A047240 -id:A047240 - cp's OEIS frontend

A113841 a(n) = a(n-1) + 2^A047240(n) for n>1, a(1)=1.

1, 3, 7, 71, 199, 455, 4551, 12743, 29127, 291271, 815559, 1864135, 18641351, 52195783, 119304647, 1193046471, 3340530119, 7635497415, 76354974151, 213793927623, 488671834567, 4886718345671, 13682811367879, 31274997412295

Offset: 1

Artur Jasinski, Jan 27 2006

Vincenzo Librandi, Table of n, a(n) for n = 1..300
Vladimir Pletser, Congruence conditions on the number of terms in sums of consecutive squared integers equal to squared integers, arXiv:1409.7969 [math.NT], 2014.
Index entries for linear recurrences with constant coefficients, signature (1, 0, 64, -64).

Cf. A099974, A112627, A080355, A080567, A099969, A099970, A099971.

CoefficientList[Series[(1 + 2 x + 4 x^2) / ((-1 + x) (-1 + 4 x) (1 + 4 x + 16 x^2)), {x, 0, 30}], x] (* Vincenzo Librandi, May 19 2013 *)
LinearRecurrence[{1,0,64,-64},{1,3,7,71},30] (* Harvey P. Dale, Nov 18 2013 *)

G.f.: x*(1+2*x+4*x^2)/((-1+x)*(-1+4*x)*(1+4*x+16*x^2)). - Vaclav Kotesovec, Nov 28 2012

a(1)=1, a(2)=3, a(3)=7, a(4)=71, a(n)=a(n-1)+64*a(n-3)-64*a(n-4). - Harvey P. Dale, Nov 18 2013

Edited with better definition and offset corrected by Omar E. Pol, Jan 08 2009

A113854 a(n) = sum(2^(A047240(i)-1), i=1..n).

Original entry on oeis.org

1, 3, 35, 99, 227, 2275, 6371, 14563, 145635, 407779, 932067, 9320675, 26097891, 59652323, 596523235, 1670265059, 3817748707, 38177487075, 106896963811, 244335917283, 2443359172835, 6841405683939, 15637498706147

Offset: 1

Artur Jasinski, Jan 27 2006

			a(2) = 2^(A047240(1)-1) + 2^(A047240(2)-1) = 2^0 + 2^1 = 3

Vincenzo Librandi, Table of n, a(n) for n = 1..300
Index entries for linear recurrences with constant coefficients, signature (1,0,64,-64).

Cf. A047240.

a = {}; s = 0; For[n = 1, n < 48, n++, If[Length[Intersection[{Mod[n, 6]}, {1, 2, 0}]] > 0, s = s + 2^(n - 1); AppendTo[a, s]]]; a
CoefficientList[Series[(1 + 2 x + 32 x^2)/((-1 + x) (-1 + 4 x) (1 + 4 x + 16 x^2)), {x, 0, 25}], x] (* Vincenzo Librandi, May 20 2013 *)
LinearRecurrence[{1,0,64,-64},{1,3,35,99},30] (* Harvey P. Dale, May 11 2025 *)

G.f.: x*(1+2*x+32*x^2)/((-1+x)*(-1+4*x)*(1+4*x+16*x^2)). - Vaclav Kotesovec, Nov 28 2012

Edited by Stefan Steinerberger, Jul 24 2007

A128205 a(n) = 2^(n-1)*A047240(n).

Original entry on oeis.org

0, 1, 4, 24, 56, 128, 384, 832, 1792, 4608, 9728, 20480, 49152, 102400, 212992, 491520, 1015808, 2097152, 4718592, 9699328, 19922944, 44040192, 90177536, 184549376, 402653184, 822083584, 1677721600, 3623878656, 7381975040, 15032385536, 32212254720

Offset: 0

Paul Barry, Feb 19 2007

-a(n) is the Hankel transform of A030662(n) = binomial(2*n,n)-1.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,0,8,-16).

Cf. A047240, A030662.

a047240[n_] := 6 Floor[n/3] + Mod[n, 3]
a128205[n_] := Map[2^(#-1) a047240[#]&, Range[0, n]]
a128205[25] (* data *) (* Hartmut F. W. Hoft, Mar 13 2017 *)
LinearRecurrence[{2,0,8,-16},{0,1,4,24},40] (* Harvey P. Dale, Feb 13 2024 *)

concat(0, Vec(x*(1 + 2*x + 16*x^2) / ((1 - 2*x)^2*(1 + 2*x + 4*x^2)) + O(x^40))) \\ Colin Barker, Mar 13 2017

a(n) = 2^(n-1)*(cos(2*Pi*n/3) + sqrt(3)*sin(2*Pi*n/3)/3 + 2n - 1);

O.g.f.: x(1+2x+16x^2)/((2x-1)^2*(4x^2+2x+1)). a(n) = 2a(n-1) + 8a(n-3) - 16a(n-4). - R. J. Mathar, Apr 28 2008

A047234 Numbers that are congruent to {0, 1, 4} mod 6.

Original entry on oeis.org

0, 1, 4, 6, 7, 10, 12, 13, 16, 18, 19, 22, 24, 25, 28, 30, 31, 34, 36, 37, 40, 42, 43, 46, 48, 49, 52, 54, 55, 58, 60, 61, 64, 66, 67, 70, 72, 73, 76, 78, 79, 82, 84, 85, 88, 90, 91, 94, 96, 97, 100, 102, 103, 106, 108, 109, 112, 114, 115, 118, 120, 121, 124

Offset: 1

N. J. A. Sloane

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

Cf. A047240, A047242.

[n : n in [0..150] | n mod 6 in [0, 1, 4]]; // Wesley Ivan Hurt, Jun 14 2016

A047234:=n->(6*n-7+cos(2*n*Pi/3)+sqrt(3)*sin(2*n*Pi/3))/3: seq(A047234(n), n=1..100); # Wesley Ivan Hurt, Jun 14 2016

Select[Range[0, 200], Mod[#, 6] == 0 || Mod[#, 6] == 1 || Mod[#, 6] == 4 &] (* Vladimir Joseph Stephan Orlovsky, Jul 07 2011 *)
LinearRecurrence[{1, 0, 1, -1}, {0, 1, 4, 6}, 100] (* Vincenzo Librandi, Jun 15 2016 *)
#+{0,1,4}&/@(6*Range[0,20])//Flatten (* Harvey P. Dale, Jul 25 2019 *)

a(n)=(n-1)\3*6+[4,0,1][n%3+1] \\ Charles R Greathouse IV, Jun 11 2015

Equals partial sums of (0, 1, 3, 2, 1, 3, 2, 1, 3, 2, ...). - Gary W. Adamson, Jun 19 2008
G.f.: x^2*(1+x)*(2*x+1)/((1+x+x^2)*(x-1)^2). - R. J. Mathar, Oct 08 2011
From Wesley Ivan Hurt, Jun 14 2016: (Start)
a(n) = a(n-1) + a(n-3) - a(n-4) for n>4.
a(n) = (6*n-7+cos(2*n*Pi/3)+sqrt(3)*sin(2*n*Pi/3))/3.
a(3k) = 6k-2, a(3k-1) = 6k-5, a(3k-2) = 6k-6. (End)
Sum_{n>=2} (-1)^n/a(n) = (3-sqrt(3))*Pi/18 + log(2)/3 + log(2+sqrt(3))/(2*sqrt(3)). - Amiram Eldar, Dec 14 2021
E.g.f.: (6 + exp(x)*(6*x - 7) + exp(-x/2)*(cos(sqrt(3)*x/2) + sqrt(3)*sin(sqrt(3)*x/2)))/3. - Stefano Spezia, Jul 26 2024

A047242 Numbers that are congruent to {0, 1, 3} mod 6.

Original entry on oeis.org

0, 1, 3, 6, 7, 9, 12, 13, 15, 18, 19, 21, 24, 25, 27, 30, 31, 33, 36, 37, 39, 42, 43, 45, 48, 49, 51, 54, 55, 57, 60, 61, 63, 66, 67, 69, 72, 73, 75, 78, 79, 81, 84, 85, 87, 90, 91, 93, 96, 97, 99, 102, 103, 105, 108, 109, 111, 114, 115, 117, 120, 121, 123

Offset: 1

N. J. A. Sloane

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

Cf. A047234, A047240, A214090.

Cf. A047261 (complement).

a047242 n = a047242_list !! n
a047242_list = elemIndices 0 a214090_list
-- Reinhard Zumkeller, Jul 06 2012

[n-1+Floor((n-1)/3)+Floor((2*n-2)/3) : n in [1..50]]; // Wesley Ivan Hurt, Dec 03 2014

A047242:=n->n-1+floor((n-1)/3)+floor((2*n-2)/3): seq(A047242(n), n=1..50); # Wesley Ivan Hurt, Dec 03 2014

Select[Range[0, 200], Mod[#, 6] == 0 || Mod[#, 6] == 1 || Mod[#, 6] == 3 &] (* Vladimir Joseph Stephan Orlovsky, Jul 07 2011 *)

Equals partial sums of (0, 1, 2, 3, 1, 2, 3, 1, 2, 3, ...). - Gary W. Adamson, Jun 19 2008
G.f.: x^2*(1+2*x+3*x^2)/((1+x+x^2)*(x-1)^2). - R. J. Mathar, Oct 08 2011
A214090(a(n)) = 0. - Reinhard Zumkeller, Jul 06 2012
a(n) = a(n-1) + a(n-3) - a(n-4), n>4. - Wesley Ivan Hurt, Dec 03 2014
a(n) = n-1 + floor((n-1)/3) + floor((2n-2)/3). - Wesley Ivan Hurt, Dec 03 2014
From Wesley Ivan Hurt, Jun 13 2016: (Start)
a(n) = (6*n-8-cos(2*n*Pi/3)+sqrt(3)*sin(2*n*Pi/3))/3.
a(3k) = 6k-3, a(3k-1) = 6k-5, a(3k-2) = 6k-6. (End)
a(n) = 2*n - 2 - sign((n-1) mod 3). - Wesley Ivan Hurt, Sep 26 2017
Sum_{n>=2} (-1)^n/a(n) = Pi/12 + log(2)/6 + log(2+sqrt(3))/(2*sqrt(3)). - Amiram Eldar, Dec 14 2021
E.g.f.: (9 + exp(x)*(6*x - 8) - exp(-x/2)*(cos(sqrt(3)*x/2) - sqrt(3)*sin(sqrt(3)*x/2)))/3. - Stefano Spezia, Jul 26 2024

A281899 a(n) = n + 6*floor(n/3).

Original entry on oeis.org

0, 1, 2, 9, 10, 11, 18, 19, 20, 27, 28, 29, 36, 37, 38, 45, 46, 47, 54, 55, 56, 63, 64, 65, 72, 73, 74, 81, 82, 83, 90, 91, 92, 99, 100, 101, 108, 109, 110, 117, 118, 119, 126, 127, 128, 135, 136, 137, 144, 145, 146, 153, 154, 155, 162, 163, 164, 171, 172, 173, 180, 181, 182, 189

Offset: 0

Bruno Berselli, Feb 06 2017

Equivalently, numbers that are congruent to {0, 1, 2} mod 9.
Also numbers m such that floor(m/3) = 3*floor(m/9).
The n-th term is 3*n, 3*n-2 or 3*n-4.
For n > 0, numbers k such that 3 | floor(k/3). - Wesley Ivan Hurt, Dec 01 2020

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

Cf. A002264.
Subsequence of A060464 and A248375.
The first differences are in A105395.
Cf. similar sequences with formula n+i*floor(n/3): A004773 (i=1), A047217 (i=2), A047240 (i=3), A047354 (i=4), A047469 (i=5), this sequence (i=6).
Cf. numbers that are congruent to {0, 1, 2} mod j: the sequences are listed in the previous row for j = 4..9, respectively.

Magma
```
[n+6*(n div 3): n in [0..70]];
```

A281899:=n->n+6*floor(n/3): seq(A281899(n), n=0..100); # Wesley Ivan Hurt, Feb 09 2017

Table[n + 6 Floor[n/3], {n, 0, 70}]
LinearRecurrence[{1,0,1,-1},{0,1,2,9},90] (* Harvey P. Dale, Feb 25 2018 *)

Maxima
```
makelist(n+6*floor(n/3), n, 0, 70);
```

a(n)=n\3*6 + n \\ Charles R Greathouse IV, Feb 07 2017

Python
```
[n+6*int(n/3) for n in range(70)]
```
Sage
```
[n+6*floor(n/3) for n in range(70)]
```

G.f.: x*(1 + x + 7*x^2)/((1 - x)^2*(1 + x + x^2)).
a(n) = a(n-1) + a(n-3) - a(n-4).
a(n) = 3*n - 2*(n mod 3). In general, n + 3*h*floor(n/3) = (h+1)*n - h*(n mod 3).
a(n) + a(n+s) = a(2*n+s-1) + 1, where s is nonnegative and not divisible by 3. Example: for s=14, a(n) + a(n+14) = a(2*n+13) + 1; for n=3, a(3) + a(17) = a(19) + 1 = 9 + 47 = 55 + 1 = 56.
a(6*k+r) = 18*k + a(r), where 0 <= r <= 5.
a(n) = 7*A002264(n) + A002264(n+1) + A002264(n+2).

A047266 Numbers that are congruent to {0, 1, 5} mod 6.

Original entry on oeis.org

0, 1, 5, 6, 7, 11, 12, 13, 17, 18, 19, 23, 24, 25, 29, 30, 31, 35, 36, 37, 41, 42, 43, 47, 48, 49, 53, 54, 55, 59, 60, 61, 65, 66, 67, 71, 72, 73, 77, 78, 79, 83, 84, 85, 89, 90, 91, 95, 96, 97, 101, 102, 103, 107, 108, 109, 113, 114, 115, 119, 120, 121, 125

Offset: 1

N. J. A. Sloane

a(n+3) is the Hankel transform of A005773(n+3). - Paul Barry, Nov 04 2008
The numbers m == 0, 2 or 10 mod 12 (the doubles of this sequence, that is, 10, 12, 14, 22, 24, 26, 34, ...) have the property that exactly 1/4 of the 3-part partitions of m form the sides of a triangle. See Mathematics Stack Exchange, 2013, link. - Ed Pegg Jr, Dec 19 2013
Row sum of a triangle where two rules build the triangle. #1 Start with the value "1" at the top of the triangle. #2 Require every "triple" to contain the values 1,2,3 (see link below). Compare with A136289 that has "3" at the apex. - Craig Knecht, Oct 17 2015
Nonnegative m such that floor(k*m^2/6) = k*floor(m^2/6), where k = 2, 3, 4 or 5. - Bruno Berselli, Dec 03 2015

Craig Knecht, Triangular row sum of A204259.
Mathematics Stack Exchange, Prove: Exactly a quarter of 3-part partitions of numbers >2 equal to 0, 2, 10 mod 12 will make a triangle, 2013.
Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).

Cf. A005773, A047240, A047242, A057078, A136289, A204259.

[n : n in [0..150] | n mod 6 in [0, 1, 5]]; // Wesley Ivan Hurt, Jun 13 2016

seq(seq(6*s+j, j=[0,1,5]), s=0..100); # Robert Israel, Dec 01 2014

Select[Range[0, 200], Mod[#, 6] == 0 || Mod[#, 6] == 1 || Mod[#, 6] == 5 &] (* Vladimir Joseph Stephan Orlovsky, Jul 07 2011 *)

concat(0, Vec(x^2*(1+4*x+x^2)/((1+x+x^2)*(x-1)^2) + O(x^100))) \\ Altug Alkan, Oct 17 2015

G.f.: x^2*(1+4*x+x^2) / ((1+x+x^2)*(x-1)^2). - R. J. Mathar, Oct 08 2011
a(n) = 2*(n-1) + A057078(n). - Robert Israel, Dec 01 2014
a(n) = a(n-1) + a(n-3) - a(n-4) for n>4. - Wesley Ivan Hurt, Nov 09 2015
From Wesley Ivan Hurt, Jun 13 2016: (Start)
a(n) = 2*n-2+cos(2*n*Pi/3)+sin(2*n*Pi/3)/sqrt(3).
a(3k) = 6k-1, a(3k-1) = 6k-5, a(3k-2) = 6k-6. (End)
Sum_{n>=2} (-1)^n/a(n) = log(2)/6 + log(2 + sqrt(3))/sqrt(3). - Amiram Eldar, Dec 14 2021

A047244 Numbers that are congruent to {0, 2, 3} mod 6.

Original entry on oeis.org

0, 2, 3, 6, 8, 9, 12, 14, 15, 18, 20, 21, 24, 26, 27, 30, 32, 33, 36, 38, 39, 42, 44, 45, 48, 50, 51, 54, 56, 57, 60, 62, 63, 66, 68, 69, 72, 74, 75, 78, 80, 81, 84, 86, 87, 90, 92, 93, 96, 98, 99, 102, 104, 105, 108, 110, 111, 114, 116, 117, 120, 122, 123

Offset: 1

N. J. A. Sloane

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

Cf. A047240, A047242.

[n : n in [0..130] | n mod 6 in [0, 2, 3]]; // Vincenzo Librandi, Oct 02 2015

A047244:=n->(6*n-7-2*cos(2*n*Pi/3))/3: seq(A047244(n), n=1..100); # Wesley Ivan Hurt, Jun 13 2016

Select[Range[0, 200], Mod[#, 6] == 0 || Mod[#, 6] == 2 || Mod[#, 6] == 3 &] (* Vladimir Joseph Stephan Orlovsky, Jul 07 2011 *)
Select[Range[0, 200], MemberQ[{0, 2, 3}, Mod[#, 6]] &] (* Vincenzo Librandi, Oct 02 2015 *)
LinearRecurrence[{1, 0, 1, -1}, {2, 3, 6, 8}, {0, 20}] (* Eric W. Weisstein, Apr 09 2018 *)
CoefficientList[Series[x (2 + x + 3 x^2)/((-1 + x)^2 (1 + x + x^2)), {x, 0, 20}], x] (* Eric W. Weisstein, Apr 09 2018 *)
Table[(6 n + Cos[2 n Pi/3] + Sqrt[3] Sin[2 n Pi/3] - 1)/3, {n, 0, 20}] (* Eric W. Weisstein, Apr 09 2018 *)

isok(n) = my(m = n % 6); (m==0) || (m==2) || (m==3); \\ Michel Marcus, Oct 02 2015

G.f.: x^2*(2+x+3*x^2) / ((1+x+x^2)*(x-1)^2). - R. J. Mathar, Oct 08 2011
From Wesley Ivan Hurt, Jun 13 2016: (Start)
a(n) = a(n-1) + a(n-3) - a(n-4) for n>4.
a(n) = (6*n-7-2*cos(2*n*Pi/3))/3.
a(3k) = 6k-3, a(3k-1) = 6k-4, a(3k-2) = 6k-6. (End)
E.g.f.: (9 + (6*x - 7)*exp(x) - 2*cos(sqrt(3)*x/2)*(cosh(x/2) - sinh(x/2)))/3. - Ilya Gutkovskiy, Jun 14 2016
Sum_{n>=2} (-1)^n/a(n) = (3-sqrt(3))*Pi/18 + log(2+sqrt(3))/(2*sqrt(3)) + log(2)/3. - Amiram Eldar, Dec 14 2021

A047245 Numbers that are congruent to {1, 2, 3} mod 6.

Original entry on oeis.org

1, 2, 3, 7, 8, 9, 13, 14, 15, 19, 20, 21, 25, 26, 27, 31, 32, 33, 37, 38, 39, 43, 44, 45, 49, 50, 51, 55, 56, 57, 61, 62, 63, 67, 68, 69, 73, 74, 75, 79, 80, 81, 85, 86, 87, 91, 92, 93, 97, 98, 99, 103, 104, 105, 109, 110, 111, 115, 116, 117, 121, 122, 123

Offset: 1

N. J. A. Sloane

a(k)^m is a term iff {a(k) is odd and m is a nonnegative integer} or {m is in A004273}. - Jerzy R Borysowicz, May 08 2023

David Lovler, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,0,1,-1).

Cf. A047240, A047244, A047258 (complement).

[2*n-3+((2*n-3) mod 3) : n in [1..100]]; // Wesley Ivan Hurt, Apr 13 2015

A047245:=n->2*n-3+((2*n-3) mod 3): seq(A047245(n), n=1..100); # Wesley Ivan Hurt, Apr 13 2015

Select[Range[0, 200], Mod[#, 6] == 1 || Mod[#, 6] == 2 || Mod[#, 6] == 3 &] (* Vladimir Joseph Stephan Orlovsky, Jul 07 2011 *)
Flatten[Table[{6n + 1, 6n + 2, 6n + 3}, {n, 0, 19}]] (* Alonso del Arte, Jul 07 2011 *)
Select[Range[0, 200], MemberQ[{1, 2, 3}, Mod[#, 6]] &] (* Vincenzo Librandi, Apr 14 2015 *)

a(n) = 3*floor((n-1)/3) + n; \\ David Lovler, Aug 03 2022

From Johannes W. Meijer, Jun 07 2011: (Start)
a(n) = ceiling(n/3) + ceiling((n-1)/3) + ceiling((n-2)/3) + 3*ceiling((n-3)/3).
G.f.: x*(1+x+x^2+3*x^3)/((x-1)^2*(x^2+x+1)). (End)
a(n) = 3*floor((n-1)/3) + n. - Gary Detlefs, Dec 22 2011
From Wesley Ivan Hurt, Apr 13 2015: (Start)
a(n) = a(n-1) + a(n-3) - a(n-4) for n>4.
a(n) = 2*n-3 + ((2*n-3) mod 3). (End)
From Wesley Ivan Hurt, Jun 13 2016: (Start)
a(n) = 2*n - 2 - cos(2*n*Pi/3) + sin(2*n*Pi/3)/sqrt(3).
a(3k) = 6k-3, a(3k-1) = 6k-4, a(3k-2) = 6k-5. (End)
Sum_{n>=1} (-1)^(n+1)/a(n) = (9-2*sqrt(3))*Pi/36 + log(2+sqrt(3))/(2*sqrt(3)) - log(2)/6. - Amiram Eldar, Dec 14 2021

A293292 Numbers with last digit less than 5 (in base 10).

Original entry on oeis.org

0, 1, 2, 3, 4, 10, 11, 12, 13, 14, 20, 21, 22, 23, 24, 30, 31, 32, 33, 34, 40, 41, 42, 43, 44, 50, 51, 52, 53, 54, 60, 61, 62, 63, 64, 70, 71, 72, 73, 74, 80, 81, 82, 83, 84, 90, 91, 92, 93, 94, 100, 101, 102, 103, 104, 110, 111, 112, 113, 114, 120, 121, 122, 123, 124, 130

Offset: 1

Bruno Berselli, Oct 05 2017

Equivalently, numbers k such that floor(k/5) = 2*floor(k/10).
After 0, partial sums of A010122 starting from the 2nd term.
The sequence differs from A007091 after a(25).
Also numbers k such that floor(k/5) is even. - Peter Luschny, Oct 05 2017

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

Cf. A010122, A239229, A257145, A293481 (complement).
Sequences of the type floor(n/d) = (10/d)*floor(n/10), where d is a factor of 10: A008592 (d=1), A197652 (d=2), this sequence (d=5), A001477 (d=10).
Sequences of the type n + r*floor(n/r): A005843 (r=1), A042948 (r=2), A047240 (r=3), A047476 (r=4), this sequence (r=5).

Magma
```
[n: n in [0..130] | n mod 10 lt 5];
```

[n: n in [0..130] | IsEven(Floor(n/5))];

Magma
```
[n+5*Floor(n/5): n in [0..70]];
```

select(k -> type(floor(k/5), even), [$0..130]); # Peter Luschny, Oct 05 2017

Table[n + 5 Floor[n/5], {n, 0, 70}]
Reap[For[k = 0, k <= 130, k++, If[Floor[k/5] == 2*Floor[k/10], Sow[k]]]][[2, 1]] (* or *) LinearRecurrence[{1, 0, 0, 0, 1, -1}, {0, 1, 2, 3, 4, 10}, 66] (* Jean-François Alcover, Oct 05 2017 *)

concat(0, Vec(x^2*(1 + x + x^2 + x^3 + 6*x^4) / ((1 - x)^2*(1 + x + x^2 + x^3 + x^4)) + O(x^70))) \\ Colin Barker, Oct 05 2017

select(k->floor(k/5) == 2*floor(k/10), vector(1000, k, k)) \\ Colin Barker, Oct 05 2017

[k for k in range(131) if (k//5) % 2 == 0] # Peter Luschny, Oct 05 2017

def A293292(n): return (n-1<<1)-(n-1)%5 # Chai Wah Wu, Oct 29 2024

[k for k in (0..130) if 2.divides(floor(k/5))] # Peter Luschny, Oct 05 2017

G.f.: x^2*(1 + x + x^2 + x^3 + 6*x^4)/((1 - x)^2*(1 + x + x^2 + x^3 + x^4)).
a(n) = a(n-1) + a(n-5) - a(n-6).
a(n) = (n-1) + 5*floor((n-1)/5) = 10*floor((n-1)/5) + ((n-1) mod 5).
a(n) = A257145(n+2) - A239229(n-1). - R. J. Mathar, Oct 05 2017
a(n) = 2n-2-((n-1) mod 5). - Chai Wah Wu, Oct 29 2024

Definition by David A. Corneth, Oct 05 2017

cp's OEIS Frontend

A113841 a(n) = a(n-1) + 2^A047240(n) for n>1, a(1)=1.

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A113854 a(n) = sum(2^(A047240(i)-1), i=1..n).

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A128205 a(n) = 2^(n-1)*A047240(n).

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A047234 Numbers that are congruent to {0, 1, 4} mod 6.

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Magma

Maple

Mathematica

PARI

Formula

A047242 Numbers that are congruent to {0, 1, 3} mod 6.

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Haskell

Magma

Maple

Mathematica

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A281899 a(n) = n + 6*floor(n/3).

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Magma

Maple

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Maxima

PARI

Python

Sage

Formula

A047266 Numbers that are congruent to {0, 1, 5} mod 6.

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