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

A cyclic Latin square is a Latin square in which row i is obtained by cyclically shifting row i-1 by d places.

Equivalently, a Latin square is cyclic if and only if each row is a cyclic permutation of the first row and each column is a cyclic permutation of the first column.

Every cyclic diagonal Latin square is a cyclic Latin square, so a(n) <= A338522(2*n+1).

Cyclic diagonal Latin squares exist only for odd orders not divisible by 3. - Andrew Howroyd, May 26 2021

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

A cyclic Latin square is a Latin square in which row i is obtained by cyclically shifting row i-1 by d places.

Equivalently, a Latin square is cyclic if and only if each row is a cyclic permutation of the first row and each column is a cyclic permutation of the first column.

There are no cyclic diagonal Latin squares of even order.

All cyclic diagonal Latin squares are pandiagonal. Conversely, all pandiagonal Latin squares are cyclic for orders 5, 7 and 11.

From Andrew Howroyd, May 01 2021: (Start)

Depending on exactly which Latin squares constitute a main class, slightly different sequences are possible. Another variation is given in A343866.

In this sequence equivalence allows for the permutation of symbols, the transpose of rows with columns, and any permutation of rows and columns that preserves the cyclic and diagonal properties. This permutation must transform every cyclic diagonal Latin square into another, but does not necessarily transform an arbitrary diagonal Latin square that is not cyclic into another diagonal Latin square.

The row (or column) permutations that satisfy this requirement form a group and are those where every k-th row (or column) is taken cyclically where k is any number that is congruent to 1 or -1 modulo every prime divisor of the order of the Latin square. (End)

A cyclic Latin square is a Latin square in which row i is obtained by cyclically shifting row i-1 by d places (see A338562, A123565 and A341585).

Cyclic diagonal Latin squares do not exist for even orders.

a(n) <= A342997(n).

All cyclic diagonal Latin squares are diagonal Latin squares, so A287647(n) <= a((n-1)/2).

A cyclic Latin square is a Latin square in which row i is obtained by cyclically shifting row i-1 by d places (see A338562, A123565 and A341585).

Cyclic diagonal Latin squares do not exist for even n.

All cyclic diagonal Latin squares are diagonal Latin squares, so a((n-1)/2) <= A287648(n).

All diagonal transversals are transversals, so a(n) <= A006717(n).

A342998 <= a(n).

Pandiagonal Latin squares exist only for odd orders not divisible by 3. All pandiagonal Latin squares for orders less than 13 are cyclic which are not counted by this sequence.

Semicyclic Latin squares are defined in the Atkin reference where the first nonzero term of this sequence is given. They are cyclic in a single direction. The direction can be horizontal or vertical or any other step such as a knights move.

Each symbol in a semicyclic Latin square occupies the same pattern of squares up to translation on the torus which in the case of a pandiagonal square is a solution to the toroidal n-queens problem.

For prime 2n+1, a(n) is a multiple of 2n+1.

From Andrew Howroyd, Nov 08 2019: (Start)

A diagonal Latin square of order n is an n X n array with every integer from 0 to n-1 in every row, every column, and both main diagonals.

Of course if even one example exists, then a(n) >= 1.

A274806 gives the number of diagonal Latin squares and A274806(6) is nonzero. This suggests that although it is not possible to have a pair of orthogonal diagonal Latin squares, a(6) should be 1 here. (End)

a(1) = 1 because there is only one (trivial) diagonal Latin square of order 1. It is orthogonal to itself, so if we allow the consideration of multiple copies of the same diagonal Latin square, we get a(1) = infinity instead.

From Eduard I. Vatutin, Mar 27 2021: (Start)

a(n) <= A287695(n) + 1.

a(p) >= A123565(p) = p-3 for all odd prime p due to existence of clique from cyclic MODLS of order p with at least A123565(p) items. It seems that for some orders p clique from cyclic MODLS can be extended by adding none cyclic DLS that are orthogonal to all cyclic DLS. (End)

a(n) <= A001438(n). - Max Alekseyev, Nov 08 2019

a(10) >= 2; a(11) >= 8; a(12) >= 4; a(13) >= 10; a(14) >= 2; a(15) >= 4. - Natalia Makarova, Sep 03 2020; updated May 30 2021

a(16) >= 14, a(17) >= 14, a(18) >= 2, a(19) >= 16, a(20) >= 2. - Natalia Makarova, Jan 08 2021

A pandiagonal Latin square is a Latin square in which the diagonal, antidiagonal and all broken diagonals and antidiagonals are transversals.

For orders n = 5, 7 and 11 all pandiagonal Latin squares are cyclic, so a(n) = A123565(2n+1) for n < 6. For n=6 (order 13), this is not true and there are 12386 inequivalent squares; of these 10 are cyclic (in all directions) and 1560 are semi-cyclic (A343867).

Pandiagonal Latin squares exist only for odd orders not divisible by 3. This is because the positions of each symbol are a solution to the toroidal n-queens problem which only has solutions for these sizes. - Andrew Howroyd, May 26 2021