cp's OEIS Frontend

Twice the odd numbers, also called singly even numbers.

Numbers having equal numbers of odd and even divisors: A001227(a(n)) = A000005(2*a(n)). - Reinhard Zumkeller, Dec 28 2003

Continued fraction for coth(1/2) = (e+1)/(e-1). The continued fraction for tanh(1/2) = (e-1)/(e+1) would be a(0) = 0, a(n) = A016825(n-1), n >= 1.

No solutions to a(n) = b^2 - c^2. - Henry Bottomley, Jan 13 2001

Sequence gives m such that 8 is the largest power of 2 dividing A003629(k)^m-1 for any k. - Benoit Cloitre, Apr 05 2002

k such that Sum_{d|k} (-1)^d = A048272(k) = 0. - Benoit Cloitre, Apr 15 2002

Also k such that Sum_{d|k} phi(d)*mu(k/d) = A007431(k) = 0. - Benoit Cloitre, Apr 15 2002

Also k such that Sum_{d|k} (d/A000005(d))*mu(k/d) = 0, k such that Sum_{d|k}(A000005(d)/d)*mu(k/d) = 0. - Benoit Cloitre, Apr 19 2002

Solutions to phi(x) = phi(x/2); primorial numbers are here. - Labos Elemer, Dec 16 2002

Together with 1, numbers that are not the leg of a primitive Pythagorean triangle. - Lekraj Beedassy, Nov 25 2003

For n > 0: complement of A107750 and A023416(a(n)-1) = A023416(a(n)) <> A023416(a(n)+1). - Reinhard Zumkeller, May 23 2005

Also the minimal value of Sum_{i=1..n+2} (p(i) - p(i+1))^2, where p(n+3) = p(1), as p ranges over all permutations of {1,2,...,n+2} (see the Mihai reference). Example: a(2)=10 because the values of the sum for the permutations of {1,2,3,4} are 10 (8 times), 12 (8 times) and 18 (8 times). - Emeric Deutsch, Jul 30 2005

Except for a(n)=2, numbers having 4 as an anti-divisor. - Alexandre Wajnberg, Oct 02 2005

A139391(a(n)) = A006370(a(n)) = A005408(n). - Reinhard Zumkeller, Apr 17 2008

Also a(n) = (n-1) + n + (n+1) + (n+2), so a(n) and -a(n) are all the integers that are sums of four consecutive integers. - Rick L. Shepherd, Mar 21 2009

The denominator in Pi/8 = 1/2 - 1/6 + 1/10 - 1/14 + 1/18 - 1/22 + .... - Mohammad K. Azarian, Oct 13 2011

This sequence gives the positive zeros of i^x + 1 = 0, x real, where i^x = exp(i*x*Pi/2). - Ilya Gutkovskiy, Aug 08 2015

Numbers k such that Sum_{j=1..k} j^3 is not a multiple of k. - Chai Wah Wu, Aug 23 2017

Numbers k such that Lucas(k) is a multiple of 3. - Bruno Berselli, Oct 17 2017

Also numbers k such that t^k == -1 (mod 5), where t is a term of A047221. - Bruno Berselli, Dec 28 2017

The even numbers form a ring, and these are the primes in that ring. Note that unique factorization into primes does not hold, since 60 = 2*30 = 6*10. - N. J. A. Sloane, Nov 11 2019

Also numbers ending with 10 in base 2. - John Keith, May 09 2022

Same as A001132 except for initial term.

Primes p such that x^2 = 2 has a solution mod p.

The primes of the form x^2 + 2xy - y^2 coincide with this sequence. These are also primes of the form u^2 - 2v^2. - Tito Piezas III, Dec 28 2008

Therefore these are composite in Z[sqrt(2)], as they can be factored as (u^2 - 2v^2)*(u^2 + 2v^2). - Alonso del Arte, Oct 03 2012

After a(1) = 2, these are the primes p such that p^4 == 1 (mod 96). - Gary Detlefs, Jan 22 2014

Also primes of the form 2v^2 - u^2. For example, 23 = 2*4^2 - 3^2. - Jerzy R Borysowicz, Oct 27 2015

Prime factors of A008865 and A028884. - Klaus Purath, Dec 07 2020

Apart from initial term(s), dimension of the space of weight 2n cuspidal newforms for Gamma_0(65).

n such that 16 is the largest power of 2 dividing A003629(k)^n - 1 for any k. - Benoit Cloitre, Mar 23 2002

Continued fraction expansion of tanh(1/4). - Benoit Cloitre, Dec 17 2002

Consider all primitive Pythagorean triples (a,b,c) with c - a = 8, sequence gives values for b. (Corresponding values for a are A078371(n), while c follows A078370(n).) - Lambert Klasen (Lambert.Klasen(AT)gmx.net), Nov 19 2004

Also numbers of the form a^2 + b^2 + c^2 + d^2, where a,b,c,d are odd integers. - Alexander Adamchuk, Dec 01 2006

If X is an n-set and Y_i (i=1,2,3) mutually disjoint 2-subsets of X then a(n-5) is equal to the number of 4-subsets of X intersecting each Y_i (i=1,2,3). - Milan Janjic, Aug 26 2007

A007814(a(n)) = 2; A037227(a(n)) = 5. - Reinhard Zumkeller, Jun 30 2012

Numbers k such that 3^k + 1 is divisible by 41. - Bruno Berselli, Aug 22 2018

Lexicographically smallest arithmetic progression of positive integers avoiding Fibonacci numbers. - Paolo Xausa, May 08 2023

From Martin Renner, May 24 2024: (Start)

Also number of points in a grid cross with equally long arms and a width of two points, e.g.:

* *

* * * *

* * * * * *

* * * * * * * * * * * * * * * * * * * *

* * * * * * * * * * * * * * * * * * * *

* * * * * *

* * * *

* *

etc. (End)

This sequence contains the n mod 8 = 5 pseudoprimes to the following modified Fermat primality criterion:

Conjecture 1: if p is an odd prime congruent to {3,5} (mod 8) then 2^((p-1)/2) mod p = p-1.

This conjecture has been tested to 10^8.

This criterion produces far fewer pseudoprimes than the 2^(n-1) mod n = 1 test and thus has a higher probability of success. The number of pseudoprimes for the two tests up to 10^k are:

10^5 5 26 19.23%

10^6 13 78 16.66%

10^7 40 228 17.54%

There are 40 terms < 10^7. If an additional constraint 3^(n-1) mod n = 1 and 5^(n-1) mod n = 1 is added, only 4 terms remain: (29341, 314821, 873181, 9863461).

This sequence appears to be a subset of A175865, A001262, A047713, A020230.

Number of terms below 10^k for k = 5..15: 5, 13, 40, 132, 369, 975, 2534, 6592, 17403, 45801, 122473. The corresponding numbers for 2^(n-1) mod n = 1: 26, 78, 228, 637, 1718, 4505, 11645, 29902, 76587, 197455, 513601. - Jens Kruse Andersen, Jul 13 2014

Also composite numbers 2n+1 with n even such that 2n+1 | 2^n+1. - Hilko Koning, Jan 27 2022

Conjecture 1 is true. With p = 2k+1 then 2^k mod (2k+1) == 2k. So 2k+1 | 2k-2^k. Prime numbers 2k+1 == +-3 (mod 8) are the prime numbers such that 2k+1 | 2^k+1 (Comments A007520). A reflection across the x-axis and +1 translation across the y-axis of the graph (2k-2^k) / (2k+1) gives the graph (2^k+1) / (2k+1). So the k values of both 2k+1 | 2k-2^k and 2k+1 | 2^k+1 are identical. - Hilko Koning, Feb 04 2022

From Klaus Purath, Jan 04 2023: (Start)

The product of two consecutive terms belongs to the sequence: a(n)*a(n+1) = a(a(n)+n) = (a(n)+n)*(a(n+1)-n-1) + 1.

a(n) is never divisible by primes given in A003629.

Each odd prime factor p divides exactly 2 out of any p consecutive terms. If a(i) and a(k) form such a pair that are divisible by p, then i + k == -6 (mod p).

The prime factors are listed in A038873 and the primes in A028886.

For n > 0, this is a proper subsequence of A079896.

Conjecture: a(n) = A079896(A265284(n-1)). -

(End)

Numbers k for which Jacobi symbol J(2,k) = -1, so 2 (as well as 2^k) is not a square mod k. - Antti Karttunen, Aug 27 2005, corrected by Jianing Song, Nov 05 2019, see also A329095.

Numbers n whose multiplicative order modulo 2^k is 2^(k - 2) for k >= 4. For k = 3, the numbers whose multiplicative order modulo 8 is 2 are in sequence A047484. - Jianing Song, Apr 29 2018

Apart from initial term(s), dimension of the space of weight 2n cuspidal newforms for Gamma_0(97).

n such that 32 is the largest power of 2 dividing A003629(k)^n-1 for any k. - Benoit Cloitre, Mar 23 2002

Continued fraction expansion of tanh(1/8). - Benoit Cloitre, Dec 17 2002

If Y and Z are 2-blocks of a (4n+1)-set X then a(n-1) is the number of 3-subsets of X intersecting both Y and Z. - Milan Janjic, Oct 28 2007

General form: (q*n+x)*q x=+1; q=2=A016825, q=3=A017197, q=4=A119413, ... x=-1; q=3=A017233, q=4=A098502, ... x=+2; q=4=A051062, ... - Vladimir Joseph Stephan Orlovsky, Feb 16 2009

a(n)*n+1 = (4n+1)^2 and a(n)*(n+1)+1 = (4n+3)^2 are both perfect squares. - Carmine Suriano, Jun 01 2014

For all positive integers n, there are infinitely many positive integers k such that k*n + 1 and k*(n+1) + 1 are both perfect squares. Except for 8, all the numbers of this sequence are the smallest integers k which are solutions for getting two perfect squares. Example: a(1) = 24 and 24 * 1 + 1 = 25 = 5^2, then 24 * (1+1) + 1 = 49 = 7^2. [Reference AMM] - Bernard Schott, Sep 24 2017

Numbers k such that 3^k + 1 is divisible by 17*193. - Bruno Berselli, Aug 22 2018

This sequence contains the n mod 8 = 3 pseudoprimes to the following modified Fermat primality criterion:

Conjecture 1: if p is a prime congruent to {3,5} mod 8 then 2^((p-1)/2) mod p = p-1.

This conjecture has been tested to 10^8.

This modified primality test has far fewer pseudoprimes than the 2^(n-1) mod n = 1 test and thus has a much higher probability of success. The number of pseudoprimes up to 10^k for the two tests are:

10^3 0 0

10^4 0 2

10^5 0 5

10^6 2 14

10^7 16 48

This sequence appears to be a subset of the composites in A175865.

The n mod 8 = 3 pseudoprimes are much rarer than the n mod 8 = 5 pseudoprimes. There are 16 terms < 10^7. If the additional constraints 3^(n-1) mod n = 1 and 5^(n-1) mod n = 1 are added, no terms remain.

Number of terms < 10^k: 0, 0, 0, 0, 0, 2, 16, 50, 132, ..., . - Robert G. Wilson v, Jul 21 2014

Number of terms < 10^k for k=5..15: 0, 2, 16, 50, 132, 341, 876, 2330, 6234, 16625, 44885. - Jens Kruse Andersen, Jul 27 2014

It appears that the terms of the sequence are also the composite numbers of A294912. - Hilko Koning, Dec 05 2019

Also composite numbers 2k+1 with k odd such that 2k+1 | 2^k+1. - Hilko Koning, Jan 27 2022

Conjecture 1 is true. With p = 2k+1 then 2^k mod (2k+1) == 2k. So 2k+1 | 2k-2^k . Prime numbers 2k+1 == +-3 (mod 8) are the prime numbers such that 2k+1 | 2^k+1 (Comments A007520). A reflection across the x-axis and +1 translation across the y-axis of the graph (2k-2^k) / (2k+1) gives the graph (2^k+1) / (2k+1). So the k values of both 2k+1 | 2k-2^k and 2k+1 | 2^k+1 are identical. - Hilko Koning, Feb 04 2022

The numbers are called Curzon numbers by Tattersall (p. 85, exercise 43).

Sequence 2*a(n)+1 apparently is A175865 (certainly it is not A003629). - Joerg Arndt, Apr 07 2013