A230089 If n is divisible by 4 then 4, if n is divisible by 2 then 2, otherwise n.

1, 2, 3, 4, 5, 2, 7, 4, 9, 2, 11, 4, 13, 2, 15, 4, 17, 2, 19, 4, 21, 2, 23, 4, 25, 2, 27, 4, 29, 2, 31, 4, 33, 2, 35, 4, 37, 2, 39, 4, 41, 2, 43, 4, 45, 2, 47, 4, 49, 2, 51, 4, 53, 2, 55, 4, 57, 2, 59, 4, 61, 2, 63, 4, 65, 2, 67, 4, 69, 2, 71, 4, 73, 2, 75, 4, 77, 2, 79, 4, 81, 2, 83, 4, 85, 2, 87, 4, 89, 2

Offset

1,2

Comments

Suggested by a puzzle in Mitteilungen der DMV, although I do not know if this solution is what they had in mind.

a(A008586(n)) = 4; a(A005408(n)) = A005408(n). - Reinhard Zumkeller, Oct 09 2013

References

Frank Lutz and Brigitte Lutz-Westphal, Eigenwillige Zahlen, Mitteilungen der DMV, 2013, Band 21, Heft 1 (p. 32).

Links

Table of n, a(n) for n=1..90.

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

Formula

Conjecture: a(n) = (3+3*(-1)^n+(-i)^n+i^n+n-(-1)^n*n)/2 where i=sqrt(-1). G.f.: -x*(4*x^5-x^4-2*x^3-2*x^2-2*x-1) / ((x-1)^2*(x+1)^2*(x^2+1)). - Colin Barker, Oct 09 2013

Maple

f:=proc(n) if (n mod 4) = 0 then 4 elif (n mod 4) = 2 then 2; else n; fi; end;

Mathematica

Table[Which[Divisible[n, 4], 4, Divisible[n, 2], 2, True, n], {n, 100}] (* or *) LinearRecurrence[{0, 1, 0, 1, 0, -1}, {1, 2, 3, 4, 5, 2}, 100] (* Harvey P. Dale, Dec 03 2017 *)

Prog

(Haskell)
a230089 n = if odd n then n else if mod n 4 == 0 then 4 else 2
-- Reinhard Zumkeller, Oct 09 2013

Crossrefs

Cf. A005843.

Sequence in context: A223490 A354749 A244734 * A081811 A304181 A034684

Adjacent sequences: A230086 A230087 A230088 * A230090 A230091 A230092

Keyword

nonn,easy

Author

N. J. A. Sloane, Oct 08 2013

Status

approved