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A059100 a(n) = n^2 + 2. 55
2, 3, 6, 11, 18, 27, 38, 51, 66, 83, 102, 123, 146, 171, 198, 227, 258, 291, 326, 363, 402, 443, 486, 531, 578, 627, 678, 731, 786, 843, 902, 963, 1026, 1091, 1158, 1227, 1298, 1371, 1446, 1523, 1602, 1683, 1766, 1851, 1938, 2027, 2118, 2211, 2306, 2403 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,1

COMMENTS

Let s(n) = Sum_{k>=1} 1/n^(2^k). Then I conjecture that the maximum element in the continued fraction for s(n) is n^2 + 2. - Benoit Cloitre, Aug 15 2002

Binomial transformation yields A081908, with A081908(0)=1 dropped. - R. J. Mathar, Oct 05 2008

a(3*n) mod 9 = 2. - Paul Curtz, Nov 07 2012

1/a(n) = R(n)/r with R(n) the n-th radius of the Pappus chain of the symmetric arbelos with semicircle radii r, r1 = r/2 = r2. See the MathWorld link for Pappus chain (there are two of them, a left and a right one. In this case these two chains are congruent). - Wolfdieter Lang, Mar 01 2013

a(n) is the number of election results for an election with n+2 candidates, say C1, C2, ..., and C(n+2), and with only two voters (each casting a single vote) that have C1 and C2 receiving the same number of votes. See link below. - Dennis P. Walsh, May 08 2013

This sequence gives the set of values such that for sequences b(k+1) = a(n)*b(k) - b(k-1), with initial values b(0) = 2, b(1) = a(n), all such sequences are invariant under this transformation: b(k) = (b(j+k) + b(j-k))/b(j), except where b(j) = 0, for all integer values of j and k, including negative values. Examples are: at n=0, b(k) = 2 for all k; at n=1, b(k) = A005248; at n=2, b(k) = 2*A001541; at n=3, b(k)= A057076; at n=4, b(k) = 2*A023039. This b(k) family are also the transformation results for all related b'(k) (i.e., those with different initial values) including non-integer values. Further, these b(k) are also the bisections of the transformations of sequences of the form G(k+1) = n * G(k) + G(k-1), and those bisections are invariant for all initial values of g(0) and g(1), including non-integer values. For n = 1 this g(k) family includes Fibonacci and Lucas, where the invariant bisection is b(k) = A005248. The applicable bisection for this transformation of g(k) is for the odd values of k, and applies for all n. Also see A000032 for a related family of sequences. - Richard R. Forberg, Nov 22 2014

LINKS

Harry J. Smith, Table of n, a(n) for n = 0..1000

Hesam Dashti, A New Upper Bound on the Length of Shortest Permutation Strings; An Algorithm for Generating Permutation Strings, arXiv:1009.5053 [math.CO], Sep 26, 2010. - Jonathan Vos Post, Sep 28 2010

Guo-Niu Han, Enumeration of Standard Puzzles

Guo-Niu Han, Enumeration of Standard Puzzles [Cached copy]

Dennis P. Walsh, Notes on a tied election

Eric Weisstein's World of Mathematics, Near-Square Prime

Eric Weisstein's World of Mathematics, Pappus chain

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

FORMULA

G.f.: (2-3x+3x^2)/(1-x)^3. - R. J. Mathar, Oct 05 2008

a(n) = ((n-2)^2 + 2*(n+1)^2)/3. - Reinhard Zumkeller, Feb 13 2009

a(n) = A000196(A156798(n) - A000290(n)). - Reinhard Zumkeller, Feb 16 2009

a(n) = 2*n+a(n-1)-1 with a(0)=2. - Vincenzo Librandi, Aug 07 2010

a(n+3) = (A166464(n+5)-A166464(n))/20. - Paul Curtz, Nov 07 2012

a(3*n+1) = 3*A056109(n). a(3*n+2) = 3*A056105(n+1). - Paul Curtz, Nov 07 2012

EXAMPLE

For n=2, a(2)=6 since there are 6 election results in a 4-candidate, 2-voter election that have candidates c1 and c2 tied. Letting <i,j> denote voter 1 voting for candidate i and voter 2 voting for candidate j, the six election results are <1,2>, <2,1>, <3,3>, <3,4>, <4,3>, and <4,4>. - Dennis P. Walsh, May 08 2013

MAPLE

with(combinat, fibonacci):seq(fibonacci(3, i)+1, i=0..49); # Zerinvary Lajos, Mar 20 2008

MATHEMATICA

Table[n^2+2, {n, 0, 50}] (* Vladimir Joseph Stephan Orlovsky, Dec 15 2008 *)

LinearRecurrence[{3, -3, 1}, {2, 3, 6}, 50] (* Vincenzo Librandi, Feb 15 2012 *)

PROG

(Sage) [lucas_number1(3, n, -2) for n in xrange(0, 50)] # Zerinvary Lajos, May 16 2009

(PARI) { for (n = 0, 1000, write("b059100.txt", n, " ", n^2+2); ) } \\ Harry J. Smith, Jun 24 2009

(Haskell)

a059100 = (+ 2) . (^ 2)

a059100_list = scanl (+) (2) [1, 3 ..]

-- Reinhard Zumkeller, Feb 09 2015

CROSSREFS

Cf. A000290, A002522, A056899. Apart from initial terms, same as A010000.

Cf. A069987, A114964 (see comment).

Cf. A008865.

Sequence in context: A034031 A121617 A157656 * A131512 A147388 A180712

Adjacent sequences:  A059097 A059098 A059099 * A059101 A059102 A059103

KEYWORD

nonn,easy

AUTHOR

Henry Bottomley, Feb 13 2001

STATUS

approved

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Last modified March 28 11:45 EDT 2017. Contains 284186 sequences.