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

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

See A105397 for definition of Fractal Jump Sequence.

a(n+2) = the digital root of the n-th centered hexagonal number (A003215). - Colin Barker, Jan 30 2015

See A105397 for definition of Fractal Jump Sequence.

Continued fraction expansion of (4+3*sqrt(2))/2. - Bruno Berselli, Nov 09 2011

Period 2: repeat [4, 8]. - Wesley Ivan Hurt, Feb 12 2017

See A105397 for definition of Fractal Jump Sequence.

From Vincenzo Librandi, Nov 13 2010: (Start)

First digit after the decimal point in the decimal expansion of (n^2 - 2)/3 (with n > 2). Examples:

for n=3, (3^2-2)/3 = 2.(3);

for n=4, (4^2-2)/3 = 4.(6);

for n=5, (5^2-2)/3 = 7.(6);

for n=6, (6^2-2)/3 = 11.(3);

for n=7, (7^2-2)/3 = 15.(6). (End)

The rules of an "Increasing Term Fractal Jump Sequence" are described in A105647.

The digit zero is omitted as it can create situations where the next term can require a 0 in the first digit.

See A105397 for definition of Fractal Jump Sequence.

See A105397 for definition of Fractal Jump Sequence.