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A067998 a(n) = n^2 - 2*n. 28
0, -1, 0, 3, 8, 15, 24, 35, 48, 63, 80, 99, 120, 143, 168, 195, 224, 255, 288, 323, 360, 399, 440, 483, 528, 575, 624, 675, 728, 783, 840, 899, 960, 1023, 1088, 1155, 1224, 1295, 1368, 1443, 1520, 1599, 1680, 1763, 1848, 1935, 2024, 2115, 2208, 2303, 2400, 2499, 2600, 2703, 2808, 2915, 3024, 3135, 3248, 3363 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,4

COMMENTS

For n != 1, numerator of 1-1/(n-1)^2. The associated denominator is A174902(n-1). - Paul Curtz, Dec 05 2010

a(n) is essentially the case 0 of the polygonal numbers. The polygonal numbers are defined as P_k(n) = Sum_{i=1..n}((k-2)*i-(k-3). Thus P_0(n) = 2*n - n^2 and a(n) = -P_0(n). - Peter Luschny, Jul 08 2011

For n >= 3, the denominator of the probability of winning the prize by switching from the initial choice of doors in a generalized Monty Hall problem with n doors: After a prize has been placed behind exactly one of the n doors at random, a contestant chooses a door. Then the host, who knows where the prize is, deliberately opens exactly one unchosen door that does not hide the prize (picked at random by the host among such doors when there is a choice) and then gives the contestant an opportunity to switch to any other door not yet opened. The numerator of this probability is n - 1 (Incidentally, GCD(n - 1, n(n - 2)) = 1). The probability of winning by switching minus the probability of winning by not switching is thus (n - 1)/(n(n - 2)) - 1/n = 1/a(n), which approaches zero as n approaches infinity, but nevertheless makes the switching strategy better for every finite n >= 3. The winning probability is 2/3 from switching in the classic 3-door Monty Hall problem; we have 3/8 and 4/15, respectively, in the 4- and 5-door generalizations. (The above analysis was independent but is consistent with the even more general "N-doors" section of the Wikipedia article, other parts of which make clear the historical importance of wording this problem as carefully as possible. See also A122774.) - Rick L. Shepherd, May 31 2014, clarified Oct 29 2015

For n > 1, a(n) is the largest integer k such that k + n^2 is a multiple of k + n. - Derek Orr, Sep 04 2014

LINKS

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Wikipedia, Monty Hall problem

FORMULA

a(n) = A005563(n-2) = A005563(-n) = A000290(n-1)-1.

G.f.: x*(3*x-1)/(1-x)^3. - Paul Barry, Mar 27 2007

E.g.f.: exp(x)*(x^2-x). - Paul Barry, Mar 27 2007

a(n) = 2*n + a(n-1) - 3 (with a(0)=0). - Vincenzo Librandi, Aug 08 2010

MAPLE

A067998:=n->n^2-2*n: seq(A067998(n), n=0..50); # Wesley Ivan Hurt, Sep 04 2014

MATHEMATICA

Table[ n^2 - 2*n, {n, 0, 60} ] (* George E. Antoniou *)

LinearRecurrence[{3, -3, 1}, {0, -1, 0}, 80] (* Vladimir Joseph Stephan Orlovsky, Feb 23 2012 *)

PROG

(PARI) a(n)=n^2-2*n;

(Haskell)

a067998 n = n * (n - 2)

a067998_list = scanl (+) 0 [-1, 1 ..]

-- Reinhard Zumkeller, Aug 26 2013

(MAGMA) [n^2-2*n : n in [0..50]]; // Wesley Ivan Hurt, Sep 04 2014

(PARI) concat(0, Vec(x*(3*x-1)/(1-x)^3 + O(x^100))) \\  Altug Alkan, Oct 30 2015

CROSSREFS

Essentially the same as A005563.

Cf. A060747 (first differences).

Cf. A000290, A174902.

Sequence in context: A131386 A132411 A005563 * A066079 A185079 A173569

Adjacent sequences:  A067995 A067996 A067997 * A067999 A068000 A068001

KEYWORD

easy,sign

AUTHOR

George E. Antoniou (george.antoniou(AT)montclair.edu), Feb 06 2002

EXTENSIONS

Edited and extended by Robert G. Wilson v, Feb 08 2002

STATUS

approved

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Last modified October 16 16:57 EDT 2018. Contains 316271 sequences. (Running on oeis4.)