

A047241


Numbers that are congruent to {1, 3} mod 6.


30



1, 3, 7, 9, 13, 15, 19, 21, 25, 27, 31, 33, 37, 39, 43, 45, 49, 51, 55, 57, 61, 63, 67, 69, 73, 75, 79, 81, 85, 87, 91, 93, 97, 99, 103, 105, 109, 111, 115, 117, 121, 123, 127, 129, 133, 135, 139, 141, 145, 147, 151, 153, 157, 159, 163, 165, 169, 171, 175, 177, 181, 183
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OFFSET

1,2


COMMENTS

Also the numbers k such that 10^p+k could possibly be prime.  Roderick MacPhee, Nov 20 2011 This statement can be written as follows. If 10^m + k = prime, for any m >= 1, then k is in this sequence. See the pink box comments by Roderick MacPhee from Dec 09 2014.  Wolfdieter Lang, Dec 09 2014
The oddindexed terms are one more than the arithmetic mean of their neighbors; the evenindexed terms are one less than the arithmetic mean of their neighbors.  Amarnath Murthy, Jul 29 2003
Partial sums are A212959.  Philippe Deléham, Mar 16 2014
12*a(n) is conjectured to be the length of the boundary after n iterations of the hexagon and square expansion shown in the link. The squares and hexagons have side length 1 in some units. The pattern is supposed to become the planar Archimedean net 4.6.12 when n > infinity.  Kival Ngaokrajang, Nov 30 2014
Positive numbers k for which 1/2 + k/3 + k^2/6 is an integer.  Bruno Berselli, Apr 12 2018


REFERENCES

L. Lovasz, J. Pelikan, K. Vesztergombi, Discrete Mathematics, Springer (2003); 14.4, p. 225.


LINKS

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
L. Lovasz, J. Pelikan, K. Vesztergombi, Discrete Mathematics, Elementary and Beyond, Springer (2003); 14.4, p. 225.
Kival Ngaokrajang, Illustration of initial terms
Index entries for linear recurrences with constant coefficients, signature (1,1,1).


FORMULA

From Paul Barry, Sep 04 2003: (Start)
O.g.f.: (1 + 2*x + 3*x^2)/((1 + x)*(1  x)^2) = (1 + 2*x + 3*x^2)/((1  x)*(1  x^2)).
E.g.f.: (6*x + 1)*exp(x)/2 + exp(x)/2;
a(n) = 3*n  5/2  (1)^n/2. (End)
a(n) = 2*floor((n1)/2) + 2*n  1.  Gary Detlefs, Mar 18 2010
a(n) = 6*n  a(n1)  8 with n > 1, a(1)=1.  Vincenzo Librandi, Aug 05 2010
a(n) = 3*n  2  ((n+1) mod 2).  Wesley Ivan Hurt, Jun 29 2013
a(1)=1, a(2)=3, a(3)=7; for n>3, a(n) = a(n1) + a(n2)  a(n3).  Harvey P. Dale, Oct 01 2013
From Benedict W. J. Irwin, Apr 13 2016: (Start)
A005408(a(n)+1) = A016813(A001651(n)),
A007310(a(n)) = A005408(A087444(n)1),
A007310(A005408(a(n)+1)) = A017533(A001651(n)). (End)


MAPLE

seq(3*k2((k+1) mod 2), k=1..100); # Wesley Ivan Hurt, Sep 28 2013


MATHEMATICA

Table[{2, 4}, {30}] // Flatten // Prepend[#, 1]& // Accumulate (* JeanFrançois Alcover, Jun 10 2013 *)
Select[Range[200], MemberQ[{1, 3}, Mod[#, 6]]&] (* or *) LinearRecurrence[{1, 1, 1}, {1, 3, 7}, 70] (* Harvey P. Dale, Oct 01 2013 *)


PROG

(Haskell)
a047241 n = a047241_list !! (n1)
a047241_list = 1 : 3 : map (+ 6) a047241_list
 Reinhard Zumkeller, Feb 19 2013
(PARI) a(n)=bitor(3*n3, 1) \\ Charles R Greathouse IV, Sep 28 2013
(Python) for n in xrange(1, 10**5):print(3*n2((n+1)%2)) # Soumil Mandal, Apr 14 2016


CROSSREFS

Cf. A047233, A056970, A007310, A047228, A047261, A047273.
Subsequence of A186422.
Union of A016921 and A016945.  Wesley Ivan Hurt, Sep 28 2013
Sequence in context: A087550 A235387 A285144 * A086515 A132222 A320634
Adjacent sequences: A047238 A047239 A047240 * A047242 A047243 A047244


KEYWORD

nonn,easy


AUTHOR

N. J. A. Sloane


EXTENSIONS

Formula corrected by Bruno Berselli, Jun 24 2010


STATUS

approved



