A047263 -id:A047263 - 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

A265187 Nonnegative m for which 2floor(m^2/11) = floor(2m^2/11).

Original entry on oeis.org

0, 1, 2, 4, 5, 6, 7, 9, 10, 11, 12, 13, 15, 16, 17, 18, 20, 21, 22, 23, 24, 26, 27, 28, 29, 31, 32, 33, 34, 35, 37, 38, 39, 40, 42, 43, 44, 45, 46, 48, 49, 50, 51, 53, 54, 55, 56, 57, 59, 60, 61, 62, 64, 65, 66, 67, 68, 70, 71, 72, 73, 75, 76, 77, 78, 79, 81, 82, 83, 84

Offset: 1

Bruno Berselli, Dec 04 2015

Also, nonnegative m not congruent to 3 or 8 (mod 11).
Integers x >= 0 satisfying k*floor(x^2/11) = floor(k*x^2/11) with k >= 0:
k = 0, 1:  x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, ...   (A001477);
k = 2:     x = 0, 1, 2, 4, 5, 6, 7, 9, 10, 11, 12, 13, ... (this sequence);
k = 3:     x = 0, 1, 5, 6, 10, 11, 12, 16, 17, 21, 22, ... (A265188);
k = 4..10: x = 0, 1, 10, 11, 12, 21, 22, 23, 32, 33, ...   (A112654);
k > 10:    x = 0, 11, 22, 33, 44, 55, 66, 77, 88, 99, ...  (A008593).
Primes in sequence: 2, 5, 7, 11, 13, 17, 23, 29, 31, 37, 43, 53, 59, ...

Bruno Berselli, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,0,1,-1).

Cf. A008593, A112654, A265188.

Cf. similar sequences provided by 2*floor(m^2/h) = floor(2*m^2/h): A005843 (h=2), A001477 (h=3,4), A008854 (h=5), A047266 (h=6), A047299 (h=7), A042965 (h=8), A060464 (h=9), A237415 (h=10), this sequence (h=11), A047263 (h=12).

[n: n in [0..100] | 2*Floor(n^2/11) eq Floor(2*n^2/11)];

Select[Range[0, 100], 2 Floor[#^2/11] == Floor[2 #^2/11] &]
Select[Range[0, 100], ! MemberQ[{3, 8}, Mod[#, 11]] &]
LinearRecurrence[{1, 0, 0, 0, 0, 0, 0, 0, 1, -1}, {0, 1, 2, 4, 5, 6, 7, 9, 10, 11}, 80]

is(n)=2*(n^2\11) == (2*n^2)\11 \\ Anders Hellström, Dec 05 2015

[n for n in (0..100) if 2*floor(n^2/11) == floor(2*n^2/11)]

G.f.: x^2*(1 + x + 2*x^2 + x^3 + x^4 + x^5 + 2*x^6 + x^7 + x^8)/((1 - x)^2*(1 + x + x^2)*(1 + x^3 + x^6)).

a(n) = a(n-1) + a(n-9) - a(n-10) for n>10.

A364123 Stolarsky-Niven numbers: numbers that are divisible by the number of 1's in their Stolarsky representation (A364121).

Original entry on oeis.org

2, 4, 6, 8, 9, 12, 14, 16, 20, 22, 24, 27, 30, 36, 38, 40, 42, 44, 48, 54, 56, 57, 60, 65, 69, 72, 75, 80, 84, 85, 90, 92, 96, 98, 100, 102, 104, 108, 112, 116, 120, 124, 126, 132, 136, 138, 145, 147, 150, 153, 155, 159, 160, 175, 180, 185, 190, 195, 196, 205

Offset: 1

Amiram Eldar, Jul 07 2023

Numbers k such that A200649(k) | k.

Fibonacci(k) + 1 is a term if k !== 3 (mod 6) (i.e., k is in A047263).

			4 is a term since its Stolarsky representation, A364121(4) = 10, has one 1 and 4 is divisible by 1.
6 is a term since its Stolarsky representation, A364121(6) = 101, has 2 1's and 6 is divisible by 2.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A047263, A200649, A364121.
Subsequences: A364124, A364125, A364126.
Similar sequences: A005349, A049445, A064150, A064438, A064481, A118363, A328208, A328212, A331085, A333426, A342726, A334308, A331728, A342426, A344341, A351714, A351719, A352089, A352107, A352320, A352342, A352508.

stol[n_] := stol[n] = If[n == 1, {}, If[n != Round[Round[n/GoldenRatio]*GoldenRatio], Join[stol[Floor[n/GoldenRatio^2] + 1], {0}], Join[stol[Round[n/GoldenRatio]], {1}]]];
stolNivQ[n_] := n > 1 && Divisible[n, Total[stol[n]]];
Select[Range[200], stolNivQ]

stol(n) = {my(phi=quadgen(5)); if(n==1, [], if(n != round(round(n/phi)*phi), concat(stol(floor(n/phi^2) + 1), [0]), concat(stol(round(n/phi)), [1])));}
isA364123(n) = n > 1 && !(n % vecsum(stol(n)));

A047252 Numbers that are congruent to {0, 1, 3, 4, 5} mod 6.

Original entry on oeis.org

0, 1, 3, 4, 5, 6, 7, 9, 10, 11, 12, 13, 15, 16, 17, 18, 19, 21, 22, 23, 24, 25, 27, 28, 29, 30, 31, 33, 34, 35, 36, 37, 39, 40, 41, 42, 43, 45, 46, 47, 48, 49, 51, 52, 53, 54, 55, 57, 58, 59, 60, 61, 63, 64, 65, 66, 67, 69, 70, 71, 72, 73, 75, 76, 77, 78, 79

Offset: 1

N. J. A. Sloane

nonn

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

Cf. A047263.

seq(floor((6*n-3)/5), n= 1..70); # Gary Detlefs, Mar 07 2010

Flatten[#+{0,1,3,4,5}&/@(6Range[0,12])]  (* Harvey P. Dale, Apr 21 2011 *)

a(n) = floor((6n-3)/5). - Gary Detlefs, Mar 07 2010
G.f.: x^2*(1+x)*(x^3+x+1) / ( (x^4+x^3+x^2+x+1)*(x-1)^2 ). - R. J. Mathar, Oct 08 2011
Sum_{n>=2} (-1)^n/a(n) = log(2+sqrt(3))/sqrt(3) - (3-2*sqrt(3))*Pi/36. - Amiram Eldar, Dec 17 2021

A239476 Number of values of k such that 2^k + (6n+3) and (6n+3)*2^k - 1 are both prime, k < 6n+3.

Original entry on oeis.org

2, 3, 5, 3, 7, 3, 1, 6, 2, 6, 6, 5, 4, 3, 2, 4, 5, 4, 1, 3, 2, 3, 3, 1, 7, 2, 2, 10, 1, 4, 1, 2, 4, 0, 3, 5, 1, 3, 4, 3, 5, 1, 5, 4, 6, 4, 2, 1, 2, 4, 4, 1, 5, 1, 4, 3, 2, 4, 3, 5, 6, 2, 6, 3, 2, 2, 2, 1, 4, 2, 1, 2, 3, 3, 4, 4, 4, 2, 3, 4, 7, 5, 2, 1, 4, 2, 1, 6, 2, 3, 2, 3, 5, 0, 5, 0, 0, 2, 2, 4, 4, 3

Offset: 0

Ilya Lopatin and Juri-Stepan Gerasimov, Mar 20 2014

nonn

Number of values of k such that 2^k + A047263(n) and (A047263(n))*2^k + 1 are both prime, k < 6n+3, where A047263(n) is complement of 6m+3 : 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...

			a(0) = 2 because
1) 2^1 + (6*0+3) = 5 and (6*0+3)*2^1 - 1 = 5 for k = 1 < (6*0+3);
2) 2^2 + (6*0+3) = 7 and (6*0+3)*2^2 - 1 = 11 for k = 2 < (6*0+3).
a(1) = 3 because
1) 2^1 + (6*1+3) = 11 and (6*1+3)*2^1 - 1 = 17 for k = 1 < (6*1+3);
2) 2^3 + (6*1+3) = 17 and (6*1+3)*2^3 - 1 = 71 for k = 3 < (6*1+3);
3) 2^7 + (6*1+3) = 137 and (6*1+3)*2^7 - 1 = 1151 for k = 7 < (6*1+3).
a(2) = 5 because
1) 2^1 + (6*2+3) = 17 and (6*2+3)*2^1 - 1 = 29 for k = 1 < (6*2+3);
2) 2^2 + (6*2+3) = 19 and (6*2+3)*2^2 - 1 = 59 for k = 2 < (6*2+3);
3) 2^4 + (6*2+3) = 31 and (6*2+3)*2^4 - 1 = 239 for k = 4 < (6*2+3);
4) 2^5 + (6*2+3) = 37 and (6*2+3)*2^5 - 1 = 479 for k = 5 < (6*2+3);
5) 2^10 + (6*2+3) = 1039 and (6*2+3)*2^10 - 1 = 15359 for k = 10 < (3*2+3).

Cf. A016945, A047263.

for(n=0, 100, m=0; for(k=0, 6*n+2, if(isprime(2^k+6*n+3) && isprime((6*n+3)*2^k-1), m++)); print1(m,", ")) \\ Colin Barker, Mar 25 2014

Offset changed to 0 by Colin Barker, Mar 25 2014

cp's OEIS Frontend

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

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A265187 Nonnegative m for which 2*floor(m^2/11) = floor(2*m^2/11).

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A364123 Stolarsky-Niven numbers: numbers that are divisible by the number of 1's in their Stolarsky representation (A364121).

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A047252 Numbers that are congruent to {0, 1, 3, 4, 5} mod 6.

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A239476 Number of values of k such that 2^k + (6n+3) and (6n+3)*2^k - 1 are both prime, k < 6n+3.

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A265187 Nonnegative m for which 2floor(m^2/11) = floor(2m^2/11).