A047389 -id:A047389 - cp's OEIS frontend

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

N. J. A. Sloane

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

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

Third row of A092260.
Cf. A004442, A008588, A010503, A016921, A016933, A016957, A016969, A019679, A115754.
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.

List([0..60], n-> 3*(1+2*n)); # G. C. Greubel, Sep 18 2019

a016945 = (+ 3) . (* 6)
a016945_list = [3, 9 ..]
-- Wesley Ivan Hurt, Sep 29 2013

[6*n+3 : n in [0..60]]; // Wesley Ivan Hurt, Sep 29 2013

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

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 *)

makelist(6*n+3, n, 0, 60); /* Wesley Ivan Hurt, Sep 29 2013 */

{a(n) = 6*n + 3} \\ Wesley Ivan Hurt, Sep 29 2013

x='x+O('x^60); Vec(3*(1+x)/(1-x)^2) \\ Altug Alkan, Apr 08 2016

[3*(1+2*n) for n in (0..60)] # G. C. Greubel, Sep 18 2019

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
From Amiram Eldar, Nov 22 2024: (Start)
Product_{n>=0} (1 - (-1)^n/a(n)) = sqrt(2)/2 (A010503).
Product_{n>=0} (1 + (-1)^n/a(n)) = sqrt(3/2) (A115754). (End)
a(n) = (n+2)^2 - (n-1)^2. - Alexander Yutkin, Mar 15 2025

A047307 Numbers that are congruent to {3, 4, 5, 6} mod 7.

Original entry on oeis.org

3, 4, 5, 6, 10, 11, 12, 13, 17, 18, 19, 20, 24, 25, 26, 27, 31, 32, 33, 34, 38, 39, 40, 41, 45, 46, 47, 48, 52, 53, 54, 55, 59, 60, 61, 62, 66, 67, 68, 69, 73, 74, 75, 76, 80, 81, 82, 83, 87, 88, 89, 90, 94, 95, 96, 97, 101, 102, 103, 104, 108, 109, 110, 111

Offset: 1

N. J. A. Sloane

Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Cf. A047288, A047389.

[n : n in [0..150] | n mod 7 in [3, 4, 5, 6]]; // Wesley Ivan Hurt, Jun 02 2016

A047307:=n->(14*n+1-3*I^(2*n)-(3-3*I)*I^(-n)-(3+3*I)*I^n)/8: seq(A047307(n), n=1..100); # Wesley Ivan Hurt, Jun 02 2016

Table[(14n+1-3*I^(2*n)-(3-3*I)*I^(-n)-(3+3*I)*I^n)/8, {n, 80}] (* Wesley Ivan Hurt, Jun 02 2016 *)
LinearRecurrence[{1,0,0,1,-1},{3,4,5,6,10},70] (* Harvey P. Dale, Dec 28 2024 *)

G.f.: x*(3+x+x^2+x^3+x^4) / ( (1+x)*(1+x^2)*(x-1)^2 ). - R. J. Mathar, Oct 25 2011
From Wesley Ivan Hurt, Jun 02 2016: (Start)
a(n) = a(n-1) + a(n-4) - a(n-5) for n>5.
a(n) = (14*n+1-3*i^(2*n)-(3-3*i)*i^(-n)-(3+3*i)*i^n)/8 where i=sqrt(-1).
a(2k) = A047288(k), a(2k-1) = A047389(k). (End)

A047362 Numbers that are congruent to {2, 3, 4, 5} mod 7.

Original entry on oeis.org

2, 3, 4, 5, 9, 10, 11, 12, 16, 17, 18, 19, 23, 24, 25, 26, 30, 31, 32, 33, 37, 38, 39, 40, 44, 45, 46, 47, 51, 52, 53, 54, 58, 59, 60, 61, 65, 66, 67, 68, 72, 73, 74, 75, 79, 80, 81, 82, 86, 87, 88, 89, 93, 94, 95, 96, 100, 101, 102, 103, 107, 108, 109, 110

Offset: 1

N. J. A. Sloane

Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Cf. A047348, A047389.

[n : n in [0..150] | n mod 7 in [2, 3, 4, 5]]; // Wesley Ivan Hurt, Jun 03 2016

A047362:=n->(14*n-7-3*(I^(2*n)+(1-I)*I^(-n)+(1+I)*I^n))/8: seq(A047362(n), n=1..100); # Wesley Ivan Hurt, Jun 03 2016

Select[Range[100], MemberQ[{2,3,4,5}, Mod[#,7]]&] (* or *) LinearRecurrence[{1,0,0,1,-1}, {2,3,4,5,9}, 60] (* Harvey P. Dale, Oct 03 2015 *)

G.f.: x*(2*x^2+3*x+2)*(x^2-x+1) / ( (1+x)*(x^2+1)*(x-1)^2 ). - R. J. Mathar, Dec 04 2011
From Wesley Ivan Hurt, Jun 03 2016: (Start)
a(n) = a(n-1) + a(n-4) - a(n-5) for n>5.
a(n) = (14*n-7-3*(i^(2*n)+(1-i)*i^(-n)+(1+i)*i^n))/8 where i=sqrt(-1).
a(2k) = A047389(k), a(2k-1) = A047348(k). (End)

A047365 Numbers that are congruent to {0, 3, 4, 5} mod 7.

Original entry on oeis.org

0, 3, 4, 5, 7, 10, 11, 12, 14, 17, 18, 19, 21, 24, 25, 26, 28, 31, 32, 33, 35, 38, 39, 40, 42, 45, 46, 47, 49, 52, 53, 54, 56, 59, 60, 61, 63, 66, 67, 68, 70, 73, 74, 75, 77, 80, 81, 82, 84, 87, 88, 89, 91, 94, 95, 96, 98, 101, 102, 103, 105, 108, 109, 110

Offset: 1

N. J. A. Sloane

Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Cf. A047345, A047389.

[n : n in [0..150] | n mod 7 in [0, 3, 4, 5]]; // Wesley Ivan Hurt, Jun 04 2016

A047365:=n->(14*n-11+I^(2*n)-(3+I)*I^(-n)-(3-I)*I^n)/8: seq(A047365(n), n=1..100); # Wesley Ivan Hurt, Jun 04 2016

Select[Range[0,100], MemberQ[{0,3,4,5}, Mod[#,7]]&] (* or *) LinearRecurrence[{1,0,0,1,-1}, {0,3,4,5,7}, 60] (* Harvey P. Dale, May 26 2012 *)

G.f.: x^2*(3+x+x^2+2*x^3) / ( (1+x)*(x^2+1)*(x-1)^2 ). - R. J. Mathar, Dec 04 2011
a(1)=0, a(2)=3, a(3)=4, a(4)=5, a(5)=7, a(n)=a(n-1)+a(n-4)-a(n-5) for n>5. - Harvey P. Dale, May 26 2012
From Wesley Ivan Hurt, Jun 04 2016: (Start)
a(n) = (14*n-11+i^(2*n)-(3+i)*i^(-n)-(3-i)*i^n)/8 where i=sqrt(-1).
a(2k) = A047389(k), a(2k-1) = A047345(k). (End)

A047366 Numbers that are congruent to {1, 3, 4, 5} mod 7.

Original entry on oeis.org

1, 3, 4, 5, 8, 10, 11, 12, 15, 17, 18, 19, 22, 24, 25, 26, 29, 31, 32, 33, 36, 38, 39, 40, 43, 45, 46, 47, 50, 52, 53, 54, 57, 59, 60, 61, 64, 66, 67, 68, 71, 73, 74, 75, 78, 80, 81, 82, 85, 87, 88, 89, 92, 94, 95, 96, 99, 101, 102, 103, 106, 108, 109, 110

Offset: 1

N. J. A. Sloane

Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Cf. A047346, A047389.

[n : n in [0..150] | n mod 7 in [1, 3, 4, 5]]; // Wesley Ivan Hurt, May 24 2016

A047366:=n->(14*n-9-I^(2*n)-(3-I)*I^(-n)-(3+I)*I^n)/8: seq(A047366(n), n=1..100); # Wesley Ivan Hurt, May 24 2016

Table[(14n-9-I^(2n)-(3-I)*I^(-n)-(3+I)*I^n)/8, {n, 80}] (* Wesley Ivan Hurt, May 24 2016 *)
Select[Range@ 120, MemberQ[{1, 3, 4, 5}, Mod[#, 7]] &] (* Michael De Vlieger, May 24 2016 *)

G.f.: x*(1+2*x+x^2+x^3+2*x^4) / ( (1+x)*(x^2+1)*(x-1)^2 ). - R. J. Mathar, Dec 04 2011
From Wesley Ivan Hurt, May 24 2016: (Start)
a(n) = a(n-1) + a(n-4) - a(n-5) for n>5.
a(n) = (14n-9-i^(2n)-(3-i)*i^(-n)-(3+i)*i^n)/8 where i=sqrt(-1).
a(2k) = A047389(k), a(2k-1) = A047346(k). (End)
E.g.f.: (8 + sin(x) - 3*cos(x) + (7*x - 4)*sinh(x) + (7*x - 5)*cosh(x))/4. - Ilya Gutkovskiy, May 25 2016

More terms from Wesley Ivan Hurt, May 24 2016

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

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A047307 Numbers that are congruent to {3, 4, 5, 6} mod 7.

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A047362 Numbers that are congruent to {2, 3, 4, 5} mod 7.

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A047365 Numbers that are congruent to {0, 3, 4, 5} mod 7.

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A047366 Numbers that are congruent to {1, 3, 4, 5} mod 7.

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