



3, 9, 15, 21, 27, 33, 39, 45, 51, 57, 63, 69, 75, 81, 87, 93, 99, 105, 111, 117, 123, 129, 135, 141, 147, 153, 159, 165, 171, 177, 183, 189, 195, 201, 207, 213, 219, 225, 231, 237, 243, 249, 255, 261, 267, 273, 279, 285, 291, 297, 303, 309, 315, 321, 327
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OFFSET

0,1


COMMENTS

Apart from initial term(s), dimension of the space of weight 2n cuspidal newforms for Gamma_0(37).
Continued fraction expansion of tanh(1/3).
If a 2set Y and a 3set Z are disjoint subsets of an nset X then a(n4) is the number of 3subsets of X intersecting both Y and Z.  Milan Janjic, Sep 08 2007
Leaves of the Odd CollatzTree: a(n) has no odd predecessors in all '3x+1' trajectories where it occurs: A139391(2*k+1) <> a(n) for all k; A082286(n)=A006370(a(n)).  Reinhard Zumkeller, Apr 17 2008
From Loren Pearson, Jul 02 2009: (Start)
Values of n in 2^n1 that produce a composite with 7 as a factor.
Their distribution in 2^n1 sequence equidistant between terms that have multiple factors of 3 (n=6,12,18,24,30,36,... where the number of factors of 3 equals [number of times 3 divides n] + 1), recognizing that all even n in the 2^n1 sequence have at least one factor of 3. Other odd n appear to be unrelated prime or semiprime composites. (End)
Let random variable X have a uniform distribution on the interval [0,c] where c is a positive constant. Then, for positive integer n, the coefficient of determination between X and X^n is (6n+3)/(n+2)^2, that is, A016945(n)/A000290(n+2). Note that the result is independent of c. For the derivation of this result, see the link in the Links section below.  Dennis P. Walsh, Aug 20 2013
Positions of 3 in A020639.  Zak Seidov, Apr 29 2015
a(n+2) gives the sum of 6 consecutive terms of A004442 starting with A004442(n).  Wesley Ivan Hurt, Apr 08 2016
Numbers k such that Fibonacci(k) mod 4 = 2.  Bruno Berselli, Oct 17 2017


REFERENCES

Friedrich L. Bauer, 'Der (ungerade) CollatzBaum', Informatik Spektrum 31 (Springer, April 2008).


LINKS

Table of n, a(n) for n=0..54.
Milan Janjic, Two Enumerative Functions
Tanya Khovanova, Recursive Sequences
Luis Manuel Rivera, Integer sequences and kcommuting permutations, arXiv preprint arXiv:1406.3081 [math.CO], 2014.
William A. Stein, Dimensions of the spaces S_k^{new}(Gamma_0(N))
William A. Stein, The modular forms database
Dennis P. Walsh, The correlation for a power curve on nonnegative support
Eric Weisstein's World of Mathematics, Collatz Problem
Index entries for sequences related to 3x+1 (or Collatz) problem
Index entries for linear recurrences with constant coefficients, signature (2,1).


FORMULA

a(n) = 3*(2*n + 1) = 3*A005408(n), odd multiples of 3.
A008615(a(n)) = n.  Reinhard Zumkeller, Feb 27 2008
A157176(a(n)) = A103333(n+1).  Reinhard Zumkeller, Feb 24 2009
a(n) = 12*n  a(n1) for n>0, a(0)=3.  Vincenzo Librandi, Nov 20 2010
G.f.: 3*(1+x)/(1x)^2.  Mario C. Enriquez, Dec 14 2016


MAPLE

seq(6*n+3, n=0..40); # Dennis P. Walsh, Aug 20 2013
A016945:=n>6*n+3; # Wesley Ivan Hurt, Sep 29 2013


MATHEMATICA

Range[3, 500, 6] (* Vladimir Joseph Stephan Orlovsky, May 26 2011 *)
Table[6 n + 3, {n, 0, 100}] (* Wesley Ivan Hurt, Sep 29 2013 *)
LinearRecurrence[{2, 1}, {3, 9}, 55] (* Ray Chandler, Jul 17 2015 *)
CoefficientList[Series[3 (x + 1)/(x  1)^2, {x, 0, 60}], x] (* Robert G. Wilson v, Dec 14 2016 *)


PROG

From Wesley Ivan Hurt, Sep 29 2013: (Start)
(Haskell)
a016945 = (+ 3) . (* 6)
a016945_list = [3, 9 ..]
(MAGMA) [6*n+3 : n in [0..100]];
(Maxima) makelist(6*n+3, n, 0, 100);
(PARI) {a(n) = 6*n + 3}
(End)
(PARI) x='x+O('x^99); Vec(3*(1+x)/(1x)^2) \\ Altug Alkan, Apr 08 2016


CROSSREFS

Third row of A092260.
Cf. A008588, A016921, A016933, A016957, A016969.
Subsequence of A061641; complement of A047263; bisection of A047241.
Cf. A000225.  Loren Pearson, Jul 02 2009
Cf. A020639.  Zak Seidov, Apr 29 2015
Cf. A004442.
Sequence in context: A030594 A032676 A228935 * A222640 A110108 A162843
Adjacent sequences: A016942 A016943 A016944 * A016946 A016947 A016948


KEYWORD

nonn,easy,changed


AUTHOR

N. J. A. Sloane


STATUS

approved



