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

Interleaves the odd squares A016754 with (1+4n^2), A053755.

Squares of positive integers (plus 1 if n is odd). - Wesley Ivan Hurt, Oct 10 2013

a(n) is the maximum total number of queens that can coexist without attacking each other on an [n+3] X [n+3] chessboard, when the lone queen is in the most vulnerable position on the board. Specifically, the lone queen will placed in any center position, facing an opponent's "army" of size a(n)-1 == A137932(n+2). - Bob Selcoe, Feb 12 2015

a(n) is also the edge chromatic number of the complement of the (n+2) X (n+2) rook graph. - Eric W. Weisstein, Jan 31 2024

The count of terms not on the principal diagonals is always even.

The last digit is the repeating pattern 0,0,0,4,8,6,4,6,8,4, which is palindromic if the leading 0's are removed, 4864684.

The sum of the last digits is 40, which is the count of the pattern times 4.

A 4 X 4 spiral is the only spiral, aside from a 0 X 0, whose count of terms that do not lie on its principal diagonals equal the count of terms that do [A137932(4) = A042948(4)] making the 4 X 4 the "perfect spiral".

Yet another property is mod(a(n), A042948(n)) = 0 iff n is even. This is a large family that includes the 4 X 4 spiral.

a(n) is the maximum number of queens of one color that can coexist without attacking one queen of the opponent's color on an [n+1] X [n+1] chessboard, when the lone queen is in the most vulnerable position on the board, i.e., on a center square. - Bob Selcoe, Feb 12 2015

Also the circumference of the (n-1) X (n-1) grid graph for n > 2. - Eric W. Weisstein, Mar 25 2018

Also the crossing number of the complete bipartite graph K_{5,n}. - Eric W. Weisstein, Sep 11 2018

Apart from initial term(s), dimension of the space of weight 2n cusp forms for Gamma_0( 10 ).

This sequence is an interleaving of A016945 with A016969. - Guenther Schrack, Nov 16 2018

In the diagram every cycle is represented by a directed graph.

After (3x + 1) the next step is (3y + 1).

After (x/2) the next step is (y/2).

A235801(n) gives the length of n-th horizontal line segment in the same diagram.

Also A004767 and A000027 interleaved.

The sequence is the interleaving of A047238 with A047241. - Guenther Schrack, Feb 12 2019

The sequence is the interleaving of A047238(n) with A016777(n-1). - Guenther Schrack, Feb 11 2019

Intersection of A000384 and A008588. Their indices are given by A047238. - Michel Marcus, Feb 27 2014

Are there any missing terms? The first 10^7 primes were examined. All these differences occur for some k < 10^5. Note that the first differences of these terms is 1, 2, 4, or 6.

From R. J. Mathar, Oct 29 2013: (Start)

This sequence of possible differences d= prime(k)^2 -q looks similar to A047238; 1 is an exception associated with the single even prime, 1=2^2-3.

[Reason: Otherwise primes are odd, squared primes are also odd, so the differences are even and therefore in the class {0,2,4} mod 6.

Furthermore primes are of the form 3n+1 or 3n+2, squared primes are of the form 9n^2+6n+1 or 9n^2+12n+4, so squared primes are of the form ==1 (mod 3).

The difference prime(k)^2-q is therefore the difference between a number ==1 (mod 3) and a number == {1,2} (mod 3) and therefore a number == {0,2} mod 3. This is never of the form 6n+4 ( == 1 mod 3). So the differences are in the class {0,2} mod 6, demonstrating that this is essentially a subsequence of A047238.]

Furthermore, differences 36, 144, 324,... of the form (6n)^2, A016910, appear in A047238 but not here, because prime(k)^2 -q=(6n)^2 is equivalent to prime(k)^2-(6n)^2 =q =(prime(k)+6n)*(prime(k)-6n), which requires an explicit factorization of the prime q. This is a contradiction if we assure that prime(k)-6n is not equal 1; if we scanned explicitly all primes up to prime(k)=10^7, for example, all (6n)^2 up to 6n<=10^7 are proved not to be in the sequence. (End)