A007679 If n mod 4 = 0 then 2^(n-1)+1 elif n mod 4 = 2 then 2^(n-1)-1 else 2^(n-1).

1, 1, 4, 9, 16, 31, 64, 129, 256, 511, 1024, 2049, 4096, 8191, 16384, 32769, 65536, 131071, 262144, 524289, 1048576, 2097151, 4194304, 8388609, 16777216, 33554431, 67108864, 134217729, 268435456, 536870911

Offset: 1

Author

N. J. A. Sloane, R. K. Guy, Simon Plouffe

Keywords

• nonn
• easy

References

• M. E. Larsen, Summa Summarum, A. K. Peters, Wellesley, MA, 2007; see p. 37. [From N. J. A. Sloane, Jan 29 2009]
• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• I. Nemes et al., How to do Monthly problems with your computer, Amer. Math. Monthly, 104 (1997), 505-519.
• Index entries for linear recurrences with constant coefficients, signature (2,-1,2).

Programs

Magma

I:=[1, 1, 4]; [n le 3 select I[n] else 2*Self(n-1)-Self(n-2)+2*Self(n-3): n in [1..40]]; // Vincenzo Librandi, May 09 2012

Maple

f:=n->2^(n-1)+cos(Pi*n/2);

Mathematica

CoefficientList[Series[(1-x+3*x^2)/((1-2*x)*(1+x^2)),{x,0,30}],x] (* Vincenzo Librandi, May 09 2012 *)
Table[Which[Mod[n,4]==0,2^(n-1)+1,Mod[n,4]==2,2^(n-1)-1,True,2^(n-1)], {n,30}] (* or *) LinearRecurrence[{2,-1,2},{1,1,4},30] (* Harvey P. Dale, May 01 2018 *)

Python

def A007679(n): return (1<Chai Wah Wu, Apr 22 2025

Formula

a(n) = 2^(n-1) + cos(n*Pi/2).

a(n) = sum(2^k*C(n-k, 2k)*n/(n-k), k=0..floor(n/3)).

a(n) = ((-i)^n+i^n+2^n)/2, where i=sqrt(-1). a(n) = 2*a(n-1)-a(n-2)+2*a(n-3). G.f.: x*(1-x+3*x^2)/((1-2*x)*(1+x^2)). [Colin Barker, May 08 2012]