by J. K. Andersen, May 15 2011
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.
+10
3
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
COMMENTS
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
EXAMPLE
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.
MATHEMATICA
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 *)
PROG
(PARI)
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)}
CROSSREFS
Cf. A038822 (number of primes between 100n and 100n+99).
Centuries whose prime pattern is the same as prime pattern in the previous century.
+10
3
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
COMMENTS
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.
FORMULA
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
Numbers k such that there are 2 primes between 100*k and 100*k + 99.
+10
2
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
COMMENTS
There are 780 possible prime patterns for centuries having 2 primes. - Tim Johannes Ohrtmann, Aug 27 2015
EXAMPLE
3020 is in this sequence because there are 2 primes between 302000 and 302099 (302009 and 302053).
PROG
(Sage)
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 (PARI) for(n=1, 1e6, if(sum(k=100*n, 100*(n+1), ispseudoprime(k))==2, print1(n", "))); \\ Charles R Greathouse IV, Feb 21 2011 (PARI) 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
CROSSREFS
Cf. A038822 (number of primes between 100n and 100n+99), A186311 (first occurrences).
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