

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
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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 nth 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(k1), with initial values b(0) = 2, b(1) = a(n), all such sequences are invariant under this transformation: b(k) = (b(j+k) + b(jk))/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 noninteger values. Further, these b(k) are also the bisections of the transformations of sequences of the form G(k+1) = n * G(k) + G(k1), and those bisections are invariant for all initial values of g(0) and g(1), including noninteger 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
GuoNiu Han, Enumeration of Standard Puzzles
GuoNiu Han, Enumeration of Standard Puzzles [Cached copy]
Dennis P. Walsh, Notes on a tied election
Eric Weisstein's World of Mathematics, NearSquare Prime
Eric Weisstein's World of Mathematics, Pappus chain
Index entries for linear recurrences with constant coefficients, signature (3,3,1).


FORMULA

G.f.: (23x+3x^2)/(1x)^3.  R. J. Mathar, Oct 05 2008
a(n) = ((n2)^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(n1)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 4candidate, 2voter 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



