A177704 -id:A177704 - cp's OEIS frontend

A001068 a(n) = floor(5*n/4), numbers that are congruent to {0, 1, 2, 3} mod 5.

0, 1, 2, 3, 5, 6, 7, 8, 10, 11, 12, 13, 15, 16, 17, 18, 20, 21, 22, 23, 25, 26, 27, 28, 30, 31, 32, 33, 35, 36, 37, 38, 40, 41, 42, 43, 45, 46, 47, 48, 50, 51, 52, 53, 55, 56, 57, 58, 60, 61, 62, 63, 65, 66, 67, 68, 70, 71, 72, 73, 75, 76, 77, 78, 80, 81, 82, 83, 85, 86, 87, 88

Offset: 0

N. J. A. Sloane

From M. F. Hasler, Oct 21 2008: (Start)
Also, for n>0, the 4th term (after [0,n,3n]) in the continued fraction expansion of arctan(1/n). (Observation by V. Reshetnikov)
Proof:
arctan(1/n) = (1/n) / (1 + (1/n)^2/( 3 + (2/n)^2/( 5 + (3/n)^2/( 7 + ...)...)
            = 1 / ( n + 1/( 3n + 4/( 5n + 9/( 7n + 25/(...)...)
            = 1 / ( n + 1/( 3n + 1/( 5n/4 + (9/4)/( 7n + 25/(...)...),
and the term added to 5n/4, (9/4)/(7n+...) = (1/4)*9/(7n+...) is less than 1/4 for all n>=2. (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Paul Erdős, Some recent problems and results in graph theory, Discr. Math., Vol. 164, No. 1-3 (1997), pp. 81-85.
Wikipedia, Continued fraction for arctangent.
R. Witula, P. Lorenc, M. Rozanski and M. Szweda, Sums of the rational powers of roots of cubic polynomials, Zeszyty Naukowe Politechniki Slaskiej, Seria: Matematyka Stosowana z. 4, Nr. kol. 1920, (2014), pp. 17-34.
Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Cf. A001622, A002265, A030308, A047203, A110255, A110256, A110257, A110258, A110259, A110260, A177704 (first differences), A249547.

[Floor(5*n/4): n in [0..80]]; // Vincenzo Librandi, Nov 13 2011

A001068:=n->floor(5*n/4); seq(A001068(k), k=0..100); # Wesley Ivan Hurt, Nov 07 2013

Table[Floor[5*n/4],{n,0,120}] (* Vladimir Joseph Stephan Orlovsky, Jan 28 2012 *)
#+{0,1,2,3}&/@(5*Range[0,20])//Flatten (* or *) Complement[Range[0,103],5*Range[20]-1] (* Harvey P. Dale, Dec 03 2023 *)

a(n)=5*n\4 /* or, cf. comment: */
a(n)=contfrac(atan(1/n))[4] \\ M. F. Hasler, Oct 21 2008

contfrac( arctan( 1/n )) = 0 + 1/( n + 1/( 3n + 1/( a(n) + 1/(...)))). - M. F. Hasler, Oct 21 2008
a(n) = Sum_{k>=0} A030308(n,k)*b(k) with b(0)=1, b(1)=2 and b(k)=5*2^(k-2) for k>1. - Philippe Deléham, Oct 17 2011.
From Bruno Berselli, Oct 17 2011:  (Start)
G.f.: x*(1+x+x^2+2*x^3)/((1+x)*(1-x)^2*(1+x^2)).
a(n) = (10*n+2*(-1)^((n-1)n/2)+(-1)^n-3)/8.
a(-n) = -A047203(n+1). (End)
From Wesley Ivan Hurt, Sep 17 2015: (Start)
a(n) = a(n-1) + a(n-4) - a(n-5) for n>4.
a(n) = n + floor(n/4). (End)
a(n) = n + A002265(n). - Robert Israel, Sep 17 2015
E.g.f.: (sin(x) + cos(x) + (5*x - 2)*sinh(x) + (5*x - 1)*cosh(x))/4. - Ilya Gutkovskiy, May 06 2016
Sum_{n>=1} (-1)^(n+1)/a(n) = log(5)/4 + sqrt(5)*log(phi)/10 + sqrt(5-2*sqrt(5))*Pi/10, where phi is the golden ratio (A001622). - Amiram Eldar, Dec 10 2021

More terms from James Sellers, Sep 19 2000

A047203 Numbers that are congruent to {0, 2, 3, 4} mod 5.

Original entry on oeis.org

0, 2, 3, 4, 5, 7, 8, 9, 10, 12, 13, 14, 15, 17, 18, 19, 20, 22, 23, 24, 25, 27, 28, 29, 30, 32, 33, 34, 35, 37, 38, 39, 40, 42, 43, 44, 45, 47, 48, 49, 50, 52, 53, 54, 55, 57, 58, 59, 60, 62, 63, 64, 65, 67, 68, 69, 70, 72, 73, 74, 75, 77, 78, 79, 80, 82, 83, 84, 85, 87, 88, 89

Offset: 1

N. J. A. Sloane

Complement of A016861. - Reinhard Zumkeller, Oct 23 2006

Melvyn B. Nathanson, On the fractional parts of roots of positive real numbers, Amer. Math. Monthly, Vol. 120, No. 5 (2013), pp. 409-429 [see p. 417].
Index entries for linear recurrences with constant coefficients, signature (1,0,0,1,-1).

Cf. A001068, A001622, A002265, A016861, A027445, A047207, A047211, A047218, A177704.

[Floor((5*n-2)/4) : n in [1..100]]; // Wesley Ivan Hurt, May 14 2016

seq(floor(5*n-2)/4), n=1..72); # Gary Detlefs, Mar 06 2010
seq(floor((15*n-5)/12), n=1..72); # Gary Detlefs, Mar 07 2010

Flatten[Table[5*n + {0, 2, 3, 4}, {n, 0, 20}]] (* T. D. Noe, Nov 12 2013 *)

a(n)=(5*n-2)\4 \\ Charles R Greathouse IV, Jun 11 2015

A027445(a(n)) mod 10 = 0. - Reinhard Zumkeller, Oct 23 2006
a(n) = floor((5n-2)/4). - Gary Detlefs, Mar 06 2010
a(n) = floor((15n-5)/12). - Gary Detlefs, Mar 07 2010
G.f.: x^2*(2+x+x^2+x^3)/((1+x)*(1+x^2)*(x-1)^2). - R. J. Mathar, Oct 08 2011
From Wesley Ivan Hurt, May 14 2016: (Start)
a(n) = a(n-1) + a(n-4) - a(n-5) for n>5.
a(n) = (10*n-7+(-1)^n+2*(-1)^((2*n+3+(-1)^n)/4))/8.
a(2n) = A047211(n), a(2n-1) = A047218(n).
a(n) = A047207(n+1) - 1.
a(n+2) = n + 2 + A002265(n) for n>0.
a(n+3)-a(n+2) = A177704(n) for n>0.
a(1-n) = - A001068(n). (End)
Sum_{n>=2} (-1)^n/a(n) = log(5)/4 + sqrt(5)*log(phi)/10 - sqrt(5-2*sqrt(5))*Pi/10, where phi is the golden ratio (A001622). - Amiram Eldar, Dec 07 2021

More terms from Reinhard Zumkeller, Oct 23 2006

A093148 a(n) = gcd(Fibonacci(n+5), Fibonacci(n+1)).

Original entry on oeis.org

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

Offset: 0

Paul Barry, Apr 02 2004

From Klaus Brockhaus, May 30 2010: (Start)
Periodic sequence: Repeat [1, 1, 1, 3].
Continued fraction expansion of (9+sqrt(165))/14.
Decimal expansion of 371/3333. (End)
Final nonzero digit of n^n in base 4. - José María Grau Ribas, Jan 19 2012

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

Cf. A000045, A061347, A167816, A177704, A178591.

[1+2*0^((n+1) mod 4) : n in [0..100]]; // Wesley Ivan Hurt, Oct 23 2014

A093148:=n->1+2*0^(n+1 mod 4): seq(A093148(n), n=0..100); # Wesley Ivan Hurt, Oct 23 2014

f[n_] := Switch[Mod[n, 4], 0, 1, 1, 1, 2, 1, 3, 3]; Array[f, 105, 0] (* Robert G. Wilson v, Jan 23 2012 *)
PadRight[{},120,{1,1,1,3}] (* Harvey P. Dale, Sep 03 2021 *)

a(n)=if(n%4==3,3,1) \\ Charles R Greathouse IV, Oct 07 2015

G.f.: (1+x+x^2+3*x^3)/(1-x^4); a(n) = 3/2-sin(Pi*n/2)-cos(Pi*n)/2.
From Klaus Brockhaus, May 30 2010: (Start)
a(n) = a(n-4) for n > 3; a(0) = a(1) = a(2) = 1, a(3) = 3.
a(n) = (3-(-1)^n+(1-(-1)^n)*i*i^n)/2 where i = sqrt(-1). (End)
a(n) = 1 + 2*0^mod(n+1, 4). - Wesley Ivan Hurt, Oct 23 2014

A236398 Period 4: repeat 1,1,2,1.

Original entry on oeis.org

1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1

Offset: 1

N. J. A. Sloane, Jan 29 2014

H. Blaine Lawson, Jr. and M.-L. Michelsohn, Spin Geometry, Princeton, p. 33.

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

Cf. A177704, A138191.

&cat [[1, 1, 2, 1]^^24]; // Bruno Berselli, Dec 10 2015

a(n)=n=n%8; if(n==3||n==7, 2, 1) \\ Charles R Greathouse IV, Jul 17 2016

a(n) = a(n-4) for n > 4. G.f.: x*(1 + x + 2*x^2 + x^3)/(1 - x^4). - Chai Wah Wu, Jun 04 2016
E.g.f.: (-2 - sin(x) + 3*sinh(x) + 2*cosh(x))/2. - Ilya Gutkovskiy, Jun 04 2016
a(n) = (5-cos(n*Pi)-2*sin(n*Pi/2))/4. - Luce ETIENNE, Feb 17 2017

Definition corrected by Paul Curtz. - N. J. A. Sloane, Oct 10 2016

A245477 Period 6: repeat [1, 1, 1, 1, 1, 2].

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1

Offset: 0

Hailey R. Olafson, Jul 23 2014

First differences of A047368. The first differences of this sequence are in A131533. - Wesley Ivan Hurt, Jul 24 2014

Binomial Transform of a(n) gives: 1, 2, 4, 8, 16, 33, 70, 149, 312, 638, 1276, 2511, ... - Wesley Ivan Hurt, Aug 13 2014

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

Cf. A024643, A047368, A097325, A130782, A131534, A172051, A177702, A177704, A177706.

[Floor((n+1)*7/6) - Floor((n)*7/6) : n in [0..100]]; // Wesley Ivan Hurt, Aug 06 2014

A:= n -> piecewise(n mod 6 = 5, 2, 1);
seq(A(n), n=0..100); # Robert Israel, Jul 23 2014

Table[2 - Sign[Mod[n + 1, 6]], {n, 0, 100}] (* Wesley Ivan Hurt, Jul 24 2014 *)
PadRight[{},120,{1,1,1,1,1,2}] (* Harvey P. Dale, Jun 02 2016 *)

a(n) = 7*(n+1)\6 - 7*n\6; \\ Michel Marcus, Jul 23 2014

[floor((n+1)*7/6) - floor((n)*7/6) for n in [0..200]]

a(n) = floor((n+1)*7/6) - floor((n)*7/6).
G.f.: 1/(1-x) + x^5/(1-x^6). - Robert Israel, Jul 23 2014
From Wesley Ivan Hurt, Jul 24 2014, Aug 06-29 2014: (Start)
  a(n) = 2 - sign((n+1) mod 6).
  a(n) = 3 - 2^sign((n+1) mod 6).
  a(n) = A172051(n) + 1.
  a(2n) = 1, a(2n+1) = A177702(n).
  Sum_{i=0..n-2} a(i) = A047368(n), n>0.
  a(n) = 1 + mod(n, 1 + mod(n-1, 3)).
  a(n) = 1 + binomial(mod(5n + 10, 6), 5). (End)
From Wesley Ivan Hurt, Jun 23 2016: (Start)
a(n) = a(n-6) for n>5.
a(n) = (7 - cos(n*Pi) + cos(n*Pi/3) - cos(2*n*Pi/3) - sqrt(3)*sin(n*Pi/3) - sqrt(3)*sin(2*n*Pi/3))/6. (End)

A177705 Decimal expansion of (3+2*sqrt(6))/5.

Original entry on oeis.org

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

Offset: 1

Klaus Brockhaus, May 11 2010

Continued fraction expansion of (3+2*sqrt(6))/5 is A177704.

			(3+2*sqrt(6))/5 = 1.57979589711327123927...

Cf. A010464, A010547, A177704.

RealDigits[(3+2Sqrt[6])/5,10,150][[1]]  (* Harvey P. Dale, Mar 14 2011 *)

Equals (6+A010547)/10. - R. J. Mathar, Feb 03 2025

A214630 a(n) is the reduced numerator of 1/4 - 1/A109043(n)^2 = (1 - 1/A026741(n)^2)/4.

Original entry on oeis.org

-1, 0, 0, 2, 3, 6, 2, 12, 15, 20, 6, 30, 35, 42, 12, 56, 63, 72, 20, 90, 99, 110, 30, 132, 143, 156, 42, 182, 195, 210, 56, 240, 255, 272, 72, 306, 323, 342, 90, 380, 399, 420, 110, 462, 483, 506, 132, 552, 575, 600, 156

Offset: 0

Paul Curtz, Jul 23 2012

The unreduced fractions are -1/0, 0/4, 0/1, 8/36, 3/16, 24/100, 2/9, 48/196, 15/64, 80/324, 6/25, ... = c(n)/A061038(n), say.
Note that c(n)=A061037(n) + (period of length 2: repeat 0, 3).
c(n) is a permutation of A198442(n). The corresponding ranks are (the 0's have been swapped for convenience) 0,2,1,6,4,10,... = A145979(n-2).
Define the following sequences, satisfying the recurrence a(n) = 2*a(n-4) - a(n-8),
e(n) = -1, 0, 0, 2, 1, 4, 1, 6, 3, 8, 2, 10, 5, ... (after -1, a permutation of A004526(n) or mix A026741(n-1), 2*n),
f(n) = 1, 2, 1, 4, 3, 6, 2, 8, 5, 10, 3, 12, 7, ..., (another permutation of A004526(n+2) or mix A026741(n+1), 2*n+2).
f(n) - e(n) = periodic of period length 4: repeat 2, 2, 1, 2.
e(n) + f(n) = 0, 2, 1, 6, 4, 10, ... = A145979(n-2).
Then c(n) = e(n)*f(n).
Note that A061038(n) - 4*c(n) = periodic of period length 4: repeat 4, 4, 1, 4.
After division (by period 2: repeat 1, 4, A010685(n)), the reduced fractions of c(n) are -1/0, 0/1 ?, 0/4 ?, 2/9, 3/16, 6/25, 2/9, 12/49, 15/64, 20/81, 6/25, ... = a(n)/b(n).
Note that a(1+4*n) + a(2+4*n) + a(3+4*n) = 2,20,56,... = A002378(1+3*n) = A194767(3*n).
A061037(n-2) - a(n-2) = 0, -3, 0, -3, 0, 3, 0, 15, 0, 33, 0, 57, ... = Fip(n-2).
Fip(n-2)/3 = 0,-1,0,-1,0,1,0,5,0,11,0,19,0,29, .... Without 0's: A165900(n) (a Fibonacci polynomial); also -A110331(n+1) (Pell numbers).
g(n) = -1, 0, 0, 1, 1, 2, 1, 3, 3, 4, ... = mix A026741(n-1), n.
h(n) = 1, 1, 1, 2, 3, 3, 2, 4, 5, 5, ... = mix A026741(n+1), n+1.
h(n) - g(n) = (period 2: repeat 2, 1, 1, 1 = A177704(n-1)).
k(n) = 1, 1, 0, 2, 3, 3, 1, 4, 5, 5, ... = mix A174239(n), n+1.
l(n) = -1, 0, 1, 1, 1, 2, 2, 3, 3, 4, ... .
k(n) - l(n) = period 4: repeat 2, 1, -1, 1.
2) By the second formula in the definition, we take first 1 - 1/A026741(n)^2.
Hence, using a convention for the first fraction, -1/0, 0/1, 0/1, 8/9, 3/4, 24/25, 8/9, 48/49, 15/16, 80/81, 24/25, ... = (A005563(n-1) - A033996(n))/A168077(n) = q(n)/A168077(n).
For a(n), we divide by 4.
Note that A214297 is the reduced numerator of  1/4 - 1/A061038(n).
Note also that A168077(n) = A026741(n)^2.

Colin Barker, 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. A000466, A002378, A131723.

m:=50; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((2*x^9+3*x^8+6*x^7+2*x^6+6*x^5+6*x^4+2*x^3-1)/((1-x)^3*(x+1)^3*(x^2+1)^3))); // G. C. Greubel, Sep 20 2018

CoefficientList[Series[(2*x^9+3*x^8+6*x^7+2*x^6+6*x^5+6*x^4+2*x^3-1)/((1-x)^3*(x+1)^3*(x^2+1)^3), {x, 0, 50}], x] (* G. C. Greubel, Sep 20 2018 *)
LinearRecurrence[{0,0,0,3,0,0,0,-3,0,0,0,1},{-1,0,0,2,3,6,2,12,15,20,6,30},60] (* Harvey P. Dale, Jul 01 2019 *)

Vec(-(2*x^9+3*x^8+6*x^7+2*x^6+6*x^5+6*x^4+2*x^3-1)/((x-1)^3*(x+ 1)^3*(x^2+1)^3) + O(x^100)) \\ Colin Barker, Jan 22 2015

a(4*n) = 4*n^2-1 = (2*n-1)*(2*n+1), a(2*n+1) = a(4*n+2) = n(n+1).
a(n)= A198442(n)/(period of length 4: repeat 1,1,4,1=A010121(n+2)).
a(n) = 3*a(n-4) - 3*a(n-8) + a(n-12). Is this the shortest possible recurrence? See A214297.
a(n+2) - a(n-2) = 0, 2, 4, 6, 2, 10, 12, 14, 4, ... = 2*A214392(n). a(-2)=a(-1)=0=a(1)=a(2).
a(n+4) - a(n-4) = 0, 4, 2, 12, 16, 20, 6, 28, 32, 36,... = 2*A188167(n). a(-4)=3=a(4), a(-3)=2=a(3).
a(n) = g(n) * h(n).
a(n) = k(n) * l(n).
G.f.: -(2*x^9+3*x^8+6*x^7+2*x^6+6*x^5+6*x^4+2*x^3-1) / ((x-1)^3*(x+1)^3*(x^2+1)^3). - Colin Barker, Jan 22 2015
From Luce ETIENNE, Apr 08 2017: (Start)
a(n) = (13*n^2-28-3*(n^2+4)*(-1)^n+3*(n^2-4)*((-1)^((2*n-1+(-1)^n)/4)+(-1)^((2*n+1-(-1)^n)/4)))/64.
a(n) = (13*n^2-28-3*(n^2+4)*cos(n*Pi)+6*(n^2-4)*cos(n*Pi/2))/64. (End)

Edited by N. J. A. Sloane, Aug 04 2012

A226276 Period 4: repeat [8, 4, 4, 4].

Original entry on oeis.org

8, 4, 4, 4, 8, 4, 4, 4, 8, 4, 4, 4, 8, 4, 4, 4, 8, 4, 4, 4, 8, 4, 4, 4, 8, 4, 4, 4, 8, 4, 4, 4, 8, 4, 4, 4, 8, 4, 4, 4, 8, 4, 4, 4, 8, 4, 4, 4, 8, 4, 4, 4, 8, 4, 4, 4, 8, 4, 4, 4, 8, 4, 4, 4, 8, 4, 4, 4, 8, 4, 4, 4, 8, 4, 4, 4, 8, 4, 4, 4, 8, 4, 4, 4

Offset: 0

Richard R. Forberg, Jun 01 2013

Old name was: A four-term repeating sequence for constructing a summation sequence from negative to positive infinity containing all primes except 2 and 5.
a(n) allows for the creation of an infinite summation sequence, s(n), extending from negative to positive infinity. (See Formula section below.) With appropriate initialization, letting "s(n+)" be the set positive s(n) values, and "s(n-)" be the absolute value of the set of negative s(n) values, the following applies:
s(n+) includes all primes of the form 4*m+1 with m>=2. Thus excluding 5.
s(n-) includes all primes of the form 4*m+3 with m>=0.
Together these include all primes (except 2 and 5) without duplication.
The primes "p(+)" within s(n+) "appear" in the form 3*p(+) within s(n-).
The primes "p(-)" within s(n-) "appear" in the form 3*p(-) within s(n+).
By using this simple repeating pattern, rather than the two well known linear formulas above, all primes (except 2 and 5) are included via a single construction mechanism, and all integers ending in the digit 5 are excluded mathematically, resulting in fewer nonprimes among the values of s(n) than there are in the combination of 4*m+1 and 4*m+3.
(NOTE: In the above "m" is not that same index as "n").
This is one of only two such repeating sequences with the property of generating a summation sequence that includes all integers ending in 1,3,7 or 9, and thus all primes except 2 and 5 (for the other see A226294). Both have the same density of primes in s(n), because both generate only 40% of the integers (in absolute value). And both presumably have the same average density of primes in positive vs. negative values of s(n).
Also, continued fraction expansion of 4 + sqrt(646)/6. - Bruno Berselli, Jun 20 2013

			s(1) = 9, s(2) = 13, s(3) = 17, s(4) = 21, s(5) = 29, s(6) = 33, s(7) = 37.
s(-1) = -3, s(-2) = -7, s(-3) = -11, s(-4) = -19, s(-5) = -23, s(-6) = -27, s(-7) = -31.

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

Cf. A002114, A002145, A121262, A177704, A226294.

&cat [[8, 4, 4, 4]^^30]; // Wesley Ivan Hurt, Jul 09 2016

seq(op([8, 4, 4, 4]), n=0..40); # Wesley Ivan Hurt, Jul 09 2016

Flatten[Table[{8, 4, 4, 4}, {20}]] (* Bruno Berselli, Jun 20 2013 *)
PadRight[{},120,{8,4,4,4}] (* Harvey P. Dale, May 11 2025 *)

a(n)=if(n%4,4,8) \\ Charles R Greathouse IV, Jul 17 2016

For generating the summation sequence s, start with s(0) = 1, and a(0) = 8.
For positive values of s(n): s(n+1) = s(n) + a(n).
For negative values of s(n): s(n-1) = s(n) - a(n-1). Here, n is negative.
All values of a(n) are positive regardless of index. For example: a(-1) = a(-2) = a(-3) = 4; a(-4) = 8. Thus the simple pattern of a(n) and the simple arithmetic for generating s(n), are maintained across the n=0 boundary, in a manner similar to extending Fibonacci numbers to negative indices.
From Bruno Berselli, Jun 20 2013: (Start)
G.f.: 4*(2+x+x^2+x^3)/((1-x)*(1+x)*(1+x^2)).
a(n) = 4 + (1 + (-1)^n)*(1 + I^(n*(n+1))). (End)
From Wesley Ivan Hurt, Jul 09 2016: (Start)
a(n) = a(n-4) for n>3.
a(n) = 5 + I^(2*n) + I^(-n) + I^n.
a(n) = 5 + cos(n*Pi) + 2*cos(n*Pi/2) + I*sin(n*Pi). (End)

Simpler name from Joerg Arndt, Jun 16 2013

A301617 Numbers not divisible by 2, 3 or 5 (A007775) with digital root 1.

Original entry on oeis.org

1, 19, 37, 73, 91, 109, 127, 163, 181, 199, 217, 253, 271, 289, 307, 343, 361, 379, 397, 433, 451, 469, 487, 523, 541, 559, 577, 613, 631, 649, 667, 703, 721, 739, 757, 793, 811, 829, 847, 883, 901, 919, 937, 973, 991, 1009, 1027, 1063, 1081, 1099

Offset: 1

Gary Croft, Mar 24 2018

Numbers == {1, 19, 37, 73} mod 90 with additive sum sequence 1{+18+18+36+18} {repeat ...}. Includes all prime numbers > 7 with digital root 1.

			1+18=19; 19+18=37; 37+36=73; 73+18=91; 91+18=109.

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

Cf. A177704, A045572.

Intersection of A007775 and A017173.

seq(seq(i+90*j,i=[1,19,37,73]),j=0..30); # Robert Israel, Mar 25 2018

LinearRecurrence[{1,0,0,1,-1},{1,19,37,73,91},50] (* Harvey P. Dale, Dec 14 2019 *)

a(n) = 1 + 18 * (n - 1 + n\4) \\ David A. Corneth, Mar 24 2018

Vec(x*(1 + 18*x + 18*x^2 + 36*x^3 + 17*x^4) / ((1 - x)^2*(1 + x)*(1 + x^2)) + O(x^60)) \\ Colin Barker, Mar 24 2018

n == {1, 19, 37, 73} mod 90.
a(n + 1) = a(n) + 18 * A177704(n + 1). - David A. Corneth, Mar 24 2018
From Colin Barker, Mar 24 2018: (Start)
G.f.: x*(1 + 18*x + 18*x^2 + 36*x^3 + 17*x^4) / ((1 - x)^2*(1 + x)*(1 + x^2)).
a(n) = a(n-1) + a(n-4) - a(n-5) for n>5.
(End)

The missing term 1081 added to the sequence by Colin Barker, Mar 24 2018

A052621 E.g.f. (2+x+x^2+x^3)/(1-x^4).

Original entry on oeis.org

2, 1, 2, 6, 48, 120, 720, 5040, 80640, 362880, 3628800, 39916800, 958003200, 6227020800, 87178291200, 1307674368000, 41845579776000, 355687428096000, 6402373705728000, 121645100408832000, 4865804016353280000

Offset: 0

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

INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 567

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

With[{nn=20},CoefficientList[Series[(2+x+x^2+x^3)/(1-x^4),{x,0,nn}],x] Range[0,nn]!] (* Harvey P. Dale, Oct 10 2023 *)

E.g.f.: -(x^3+x^2+x+2)/(-1+x^4)
Recurrence: {a(1)=1, a(3)=6, a(2)=2, a(0)=2, (-n^4-35*n^2-50*n-24-10*n^3)*a(n)+a(n+4)=0}
Sum(1/4*(_alpha^3+_alpha^2+2*_alpha+1)*_alpha^(-1-n), _alpha=RootOf(-1+_Z^4))*n!
2n! if n is 0 mod 4, n! otherwise.
a(n)=n!*A177704(n+3). - R. J. Mathar, Jun 03 2022

cp's OEIS Frontend

A001068 a(n) = floor(5*n/4), numbers that are congruent to {0, 1, 2, 3} mod 5.

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A047203 Numbers that are congruent to {0, 2, 3, 4} mod 5.

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A093148 a(n) = gcd(Fibonacci(n+5), Fibonacci(n+1)).

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A236398 Period 4: repeat 1,1,2,1.

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A245477 Period 6: repeat [1, 1, 1, 1, 1, 2].

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A177705 Decimal expansion of (3+2*sqrt(6))/5.

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A214630 a(n) is the reduced numerator of 1/4 - 1/A109043(n)^2 = (1 - 1/A026741(n)^2)/4.

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