A060747 a(n) = 2*n - 1.

-1, 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99, 101, 103, 105, 107, 109, 111, 113, 115, 117, 119, 121, 123, 125, 127, 129, 131, 133, 135, 137, 139, 141, 143, 145, 147, 149, 151

Offset: 0

Author

Henry Bottomley, Apr 26 2001

Keywords

• sign
• easy

Comments

If you put n red balls and n blue balls in a bag and draw them one by one without replacement, the probability of never having drawn equal numbers of the two colors before the final ball is drawn is 1/a(n) unsigned.

abs(a(n)) = 2n - 1 + 2*0^n. It has A048495 as binomial transform. - Paul Barry, Jun 09 2003

For n >= 1, a(n) = numbers k such that arithmetic mean of the first k positive integers is an integer. A040001(a(n)) = 1. See A145051 and A040001. - Jaroslav Krizek, May 28 2010

From Jaroslav Krizek, May 28 2010: (Start)

For n >= 1, a(n) = corresponding values of antiharmonic means to numbers from A016777 (numbers k such that antiharmonic mean of the first k positive integers is an integer).

a(n) = A000330(A016777(n)) / A000217(A016777(n)) = A146535(A016777(n)+1). (End)

Links

• Tanya Khovanova, Recursive Sequences
• Index entries for linear recurrences with constant coefficients, signature (2,-1).

Crossrefs

Cf. A000217, A000330, A000984, A001477, A002420, A005408, A005843, A016777, A023443, A040001, A048495, A078008, A145051, A146535, A161680.

Programs

Haskell

a060747 = subtract 1 . (* 2)
a060747_list = [-1, 1 ..] -- Reinhard Zumkeller, Jul 05 2015
-- Reinhard Zumkeller, Jul 05 2015

Mathematica

Table[2*n - 1, {n, 0, 200}] (* Vladimir Joseph Stephan Orlovsky, Feb 16 2012 *)
LinearRecurrence[{2,-1},{-1,1},80] (* Harvey P. Dale, Mar 27 2020 *)

PARI

a(n)=2*n-1 \\ Charles R Greathouse IV, Sep 24 2015

Formula

a(n) = A005408(n)-2 = A005843(n)-1 = -A000984(n)/A002420(n) = A001477(n)+A023443(n).

G.f.: (3*x - 1)/(1 - x)^2.

Abs(a(n)) = Sum_{k=0..n} (A078008(k) mod 4). - Paul Barry, Mar 12 2004

E.g.f.: exp(x)*(2*x-1). - Paul Barry, Mar 31 2007

a(n) = 2*a(n-1) - a(n-2); a(0)=-1, a(1)=1. - Philippe Deléham, Nov 03 2008

a(n) = 4*n - a(n-1) - 4 for n>0, with a(0)=-1. - Vincenzo Librandi, Aug 07 2010

a(n) = A161680(A005843(n))/n for n > 0. - Stefano Spezia, Feb 14 2025