A064680 -id:A064680 - cp's OEIS frontend

A174012 a(n) = 3 * A064680(n).

0, 6, 3, 18, 6, 30, 9, 42, 12, 54, 15, 66, 18, 78, 21, 90, 24, 102, 27, 114, 30, 126, 33, 138, 36, 150, 39, 162, 42, 174, 45, 186, 48, 198, 51, 210, 54, 222, 57, 234, 60, 246, 63, 258, 66, 270, 69, 282, 72, 294, 75, 306, 78, 318, 81, 330, 84, 342, 87, 354, 90, 366, 93, 378, 96

Offset: 0

Paul Curtz, Mar 05 2010

Cf. A008585, A017593, A064680, A080425, A165988.

a(n) = A064680(3*n), similar to A165988.
a(n) mod 9 = 3*A080425(n) (period length 3).
a(2n+1) = A017593(n).
a(2n) = A008585(n).

a(0) = 0 prepended by Georg Fischer, Jul 01 2020

A042968 Numbers not divisible by 4.

Original entry on oeis.org

1, 2, 3, 5, 6, 7, 9, 10, 11, 13, 14, 15, 17, 18, 19, 21, 22, 23, 25, 26, 27, 29, 30, 31, 33, 34, 35, 37, 38, 39, 41, 42, 43, 45, 46, 47, 49, 50, 51, 53, 54, 55, 57, 58, 59, 61, 62, 63, 65, 66, 67, 69, 70, 71, 73, 74, 75, 77, 78, 79, 81, 82, 83, 85, 86, 87, 89, 90, 91, 93, 94, 95, 97, 98, 99, 101, 102

Offset: 1

N. J. A. Sloane, Dec 11 1999

Equivalently, numbers whose square part is odd. Cf. A028982. - Peter Munn, Jul 14 2020
More generally the sequence of numbers not divisible by some fixed integer m >= 2 is given by a(n,m) = 1 + n + floor(n/(m-1)). - Benoit Cloitre, Jul 11 2009
Also a(n,m) = floor((m*n-1)/(m-1)) [with offset 1]. - Gary Detlefs, May 14 2011
Numbers not having more even than odd divisors: A048272(a(n)) >= 0. - Reinhard Zumkeller, Jan 21 2012
Extending the comments of Benoit Cloitre (Jul 11 2009) and Gary Detlefs (May 14 2011), the g.f. is A(m,x) = (1-x^m) / ((1-x^(m-1))*(1-x)^2) where m >= 2 is fixed. - Werner Schulte, Apr 26 2018

			G.f. = 1 + 2*x + 3*x^2 + 5*x^3 + 6*x^4 + 7*x^5 + 9*x^6 + 10*x^7 + 11*x^8 + ... - _Michael Somos_, Jun 17 2018

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

Cf. A001651, A001935, A028982, A070048, A042965, A064680.
Cf. A071619 (partial sums); A008586 (complement).
Numbers that are congruent to {k0,k1,k2} mod 4: A004772, A004773, A042965, a(n).

a042968 = (`div` 3) . (subtract 1) . (* 4)
a042968_list = filter ((/= 0) . (`mod` 4)) [1..]
-- Reinhard Zumkeller, Sep 02 2012

[n+1+Floor(n/3): n in [0..80]]; // Vincenzo Librandi, Aug 03 2015

seq(n+floor((n-1)/3), n=1..80); # Muniru A Asiru, Feb 17 2019

Select[Table[n,{n,200}], Mod[#,4] != 0&] (* Vladimir Joseph Stephan Orlovsky, Feb 18 2011 *)
LinearRecurrence[{1,0,1,-1},{1,2,3,5},80]  (* or *) Drop[Range[110],{4,-1,4}] (* Harvey P. Dale, Jan 07 2023 *)

PARI
```
{a(n) = 1 + n + n\3};
```

def A042968(n): return n+(n-1)//3 # Chai Wah Wu, Apr 13 2025

[1+n+floor(n/3) for n in (0..80)] # G. C. Greubel, Feb 17 2019

a(n) = a(n-1) + a(n-3) - a(n-4).
a(n) = a(n-3) + 4, with a(1) = 1.
G.f.: x * (1+x) * (1+x^2) / ( (1+x+x^2)*(1-x)^2 ). - Michael Somos, Jan 12 2000
A064680(A064680(a(n))) = a(n). - Reinhard Zumkeller, Oct 19 2001
Nearest integer to (Sum_{k>n} 1/k^4)/(Sum_{k>n} 1/k^5). - Benoit Cloitre, Jun 12 2003
a(n) = n + 1 + floor(n/3). - Benoit Cloitre, Jul 11 2009
a(n) = floor((4*n+3)/3). - Gary Detlefs, May 14 2011
A214546(a(n)) >= 0 for n > 0. - Reinhard Zumkeller, Jul 20 2012
a(n) = 2*n - ceiling(2*n/3) + 1. - Arkadiusz Wesolowski, Sep 21 2012
Sum_{k=0..n} a(n) = A071619(n+1). - L. Edson Jeffery, Jul 30 2014
The g.f. A(x) satisfies x*A(x)^2 = (B(x)/x)^2 + (B(x)/x), where B(x) is the o.g.f. of A042965. - Peter Bala, Apr 12 2017
a(n) = (12*n + 6 + 3*cos(2*n*Pi/3) + sqrt(3)*sin(2*n*Pi/3))/9. - Wesley Ivan Hurt, Sep 30 2017
Euler transform of length 4 sequence [2, 0, 1, -1]. - Michael Somos, Jun 17 2018
a(n) = -a(-1-n) for all n in Z. - Michael Somos, Jun 17 2018
E.g.f.: (2/3)*exp(x)*(1 + 2*x) + (1/9)*exp(-x/2)*(3*cos(sqrt(3)*x/2) + sqrt(3)*sin(sqrt(3)*x/2)). - Stefano Spezia, Nov 16 2019
a(n) = (12*n + 6 + w^(2*n)*(w + 2) - w^n*(w - 1))/9 where w = (-1 + sqrt(-3))/2. - Guenther Schrack, Jun 07 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = (2*sqrt(2)-1)*Pi/8. - Amiram Eldar, Dec 05 2021

Edited by Peter Munn, Nov 16 2019

I restored my original (1999) definition and offset, which in the intervening 21 years had been lost. - N. J. A. Sloane, Jun 12 2021

A168077 a(2n) = A129194(2n)/2; a(2n+1) = A129194(2n+1).

Original entry on oeis.org

0, 1, 1, 9, 4, 25, 9, 49, 16, 81, 25, 121, 36, 169, 49, 225, 64, 289, 81, 361, 100, 441, 121, 529, 144, 625, 169, 729, 196, 841, 225, 961, 256, 1089, 289, 1225, 324, 1369, 361, 1521, 400, 1681, 441, 1849, 484, 2025, 529, 2209, 576, 2401, 625, 2601

Offset: 0

Paul Curtz, Nov 18 2009

From Paul Curtz, Mar 26 2011: (Start)
Successive A026741(n) * A026741(n+p):
  p=0:  0, 1,  1,  9,  4, 25,  9,     a(n),
  p=1:  0, 1,  3,  6, 10, 15, 21,  A000217,
  p=2:  0, 3,  2, 15,  6, 35, 12,  A142705,
  p=3:  0, 2,  5,  9, 14, 20, 27,  A000096,
  p=4:  0, 5,  3, 21,  8, 45, 15,  A171621,
  p=5:  0, 3,  7, 12, 18, 25, 33,  A055998,
  p=6:  0, 7,  4, 27, 10, 55, 18,
  p=7:  0, 4,  9, 15, 22, 30, 39,  A055999,
  p=8:  0, 9,  5, 33, 12, 65, 21,  (see A061041),
  p=9:  0, 5, 11, 18, 26, 35, 45,  A056000. (End)
The moment generating function of p(x, m=2, n=1, mu=2) = 4*x*E(x, 2, 1), see A163931 and A274181, is given by M(a) = (-4 * log(1-a) - 4 * polylog(2, a))/a^2. The series expansion of M(a) leads to the sequence given above. - Johannes W. Meijer, Jul 03 2016
Multiplicative because both A129194 and A040001 are. - Andrew Howroyd, Jul 26 2018

G. C. Greubel, Table of n, a(n) for n = 0..1000
John M. Campbell, An Integral Representation of Kekulé Numbers, and Double Integrals Related to Smarandache Sequences, arXiv preprint arXiv:1105.3399 [math.GM], 2011.
Index entries for linear recurrences with constant coefficients, signature (0,3,0,-3,0,1).

Cf. A040001, A168068, A181318.

I:=[0,1,1,9,4,25]; [n le 6 select I[n] else 3*Self(n-2)-3*Self(n-4)+Self(n-6): n in [1..60]]; // Vincenzo Librandi, Jul 10 2016

a := proc(n): n^2*(5-3*(-1)^n)/8 end: seq(a(n), n=0..46); # Johannes W. Meijer, Jul 03 2016

LinearRecurrence[{0,3,0,-3,0,1},{0,1,1,9,4,25},60] (* Harvey P. Dale, May 14 2011 *)
f[n_] := Numerator[(n/2)^2]; Array[f, 60, 0] (* Robert G. Wilson v, Dec 18 2012 *)
CoefficientList[Series[x(1+x+6x^2+x^3+x^4)/((1-x)^3(1+x)^3), {x,0,60}], x] (* Vincenzo Librandi, Jul 10 2016 *)

concat(0, Vec(x*(1+x+6*x^2+x^3+x^4)/((1-x)^3*(1+x)^3) + O(x^60))) \\ Altug Alkan, Jul 04 2016

a(n) = lcm(4, n^2)/4; \\ Andrew Howroyd, Jul 26 2018

(x*(1+x+6*x^2+x^3+x^4)/(1-x^2)^3).series(x, 60).coefficients(x, sparse=False) # G. C. Greubel, Feb 20 2019

From R. J. Mathar, Jan 22 2011: (Start)
G.f.: x*(1 + x + 6*x^2 + x^3 + x^4) / ((1-x)^3*(1+x)^3).
a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6).
a(n) = n^2*(5 - 3*(-1)^n)/8. (End)
a(n) = A026741(n)^2.
a(2*n) = A000290(n); a(2*n+1) = A016754(n).
a(n) - a(n-4) = 4*A064680(n+2). - Paul Curtz, Mar 27 2011
4*a(n) = A061038(n) * A010121(n+2) = A109043(n)^2, n >= 2. - Paul Curtz, Apr 07 2011
a(n) = A129194(n) / A040001(n). - Andrew Howroyd, Jul 26 2018
From Peter Bala, Feb 19 2019: (Start)
a(n) = numerator(n^2/(n^2 + 4)) = n^2/(gcd(n^2,4)) = (n/gcd(n,2))^2.
a(n) = n^2/b(n), where b(n) = [1, 4, 1, 4, ...] is a purely periodic sequence of period 2. Thus a(n) is a quasi-polynomial in n.
O.g.f.: x*(1 + x)/(1 - x)^3 - 3*x^2*(1 + x^2)/(1 - x^2)^3.
Cf. A181318. (End)
From Werner Schulte, Aug 30 2020: (Start)
Multiplicative with a(2^e) = 2^(2*e-2) for e > 0, and a(p^e) = p^(2*e) for prime p > 2.
Dirichlet g.f.: zeta(s-2) * (1 - 3/2^s).
Dirichlet convolution with A259346 equals A000290.
Sum_{n>0} 1/a(n) = Pi^2 * 7 / 24. (End)
Sum_{k=1..n} a(k) ~ (5/24) * n^3. - Amiram Eldar, Nov 28 2022

A188134 a(4n) = n, a(1+2n) = 4+8n, a(2+4n) = 2+4*n.

Original entry on oeis.org

0, 4, 2, 12, 1, 20, 6, 28, 2, 36, 10, 44, 3, 52, 14, 60, 4, 68, 18, 76, 5, 84, 22, 92, 6, 100, 26, 108, 7, 116, 30, 124, 8, 132, 34, 140, 9, 148, 38, 156, 10, 164, 42, 172, 11, 180, 46, 188, 12, 196, 50, 204, 13, 212, 54, 220, 14, 228, 58, 236, 15, 244, 62

Offset: 0

Paul Curtz, Mar 21 2011

G. C. Greubel, Table of n, a(n) for n = 0..5000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,2,0,0,0,-1).

Cf. A016825, A017113, A060819, A061037, A064680, A146160, A176895, A177499.

[(64-3*(1+(-1)^n)*(9+(-1)^(n div 2)))*n/16 : n in [0..80]]; // Wesley Ivan Hurt, Jul 06 2016

A188134:=n->8*n/(11 + 9*cos(Pi*n) + 12*cos(n*Pi/2)): seq(A188134(n), n=0..100); # Wesley Ivan Hurt, Jul 06 2016

Table[8 n/(11 + 9 Cos[Pi*n] + 12 Cos[n*Pi/2]), {n, 0, 80}] (* Wesley Ivan Hurt, Jul 06 2016 *)
CoefficientList[Series[x*(4+2*x+12*x^2+x^3+12*x^4+2*x^5+4*x^6)/(1-x^4)^2, {x, 0, 50}], x] (* G. C. Greubel, Sep 20 2018 *)
LinearRecurrence[{0,0,0,2,0,0,0,-1},{0,4,2,12,1,20,6,28},70] (* Harvey P. Dale, Aug 14 2019 *)

x='x+O('x^50); concat([0], Vec(x*(4+2*x+12*x^2+x^3+12*x^4+ 2*x^5 +4*x^6)/(1-x^4)^2)) \\ G. C. Greubel, Sep 20 2018

a(n) = 2*a(n-4) - a(n-8) for n>7.
a(n) = A176895(n) * A060819(n).
a(n) = (4*A061037(n+2))/(n+4).
a(n) = 4*n / A146160(n).
a(2*n) = A064680(n).
a(1+2*n) = A017113(n).
a(4*n) = a(-4+4*n) +  1.
a(1+4*n) = a(-3+4*n) + 16.
a(2+4*n) = a(-2+4*n) +  4.
a(3+4*n) = a(-1+4*n) + 16.  See A177499.
From Bruno Berselli, Mar 22 2011: (Start)
G.f.: x*(4+2*x+12*x^2+x^3+12*x^4+2*x^5+4*x^6)/(1-x^4)^2.
a(n) = (64-3*(1+(-1)^n)*(9+i^n))*n/16 with i=sqrt(-1).
a(n)/a(n-4) = n/(n-4) for n>4. (End)
a(n) = 8*n/(11 + 9*cos(Pi*n) + 12*cos(n*Pi/2)). - Wesley Ivan Hurt, Jul 06 2016
a(n) = lcm(4,n)/gcd(4,n). - R. J. Mathar, Feb 12 2019
Sum_{k=1..n} a(k) ~ (37/32)*n^2. - Amiram Eldar, Oct 07 2023

A352544 a(n) = n/2 if n is even, otherwise n + A004186(n), where A004186 = arrange digits in decreasing order.

Original entry on oeis.org

0, 2, 1, 6, 2, 10, 3, 14, 4, 18, 5, 22, 6, 44, 7, 66, 8, 88, 9, 110, 10, 42, 11, 55, 12, 77, 13, 99, 14, 121, 15, 62, 16, 66, 17, 88, 18, 110, 19, 132, 20, 82, 21, 86, 22, 99, 23, 121, 24, 143, 25, 102, 26, 106, 27, 110, 28, 132, 29, 154, 30, 122, 31, 126, 32, 130, 33, 143, 34, 165, 35, 142, 36, 146, 37, 150, 38, 154, 39, 176, 40, 162, 41, 166, 42, 170, 43, 174, 44, 187, 45

Offset: 0

Eric Angelini and M. F. Hasler, Mar 20 2022

A variant of the Collatz (3x+1) map A006370. See A352540 - A352543 for more about iterations of this map.

Eric Angelini, Divide by 2 or add the biggest anagram, math-fun discussion list, Mar 20 2022

Cf. A004186.
Coincides with A064680 (half if even, double if odd) for n < 13.
Cf. A352540 (initial values with infinite orbit under A352544), A352541 (number of iterations of A352544 to reach a value for the second time), A352542 (orbit of 89 under A352544), A352543 (numbers that end in a loop of length > 3 under A352544), A352545 (representatives of loops of length > 3).

apply( {A352544(n)=if(n%2,n+A004186(n),n\2)}, [0..90]) \\ with A004186(n)=fromdigits(vecsort(digits(n),,4))

A283442 a(n) = lcm(n,5) / gcd(n,5).

Original entry on oeis.org

0, 5, 10, 15, 20, 1, 30, 35, 40, 45, 2, 55, 60, 65, 70, 3, 80, 85, 90, 95, 4, 105, 110, 115, 120, 5, 130, 135, 140, 145, 6, 155, 160, 165, 170, 7, 180, 185, 190, 195, 8, 205, 210, 215, 220, 9, 230, 235, 240, 245, 10, 255, 260, 265, 270, 11, 280, 285, 290

Offset: 0

Colin Barker, Mar 07 2017

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,2,0,0,0,0,-1).

Cf. A064680, A130724, A188134, A283443, A283444.

Cf. A109009, A109046.

Table[LCM[n, 5] / GCD[n, 5], {n,0,58}] (* Indranil Ghosh, Mar 08 2017 *)
LinearRecurrence[{0,0,0,0,2,0,0,0,0,-1},{0,5,10,15,20,1,30,35,40,45},60] (* Harvey P. Dale, Aug 11 2019 *)

concat(0, Vec(x*(5 + 10*x + 15*x^2 + 20*x^3 + x^4 + 20*x^5 + 15*x^6 + 10*x^7 + 5*x^8) / ((1 - x)^2*(1 + x + x^2 + x^3 + x^4)^2) + O(x^100)))

{for (n=0, 58, print1((lcm(n, 5) / gcd(n, 5)),", "))}; \\ Indranil Ghosh, Mar 08 2017

G.f.: x*(5 + 10*x + 15*x^2 + 20*x^3 + x^4 + 20*x^5 + 15*x^6 + 10*x^7 + 5*x^8) / ((1 - x)^2*(1 + x + x^2 + x^3 + x^4)^2).
a(n) = A109046(n) / A109009(n).
a(n) = 2*a(n-5) - a(n-10) for n>9.
Sum_{k=1..n} a(k) ~ (101/50)*n^2. - Amiram Eldar, Oct 07 2023

A191662 a(n) = n! / A000034(n-1).

Original entry on oeis.org

1, 1, 6, 12, 120, 360, 5040, 20160, 362880, 1814400, 39916800, 239500800, 6227020800, 43589145600, 1307674368000, 10461394944000, 355687428096000, 3201186852864000, 121645100408832000, 1216451004088320000, 51090942171709440000, 562000363888803840000

Offset: 1

Paul Curtz, Jun 10 2011

The a(n) are the denominators in the formulas of the k-dimensional square pyramidal numbers:
A005408 = (2*n+1)/1                       = 1, 3,  5,   7,   9, ... (k=1)
A000290 = (n^2)/1                         = 1, 4,  9,  16,  25, ... (k=2)
A000330 = n*(n+1)*(2*n+1)/6               = 1, 5, 14,  30,  55, ... (k=3)
A002415 = (n^2)*(n^2-1)/12                = 1, 6, 20,  50, 105, ... (k=4)
A005585 = n*(n+1)*(n+2)*(n+3)*(2*n+3)/120 = 1, 7, 27,  77, 182, ... (k=5)
A040977 = (n^2)*(n^2-1)*(n^2-4)/360       = 1, 8, 35, 112, 294, ... (k=6)
A050486 (k=7), A053347 (k=8), A054333 (k=9), A054334 (k=10), A057788 (k=11).
The first superdiagonal of this array appears in A029651. - Paul Curtz, Jul 04 2011
The general formula for the k-dimensional square pyramidal numbers is (2*n+k)*binomial(n+k-1,k-1)/k, k >= 1, n >= 0, see A097207. - Johannes W. Meijer, Jun 22 2011

Cf. A000142, A009445, A002674, A064680.
Cf. A000290, A000330, A002415, A005408, A005585, A029651, A040977, A050486, A053347, A054333, A054334, A057788.
Cf. A052612.

A191662:= proc(n): n!/A000034(n-1) end: A000034 := proc(n) op((n mod 2)+1, [1, 2]) ; end proc: seq(A191662(n),n=1..17); # Johannes W. Meijer, Jun 22 2011

Array[If[EvenQ[#],#!/2,#!]&,20] (* Harvey P. Dale, Mar 14 2014 *)

a(2*n-1) = (2*n-1)!, a(2*n) = (2*n)!/2.
a(n+1) = A064680(n+1) * a(n).
From Amiram Eldar, Jul 06 2022: (Start)
Sum_{n>=1} 1/a(n) = sinh(1) + 2*cosh(1) - 2.
Sum_{n>=1} (-1)^(n+1)/a(n) = sinh(1) - 2*cosh(1) + 2. (End)
D-finite with recurrence: a(n) - (n-1)*n*a(n-2) = 0 for n >= 3 with a(1)=a(2)=1. - Georg Fischer, Nov 25 2022
a(n) = A052612(n)/2 for n >= 1. - Alois P. Heinz, Sep 05 2023

More terms from Harvey P. Dale, Mar 14 2014

A283443 a(n) = lcm(n,6) / gcd(n,6).

Original entry on oeis.org

0, 6, 3, 2, 6, 30, 1, 42, 12, 6, 15, 66, 2, 78, 21, 10, 24, 102, 3, 114, 30, 14, 33, 138, 4, 150, 39, 18, 42, 174, 5, 186, 48, 22, 51, 210, 6, 222, 57, 26, 60, 246, 7, 258, 66, 30, 69, 282, 8, 294, 75, 34, 78, 318, 9, 330, 84, 38, 87, 354, 10, 366, 93, 42

Offset: 0

Colin Barker, Mar 07 2017

If n == 2 or 4 (mod 6) then a(n) = 3*n/2; if n == 3 (mod 6) then a(n) = 2*n/3; if n == 1 or 5 (mod 6) then a(n) = 6*n; otherwise, a(n) = n/6. Examples: n = 50 = 6*8+2, a(50) = 3*50/2 = 75; n = 23 = 6*3+5, a(23) = 6*23 = 138. - Bruno Berselli, Mar 08 2017

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,2,0,0,0,0,0,-1).

Cf. A064680, A130724, A188134, A283442, A283444.

Table[LCM[n, 6] / GCD[n, 6], {n,0,63}] (* Indranil Ghosh, Mar 08 2017 *)

concat(0, Vec(x*(6 + 3*x + 2*x^2 + 6*x^3 + 30*x^4 + x^5 + 30*x^6 + 6*x^7 + 2*x^8 + 3*x^9 + 6*x^10) / ((1 - x)^2*(1 + x)^2*(1 - x + x^2)^2*(1 + x + x^2)^2) + O(x^100)))

{for (n=0, 63, print1((lcm(n, 6) / gcd(n, 6)),", "))}; \\ Indranil Ghosh, Mar 08 2017

G.f.: x*(6 + 3*x + 2*x^2 + 6*x^3 + 30*x^4 + x^5 + 30*x^6 + 6*x^7 + 2*x^8 + 3*x^9 + 6*x^10) / ((1 - x)^2*(1 + x)^2*(1 - x + x^2)^2*(1 + x + x^2)^2).
a(n) = 2*a(n-6) - a(n-12) for n>11.
a(n) = A109047(n)/A089128(n). - R. J. Mathar, Feb 12 2019
Sum_{k=1..n} a(k) ~ (95/72)*n^2. - Amiram Eldar, Oct 07 2023

A191567 Four interlaced 2nd order polynomials: a(4k) = k(1+2k); a(1+2k) = 2(1+2k)(3+2k); a(2+4k) = 4(1+k)(1+2k).

Original entry on oeis.org

0, 6, 4, 30, 3, 70, 24, 126, 10, 198, 60, 286, 21, 390, 112, 510, 36, 646, 180, 798, 55, 966, 264, 1150, 78, 1350, 364, 1566, 105, 1798, 480, 2046, 136, 2310, 612, 2590, 171, 2886, 760, 3198, 210, 3526, 924, 3870, 253, 4230, 1104, 4606, 300, 4998, 1300, 5406, 351

Offset: 0

Paul Curtz, Jun 12 2011

a(n) = T(0,n) and differences T(n,k) = T(n-1,k+1) - T(n-1,k) define the array
0,   6,  4,  30,    3,  70,   24,  126,   10,  198,   60,  286,   21,  390,  ..
6,  -2, 26, -27,   67, -46,  102, -116,  188, -138,  226, -265,  369, -278, ..
-8, 28 -53,  94, -113, 148, -218,  304, -326,  364, -491,  634, -647,  676, ...
T(3,n) mod 9 is the sequence 1, 1, 1, 4, 4, 4, 7, 7, 7, 4, 4, 4 (and periodically repeated with period 12).
A064680(2+n) divides a(n), where b(n) = a(n)/A064680(2+n) = 0, 1, 2, 3, 1, 5, 6, 7, 2,... for n>=0, obeys b(4*k) = k and has recurrence b(n) = 2*b(n-4) - b(n-8).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,3,0,0,0,-3,0,0,0,1).

Cf. A014105, A000466, A000384, A177427.

a:=[0,6,4,30,3,70,24,126,10,198,60,286];; for n in [13..60] do a[n]:= 3*a[n-4]-3*a[n-8]+a[n-12]; od; a; # G. C. Greubel, Feb 26 2019

I:=[0,6,4,30,3,70,24,126,10,198,60,286]; [n le 12 select I[n] else 3*Self(n-4)-3*Self(n-8)+Self(n-12): n in [1..60]]; // Vincenzo Librandi, Apr 23 2017

Table[Which[OddQ@ n, 2 (1 + 2 #) (3 + 2 #) &[(n - 1)/2], Mod[n, 4] == 0, # (1 + 2 #) &[n/4], True, 4 (1 + #) (1 + 2 #) &[(n - 2)/4]], {n, 0, 60}] (* or *)
CoefficientList[Series[x(6 +4x +30x^2 +3x^3 +52x^4 +12x^5 +36x^6 +x^7 +6x^8 -2x^10)/((1-x)^3*(1+x)^3*(1+x^2)^3), {x, 0, 60}], x] (* Michael De Vlieger, Apr 22 2017 *)
LinearRecurrence[{0,0,0,3,0,0,0,-3,0,0,0,1}, {0,6,4,30,3,70,24,126,10,198,60, 286}, 80] (* Vincenzo Librandi, Apr 23 2017 *)

m=60; v=concat([0,6,4,30,3,70,24,126,10,198,60,286], vector(m-12)); for(n=13, m, v[n]=3*v[n-4]-3*v[n-8]+v[n-12]); v \\ G. C. Greubel, Feb 26 2019

(x*(6+4*x+30*x^2+3*x^3+52*x^4+12*x^5+36*x^6+x^7+6*x^8-2*x^10)/((1-x)^3 *(1+x)^3*(1+x^2)^3 )).series(x, 60).coefficients(x, sparse=False) # G. C. Greubel, Feb 26 2019

a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12).
a(n) = A061037(n+2) + A181318(n). - Paul Curtz, Jul 19 2011
a(n) = A060819(n) * A145979(n). - Paul Curtz, Sep 06 2011
G.f.: x*(6+4*x+30*x^2+3*x^3+52*x^4+12*x^5+36*x^6+x^7+6*x^8-2*x^10) /( (1-x)^3 *(1+x)^3 *(1+x^2)^3 ). - R. J. Mathar, Jun 17 2011
Let BEB(n) = a(n)/A061038(n+2) = A060819(n)/A145979(n). Then (BEB(n))^2 = A181318(n)/A061038(n+2) = BEB(n) - A061037(n+2)/A061038(n+2). - Paul Curtz, Jul 19 2011, index corrected by R. J. Mathar, Sep 09 2011
From Luce ETIENNE, Apr 18 2017: (Start)
a(n) = n*(n + 2)*(37 - 27*(-1)^n - 3*((-1)^((2*n + 1 - (-1)^n)/4) + (-1)^((2*n - 1 + (-1)^n)/4)))/32.
a(n) = n*(n+2)*(37-27*cos(n*Pi) - 6*cos(n*Pi/2))/32.
a(n) = n*(n + 2)*(37 - 27*(-1)^n - 3*(i^n + (-i)^n))/32, where i=sqrt(-1). (End)

A283444 a(n) = lcm(n,7) / gcd(n,7).

Original entry on oeis.org

0, 7, 14, 21, 28, 35, 42, 1, 56, 63, 70, 77, 84, 91, 2, 105, 112, 119, 126, 133, 140, 3, 154, 161, 168, 175, 182, 189, 4, 203, 210, 217, 224, 231, 238, 5, 252, 259, 266, 273, 280, 287, 6, 301, 308, 315, 322, 329, 336, 7, 350, 357, 364, 371, 378, 385, 8, 399

Offset: 0

Colin Barker, Mar 07 2017

Similar to row 7 of A059897. Apart from the extra a(0) = 0, differs first at a(49) = 7 <> 343 = A059897(7,49). Note that a(1) = 7 also, whereas all rows of A059897 are permutations of the positive integers. - Peter Munn, Jan 16 2020

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,2,0,0,0,0,0,0,-1).

Cf. A059897, A064680, A130724, A188134, A283442, A283443.

Table[LCM[n, 7] / GCD[n, 7], {n,0,57}] (* Indranil Ghosh, Mar 08 2017 *)
LinearRecurrence[{0,0,0,0,0,0,2,0,0,0,0,0,0,-1},{0,7,14,21,28,35,42,1,56,63,70,77,84,91},60] (* Harvey P. Dale, Apr 05 2018 *)

concat(0, Vec(x*(7 + 14*x + 21*x^2 + 28*x^3 + 35*x^4 + 42*x^5 + x^6 + 42*x^7 + 35*x^8 + 28*x^9 + 21*x^10 + 14*x^11 + 7*x^12) / ((1 - x)^2*(1 + x + x^2 + x^3 + x^4 + x^5 + x^6)^2) + O(x^100)))

{for (n=0, 57, print1((lcm(n, 7) / gcd(n, 7)),", "))}; \\ Indranil Ghosh, Mar 08 2017

G.f.: x*(7 + 14*x + 21*x^2 + 28*x^3 + 35*x^4 + 42*x^5 + x^6 + 42*x^7 + 35*x^8 + 28*x^9 + 21*x^10 + 14*x^11 + 7*x^12) / ((1 - x)^2*(1 + x + x^2 + x^3 + x^4 + x^5 + x^6)^2).
a(n) = 2*a(n-7) - a(n-14) for n > 13.
a(n) = 7^(-(m^6 - 21*m^5 + 175*m^4 - 735*m^3 + 1624*m^2 - 1764*m + 360)/360)*n where m = (n mod 7). - Luce ETIENNE, Nov 18 2019
Sum_{k=1..n} a(k) ~ (295/98)*n^2. - Amiram Eldar, Oct 07 2023

cp's OEIS Frontend

A174012 a(n) = 3 * A064680(n).

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A042968 Numbers not divisible by 4.

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A168077 a(2n) = A129194(2n)/2; a(2n+1) = A129194(2n+1).

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A188134 a(4*n) = n, a(1+2*n) = 4+8*n, a(2+4*n) = 2+4*n.

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A352544 a(n) = n/2 if n is even, otherwise n + A004186(n), where A004186 = arrange digits in decreasing order.

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A283442 a(n) = lcm(n,5) / gcd(n,5).

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A191662 a(n) = n! / A000034(n-1).

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A188134 a(4n) = n, a(1+2n) = 4+8n, a(2+4n) = 2+4*n.

A191567 Four interlaced 2nd order polynomials: a(4k) = k(1+2k); a(1+2k) = 2(1+2k)(3+2k); a(2+4k) = 4(1+k)(1+2k).