A144437 -id:A144437 - cp's OEIS frontend

A168615 Inverse binomial transform of A169609, or of A144437 preceded by 1.

1, 2, -2, 0, 6, -18, 36, -54, 54, 0, -162, 486, -972, 1458, -1458, 0, 4374, -13122, 26244, -39366, 39366, 0, -118098, 354294, -708588, 1062882, -1062882, 0, 3188646, -9565938, 19131876, -28697814, 28697814, 0, -86093442, 258280326, -516560652

Offset: 0

Paul Curtz, Dec 01 2009

sign

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-3,-3).

Cf. A169609, A144437, A000748, A057083, A057682.

[ n le 2 select n else n eq 3 select -2 else -3*Self(n-1)-3*Self(n-2): n in [1..37] ]; // Klaus Brockhaus, Dec 03 2009

Join[{1,2,-2}, LinearRecurrence[{-3, -3}, {0, 6}, 25]] (* G. C. Greubel, Jul 27 2016 *)
LinearRecurrence[{-3,-3},{1,2,-2},40] (* Harvey P. Dale, Jul 21 2024 *)

a(n) = -3*a(n-1) - 3*a(n-2) for n > 2; a(0) = 1, a(1) = 2, a(2) = -2.
a(n) = 2*A123877(n-1), n>0.
G.f.: 1+2*x*(1+2*x)/(1+3*x+3*x^2).
a(6*m + 3) = 0, m>=0. - G. C. Greubel, Jul 27 2016

Edited and extended by Klaus Brockhaus, Dec 03 2009

A163979 a(n) = n*(n-1) + A144437(n+2).

Original entry on oeis.org

1, 3, 5, 7, 15, 23, 31, 45, 59, 73, 93, 113, 133, 159, 185, 211, 243, 275, 307, 345, 383, 421, 465, 509, 553, 603, 653, 703, 759, 815, 871, 933, 995, 1057, 1125, 1193, 1261, 1335, 1409, 1483, 1563, 1643, 1723, 1809, 1895, 1981, 2073, 2165, 2257, 2355, 2453, 2551

Offset: 0

Paul Curtz, Aug 07 2009

First differences are 2, 2, 2, 8, 8, 8, 14, 14, 14, 20, 20, 20,... (triplicated A016933).

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

LinearRecurrence[{2,-1,1,-2,1},{1,3,5,7,15},60]  (* or *) CoefficientList[ Series[-(1+x+5x^4-x^3)/((1+x+x^2)(x-1)^3), {x,0,60}],x]  (* Harvey P. Dale, Apr 20 2011 *)

x='x+O('x^50); Vec((1+x-x^3+5*x^4)/((1+x+x^2)*(1-x)^3)) \\ G. C. Greubel, Aug 24 2017

a(n) = A002378(n-1) + A144437(n+2).
a(n) = 2*a(n-1) - a(n-2) + a(n-3) - 2*a(n-4) + a(n-5).
G.f.: (1 +x -x^3 +5*x^4)/( (1 +x +x^2)*(1 -x)^3 ).
E.g.f.: (1/3)*((7+3*x^2)*exp(x) - 4*exp(-x/2)*cos(sqrt(3)*x/2)). - G. C. Greubel, Aug 24 2017

Edited and extended by R. J. Mathar, Aug 12 2009

A051176 If n mod 3 = 0 then n/3 else n.

Original entry on oeis.org

0, 1, 2, 1, 4, 5, 2, 7, 8, 3, 10, 11, 4, 13, 14, 5, 16, 17, 6, 19, 20, 7, 22, 23, 8, 25, 26, 9, 28, 29, 10, 31, 32, 11, 34, 35, 12, 37, 38, 13, 40, 41, 14, 43, 44, 15, 46, 47, 16, 49, 50, 17, 52, 53, 18, 55, 56, 19, 58, 59, 20, 61, 62, 21, 64, 65, 22, 67

Offset: 0

N. J. A. Sloane

Numerator of n/3. - Wesley Ivan Hurt, Jul 18 2014

			G.f. = x + 2*x^2 + x^3 + 4*x^4 + 5*x^5 + 2*x^6 + 7*x^7 + 8*x^8 + 3*x^9 + ...

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

Cf. A026741, A051176, A060819, A060791, A060789 for n / GCD(n,k) for k=2..6. See also A106608 thru A106612 (k = 7 thru 11), A051724 (k = 12), A106614 thru A106621 (k = 13 thru 20).

Cf. A109044, A011655, A144437, A167192.

a051176 n = if m == 0 then n' else n  where (n',m) = divMod n 3
-- Reinhard Zumkeller, Aug 27 2012

[Numerator(n/3): n in [0..70]]; // G. C. Greubel, Feb 19 2019

A051176:=n->numer(n/3); seq(A051176(n), n=0..100); # Wesley Ivan Hurt, Jul 18 2014

If[Divisible[#,3],#/3,#]&/@Range[0,70] (* Harvey P. Dale, Feb 07 2011 *)
a[n_] := Numerator[n/3]; Array[a, 100, 0] (* Wesley Ivan Hurt, Jul 18 2014 *)
LinearRecurrence[{0,0,2,0,0,-1},{0,1,2,1,4,5},70] (* Harvey P. Dale, Jul 12 2025 *)

a(n) = if (n % 3, n, n/3); \\ Michel Marcus, Feb 02 2016

[numerator(n/3) for n in range(70)] # G. C. Greubel, Feb 19 2019

a(n) = n / gcd(n,3).
G.f.: x*(1+2*x+x^2+2*x^3+x^4)/(1-x^3)^2 = x*(1+2*x+x^2+2*x^3+x^4) / ( (x-1)^2*(1+x+x^2)^2 ). - Len Smiley, Apr 30 2001
Multiplicative with a(3^e) = 3^(e-1), a(p^e) = p^e otherwise. - Mitch Harris, Jun 09 2005
a(n) = A167192(n+3, 3). - Reinhard Zumkeller, Oct 30 2009
From R. J. Mathar, Apr 18 2011: (Start)
a(n) = A109044(n)/3.
Dirichlet g.f.: zeta(s-1)*(1-2/3^s). (End)
a(n) = n/3 * (1 + 2*A011655(n)) = n*A144437(n)/3. - Timothy Hopper, Feb 23 2017
G.f.: x /(1 - x)^2 - 2 * x^3/(1 - x^3)^2. - Michael Somos, Mar 05 2017
a(n) = a(-n) for all n in Z. - Michael Somos, Mar 05 2017
a(n) = n*(7 - 4*cos((2*Pi*n)/3)) / 9. - Colin Barker, Mar 05 2017
Sum_{k=1..n} a(k) ~ (7/18) * n^2. - Amiram Eldar, Nov 25 2022
Sum_{n>=1} (-1)^(n+1)/a(n) = 5*log(2)/3. - Amiram Eldar, Sep 08 2023

A069705 a(n) = 2^n mod 7.

Original entry on oeis.org

1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4

Offset: 0

Jon Perry, Jan 14 2003

Periodic sequence with period [1,2,4]. - Philippe Deléham, Sep 25 2006
From Klaus Brockhaus, May 23 2010: (Start)
Continued fraction expansion of (11 + sqrt(229))/18.
Decimal expansion of 124/999. (End)

			a(4)=16 mod 7=2, a(5)=32 mod 7=4, a(6)=64 mod 7=1.

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

Cf. A000079, A050985.
Cf. A178233 (decimal expansion of (11+sqrt(229))/18).
Cf. A052901, A144437.

List([0..83],n->PowerMod(2,n,7)); # Muniru A Asiru, Jan 31 2019

[Modexp(2, n, 7): n in [0..100]]; // Vincenzo Librandi, Mar 25 2016

A069705 := proc(n) op((n mod 3)+1,[1,2,4]) ; end proc: # R. J. Mathar, Feb 05 2011

PowerMod[2,Range[0,110],7] (* or *) PadRight[{},110,{1,2,4}] (* Harvey P. Dale, Mar 28 2015 *)

a(n)=2^(n%3)%7 \\ Charles R Greathouse IV, Jun 11 2015

a(n) = lift(Mod(2, 7)^n); \\ Altug Alkan, Mar 25 2016

[power_mod(2,n,7) for n in range(0, 105)] # Zerinvary Lajos, Jun 07 2009

n=0 mod 3 -> a(n)=1 n=1 mod 3 -> a(n)=2 n=2 mod 3 -> a(n)=4.
a(n) = 2^(n mod 3). - Paul Barry, Oct 06 2003
a(n) = cubefree part of 2^n = A000079(A050985(n)). - Artur Jasinski, Oct 15 2008
From R. J. Mathar, Apr 13 2010: (Start)
a(n) = a(n-3).
G.f.: (1+2*x+4*x^2)/((1-x) * (1+x+x^2)). (End)
a(n) = (7+5*cos(2*(n+1)*Pi/3)-sqrt(3)*sin(2*(n+1)*Pi/3))/3. - Wesley Ivan Hurt, Oct 01 2017
From Nicolas Bělohoubek, Nov 11 2021: (Start)
a(n) = 8/(a(n-2)*a(n-1)).
a(n) = 7 - a(n-2) - a(n-1). See also A052901 or A144437. (End)
a(n) = n + 1 + floor((n+1)/3) - 4*floor(n/3). - Ridouane Oudra, Sep 25 2024
E.g.f.: (7*exp(x) - 2*exp(-x/2)*(2*cos(sqrt(3)*x/2) + sqrt(3)*sin(sqrt(3)*x/2)))/3. - Stefano Spezia, Sep 27 2024

A169609 Period 3: repeat [1, 3, 3].

Original entry on oeis.org

1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3

Offset: 0

Klaus Brockhaus, Dec 03 2009

Interleaving of A000012, A010701 and A010701.
Also continued fraction expansion of (5+sqrt(65))/10 = 1.3062257748....
Also decimal expansion of 133/999.
a(n) = A144437(n) for n > 0.
Unsigned version of A154595.
Binomial transform of A168615.
Inverse binomial transform of A168673.
Essentially first differences of A047347.

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

Cf. A000012 (all 1's sequence), A010701 (all 3's sequence), A144437 (repeat 3, 3, 1), A154595 (repeat 1, 3, 3, -1, -3, -3), A168615, A168673, A047347 (congruent to {0, 1, 4} mod 7), A010684 (repeat 1, 3).
Cf. A171419 (decimal expansion of (5+sqrt(65))/10).
Cf. A146094.

[ n mod 3 eq 0 select 1 else 3: n in [0..104] ];

&cat [[1, 3, 3]^^30]; // Wesley Ivan Hurt, Jul 02 2016

seq(op([1, 3, 3]), n=0..50); # Wesley Ivan Hurt, Jul 02 2016

PadRight[{},120,{1,3,3}] (* or *) LinearRecurrence[{0,0,1},{1,3,3},120] (* Harvey P. Dale, Apr 29 2015 *)

a(n) = a(n-3) for n > 2, with a(0) = 1, a(1) = 3, a(2) = 3.
G.f.: (1+3*x+3*x^2)/(1-x^3).
a(n) = (7/3)+(2/3)*cos((2*Pi/3)*(n+1))-(2*sqrt(3)/3)*sin((2*Pi/3)*(n+1)). [Richard Choulet, Mar 15 2010]
a(n) = a(n-a(n-2)) for n>=2. Example: a(5) = a(5-a(3)) = a(5-a(3-a(1))) = a(5-a(3-3)) = a(5-a(0)) = a(5-1) = a(4) = a(4-a(2)) = a(4-3) = a(1) = 3. [Richard Choulet, Mar 15 2010; edited by Klaus Brockhaus, Nov 21 2010]
a(n) = 1 + 2*sgn(n mod 3). - Wesley Ivan Hurt, Jul 02 2016
a(n) = 3/gcd(n,3). - Wesley Ivan Hurt, Jul 11 2016

Keywords cofr, cons added by Klaus Brockhaus, Apr 20 2010

Minor edits, crossref added by Klaus Brockhaus, May 03 2010

A052901 Periodic with period 3: a(3n)=3, a(3n+1)=a(3n+2)=2.

Original entry on oeis.org

3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2

Offset: 0

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Continued fraction expansion of (15 + sqrt(365))/10 = A176979. - Klaus Brockhaus, Apr 30 2010
First differences of A047390. - Tom Edgar, Jul 17 2014
Also decimal expansion of 322/999. - Nicolas Bělohoubek, Nov 11 2021

INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 878
Index entries for linear recurrences with constant coefficients, signature (0,0,1).

Cf. A176979 (decimal expansion of (15+sqrt(365))/10).

Cf. A208131 (partial products).

a052901 n = a052901_list !! n
a052901_list = cycle [3,2,2]  -- Reinhard Zumkeller, Apr 08 2012

spec := [S,{S=Union(Sequence(Z),Sequence(Z),Sequence(Prod(Z,Z,Z)))},unlabeled]: seq(combstruct[count](spec,size=n), n=0..20);

PadRight[{},110,{3,2,2}] (* Harvey P. Dale, Mar 19 2013 *)
LinearRecurrence[{0, 0, 1},{3, 2, 2},105] (* Ray Chandler, Aug 25 2015 *)

Vec((2*x^2+2*x+3)/(1-x^3)+O(x^99)) \\ Charles R Greathouse IV, Apr 08 2012

G.f.: (2*x^2 + 2*x + 3)/(1-x^3).
a(n) = Sum((1/3)*(2*alpha^2 + 3*alpha + 2)*alpha^(-1-n), where alpha = RootOf(-1+x^3)).
a(n) = ceiling(7*(n+1)/3) - ceiling(7*n/3). - Tom Edgar, Jul 17 2014
From Nicolas Bělohoubek, Nov 11 2021: (Start)
a(n) = 12/(a(n-2)*a(n-1)).
a(n) = 7 - a(n-2) - a(n-1). See also A069705 or A144437. (End)

More terms from James Sellers, Jun 06 2000

A167192 Triangle read by rows: T(n,k) = (n-k)/gcd(n,k), 1 <= k <= n.

Original entry on oeis.org

0, 1, 0, 2, 1, 0, 3, 1, 1, 0, 4, 3, 2, 1, 0, 5, 2, 1, 1, 1, 0, 6, 5, 4, 3, 2, 1, 0, 7, 3, 5, 1, 3, 1, 1, 0, 8, 7, 2, 5, 4, 1, 2, 1, 0, 9, 4, 7, 3, 1, 2, 3, 1, 1, 0, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 11, 5, 3, 2, 7, 1, 5, 1, 1, 1, 1, 0, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 13, 6, 11, 5, 9, 4, 1, 3, 5, 2, 3

Offset: 1

Reinhard Zumkeller, Oct 30 2009

			The triangle T(n,k) begins:
n\k   1   2   3   4  5  6  7  8  9 10 11 12 13  14  15 ...
1:    0
2:    1   0
3:    2   1   0
4:    3   1   1   0
5:    4   3   2   1  0
6:    5   2   1   1  1  0
7:    6   5   4   3  2  1  0
8:    7   3   5   1  3  1  1  0
9:    8   7   2   5  4  1  2  1  0
10:   9   4   7   3  1  2  3  1  1  0
11:  10   9   8   7  6  5  4  3  2  1  0
12:  11   5   3   2  7  1  5  1  1  1  1  0
13:  12  11  10   9  8  7  6  5  4  3  2  1  0
14:  13   6  11   5  9  4  1  3  5  2  3  1  1   0
15:  14  13   4  11  2  3  8  7  2  1  4  1  2   1   0
- _Wolfdieter Lang_, Feb 20 2013

Indranil Ghosh, Rows 1..120 of triangle, flattened

Cf. A054531, A112543, A164306.

Flatten[Table[(n-k)/GCD[n,k],{n,20},{k,n}]] (* Harvey P. Dale, Nov 27 2015 *)

for(n=1,10, for(k=1,n, print1((n-k)/gcd(n,k), ", "))) \\ G. C. Greubel, Sep 13 2017

T(n,k) = (n-k)/gcd(n,k), 1 <= k <= n.
T(n,k) = A025581(n,k)/A050873(n,k);
T(n,1) = A001477(n-1);
T(n,2) = A026741(n-2) for n > 1;
T(n,3) = A051176(n-3) for n > 2;
T(n,4) = A060819(n-4) for n > 4;
T(n,n-3) = A144437(n) for n > 3;
T(n,n-2) = A000034(n) for n > 2;
T(n,n-1) = A000012(n);
T(n,n) = A000004(n).

A143025 Period length 4: repeat [1, 8, 2, 8].

Original entry on oeis.org

1, 8, 2, 8, 1, 8, 2, 8, 1, 8, 2, 8, 1, 8, 2, 8, 1, 8, 2, 8, 1, 8, 2, 8, 1, 8, 2, 8, 1, 8, 2, 8, 1, 8, 2, 8, 1, 8, 2, 8, 1, 8, 2, 8, 1, 8, 2, 8, 1, 8, 2, 8, 1, 8, 2, 8, 1, 8, 2, 8, 1, 8, 2, 8, 1, 8, 2, 8, 1, 8, 2, 8, 1, 8, 2, 8, 1, 8, 2, 8, 1, 8, 2, 8, 1, 8

Offset: 0

Paul Curtz, Oct 13 2008

Numerator of 1/n^2-1/(3n)^2 if n>0.
This can be generated from the transitions between principal quantum numbers n and 3n in the Hydrogen series: A005563(2), A061037(6), A061039(9), A061041(12), A061043(15), A061045(18), A061047(21), A061049(24),... (The mention of A005563(2) is somewhat a fluke to maintain the periodic pattern.)
Related to the continued fraction of (12*sqrt(55)-72)/19 = 0.89444115.. = 0+1/(1+1/(8+1/(2+...))). - R. J. Mathar, Jun 27 2011

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

Cf. A045944, A144437.

&cat [[1, 8, 2, 8]^^30]; // Wesley Ivan Hurt, Jul 10 2016

seq(op([1, 8, 2, 8]), n=0..50); # Wesley Ivan Hurt, Jul 10 2016

PadRight[{}, 120, {1, 8, 2, 8}] (* Harvey P. Dale, Jul 01 2015 *)

a(n)=[1,8,2,8][n%4+1] \\ Charles R Greathouse IV, Jun 02 2011

a(n+4) = a(n).
G.f.: (1+8*x+2*x^2+8*x^3)/(1-x^4).
From Wesley Ivan Hurt, Jul 10 2016: (Start)
a(n) = (19 - 13*I^(2*n) - I^(-n) - I^n)/4, where I = sqrt(-1).
a(n) = (19 - 2*cos(n*Pi/2) - 13*cos(n*Pi))/4. (End)

Partially edited by R. J. Mathar, Dec 10 2008

A147560 a(n) = 4*A046162(n+1).

Original entry on oeis.org

0, 4, 16, 12, 64, 100, 48, 196, 256, 108, 400, 484, 192, 676, 784, 300, 1024, 1156, 432, 1444, 1600, 588, 1936, 2116, 768, 2500, 2704, 972, 3136, 3364, 1200, 3844, 4096, 1452, 4624, 4900, 1728, 5476, 5776, 2028, 6400, 6724, 2352, 7396, 7744, 2700, 8464

Offset: 0

Paul Curtz, Nov 07 2008

G. C. Greubel, Table of n, a(n) for n = 0..5000

Companion to A144437.
Cf. A046162.
Cf. A171522. [R. J. Mathar, Dec 15 2009]

[4*Numerator(n^2/(n^2+3*n+3)): n in [0..70]]; // G. C. Greubel, Oct 27 2022

A046162 := proc(n) (n-1)^2/(n^2+n+1) ; numer(%) ; end proc: A147560 := proc(n) 4*A046162(n+1) ; end proc: seq(A147560(n),n=0..70) ; # R. J. Mathar, Dec 15 2009

a[n_] := 4 * Numerator[n^2/(n^2 + 3*n + 3)]; Array[a, 50, 0] (* Amiram Eldar, Aug 14 2022 *)

[4*numerator(n^2/(n^2 +3*n +3)) for n in range(71)] # G. C. Greubel, Oct 27 2022

a(n) = 4*numerator(n^2/(n^2 + 3*n + 3)).
Sum_{n>=1} 1/a(n) = 11*Pi^2/216. - Amiram Eldar, Aug 14 2022
G.f.: 4*x*(1 + 4*x + 3*x^2 + 13*x^3 + 13*x^4 + 3*x^5 + 4*x^6 + x^7)/(1-x^3)^3. - G. C. Greubel, Oct 27 2022

More terms from R. J. Mathar, Dec 15 2009

A171417 Decimal expansion of (5+sqrt(65))/4.

Original entry on oeis.org

3, 2, 6, 5, 5, 6, 4, 4, 3, 7, 0, 7, 4, 6, 3, 7, 4, 1, 3, 0, 9, 1, 6, 5, 3, 3, 0, 7, 5, 7, 5, 9, 4, 2, 7, 8, 2, 7, 8, 3, 5, 9, 9, 0, 7, 6, 4, 0, 2, 1, 4, 3, 3, 4, 6, 9, 8, 4, 1, 4, 8, 0, 9, 7, 3, 1, 5, 9, 6, 8, 7, 3, 7, 7, 5, 6, 4, 2, 2, 0, 5, 0, 7, 4, 0, 0, 3, 8, 5, 6, 6, 6, 7, 9, 3, 0, 7, 6, 6, 0, 9, 0, 9, 3, 6

Offset: 1

Klaus Brockhaus, Dec 08 2009

Continued fraction expansion of (5+sqrt(65))/4 is A144437.

			(5+sqrt(65))/4 = 3.26556443707463741309....

Cf. A010517 (decimal expansion of sqrt(65)), A144437 (repeat 3, 3, 1).

RealDigits[(5+Sqrt[65])/4,10,120][[1]] (* Harvey P. Dale, Jun 17 2011 *)

cp's OEIS Frontend

A168615 Inverse binomial transform of A169609, or of A144437 preceded by 1.

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A163979 a(n) = n*(n-1) + A144437(n+2).

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A051176 If n mod 3 = 0 then n/3 else n.

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A069705 a(n) = 2^n mod 7.

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A169609 Period 3: repeat [1, 3, 3].

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A052901 Periodic with period 3: a(3n)=3, a(3n+1)=a(3n+2)=2.

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