%I A011943
%S A011943 1,7,97,1351,18817,262087,3650401,50843527,708158977,9863382151,
%T A011943 137379191137,1913445293767,26650854921601,371198523608647,
%U A011943 5170128475599457,72010600134783751,1002978273411373057
%N A011943 Numbers n such that any group of n consecutive integers has integral 
               standard deviation {viz. A011944(n)}.
%C A011943 If k is in the sequence, then it has successor 7*k + 4*sqrt{3*(k^2 - 
               1)}. - Lekraj Beedassy (blekraj(AT)yahoo.com), Jun 28 2002
%C A011943 Chebyshev's polynomials T(n,x) evaluated at x=7.
%C A011943 a(n+1) give all (nontrivial) solutions of Pell equation a(n+1)^2 - 48*b(n+1)^2 
               = +1 with b(n+1)=A007655(n+2), n>=0.
%C A011943 Also numbers x of the form 3k+1 such that x^2 = 3n^2+1. Also solutions 
               of x in x^2 - 3*y^2 = 1 in A001075 if x = 3k+1 k=1,2,... - Cino Hilliard 
               (hillcino368(AT)gmail.com), Mar 05 2005
%C A011943 Equals sqrt(12*A011944(n)^2 + 1).
%C A011943 In addition to having integral standard deviation, these n consecutive 
               integers also have integral mean. This question was posed by Jim 
               Delany of Cal Poly in 1989. The solution appeared in the American 
               Mathematical Monthly Vol. 97, No. 5, (May, 1990), pp. 432 as problem 
               E3302. - Ronald S. Tiberio (chuck_tiberio(AT)wellesley.k12.ma.us), 
               Jun 23 2008
%C A011943 Lebl and Lichtblau give the formula a(d) = ((7+4*sqrt(3))^d + (7-4*sqrt(3))^d)/
               2 in Theorem 1.2(iii), p.4. Abstract: Let p(x,y) be a polynomial 
               of degree d with N positive coefficients and no negative coefficients, 
               such that p=1 when x+y=1. It is known that the sharp estimate d <= 
               2N-3 holds. In this paper we study the p that minimize N and we give 
               complete classification of these polynomials up to d=17 by computational 
               methods. We use a linear algebra approach and a mixed linear programming 
               approach. The question is motivated by a question in CR geometry. 
               In particular, a complete classification of polynomials minimizing 
               N is an important first step in the complete classification of CR 
               maps of spheres in different dimensions. - Jonathan Vos Post (jvospost3(AT)gmail.com), 
               Aug 05 2008
%D A011943 E. K. Lloyd "The standard deviation of 1, 2, .., n, Pell's equation and 
               rational triangles", preprint.
%H A011943 Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/
               RecursiveSequences.html">Recursive Sequences</a>
%H A011943 Jiri Lebl and Daniel Lichtblau, <a href="http://arxiv.org/pdf/0808.0284">
               Uniqueness of certain polynomials constant on a hyperplane</a>
%H A011943 <a href="Sindx_Ch.html#Cheby">Index entries for sequences related to 
               Chebyshev polynomials.</a>
%H A011943 <a href="Sindx_Rea.html#recLCC">Index entries for sequences related to 
               linear recurrences with constant coefficients</a>
%F A011943 a(m) = 14a(m-1) - a(m-2).
%F A011943 a(n) ~ (1/2)*(2 + sqrt(3))^(2*n) - Joe Keane (jgk(AT)jgk.org), May 15 
               2002
%F A011943 a(n) = T(n, 7) = (S(n, 14)-S(n-2, 14))/2 = T(2*n, 2) with S(n, x) := 
               U(n, x/2) and T(n, x), resp. U(n, x), are Chebyshev's polynomials 
               of the first, resp. second, kind. See A053120 and A049310. S(-2, 
               x) := -1, S(-1, x) := 0, S(n, 14)=A007655(n+2).
%F A011943 a(n) = ((7+4*sqrt(3))^n + (7-4*sqrt(3))^n)/2.
%F A011943 a(n) = sqrt(48*A007655(n+1)^2 + 1).
%F A011943 G.f.: (1-7*x)/(1-14*x+x^2).
%F A011943 a(n) = Cosh[2n*ArcSinh[Sqrt[3]]] - Herbert Kociemba (kociemba(AT)t-online.de), 
               Apr 24 2008
%F A011943 a(n) = ((7+4*sqrt(3))^n + (7-4*sqrt(3))^n)/2. - Jonathan Vos Post (jvospost3(AT)gmail.com), 
               Aug 05 2008
%o A011943 (PARI) a(n)=if(n<0,0,subst(poltchebi(n),x,7))
%o A011943 (PARI) g(n) = forstep(x=1,n,3,y=(x^2-1)/3;if(issquare(y),print1(x","))) 
               (Hilliard)
%Y A011943 a(n)=A001075(2n)
%Y A011943 Cf. A007654, A011944.
%Y A011943 Sequence in context: A093172 A074110 A155644 this_sequence A083083 A022007 
               A058805
%Y A011943 Adjacent sequences: A011940 A011941 A011942 this_sequence A011944 A011945 
               A011946
%K A011943 nonn,easy
%O A011943 1,2
%A A011943 E. K. Lloyd
%E A011943 Better description from Lekraj Beedassy (blekraj(AT)yahoo.com), Jun 27 
               2002
%E A011943 Chebyshev comments from W. Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de), 
               Nov 08 2002