Displaying 1-10 of 44 results found.
430646, 491572, 505572, 468318, 664338, 623962, 672132, 650628, 426224, 395410, 749622, 470874, 440004, 225336, 206090, 337014, 358670, 120306, 182388, 152680, 44666, 383260, 503380, 245786, 360250, 336066, 325314, 25308, 53278, -405460, -314318, -789560
COMMENTS
If we analyze the b-files of A076336 and A101036, we can see that the motivation of this sequence is oscillation similar graph of it. Since both sequences ( A076336, A101036) contain the families of congruences, there are repeated values in this sequence. For the first 15000 terms, the most repeated values -376924 and -2843318 appear 28 times in this sequence.
254695787, 265880597, 748135097, 758012237, 785868467, 792874337, 804059147, 806930417, 869860337, 893537627, 896408897, 949461677, 1501696307, 1556312687, 1567497497, 1602359597, 1647098837, 1668160787, 1697536277, 1698843947, 1757639267, 1826055797
COMMENTS
This sequence is an example for the relation between A076336 and A101036. See A271110 for the motivation of "376924" that sequence focuses on.
EXAMPLE
254695787 is a term because A076336(1714) = A101036(1714) + 2*2*17*23*241 = 254318863 + 2*2*17*23*241 = 254695787
332252059, 341679859, 412107289, 479216149, 487082389, 530260069, 557509819, 568694629, 579879439, 586184119, 621300109, 1158170989, 1161489559, 1584950779, 1709545249, 1717411489, 1720730059, 1739781109, 1775092549, 1782958789, 1794143599, 1795705159
COMMENTS
This sequence is an example for the relation between A076336 and A101036. See A271110 for the motivation of "2843318" that sequence focuses on.
EXAMPLE
332252059 is a term because A076336(2208) = A101036(2208) + 2*17*241*347 = 329408741 + 2*17*241*347 = 332252059
a(n) is the smallest cardinality of all covering sets associated with Riesel number A101036(n).
+20
0
6, 6, 7, 7, 6, 7, 6, 6, 6, 6, 7, 6, 6, 7, 7, 6
COMMENTS
The condition for choosing a covering set is necessary as there are Riesel numbers with more than one covering set, see A263392.
EXAMPLE
n | Riesel number | Covering set | a(n)
--------------------------------------------------------
1 | 509203 | {3, 5, 7, 13, 17, 241} | 6
2 | 762701 | {3, 5, 7, 13, 17, 241} | 6
3 | 777149 | {3, 5, 7, 13, 19, 37, 73} | 7
4 | 790841 | {3, 5, 7, 13, 19, 37, 73} | 7
5 | 992077 | {3, 5, 7, 13, 17, 241} | 6
6 | 1106681 | {3, 5, 7, 13, 19, 37, 73} | 7
7 | 1247173 | {3, 5, 7, 13, 17, 241} | 6
8 | 1254341 | {3, 5, 7, 13, 17, 241} | 6
9 | 1330207 | {3, 5, 7, 13, 17, 241} | 6
10 | 1330319 | {3, 5, 7, 13, 17, 241} | 6
11 | 1715053 | {3, 5, 7, 13, 19, 37, 73} | 7
12 | 1730653 | {3, 5, 7, 13, 17, 241} | 6
13 | 1730681 | {3, 5, 7, 13, 17, 241} | 6
14 | 1744117 | {3, 5, 7, 13, 19, 73, 109} | 7
15 | 1830187 | {3, 5, 7, 13, 37, 73, 109} | 7
16 | 1976473 | {3, 5, 7, 13, 17, 241} | 6
(Provable) Sierpiński numbers: odd numbers n such that for all k >= 1 the numbers n*2^k + 1 are composite.
+10
73
78557, 271129, 271577, 322523, 327739, 482719, 575041, 603713, 903983, 934909, 965431, 1259779, 1290677, 1518781, 1624097, 1639459, 1777613, 2131043, 2131099, 2191531, 2510177, 2541601, 2576089, 2931767, 2931991, 3083723, 3098059, 3555593, 3608251
COMMENTS
"Provable" in the definition means provable by any method (whether using a covering set or not). - N. J. A. Sloane, Aug 03 2024 It is only a conjecture that this sequence is complete up to 3000000 - there may be missing terms.
It is conjectured that 78557 is the smallest Sierpiński number. - T. D. Noe, Oct 31 2003 Sierpiński numbers may be proved by exhibiting a periodic sequence p of prime divisors with p(k) | n*2^k+1 and disproved by finding a prime n*2^k+1. It is conjectured by some people that numbers that cannot be proved to be Sierpiński by this method are non-Sierpiński. However, some numbers resist both proof and disproof. - David W. Wilson, Jan 17 2005 [Edited by N. J. A. Sloane, Aug 03 2024] Sierpiński showed that this sequence is infinite.
There are four related sequences that arise in this context:
S1: Numbers n such that n*2^k + 1 is composite for all k (this sequence)
S2: Odd numbers n such that 2^k + n is composite for all k (apparently it is conjectured that S1 and S2 are the same sequence)
S3: Numbers n such that n*2^k + 1 is prime for all k (empty)
S4: Numbers n such that 2^k + n is prime for all k (empty)
The following argument, due to Michael Reid, attempts to show that S3 and S4 are empty: If p is a prime divisor of n + 1, then for k = p - 1, the term (either n*2^k + 1 or 2^k + n) is a multiple of p (and also > p, so not prime). [However, David McAfferty points that for the case S3, this argument fails if p is of the form 2^m-1. So it may only be a conjecture that the set S3 is empty. - N. J. A. Sloane, Jun 27 2021] a(1) = 78557 is also the smallest odd n for which either n^p*2^k + 1 or n^p + 2^k is composite for every k > 0 and every prime p greater than 3. - Arkadiusz Wesolowski, Oct 12 2015 n = 4008735125781478102999926000625 = ( A213353(1))^4 is in this sequence but is thought not to satisfy the conjecture mentioned by David W. Wilson above. For this multiplier, all n*2^(4m + 2) + 1 are composite by an Aurifeuillean factorization. Only the remaining cases, n*2^k + 1 where k is not 2 modulo 4, are covered by a finite set of primes (namely {3, 17, 97, 241, 257, 673}). See Izotov link for details (although with another prime set). - Jeppe Stig Nielsen, Apr 14 2018 Conjecture: if S is a (provable) Sierpiński number, then there exists a prime P such that S^p is also a Sierpiński number for every prime p > P. - Thomas Ordowski, Jul 12 2022 Problem: are there odd numbers K such that K - 2^m is a Sierpiński number for every 1 < 2^m < K? If so, then all positive values of (K - 2^m)*2^n + 1 are composite. Also, by the dual Sierpiński conjecture, K - 2^m + 2^n is composite for every 1 < 2^m < K and for every n > 0. Note that, by the dual Sierpiński conjecture, if p is an odd prime and 1 < 2^m < p, then there exists n such that (p - 2^m)*2^n + 1 is prime. So if such a number K exists, it must be composite. - Thomas Ordowski, Jul 20 2022 1) The above Conjecture is true for Sierpiński numbers provable by a "covering set", with P equal to the largest prime factor of the elements of that set*, according to the explanation from Michael Filaseta posted Jul 12 2022 on the SeqFan mailing list, cf. links. (*More generally: for S^p with any p coprime to all elements of the covering set, but not necessarily prime.)
2) Wilson's comment from 2005 (also the first part, not only the conjecture) is misleading if not wrong because there are provable Sierpiński numbers for which a covering set is not known (maybe even believed not to exist), as explained by Nielsen in his above comment from 2018. (End)
REFERENCES
R. K. Guy, Unsolved Problems in Number Theory, Section B21.
C. A. Pickover, The Math Book, Sterling, NY, 2009; see p. 420.
P. Ribenboim, The Book of Prime Number Records, 2nd. ed., 1989, p. 282.
LINKS
T. D. Noe and Arkadiusz Wesolowski, Table of n, a(n) for n = 1..15000 (T. D. Noe supplied 13394 terms which came from McLean. a(1064), a(7053), and a(13397)-a(15000) from Arkadiusz Wesolowski.) Michael Filaseta, quoted by T. Ordowski, Re: Is it true? SeqFan mailing list, Jul 12 2022 Payam Samidoost, 4847 [Broken link?] Payam Samidoost, 4847 [Cached copy]
Odd numbers that are not of the form p + 2^a + 2^b, a, b > 0, p prime.
+10
51
1, 3, 5, 6495105, 848629545, 1117175145, 2544265305, 3147056235, 3366991695, 3472109835, 3621922845, 3861518805, 4447794915, 4848148485, 5415281745, 5693877405, 6804302445, 7525056375, 7602256605, 9055691835, 9217432215
COMMENTS
Crocker shows that this sequence is infinite.
All members above 5 found so far (up to 2.5 * 10^11) are divisible by 255 = 3 * 5 * 17, and many are divisible by 257. I conjecture that all members of this sequence greater than 5 are divisible by 255. This implies that all odd numbers (greater than 7) are the sum of a prime and at most three positive powers of two.
Pan shows that, for every c > 1, a(n) << x^c. More specifically, there are constants C,D > 0 such that there are at least Dx/exp(C log x log log log log x/log log log x) members of this sequence up to x. - Charles R Greathouse IV, Apr 11 2016 All terms > 5 are numbers k > 3 such that k - 2^n is a de Polignac number ( A006285) for every n > 0 with 2^n < k. Are there numbers K such that |K - 2^n| is a Riesel number ( A101036) for every n > 0? If so, ||K - 2^n| - 2^m| is composite for every pair m,n > 0, by the dual Riesel conjecture. - Thomas Ordowski, Jan 06 2024
EXAMPLE
Prime factorization of terms:
F_0 = 3, F_1 = 5, F_2 = 17, F_3 = 257 are Fermat numbers (cf. A000215) 6495105 = 3 * 5 * 17 * 25471
848629545 = 3 * 5 * 17 * 461 * 7219
1117175145 = 3 * 5 * 17 * 257 * 17047
2544265305 = 3^2 * 5 * 17 * 257 * 12941
3147056235 = 3^2 * 5 * 17 * 257 * 16007
3366991695 = 3 * 5 * 17 * 83 * 257 * 619
3472109835 = 3 * 5 * 17 * 257 * 52981
3621922845 = 3 * 5 * 17^2 * 257 * 3251
3861518805 = 3^3 * 5 * 17 * 257 * 6547
4447794915 = 3^3 * 5 * 17 * 257 * 7541
4848148485 = 3^4 * 5 * 17 * 704161
5415281745 = 3 * 5 * 17 * 21236399
5693877405 = 3^2 * 5 * 17 * 257 * 28961
6804302445 = 3^2 * 5 * 17 * 53 * 257 * 653
7525056375 = 3^2 * 5^3 * 17 * 257 * 1531
7602256605 = 3 * 5 * 17 * 257 * 311 * 373
9055691835 = 3 * 5 * 17 * 257 * 138181
9217432215 = 3^2 * 5 * 17 * 173 * 257 * 271
PROG
(PARI) is(n)=if(n%2==0, return(0)); for(a=1, log(n)\log(2), for(b=1, a, if(isprime(n-2^a-2^b), return(0)))); 1 \\ Charles R Greathouse IV, Nov 27 2013 (Python)
from itertools import count, islice
from sympy import isprime
def A156695_gen(startvalue=1): # generator of terms >= startvalue for n in count(max(startvalue+(startvalue&1^1), 1), 2):
l = n.bit_length()-1
for a in range(l, 0, -1):
c = n-(1<<a)
for b in range(min(a, l-1), 0, -1):
if isprime(c-(1<<b)):
break
else:
continue
break
else:
yield n
Riesel numbers: odd numbers n such that for all k >= 1 the numbers n*2^k - 1 are composite.
+10
32
COMMENTS
509203 has been proved to be a member of the sequence, and is conjectured to be the smallest member. However, as of 2009, there are still several smaller numbers which are candidates and have not yet been ruled out (see links).
Riesel numbers are proved by exhibiting a periodic sequence p of prime divisors with p(k) | n*2^k-1 and disproved by finding prime n*2^k-1. It is conjectured that numbers that cannot be proved Riesel in this way are non-Riesel. However, some numbers resist both proof and disproof.
Others conjecture the opposite: that there are infinitely many Riesel numbers that do not arise from a covering system, see A101036. The word "odd" is needed in the definition because otherwise for any term n, all numbers n*2^m, m >= 1, would also be Riesel numbers, but we don't want them in this sequence (as is manifest from A101036). Since 1 and 3 obviously are not in this sequence, for any n in this sequence n-1 is an even number > 2 and therefore composite, so one could replace "k >= 1" equivalently by "k >= 0". - M. F. Hasler, Aug 20 2020 Named after the Swedish mathematician Hans Ivar Riesel (1929-2014). - Amiram Eldar, Apr 02 2022
REFERENCES
R. K. Guy, Unsolved Problems in Number Theory, Section B21.
Paulo Ribenboim, The Book of Prime Number Records, 2nd ed., 1989, p. 282.
LINKS
Hans Riesel, Some large prime numbers. Translated from the Swedish original (Några stora primtal, Elementa 39 (1956), pp. 258-260) by Lars Blomberg.
KEYWORD
nonn,bref,hard,more,changed
EXTENSIONS
Normally we require at least four terms but we will make an exception for this sequence in view of its importance. - N. J. A. Sloane, Nov 07 2002. See A101036 for the most likely extension. Definition corrected ("odd" added) by M. F. Hasler, Aug 23 2020
Riesel problem: start with n; repeatedly double and add 1 until reaching a prime. Sequence gives number of steps to reach a prime or 0 if no prime is ever reached.
+10
29
1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 3, 4, 1, 1, 2, 2, 1, 2, 1, 1, 4, 1, 3, 2, 1, 3, 4, 1, 1, 2, 2, 1, 2, 1, 1, 2, 3, 1, 2, 1, 7, 24, 1, 3, 4, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 3, 12, 2, 3, 4, 2, 1, 4, 1, 5, 2, 1, 1, 2, 4, 7, 2552, 1, 1, 2, 2, 1, 4, 3, 1, 2, 1, 5, 6, 1, 23, 4, 1, 1, 2, 3, 3, 2, 1, 1, 4, 1, 1
COMMENTS
a(n) is the smallest m >= 1 such that (n+1)*2^m - 1 is prime (or 0 if no such prime exists).
It is conjectured that n = 509203 is the smallest Riesel number, i.e., n*2^k - 1 is composite for every k>0. - Robert G. Wilson v, Mar 01 2015. [This would imply that a(509203) is the first zero term in this sequence. - N. J. A. Sloane, Jul 31 2024] Both the Ballinger-Keller and Prime Wiki links assert that 104917*2^340181-1 is prime, but leave open the possibility that there is an m < 340181 which makes 104917*2^m - 1 a prime.
This question was finally settled by Lucas A. Brown on Jul 31 2024, who showed that m = 340181 is the smallest value that gives a prime. This implies that a(104917) = 340181. Brown used a Python program (see below), with BPSW for the primality testing and gmpy2 to handle the arithmetic. The program was started on Jul 30 2024 and finished on Jul 31 2024.
He reports that it took about 15 hours in wall-clock time, and used 24 threads running in parallel. (End)
LINKS
Hans Riesel, Some large prime numbers. Translated from the Swedish original (Några stora primtal, Elementa 39 (1956), pp. 258-260) by Lars Blomberg.
FORMULA
If a(n) = 0, then a(2n+1) is also 0. Conjecture: If a(n) = 1, then a(2n+1) is not 0. - Jeppe Stig Nielsen, Feb 12 2023
EXAMPLE
For n=4; the smallest m>=1 such that (4+1)*2^m-1 is prime is m=2: 5*2^2-1=19 (prime). - Jaroslav Krizek, Feb 13 2011
MAPLE
local twox1, k ;
twox1 := 2*n+1 ;
k := 1;
while not isprime(twox1) do
twox1 := 2*twox1+1 ;
k := k+1 ;
end do:
return k;
MATHEMATICA
a[n_] := Block[{s=n, c=1}, While[ ! PrimeQ[2*s+1], s = 2*s+1; c++]; c]; Table[ a[n], {n, 1, 99} ] (* Jean-François Alcover, Feb 06 2012, after Pari *) a[n_] := Block[{k = 1}, While[ !PrimeQ[2^k (n + 1) - 1], k++]; k]; Array[a, 100] (* Robert G. Wilson v, Feb 14 2015 *) (* Corrected by Paolo Xausa, Jul 30 2024 *)
PROG
(PARI) a(n)=if(n<0, 0, s=n; c=1; while(isprime(2*s+1)==0, s=2*s+1; c++); c)
(Python, designed specifically for n = 104917)
#! /usr/bin/env python3
from labmath import primegen, isprime, mpz, count
from multiprocessing import Pool
primes = list(primegen(1000000))
def test(n):
for p in primes:
if (104917 * pow(2, n, p)) % p == 1:
return (n, False)
return (n, isprime(104917 * mpz(2)**n - 1, tb=[]))
with Pool(24) as P:
for (n, result) in P.imap(test, count()):
print('\b'*80, n, end='', flush=True)
if result:
Riesel problem: a(n) = smallest m >= 0 such that n*2^m-1 is prime, or -1 if no such prime exists.
+10
25
2, 1, 0, 0, 2, 0, 1, 0, 1, 1, 2, 0, 3, 0, 1, 1, 2, 0, 1, 0, 1, 1, 4, 0, 3, 2, 1, 3, 4, 0, 1, 0, 2, 1, 2, 1, 1, 0, 3, 1, 2, 0, 7, 0, 1, 3, 4, 0, 1, 2, 1, 1, 2, 0, 1, 2, 1, 3, 12, 0, 3, 0, 2, 1, 4, 1, 5, 0, 1, 1, 2, 0, 7, 0, 1, 1, 2, 2, 1, 0, 3, 1, 2, 0, 5, 6, 1, 23, 4, 0, 1, 2, 3, 3, 2, 1, 1, 0, 1, 1, 10, 0, 3
LINKS
Hans Riesel, Some large prime numbers. Translated from the Swedish original (Några stora primtal, Elementa 39 (1956), pp. 258-260) by Lars Blomberg.
PROG
(Haskell)
a040081 = length . takeWhile ((== 0) . a010051) .
iterate ((+ 1) . (* 2)) . (subtract 1)
(PARI) a(n)=for(k=0, 2^16, if(ispseudoprime(n*2^k-1), return(k))) \\ Eric Chen, Jun 01 2015 (Python)
from sympy import isprime
def a(n):
m = 0
while not isprime(n*2**m - 1): m += 1
return m
Riesel problem: start with n; repeatedly double and add 1 until reach a prime. Sequence gives a(n) = prime reached, or 0 if no prime is ever reached.
+10
23
3, 5, 7, 19, 11, 13, 31, 17, 19, 43, 23, 103, 223, 29, 31, 67, 71, 37, 79, 41, 43, 367, 47, 199, 103, 53, 223, 463, 59, 61, 127, 131, 67, 139, 71, 73, 151, 311, 79, 163, 83, 5503, 738197503, 89, 367, 751, 191, 97, 199, 101, 103, 211, 107, 109, 223, 113, 463
COMMENTS
Equivalently, a(n) = smallest prime of form (n+1)*2^k-1 for k >= 1, or 0 if no such prime exists.
a(509202) = 0 (i.e. never reaches a prime) - Chris Nash (chris_nash(AT)hotmail.com). (Of course this is related to the entry 509203 of A076337.) a(73) is a 771-digit prime reached after 2552 iterations - Warut Roonguthai. This was proved to be a prime by Paul Jobling (Paul.Jobling(AT)WhiteCross.com) using PrimeForm and by Ignacio Larrosa Cañestro using Titanix (http://www.znz.freesurf.fr/pages/titanix.html). [Oct 30 2000]
LINKS
Hans Riesel, Some large prime numbers. Translated from the Swedish original (Några stora primtal, Elementa 39 (1956), pp. 258-260) by Lars Blomberg.
EXAMPLE
a(4)=19 because 4 -> 9 (composite) -> 19 (prime).
MATHEMATICA
Table[NestWhile[2#+1&, 2n+1, !PrimeQ[#]&], {n, 60}] (* Harvey P. Dale, May 08 2011 *) (* Will run for ever if a(n) = 0. - N. J. A. Sloane, Jul 29 2024 *)
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