cp's OEIS Frontend

Numbers n such that phi(4n) = phi(3n). - Benoit Cloitre, Aug 06 2003

Or, numbers relatively prime to 2 and 3, or coprime to 6, or having only prime factors >= 5; also known as 5-rough numbers. (Edited by M. F. Hasler, Nov 01 2014: merged with comments from Zak Seidov, Apr 26 2007 and Michael B. Porter, Oct 09 2009)

Apart from initial term(s), dimension of the space of weight 2n cuspidal newforms for Gamma_0( 38 ).

Numbers k such that k mod 2 = 1 and (k+1) mod 3 <> 1. - Klaus Brockhaus, Jun 15 2004

Also numbers n such that the sum of the squares of the first n integers is divisible by n, or A000330(n) = n*(n+1)*(2*n+1)/6 is divisible by n. - Alexander Adamchuk, Jan 04 2007

Numbers n such that the sum of squares of n consecutive integers is divisible by n, because A000330(m+n) - A000330(m) = n*(n+1)*(2*n+1)/6 + n*(m^2+n*m+m) is divisible by n independent of m. - Kaupo Palo, Dec 10 2016

A126759(a(n)) = n + 1. - Reinhard Zumkeller, Jun 16 2008

Terms of this sequence (starting from the second term) are equal to the result of the expression sqrt(4!*(k+1) + 1) - but only when this expression yields integral values (that is when the parameter k takes values, which are terms of A144065). - Alexander R. Povolotsky, Sep 09 2008

For n > 1: a(n) is prime if and only if A075743(n-2) = 1; a(2*n-1) = A016969(n-1), a(2*n) = A016921(n-1). - Reinhard Zumkeller, Oct 02 2008

A156543 is a subsequence. - Reinhard Zumkeller, Feb 10 2009

Numbers n such that ChebyshevT(x, x/2) is not an integer (is integer/2). - Artur Jasinski, Feb 13 2010

If 12*k + 1 is a perfect square (k = 0, 2, 4, 10, 14, 24, 30, 44, ... = A152749) then the square root of 12*k + 1 = a(n). - Gary Detlefs, Feb 22 2010

A089128(a(n)) = 1. Complement of A047229(n+1) for n >= 1. See A164576 for corresponding values A175485(a(n)). - Jaroslav Krizek, May 28 2010

Cf. property described by Gary Detlefs in A113801 and in Comment: more generally, these numbers are of the form (2*h*n+(h-4)*(-1)^n-h)/4 (with h, n natural numbers), therefore ((2*h*n+(h-4)*(-1)^n-h)/4)^2-1 == 0 (mod h); in this case, a(n)^2 - 1 == 0 (mod 6). Also a(n)^2 - 1 == 0 (mod 12). - Bruno Berselli, Nov 05 2010 - Nov 17 2010

Numbers n such that ( Sum_{k = 1..n} k^14 ) mod n = 0. (Conjectured) - Gary Detlefs, Dec 27 2011

From Peter Bala, May 02 2018: (Start)

The above conjecture is true. Apply Ireland and Rosen, Proposition 15.2.2. with m = 14 to obtain the congruence 6*( Sum_{k = 1..n} k^14 )/n = 7 (mod n), true for all n >= 1. Suppose n is coprime to 6, then 6 is a unit in Z/nZ, and it follows from the congruence that ( Sum_{k = 1..n} k^14 )/n is an integer. On the other hand, if either 2 divides n or 3 divides n then the congruence shows that ( Sum_{k = 1..n} k^14 )/n cannot be integral. (End)

A126759(a(n)) = n and A126759(m) < n for m < a(n). - Reinhard Zumkeller, May 23 2013

(a(n-1)^2 - 1)/24 = A001318(n), the generalized pentagonal numbers. - Richard R. Forberg, May 30 2013

Numbers k for which A001580(k) is divisible by 3. - Bruno Berselli, Jun 18 2014

Numbers n such that sigma(n) + sigma(2n) = sigma(3n). - Jahangeer Kholdi and Farideh Firoozbakht, Aug 15 2014

a(n) are values of k such that Sum_{m = 1..k-1} m*(k-m)/k is an integer. Sums for those k are given by A062717. Also see Detlefs formula below based on A062717. - Richard R. Forberg, Feb 16 2015

a(n) are exactly those positive integers m such that the sequence b(n) = n*(n + m)*(n + 2*m)/6 is integral, and also such that the sequence c(n) = n*(n + m)*(n + 2*m)*(n + 3*m)/24 is integral. Cf. A007775. - Peter Bala, Nov 13 2015

Along with 2, these are the numbers k such that the k-th Fibonacci number is coprime to every Lucas number. - Clark Kimberling, Jun 21 2016

This sequence is the Engel expansion of 1F2(1; 5/6, 7/6; 1/36) + 1F2(1; 7/6, 11/6; 1/36)/5. - Benedict W. J. Irwin, Dec 16 2016

The sequence a(n), n >= 4 is generated by the successor of the pair of polygonal numbers {P_s(4) + 1, P_(2*s - 1)(3) + 1}, s >= 3. - Ralf Steiner, May 25 2018

The asymptotic density of this sequence is 1/3. - Amiram Eldar, Oct 18 2020

Also, the only vertices in the odd Collatz tree A088975 that are branch values to other odd nodes t == 1 (mod 2) of A005408. - Heinz Ebert, Apr 14 2021

From Flávio V. Fernandes, Aug 01 2021: (Start)

For any two terms j and k, the product j*k is also a term (the same property as p^n and smooth numbers).

From a(2) to a(phi(A033845(n))), or a((A033845(n))/3), the terms are the totatives of the A033845(n) itself. (End)

Also orders n for which cyclic and semicyclic diagonal Latin squares exist (see A123565 and A342990). - Eduard I. Vatutin, Jul 11 2023

If k is in the sequence, then k*2^m + 3 is also in the sequence, for all m > 0. - Jules Beauchamp, Aug 29 2024

Polya proved (see Ahrens) that the number of solutions to this problem for an m X m board is > 0 iff m is coprime to 6. - Jonathan Vos Post, Feb 20 2005

A strong complete mapping of a cyclic group (Z_n,+) is a permutation f(x) of Z_n such that f(0)=0 and that f(x)-x and f(x)+x are both permutations.

a(n) is the number of solutions of the toroidal n-queen problem (A007705) with 2n+1 queens and one queen in the top-left corner.

Also a(n) is the number of horizontally or vertically semicyclic diagonal Latin squares of order 2n+1 with the first row in ascending order. Horizontally semicyclic diagonal Latin square is a square where each row r(i) is a cyclic shift of the first row r(0) by some value d(i) (see example). Vertically semicyclic diagonal Latin square is a square where each column c(i) is a cyclic shift of the first column c(0) by some value d(i). Cyclic diagonal Latin squares (see A123565) fall under the definition of vertically and horizontally semicyclic diagonal Latin squares simultaneously, in this type of squares each row r(i) is obtained from the previous one r(i-1) using cyclic shift by some value d. Definition from A343867 includes this type of squares but not only it. - Eduard I. Vatutin, Jan 25 2022

The Latin squares considered here are diagonal Latin squares that are isomorphic to cyclic Latin squares. They are can be obtained from cyclic Latin squares (see A338522) by diagonalization (getting a corresponding pair of transversals and placing them on the diagonals, see article). These Latin squares have some interesting properties, for example, there are a large number of diagonal transversals.

A horizontally semicyclic diagonal Latin square is a square where each row r(i) is a cyclic shift of the first row r(0) by some value d(i) (see example).

A horizontally semicyclic diagonal Latin square is a square where each row r(i) is a cyclic shift of the first row r(0) by some value d(i) (see example). Similarly, a vertically semicyclic diagonal Latin square is a square where each column c(i) is a cyclic shift of the first column c(0) by some value d(i).

A horizontally semicyclic diagonal Latin square is a square where each row r(i) is a cyclic shift of the first row r(0) by some value d(i) (see example). A vertically semicyclic diagonal Latin square is a square where each column c(i) is a cyclic shift of the first column c(0) by some value d(i). Cyclic diagonal Latin squares (see A338562) fall under the definition of vertically and horizontally semicyclic diagonal Latin squares simultaneously, in this type of squares each row r(i) is obtained from the previous one r(i-1) using cyclic shift by some value d.

Semicyclic diagonal Latin squares do not exist for even orders n.