

A007310


Numbers congruent to 1 or 5 mod 6.


228



1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53, 55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 91, 95, 97, 101, 103, 107, 109, 113, 115, 119, 121, 125, 127, 131, 133, 137, 139, 143, 145, 149, 151, 155, 157, 161, 163, 167, 169, 173, 175
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OFFSET

1,2


COMMENTS

Or, numbers relatively prime to 2 and 3, or coprime to 6, or having only prime factors >= 5; also known as 5rough 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
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
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
Cf. property described by Gary Detlefs in A113801 and in Comment: more generally, these numbers are of the form (2*h*n+(h4)*(1)^nh)/4 (with h, n natural numbers), therefore ((2*h*n+(h4)*(1)^nh)/4)^21 == 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
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)
a(n) are values of k such that Sum_{m = 1..k1} m*(km)/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 kth 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
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)


REFERENCES

K. Ireland and M. Rosen, A Classical Introduction to Modern Number Theory, SpringerVerlag, 1980.


LINKS



FORMULA

a(n) = 3*n  1  (n mod 2).  Zak Seidov, Jan 18 2006
a(1) = 1 then alternatively add 4 and 2. a(1) = 1, a(n) = a(n1) + 3 + (1)^n.  Zak Seidov, Mar 25 2006
1 + 1/5^2 + 1/7^2 + 1/11^2 + ... = Pi^2/9 [Jolley].  Gary W. Adamson, Dec 20 2006
For n >= 3 a(n) = a(n2) + 6.  Zak Seidov, Apr 18 2007
Expand (x+x^5)/(1x^6) = x + x^5 + x^7 + x^11 + x^13 + ...
O.g.f.: x*(1+4*x+x^2)/((1+x)*(1x)^2). (End)
a(n) = 6*floor(n/2) + (1)^(n+1).  Gary Detlefs, Dec 29 2011
a(n) = 2*n + 1 + 2*floor((n2)/2) = 2*n  1 + 2*floor(n/2), leading to the o.g.f. given by R. J. Mathar above.  Wolfdieter Lang, Jan 20 2012
1  1/5^3 + 1/7^3  1/11^3 +  ... = Pi^3*sqrt(3)/54 (L. Euler).  Philippe Deléham, Mar 09 2013
a(n) = 2*floor(3*n/2)  1.
a(k*n) = k*a(n) + (4*k + (1)^k  3)/2 for k>0 and odd n, a(k*n) = k*a(n) + k  1 for even n. Some special cases:
k=2: a(2*n) = 2*a(n) + 3 for odd n, a(2*n) = 2*a(n) + 1 for even n;
k=3: a(3*n) = 3*a(n) + 4 for odd n, a(3*n) = 3*a(n) + 2 for even n;
k=4: a(4*n) = 4*a(n) + 7 for odd n, a(4*n) = 4*a(n) + 3 for even n;
k=5: a(5*n) = 5*a(n) + 8 for odd n, a(5*n) = 5*a(n) + 4 for even n, etc. (End)
a(2*m) = 6*m  1, m >= 1; a(2*m + 1) = 6*m + 1, m >= 0.  Ralf Steiner, May 17 2018


EXAMPLE

G.f. = x + 5*x^2 + 7*x^3 + 11*x^4 + 13*x^5 + 17*x^6 + 19*x^7 + 23*x^8 + ...


MAPLE

seq(seq(6*i+j, j=[1, 5]), i=0..100); # Robert Israel, Sep 08 2014


MATHEMATICA

Select[Range[200], MemberQ[{1, 5}, Mod[#, 6]] &] (* Harvey P. Dale, Aug 27 2013 *)
Table[2*Floor[3*n/2]  1, {n, 1000}] (* Mikk Heidemaa, Feb 11 2016 *)


PROG

(PARI) \\ given an element from the sequence, find the next term in the sequence.
(Sage) [i for i in range(150) if gcd(6, i) == 1] # Zerinvary Lajos, Apr 21 2009
(Haskell)
a007310 n = a007310_list !! (n1)
a007310_list = 1 : 5 : map (+ 6) a007310_list
(GAP) Filtered([1..150], n>n mod 6=1 or n mod 6=5); # Muniru A Asiru, Dec 19 2018
(Python)


CROSSREFS

Cf. A000330, A001580, A032528 (partial sums), A038509 (subsequence of composites), A047209, A047336, A047522, A056020, A084967, A090771, A091998, A144065, A175885A175887.
Row 3 of A260717 (without the initial 1).


KEYWORD

nonn,easy


AUTHOR

C. Christofferson (Magpie56(AT)aol.com)


STATUS

approved



