A016945 a(n) = 6*n+3.

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

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 2-set Y and a 3-set Z are disjoint subsets of an n-set X then a(n-4) is the number of 3-subsets of X intersecting both Y and Z. - Milan Janjic, Sep 08 2007

Leaves of the Odd Collatz-Tree: 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

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

Also numbers k such that t^k == -1 (mod 7), where t is a member of A047389. - Bruno Berselli, Dec 28 2017

Links

Table of n, a(n) for n=0..54.

Milan Janjic, Two Enumerative Functions.

Friedrich L. Bauer, Der (ungerade) Collatz-Baum, Informatik Spektrum 31 (Springer, April 2008), pp. 379-384.

Tanya Khovanova, Recursive Sequences.

Luis Manuel Rivera, Integer sequences and k-commuting 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(n-1) for n>0, a(0)=3. - Vincenzo Librandi, Nov 20 2010

G.f.: 3*(1+x)/(1-x)^2. - Mario C. Enriquez, Dec 14 2016

E.g.f.: 3*(1 + 2*x)*exp(x). - G. C. Greubel, Sep 18 2019

Sum_{n>=0} (-1)^n/a(n) = Pi/12 (A019679). - Amiram Eldar, Dec 10 2021

Maple

seq(6*n+3, n=0..60); # Dennis P. Walsh, Aug 20 2013
A016945:=n->6*n+3; # Wesley Ivan Hurt, Sep 29 2013

Mathematica

Range[3, 350, 6] (* Vladimir Joseph Stephan Orlovsky, May 26 2011 *)
Table[6n+3, {n, 0, 60}] (* Wesley Ivan Hurt, Sep 29 2013 *)
LinearRecurrence[{2, -1}, {3, 9}, 55] (* Ray Chandler, Jul 17 2015 *)
CoefficientList[Series[3(1+x)/(1-x)^2, {x, 0, 60}], x] (* Robert G. Wilson v, Dec 14 2016 *)

Prog

(Haskell)
a016945 = (+ 3) . (* 6)
a016945_list = [3, 9 ..]
-- Wesley Ivan Hurt, Sep 29 2013
(Magma) [6*n+3 : n in [0..60]]; // Wesley Ivan Hurt, Sep 29 2013
(Maxima) makelist(6*n+3, n, 0, 60); /* Wesley Ivan Hurt, Sep 29 2013 */
(PARI) {a(n) = 6*n + 3} \\ Wesley Ivan Hurt, Sep 29 2013
(PARI) x='x+O('x^60); Vec(3*(1+x)/(1-x)^2) \\ Altug Alkan, Apr 08 2016
(Sage) [3*(1+2*n) for n in (0..60)] # G. C. Greubel, Sep 18 2019
(GAP) List([0..60], n-> 3*(1+2*n)); # G. C. Greubel, Sep 18 2019

Crossrefs

Third row of A092260.

Cf. A004442, A008588, A016921, A016933, A016957, A016969, A019679.

Subsequence of A061641; complement of A047263; bisection of A047241.

Cf. A000225. - Loren Pearson, Jul 02 2009

Cf. A020639. - Zak Seidov, Apr 29 2015

Odd numbers in A355200.

Sequence in context: A032676 A366849 A228935 * A222640 A110108 A162843

Adjacent sequences: A016942 A016943 A016944 * A016946 A016947 A016948

Keyword

nonn,easy

Author

N. J. A. Sloane

Status

approved