OFFSET
1,2
COMMENTS
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)
(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
REFERENCES
K. Ireland and M. Rosen, A Classical Introduction to Modern Number Theory, Springer-Verlag, 1980.
LINKS
Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Andreas Enge, William Hart, and Fredrik Johansson, Short addition sequences for theta functions, arXiv:1608.06810 [math.NT], 2016-2018.
B. W. J. Irwin, Constants Whose Engel Expansions are the k-rough Numbers.
L. B. W. Jolley, Summation of Series, Dover, 1961
Cedric A. B. Smith, Prime factors and recurring duodecimals, Math. Gaz. 59 (408) (1975) 106-109.
William A. Stein's The Modular Forms Database, PARI-readable dimension tables for Gamma_0(N).
Eric Weisstein's World of Mathematics, Rough Number.
Eric Weisstein's World of Mathematics, Pi Formulas. [Jaume Oliver Lafont, Oct 23 2009]
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).
FORMULA
a(n) = (6*n + (-1)^n - 3)/2. - Antonio Esposito, Jan 18 2002
a(n) = a(n-1) + a(n-2) - a(n-3), n >= 4. - Roger L. Bagula
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(n-1) + 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(n-2) + 6. - Zak Seidov, Apr 18 2007
From R. J. Mathar, May 23 2008: (Start)
Expand (x+x^5)/(1-x^6) = x + x^5 + x^7 + x^11 + x^13 + ...
O.g.f.: x*(1+4*x+x^2)/((1+x)*(1-x)^2). (End)
a(n) = 6*floor(n/2) - 1 + 2*(n mod 2). - Reinhard Zumkeller, Oct 02 2008
1 + 1/5 - 1/7 - 1/11 + + - - ... = Pi/3 = A019670 [Jolley eq (315)]. - Jaume Oliver Lafont, Oct 23 2009
a(n) = ( 6*A062717(n)+1 )^(1/2). - Gary Detlefs, Feb 22 2010
a(n) = 6*A000217(n-1) + 1 - 2*Sum_{i=1..n-1} a(i), with n > 1. - Bruno Berselli, Nov 05 2010
a(n) = 6*n - a(n-1) - 6 for n>1, a(1) = 1. - Vincenzo Librandi, Nov 18 2010
Sum_{n >= 1} (-1)^(n+1)/a(n) = A093766 [Jolley eq (84)]. - R. J. Mathar, Mar 24 2011
a(n) = 6*floor(n/2) + (-1)^(n+1). - Gary Detlefs, Dec 29 2011
a(n) = 3*n + ((n+1) mod 2) - 2. - Gary Detlefs, Jan 08 2012
a(n) = 2*n + 1 + 2*floor((n-2)/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 + 1/7 - 1/11 + - ... = Pi*sqrt(3)/6 = A093766 (L. Euler). - Philippe Deléham, Mar 09 2013
1 - 1/5^3 + 1/7^3 - 1/11^3 + - ... = Pi^3*sqrt(3)/54 (L. Euler). - Philippe Deléham, Mar 09 2013
gcd(a(n), 6) = 1. - Reinhard Zumkeller, Nov 14 2013
a(n) = sqrt(6*n*(3*n + (-1)^n - 3)-3*(-1)^n + 5)/sqrt(2). - Alexander R. Povolotsky, May 16 2014
a(n) = 3*n + 6/(9*n mod 6 - 6). - Mikk Heidemaa, Feb 05 2016
From Mikk Heidemaa, Feb 11 2016: (Start)
a(n) = 2*floor(3*n/2) - 1.
a(n) = A047238(n+1) - 1. (suggested by Michel Marcus) (End)
E.g.f.: (2 + (6*x - 3)*exp(x) + exp(-x))/2. - Ilya Gutkovskiy, Jun 18 2016
From Bruno Berselli, Apr 27 2017: (Start)
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)
From Antti Karttunen, May 20 2017: (Start)
a((5*n)-3) = A255413(n).
A126760(a(n)) = n. (End)
a(2*m) = 6*m - 1, m >= 1; a(2*m + 1) = 6*m + 1, m >= 0. - Ralf Steiner, May 17 2018
From Amiram Eldar, Nov 22 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = sqrt(3) (A002194).
Product_{n>=2} (1 + (-1)^n/a(n)) = Pi/3 (A019670). (End)
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 *)
a[n_] := (6 n + (-1)^n - 3)/2; a[rem156, 60] (* Robert G. Wilson v, May 26 2014 from a suggestion by N. J. A. Sloane *)
Flatten[Table[6n + {1, 5}, {n, 0, 24}]] (* Alonso del Arte, Feb 06 2016 *)
Table[2*Floor[3*n/2] - 1, {n, 1000}] (* Mikk Heidemaa, Feb 11 2016 *)
PROG
(PARI) isA007310(n) = gcd(n, 6)==1 \\ Michael B. Porter, Oct 09 2009
(PARI) A007310(n)=n\2*6-(-1)^n \\ M. F. Hasler, Oct 31 2014
(PARI) \\ given an element from the sequence, find the next term in the sequence.
nxt(n) = n + 9/2 - (n%6)/2 \\ David A. Corneth, Nov 01 2016
(Sage) [i for i in range(150) if gcd(6, i) == 1] # Zerinvary Lajos, Apr 21 2009
(Haskell)
a007310 n = a007310_list !! (n-1)
a007310_list = 1 : 5 : map (+ 6) a007310_list
-- Reinhard Zumkeller, Jan 07 2012
(Magma) [n: n in [1..250] | n mod 6 in [1, 5]]; // Vincenzo Librandi, Feb 12 2016
(GAP) Filtered([1..150], n->n mod 6=1 or n mod 6=5); # Muniru A Asiru, Dec 19 2018
(Python)
def A007310(n): return (n+(n>>1)<<1)-1 # Chai Wah Wu, Oct 10 2023
CROSSREFS
A005408 \ A016945. Union of A016921 and A016969; union of A038509 and A140475. Essentially the same as A038179. Complement of A047229. Subsequence of A186422.
Cf. A000330, A001580, A002194, A019670, A032528 (partial sums), A038509 (subsequence of composites), A047209, A047336, A047522, A056020, A084967, A090771, A091998, A144065, A175885-A175887.
For k-rough numbers with other values of k, see A000027, A005408, A007775, A008364-A008366, A166061, A166063.
Cf. A126760 (a left inverse).
Row 3 of A260717 (without the initial 1).
Cf. A105397 (first differences).
KEYWORD
nonn,easy
AUTHOR
C. Christofferson (Magpie56(AT)aol.com)
STATUS
approved