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If k is in the sequence and m=3^(k-1) then m is a term of A033632 (phi(sigma(m)) = sigma(phi(m))), so 3^(A028491-1) is a subsequence of A033632. For example since 9551 is in the sequence, phi(sigma(3^9550)) = sigma(phi(3^9550)). - Farideh Firoozbakht, Feb 09 2005

Salas lists these, except 3, in "Open Problems" p. 6 [March 2012], and proves that the Cantor primes > 3 are exactly the prime-valued cyclotomic polynomials of the form Phi_s(3^{s^j}) == 1 (mod 4).

Also, k such that 3^k-1 is a semiprime - see also A080892. - M. F. Hasler, Mar 19 2013

From Markus Voege, Nov 24 2009: (Start)

On the difference between this sequence and A038147:

The first term that differs is for n=6; for all subsequent terms, the number of polyhexes is larger than the number of planar polyhexes.

If I recall correctly, polyhexes are clusters of regular hexagons that are joined at the edges and are LOCALLY embeddable in the hexagonal lattice.

"Planar polyhexes" are polyhexes that are GLOBALLY embeddable in the honeycomb lattice.

Example: (Planar) polyhex with 6 cells (x) and a hole (O):

.. x x

. x O x

.. x x

Polyhex with 6 cells that is cut open (I):

.. xIx

. x O x

.. x x

This polyhex is not globally embeddable in the honeycomb lattice, since adjacent cells of the lattice must be joined. But it can be embedded locally everywhere. It is a start of a spiral. For n>6 the spiral can be continued so that the cells overlap.

Illegal configuration with cut (I):

.. xIx

. x x x

.. x x

This configuration is NOT a polyhex since the vertex at

.. xIx

... x

is not embeddable in the honeycomb lattice.

One has to keep in mind that these definitions are inspired by chemistry. Hence, potential molecules are often the motivation for these definitions. Think of benzene rings that are fused at a C-C bond.

The (planar) polyhexes are "free" configurations, in contrast to "fixed" configurations as in A001207 = Number of fixed hexagonal polyominoes with n cells.

A000228 (planar polyhexes) and A001207 (fixed hexagonal polyominoes) differ only by the attribute "free" vs. "fixed," that is, whether the different orientations and reflections of an embedding in the lattice are counted.

The configuration

. x x .... x

.. x .... x x

is counted once as free and twice as fixed configurations.

Since most configurations have no symmetry, (A001207 / A000228) -> 12 for n -> infinity. (End)

Here two polycubes that differ by reflection are considered different. - Joerg Arndt, Apr 26 2023

Number of oriented polyominoes with n cubical cells of the regular tiling with Schläfli symbol {4,3,4}. For oriented polyominoes, chiral pairs are counted as two. - Robert A. Russell, Mar 21 2024

Probably finite.

The persistence of a number (A031346) is the number of times you need to multiply the digits together before reaching a single digit.

From David A. Corneth, Sep 23 2016: (Start)

For n > 1, the digit 0 doesn't occur. Therefore the digit 1 doesn't occur and all terms have digits in nondecreasing order.

a(n) consists of at most one three and at most one two but not both. If they contain both, they could be replaced with a single digit 6 giving a lesser number. Two threes can be replaced with a 9. Similarily, there's at most one four and one six but not both. Two sixes can be replaced with 49. A four and a six can be replaced with a three and an eight. For n > 2, an even number and a five don't occur together.

Summarizing, a term a(n) for n > 2 consists of 7's, 8's and 9's with a prefix of one of the following sets of digits: {{}, {2}, {3}, {4}, {6}, {2,6}, {3,5}, {5, 5,...}} [Amended by Kohei Sakai, May 27 2017]

No more up to 10^200. (End)

From Benjamin Chaffin, Sep 29 2016: (Start)

Let p(n) be the product of the digits of n, and P(n) be the multiplicative persistence of n. Any p(n) > 1 must have only prime factors from one of the two sets {2,3,7} or {3,5,7}. The following are true of all p(n) < 10^20000:

The largest p(n) with P(p(n))=10 is 2^4 * 3^20 * 7^5. The only other such p(n) known is p(a(11))=2^19 * 3^4 * 7^6.

The largest p(n) with P(p(n))=9 is 2^33 * 3^3 (12 digits).

The largest p(n) with P(p(n))=8 is 2^9 * 3^5 * 7^8 (12 digits).

The largest p(n) with P(p(n))=7 is 2^24 * 3^18 (16 digits).

The largest p(n) with P(p(n))=6 is 2^24 * 3^6 * 7^6 (16 digits).

The largest p(n) with P(p(n))=5 is 2^35 * 3^2 * 7^6 (17 digits).

The largest p(n) with P(p(n))=4 is 2^59 * 3^5 * 7^2 (22 digits).

The largest p(n) with P(p(n))=3 is 2^4 * 3^17 * 7^38 (42 digits).

The largest p(n) with P(p(n))=2 is 2^25 * 3^227 * 7^28 (140 digits).

All p(n) between 10^140 and 10^20000 have a persistence of 1, meaning they contain a 0 digit. (End)

Benjamin Chaffin's comments imply that there are no more terms up to 10^20585. For every number N between 10^200 with 10^20585 with persistence greater than 1, the product of the digits of N is between 10^140 and 10^20000, and each of these products has a persistence of 1. - David Radcliffe, Mar 22 2019

From A.H.M. Smeets, Nov 16 2018: (Start)

Let p_10(n) be the product of the digits of n in base 10. We can define an equivalence relation DP_10 on n by n DP_10 m if and only if p_10(n) = p_10(m); the name DP_b for the equivalence relation stands for "digits product for representation in base b". A number n is called the class representative number of class n/DP_10 if and only if p_10(n) = p_10(m), m >= n; i.e., if it is the smallest number of that class; it is also called the reduced number.

For any multiplicative persistence, except multiplicative persistence 2, the set of class representative numbers with that multiplicative persistence is conjectured to be finite.

Each class representative number represents an infinite set of numbers with the same multiplicative persistence.

For multiplicative persistence 2, only the set of class representative numbers which end in the digit zero is infinite. The table of numbers of class representative numbers of different multiplicative persistence (mp) is given by:

final digit

mp total 0 1 2 3 4 5 6 7 8 9

====================================================

0 10 1 1 1 1 1 1 1 1 1 1

1 10 1 1 1 1 1 1 1 1 1 1

2 inf inf 0 4 0 1 1 5 0 7 0

3 12199 12161 0 8 0 3 3 8 0 16 0

4 408 342 0 14 0 5 4 19 0 24 0

5 151 88 0 9 0 1 3 37 0 13 0

6 41 24 0 1 0 0 0 14 0 2 0

7 13 9 0 0 0 0 0 4 0 0 0

8 8 7 0 0 0 0 0 1 0 0 0

9 5 5 0 0 0 0 0 0 0 0 0

10 2 2 0 0 0 0 0 0 0 0 0

11 2 2 0 0 0 0 0 0 0 0 0

It is observed from this that for the reduced numbers with multiplicative persistence 1, the primes 11, 13, 17 and 19, will not occur in any trajectory of another (larger) number; i.e., all numbers represented by the reduced numbers 11, 13, 17 and 19 have a prime factor of at least 11 (conjectured from the observations).

Example for numbers represented by the reduced number 19: 91 = 7*13, 133 = 7*19, 313 is prime, 331 is prime, 119 = 7*17, 191 is prime, 911 is prime, 1133 = 11*103, 1313 = 13*101, 1331 = 11^3, 3113 = 11*283, 3131 = 31*101 and 3311 = 7*11*43.

In fact all trajectories can be projected to a trajectory in one of the ten trees with reduced numbers with roots 0..9, and the numbers represented by the reduced number of each leaf have a prime factor of at least 11 (as conjectured from the observations).

Example of the trajectory of 277777788888899 (see A121111) in the tree of reduced numbers (the unreduced numbers are given between brackets): 277777788888899 -> 3778888999 (4996238671872) -> 26888999 (438939648) -> 2677889 (4478976) -> 68889 (338688) -> 6788 (27648) -> 2688 (2688) -> 678 (768) -> 69 (336) -> 45 (54) -> 10 (20) -> 0. (End)

From Tim Peters, Sep 19 2023: (Start)

New lower bound: if a(12) exists, it must be > 2.67*10^30000. It continues to be the case that the digit products for all candidates with at least 20000 digits (roughly where the last long run reported here stopped) contain a zero digit, so the candidates all have persistence 2. More, the digit products all contain at least one zero in their last 306 digits. An extreme is the digit product 2^13802 * 3^16807 * 7^1757. That has 13659 decimal digits, 1335 of which are zeros. It ends with a zero followed by 305 nonzero digits. So to confirm that the large candidates with no more than 30000 digits have persistence 2, it would suffice to compute digit products modulo 10^306.

Note: by "candidate" I mean a digit string matching one of these eight (pairwise disjoint) simple regular expressions. Each such string gives the smallest integer with its digit product (and viewing the empty string as having digit product 1), and their union covers all digit products that don't end with a zero.

7* 8* 9*

2 7* 8* 9*

3 7* 8* 9*

4 7* 8* 9*

5 5* 7* 9*

6 7* 8* 9*

26 7* 8* 9*

35 5* 7* 9*

There are (8*N^2 + 13*N + 6)*(N + 1)/6 such strings with no more than N digits. A long computer run checked N=30000, a bit over 36*10^12 candidates. The smallest candidate with more than 30000 digits is > 2.67*10^30000, which is the smallest remaining possibility for a(12). (End)

"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

From M. F. Hasler, Jul 23 2022: (Start)

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)

If holes are not allowed, we get A070765. - Joseph Myers, Apr 20 2009

It is a consequence of Madras's 1999 pattern theorem that almost all polyiamonds have holes, i.e., lim_{n->oo} A070765(n)/A000577(n) = 0. - Johann Peters, Jan 06 2024

a(n) is the smallest number m such that pi(m)=d(m)^n, where d(m) is number of positive divisors of m (the proof is easy). - Farideh Firoozbakht, Jun 06 2005

a(6) is often quoted as 102981500, but this is incorrect.

All terms are primes.

No other terms less than 10^5. - Robert Price, May 28 2012

No other terms less than 10^6. - Jon Grantham, Jul 29 2023