A164115 -id:A164115 - cp's OEIS frontend

A164116 Expansion of (1 - x) * (1 - x^4) / (1 - x^5) in powers of x.

1, -1, 0, 0, -1, 2, -1, 0, 0, -1, 2, -1, 0, 0, -1, 2, -1, 0, 0, -1, 2, -1, 0, 0, -1, 2, -1, 0, 0, -1, 2, -1, 0, 0, -1, 2, -1, 0, 0, -1, 2, -1, 0, 0, -1, 2, -1, 0, 0, -1, 2, -1, 0, 0, -1, 2, -1, 0, 0, -1, 2, -1, 0, 0, -1, 2, -1, 0, 0, -1, 2, -1, 0, 0, -1, 2, -1, 0, 0, -1, 2, -1, 0, 0, -1, 2, -1, 0, 0, -1, 2, -1, 0, 0, -1, 2, -1, 0, 0, -1, 2, -1, 0, 0, -1

Offset: 0

Michael Somos, Aug 10 2009

This sequence with a(0) replaced by 2 appears, together with three other sequences, in the formula 2*exp(2*Pi*n*I/5) = 2*T(n,x) + S(n-1,x)*sqrt(2+phi)*I, with x = (phi-1)/2 and I = sqrt(-1), where phi = (1+sqrt(5))/2 (golden section) after reduction of powers using phi^2 = phi+1. T and S are Chebyshev polynomials from A053120 and A049310. This results in 2*exp(2*Pi*n*I/5) = (A(n) + B(n)*phi) + (C(n) + D(n)*phi)*sqrt(2+phi)*I, with A(n) = a(n+5), B(n) = A080891(n), C(n) = A156174(n+4) and D(n) = A010891(n+3) for n >= 0. - Wolfdieter Lang, Feb 26 2014

			G.f. = 1 - x - x^4 + 2*x^5 - x^6 - x^9 + 2*x^10 - x^11 - x^14 + 2*x^15 - x^16 + ...
exp(2*Pi*3*I/5) = (0 - phi) + (1 - phi)*sqrt(2+phi)*I, with phi = (1+sqrt(5))/2. - _Wolfdieter Lang_, Feb 26 2014

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

Cf. A164115, A164118.

m:=50; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((1 - x^4)/(1+x+x^2+x^3+x^4))); // G. C. Greubel, Sep 22 2018

CoefficientList[Series[(1-x)(1-x^4)/(1-x^5),{x,0,110}],x] (* Harvey P. Dale, Sep 25 2013 *)

{a(n) = -(n==0) + [2, -1, 0, 0, -1][n%5 + 1]};

Euler transform of length 5 sequence [-1, 0, 0, -1, 1].
a(n) = a(-n) for all n in Z. a(n+5) = a(n) unless n=0 or n=-5.
G.f.: (1 - x^4)/(1 + x + x^2 + x^3 + x^4).
a(n) = (-1)^n * A164118(n). Convolution inverse of A164115.
a(n) = 2*0^mod(n,5) - 0^n - mod(mod(n+2,5),2). - Wesley Ivan Hurt, Apr 28 2015

A177704 Period 4: repeat [1, 1, 1, 2].

Original entry on oeis.org

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

Offset: 0

Klaus Brockhaus, May 11 2010

Continued fraction expansion of (3 + 2*sqrt(6))/5.
Decimal expansion of 1112/9999.
a(n) = A164115(n + 1) = (-1)^(n + 1) * A164117(n + 1) = A138191(n + 3) = A107453(n + 5).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
David Nacin, Van der Laan Sequences and a Conjecture on Padovan Numbers, J. Int. Seq., Vol. 26 (2023), Article 23.1.2.
Index entries for linear recurrences with constant coefficients, signature (0,0,0,1).

Cf. A001068, A107453, A138191, A164115, A164117, A177705, A219977, A236398.

Magma
```
&cat[ [1, 1, 1, 2]: k in [1..27] ];
```

A177704:=n->floor((n+1)*5/4) - floor(n*5/4): seq(A177704(n), n=0..100); # Wesley Ivan Hurt, Jun 15 2016

Table[Floor[(n + 1)*5/4] - Floor[n*5/4], {n, 0, 100}] (* Wesley Ivan Hurt, Jun 15 2016 *)
LinearRecurrence[{0, 0, 0, 1}, {1, 1, 1, 2}, 100] (* Vincenzo Librandi, Jun 16 2016 *)

a(n) = if(n%4==3, 2, 1) \\ Felix Fröhlich, Jun 15 2016

a(n) = (5-(-1)^n + i*i^n-i*(-i)^n)/4 where i = sqrt(-1).
a(n) = a(n-4) for n > 3; a(0) = 1, a(1) = 1, a(2) = 1, a(3) = 2.
G.f.: (1+x+x^2+2*x^3)/(1-x^4).
a(n) = 1 + (1-(-1)^n) * (1+i^(n+1))/4 where i = sqrt(-1). - Bruno Berselli, Apr 05 2011
a(n) = 5/4 - sin(Pi*n/2)/2 - (-1)^n/4. - R. J. Mathar, Oct 08 2011
a(n) = floor((n+1)*5/4) - floor(n*5/4). - Hailey R. Olafson, Jul 23 2014
From Wesley Ivan Hurt, Jun 15 2016: (Start)
a(n+3) - a(n+2) = A219977(n).
Sum_{i=0..n-1} a(i) = A001068(n). (End)
E.g.f.: (-sin(x) + 3*sinh(x) + 2*cosh(x))/2. - Ilya Gutkovskiy, Jun 15 2016

A186646 Every fourth term of the sequence of natural numbers 1,2,3,4,... is halved.

Original entry on oeis.org

1, 2, 3, 2, 5, 6, 7, 4, 9, 10, 11, 6, 13, 14, 15, 8, 17, 18, 19, 10, 21, 22, 23, 12, 25, 26, 27, 14, 29, 30, 31, 16, 33, 34, 35, 18, 37, 38, 39, 20, 41, 42, 43, 22, 45, 46, 47, 24, 49, 50, 51, 26, 53, 54, 55, 28, 57, 58, 59, 30, 61, 62, 63, 32, 65, 66, 67, 34, 69, 70, 71, 36, 73, 74, 75, 38, 77, 78, 79, 40, 81, 82, 83, 42, 85, 86, 87, 44, 89, 90, 91, 46, 93, 94, 95, 48, 97, 98, 99

Offset: 1

R. J. Mathar, Feb 25 2011

a(n) is the length of the period of the sequence k^2 mod n, k=1,2,3,4,..., i.e., the length of the period of A000035 (n=2), A011655 (n=3), A000035 (n=4), A070430 (n=5), A070431 (n=6), A053879 (n=7), A070432 (n=8), A070433 (n=9), A008959 (n=10), A070434 (n=11), A070435 (n=12) etc.
From Franklin T. Adams-Watters, Feb 24 2011: (Start)
Clearly if gcd(n,m) = 1, a(nm) = lcm(a(n),a(m)), so it suffices to establish this for prime powers.
If p is a prime, the period must divide p, but k^2 mod p is not constant, so a(p) = p.
a(p^e), e > 1, must be divisible by a(p^(e-1)), and must divide p^e. If p != 2, (p^(e-1)+1)^2 =  p^(2e-2)+2p^(e-1)+1 == 2p^(e-1)+1 (mod p^2), so a(p^e) != p^(e-1); it must then be e.
By inspection, a(4) = 2 and a(8) = 4.
This leaves a(2^e), e > 3. But then (2^(e-2)+1)^2 = 2^(2e-4)+2^(e-1)+1 == 2^(e-1)+1 (mod 2^e), so a(n) > 2^(e-2). On the other hand, (2^(e-1)+c)^2 = 2^(2e-2)+c2^e+c^2 == c^2 (mod 2^e). Hence the period is 2^(e-1). (End)

Index entries for linear recurrences with constant coefficients, signature (0,0,0,2,0,0,0,-1).

Cf. A000224 (size of the set of moduli of k^2 mod n), A019554, A060819, A061037, A090129, A142705, A164115, A283971.

A186646 := proc(n) if n mod 4 = 0 then n/2 ; else n ; end if; end proc ;

Flatten[Table[{n,n+1,n+2,(n+3)/2},{n,1,101,4}]] (* or *) LinearRecurrence[ {0,0,0,2,0,0,0,-1},{1,2,3,2,5,6,7,4},100] (* Harvey P. Dale, May 30 2014 *)
Table[n (7 - (-1)^n - 2 Cos[n Pi/2])/8, {n, 100}] (* Federico Provvedi , Jan 02 2018 *)

a(n)=if(n%4,n,n/2) \\ Charles R Greathouse IV, Oct 16 2015

def A186646(n): return n if n&3 else n>>1 # Chai Wah Wu, Jan 10 2023

a(n) = 2*a(n-4) - a(n-8).
a(4n) = 2n; a(4n+1) = 4n+1; a(4n+2) = 4n+2; a(4n+3) = 4n+3.
a(n) = n/A164115(n).
G.f.: x*(1 + 2*x + 3*x^2 + 2*x^3 + 3*x^4 + 2*x^5 + x^6) / ( (x-1)^2*(1+x)^2*(x^2+1)^2 ).
Dirichlet g.f.: (1-2/4^s)*zeta(s-1).
A019554(n) | a(n). - Charles R Greathouse IV, Feb 24 2011
a(n) = n*(7 - (-1)^n - (-i)^n - i^n)/8, with i=sqrt(-1). - Bruno Berselli, Feb 25 2011
Multiplicative with a(p^e)=2^e if p=2 and e<=1; a(p^e)=2^(e-1) if p=2 and e>=2; a(p^e)=p^e otherwise. - David W. Wilson, Feb 26 2011
a(n) * A060819(n+2) = A142705(n+1) = A061037(2n+2). - Paul Curtz, Mar 02 2011
a(n) = n - (n/2)*floor(((n-1) mod 4)/3). - Gary Detlefs, Apr 14 2013
a(2^n) = A090129(n+1). - R. J. Mathar, Oct 09 2014
a(n) = n*(7 - (-1)^n - 2*cos(n*Pi/2))/8. - Federico Provvedi, Jan 02 2018
E.g.f.: (1/4)*x*(4*cosh(x) + sin(x) + 3*sinh(x)). - Stefano Spezia, Jan 26 2020
Sum_{k=1..n} a(k) ~ (7/16) * n^2. - Amiram Eldar, Nov 28 2022

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A164116 Expansion of (1 - x) * (1 - x^4) / (1 - x^5) in powers of x.

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A177704 Period 4: repeat [1, 1, 1, 2].

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A186646 Every fourth term of the sequence of natural numbers 1,2,3,4,... is halved.

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