cp's OEIS Frontend

Same as Pisot sequence L(2,3).

Length of the continued fraction for Sum_{k=0..n} 1/3^(2^k). - Benoit Cloitre, Nov 12 2003

See also A004119 for a(n) = 2a(n-1)-1 with first term = 1. - Philippe Deléham, Feb 20 2004

From the second term on (n>=1), in base 2, these numbers present the pattern 1000...0001 (with n-1 zeros), which is the "opposite" of the binary 2^n-2: (0)111...1110 (cf. A000918). - Alexandre Wajnberg, May 31 2005

Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=5, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n-1)=(-1)^(n-1)* charpoly(A,3). - Milan Janjic, Jan 27 2010

First differences of A006127. - Reinhard Zumkeller, Apr 14 2011

The odd prime numbers in this sequence form A019434, the Fermat primes. - David W. Wilson, Nov 16 2011

Pisano period lengths: 1, 1, 2, 1, 4, 2, 3, 1, 6, 4, 10, 2, 12, 3, 4, 1, 8, 6, 18, 4, ... . - R. J. Mathar, Aug 10 2012

Is the mentioned Pisano period lengths (see above) the same as A007733? - Omar E. Pol, Aug 10 2012

Only positive integers that are not 1 mod (2k+1) for any k>1. - Jon Perry, Oct 16 2012

For n >= 1, a(n) is the total length of the segments of the Hilbert curve after n iterations. - Kival Ngaokrajang, Mar 30 2014

Frénicle de Bessy (1657) proved that a(3) = 9 is the only square in this sequence. - Charles R Greathouse IV, May 13 2014

a(n) is the number of distinct possible sums made with at most two elements in {1,...,a(n-1)} for n > 0. - Derek Orr, Dec 13 2014

For n > 0, given any set of a(n) lattice points in R^n, there exist 2 distinct members in this set whose midpoint is also a lattice point. - Melvin Peralta, Jan 28 2017

Also the number of independent vertex sets, irredundant sets, and vertex covers in the (n+1)-star graph. - Eric W. Weisstein, Aug 04 and Sep 21 2017

Also the number of maximum matchings in the 2(n-1)-crossed prism graph. - Eric W. Weisstein, Dec 31 2017

Conjecture: For any integer n >= 0, a(n) is the permanent of the (n+1) X (n+1) matrix with M(j, k) = -floor((j - k - 1)/(n + 1)). This conjecture is inspired by the conjecture of Zhi-Wei Sun in A036968. - Peter Luschny, Sep 07 2021

Although this is a list, and lists normally have offset 1, it seems better to make an exception in this case. - N. J. A. Sloane, Mar 13 2010

The subsequence 0,1,2,3,4 gives the known values of n such that 2^(2^n)+1 is a prime (see A019434, the Fermat primes). - N. J. A. Sloane, Jun 16 2010

Also: The identity map, defined on the set of nonnegative integers. The restriction to the positive integers yields the sequence A000027. - M. F. Hasler, Nov 20 2013

The number of partitions of 2n into exactly 2 parts. - Colin Barker, Mar 22 2015

The number of orbits of Aut(Z^7) as function of the infinity norm n of the representative lattice point of the orbit, when the cardinality of the orbit is equal to 8960 or 168.- Philippe A.J.G. Chevalier, Dec 29 2015

Partial sums give A000217. - Omar E. Pol, Jul 26 2018

First differences are A000012 (the "all 1's" sequence). - M. F. Hasler, May 30 2020

See A061579 for the transposed infinite square matrix, or triangle with rows reversed. - M. F. Hasler, Nov 09 2021

This is the unique sequence (a(n)) that satisfies the inequality a(n+1) > a(a(n)) for all n in N. This simple and surprising result comes from the 6th problem proposed by Bulgaria during the second day of the 19th IMO (1977) in Belgrade (see link and reference). - Bernard Schott, Jan 25 2023

It is conjectured that just the first 5 numbers in this sequence are primes.

An infinite coprime sequence defined by recursion. - Michael Somos, Mar 14 2004

For n>0, Fermat numbers F(n) have digital roots 5 or 8 depending on whether n is even or odd (Koshy). - Lekraj Beedassy, Mar 17 2005

This is the special case k=2 of sequences with exact mutual k-residues. In general, a(1)=k+1 and a(n)=min{m | m>a(n-1), mod(m,a(i))=k, i=1,...,n-1}. k=1 gives Sylvester's sequence A000058. - Seppo Mustonen, Sep 04 2005

For n>1 final two digits of a(n) are periodically repeated with period 4: {17, 57, 37, 97}. - Alexander Adamchuk, Apr 07 2007

For 1 < k <= 2^n, a(A007814(k-1)) divides a(n) + 2^k. More generally, for any number k, let r = k mod 2^n and suppose r != 1, then a(A007814(r-1)) divides a(n) + 2^k. - T. D. Noe, Jul 12 2007

From Daniel Forgues, Jun 20 2011: (Start)

The Fermat numbers F_n are F_n(a,b) = a^(2^n) + b^(2^n) with a = 2 and b = 1.

For n >= 2, all factors of F_n = 2^(2^n) + 1 are of the form k*(2^(n+2)) + 1 (k >= 1).

The products of distinct Fermat numbers (in their binary representation, see A080176) give rows of Sierpiński's triangle (A006943). (End)

Let F(n) be a Fermat number. For n > 2, F(n) is prime if and only if 5^((F(n)-1)/4) == sqrt(F(n)-1) (mod F(n)). - Arkadiusz Wesolowski, Jul 16 2011

Conjecture: let the smallest prime factor of Fermat number F(n) be P(F(n)). If F(n) is composite, then P(F(n)) < 3*2^(2^n/2 - n - 2). - Arkadiusz Wesolowski, Aug 10 2012

The Fermat primes are not Brazilian numbers, so they belong to A220627, but the Fermat composites are Brazilian numbers so they belong to A220571. For a proof, see Proposition 3 page 36 on "Les nombres brésiliens" in Links. - Bernard Schott, Dec 29 2012

It appears that this sequence is generated by starting with a(0)=3 and following the rule "Write in binary and read in base 4". For an example of "Write in binary and read in ternary", see A014118. - John W. Layman, Jul 30 2013

Conjecture: the numbers > 5 in this sequence, i.e., 2^2^k + 1 for k>1, are exactly the numbers n such that (n-1)^4-1 divides 2^(n-1)-1. - M. F. Hasler, Jul 24 2015

It is conjectured that this sequence is infinite, but this has never been proved.

An equivalent description: primes of form P = (p1*p2*...*pm)^k + 1 where p1..pm are primes and k > 1, since then k must be even for P to be prime.

Also prime = p(n) if A054269(n) = 1, i.e., quotient-cycle-length = 1 in continued fraction expansion of sqrt(p). - Labos Elemer, Feb 21 2001

Also primes p such that phi(p) is a square.

Also primes of form x*y + z, where x, y and z are three successive numbers. - Giovanni Teofilatto, Jun 05 2004

It is a result that goes back to Mirsky that the set of primes p for which p-1 is squarefree has density A, where A = A005596 denotes the Artin constant. More precisely, Sum_{p <= x} mu(p-1)^2 = A*x/log x + o(x/log x) as x tends to infinity. Conjecture: Sum_{p <= x, mu(p-1)=1} 1 = (A/2)*x/log x + o(x/log x) and Sum_{p <= x, mu(p-1)=-1} 1 = (A/2)*x/log x + o(x/log x). - Pieter Moree (moree(AT)mpim-bonn.mpg.de), Nov 03 2003

Also primes of the form x^y + 1, where x > 0, y > 1. Primes of the form x^y - 1 (x > 0, y > 1) are the Mersenne primes listed in A000668(n) = {3, 7, 31, 127, 8191, 131071, 524287, 2147483647, ...}. - Alexander Adamchuk, Mar 04 2007

With the exception of the first two terms {2,5}, the continued fraction (1 + sqrt(p))/2 has period 3. - Artur Jasinski, Feb 03 2010

With the exception of the first term {2}, congruent to 1 (mod 4). - Artur Jasinski, Mar 22 2011

With the exception of the first two terms, congruent to 1 or 17 (mod 20). - Robert Israel, Oct 14 2014

From Bernard Schott, Mar 22 2019: (Start)

These primes are the primitive terms which generate the sequence of integers with only one prime factor and whose Euler's totient is a square: A054755. So this sequence is a subsequence of A054755 and of A039770. Additionally, the terms of this sequence also have a square cototient, so this sequence is a subsequence of A063752 and A054754.

If p prime = n^2 + 1, phi(p) = n^2 and cototient(p) = 1^2.

Except for 3, the four Fermat primes in A019434 {5, 17, 257, 65537}, belong to this sequence; with F_k = 2^(2^k) + 1, phi(F_k) = (2^(2^(k-1)))^2.

See the file "Subfamilies and subsequences" (& I) in A039770 for more details, proofs with data, comments, formulas and examples. (End)

In this sequence, primes ending with 7 seem to appear twice as often as primes ending with 1. This is because those with 7 come from integers ending with 4 or 6, while those with 1 come only from integers ending with 0 (see De Koninck & Mercier reference). - Bernard Schott, Nov 29 2020

The set of odd primes p for which every elliptic curve of the form y^2 = x^3 + d*x has order p-1 over GF(p) for those d with (d,p)=1 and d a fourth power modulo p. - Gary Walsh, Sep 01 2021 [edited, Gary Walsh, Apr 26 2025]

Also the rows of both triangles A237270 and A237593 interleaved.

Also, irregular triangle read by rows in which T(n,k) is the area of the k-th region (from left to right in ascending diagonal) of the n-th symmetric set of regions (from the top to the bottom in descending diagonal) in the two-dimensional diagram of the perspective view of the infinite stepped pyramid described in A245092 (see the diagram in the Links section).

The diagram of the symmetric representation of sigma is also the top view of the pyramid, see Links section. For more information about the diagram see also A237593 and A237270.

The number of cubes at the n-th level is also A024916(n), the sum of all divisors of all positive integers <= n.

Note that this pyramid is also a quarter of the pyramid described in A244050. Both pyramids have infinitely many levels.

Odd-indexed rows are also the rows of the irregular triangle A237270.

Even-indexed rows are also the rows of the triangle A237593.

Lengths of the odd-indexed rows are in A237271.

Lengths of the even-indexed rows give 2*A003056.

Row sums of the odd-indexed rows gives A000203, the sum of divisors function.

Row sums of the even-indexed rows give the positive even numbers (see A005843).

Row sums give A245092.

From the front view of the stepped pyramid emerges a geometric pattern which is related to A001227, the number of odd divisors of the positive integers.

The connection with the odd divisors of the positive integers is as follows: A261697 --> A261699 --> A237048 --> A235791 --> A237591 --> A237593 --> A237270 --> this sequence.

The sequence is a Lucas sequence V(P,Q) with P = 5 and Q = 4, so if n is a prime number, then V_n(5,4) - 5 is divisible by n. The smallest pseudoprime q which divides V_q(5,4) - 5 is 15.

Also the edge cover number of the (n+1)-Sierpinski tetrahedron graph. - Eric W. Weisstein, Sep 20 2017

First bisection of A000051, A049332, A052531 and A014551. - Klaus Purath, Sep 23 2020

The only primes of binary weight 2 are the Fermat primes (only five are known: 11, 101, 10001, 100000001, 10000000000000001); the repunits base 2 are the Mersenne primes. - Daniel Forgues, Nov 07 2011

a(n) is the rank of prime(n) in the base-2 dominance order on the natural numbers. - Tom Edgar, Mar 25 2014

From Antti Karttunen, Apr 07 2020: (Start)

Also the least number of iterations of nondeterministic map k -> k-(k/p) needed to reach a power of 2, when any prime factor p of k can be used. The minimal length path to the nearest power of 2 (= 2^A064415(n)) is realized whenever one uses any of the A005087(k) distinct odd prime factors of the current k, at any step of the process. For example, this could be done by iterating with the map k -> k-(k/A078701(k)), i.e., by using the least odd prime factor of k (instead of the largest prime).

Proof: Viewing the prime factorization of changing k as a multiset ("bag") of primes, we see that liquefying any odd prime p with step p -> (p-1) brings at least one more 2 to the bag, while applying p -> (p-1) to any 2 just removes it from the bag, but gives nothing back. Thus the largest (and thus also the nearest) power of 2 is reached by eliminating - step by step - all odd primes from the bag, but none of 2's, and it doesn't matter in which order this is done.

The above implies also that the sequence is totally additive, which also follows because both A064097 and A064415 are. That A064097(n) = A329697(n) + A054725(n) for all n > 1 can be also seen by comparing the initial conditions and the recursion formulas of these three sequences.

For any n, A333787(n) is either the nearest power of 2 reached (= 2^A064415(n)), or occurs on some of the paths from n to there.

(End)

A003401 gives the numbers k where a(k) = A005087(k). See also A336477. - Antti Karttunen, Mar 16 2021

The definition is given by Guy: a prime p is in class 1- if the only prime divisors of p - 1 are 2 or 3; and p is in class r- if every prime factor of p - 1 is in some class <= r- - 1, with equality for at least one prime factor. - N. J. A. Sloane, Sep 22 2012

See A005105 for the definition of class r+ primes.

Gleason, p. 191: a regular polygon of n sides can be constructed by ruler, compass and angle-trisector iff n = 2^r * 3^s * p_1 * p_2 * ... * p_k, where p_1, p_2, ..., p_k are distinct elements of this sequence and > 3.

Sequence gives primes solutions to p == +1 (mod phi(p-1)). - Benoit Cloitre, Feb 22 2002

These are the primes p for which p-1 is 3-smooth. Primes for which either p+1 or p-1 have many small factors are more easily proved prime, so most of the largest primes found have this property. - Michael B. Porter, Feb 19 2013

For terms p > 3, omega(p-1) = 3 - p mod 3. Consider terms > 3. Clearly, t > 0. If p == 1 mod 3, u > 0: hence omega(p-1) = 2 because p-1 has two prime factors. If p == 2 mod 3, u = 0: hence omega(p-1) = 1 because p-1 is a power of 2. The latter case corresponds to terms that are Fermat primes > 3. Similar arguments demonstrate the converse, that for p > 3, if omega(p-1) = 3 - p mod 3, p is a term. - Chris Boyd, Mar 22 2014

The subset of A055600 which are prime. - Robert G. Wilson v, Jul 19 2014

Named after the American mathematician James Pierpont (1866-1938). - Amiram Eldar, Jun 09 2021