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A011943 Numbers n such that any group of n consecutive integers has integral standard deviation {viz. A011944(n)}. 22
1, 7, 97, 1351, 18817, 262087, 3650401, 50843527, 708158977, 9863382151, 137379191137, 1913445293767, 26650854921601, 371198523608647, 5170128475599457, 72010600134783751, 1002978273411373057, 13969685227624439047, 194572614913330773601, 2710046923559006391367 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

If k is in the sequence, then it has successor 7*k + 4*sqrt{3*(k^2 - 1)}. - Lekraj Beedassy, Jun 28 2002

Chebyshev's polynomials T(n,x) evaluated at x=7.

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.

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, Mar 05 2005

Equals sqrt(12*A011944(n)^2 + 1).

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

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. - Jonathan Vos Post, Aug 05 2008

LINKS

Robert Israel, Table of n, a(n) for n = 1..788

R. K. Guy, Letter to N. J. A. Sloane concerning A001075, A011943, A094347 [Scanned and annotated letter, included with permission]

Tanya Khovanova, Recursive Sequences

Jiri Lebl and Daniel Lichtblau, Uniqueness of certain polynomials constant on a hyperplane, arXiv:0808.0284 [math.CV], 2008-2010.

E. Keith Lloyd, The Standard Deviation of 1, 2,..., n: Pell's Equation and Rational Triangles, Math. Gaz. vol 81 (1997), 231-243.

Index entries for sequences related to Chebyshev polynomials.

Index entries for linear recurrences with constant coefficients, signature (14,-1).

FORMULA

a(m) = 14a(m-1) - a(m-2).

a(n) ~ (1/2)*(2 + sqrt(3))^(2*n). - Joe Keane (jgk(AT)jgk.org), May 15 2002

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).

a(n) = ((7+4*sqrt(3))^n + (7-4*sqrt(3))^n)/2.

a(n) = sqrt(48*A007655(n+1)^2 + 1).

G.f.: (1-7*x)/(1-14*x+x^2).

a(n) = cosh(2n*arcsinh(sqrt(3))). - Herbert Kociemba, Apr 24 2008

MAPLE

seq(orthopoly[T](n, 7), n = 0..50); # Robert Israel, Jun 02 2015

MATHEMATICA

LinearRecurrence[{14, -1}, {1, 7}, 30] (* Harvey P. Dale, Dec 16 2013 *)

a[n_]:=1/2((7-4 Sqrt[3])^n+(7+4 Sqrt[3])^n); Table[a[n] // Simplify, {n, 0, 20}] (* Gerry Martens, May 30 2015 *)

PROG

(PARI) a(n)=if(n<0, 0, subst(poltchebi(n), x, 7))

(PARI) g(n) = forstep(x=1, n, 3, y=(x^2-1)/3; if(issquare(y), print1(x", "))) \\ Cino Hilliard, Mar 05 2005

(MAGMA) I:=[1, 7]; [n le 2 select I[n] else 14*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Apr 19 2015

CROSSREFS

a(n) = A001075(2n).

Row 2 of array A188644

Cf. A007654, A011944, A102344.

Sequence in context: A155644 A243867 A232290 * A218669 A188441 A178808

Adjacent sequences:  A011940 A011941 A011942 * A011944 A011945 A011946

KEYWORD

nonn,easy

AUTHOR

E. K. Lloyd

EXTENSIONS

Better description from Lekraj Beedassy, Jun 27 2002

Chebyshev comments from Wolfdieter Lang, Nov 08 2002

More terms from Vincenzo Librandi, Apr 19 2015

STATUS

approved

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Last modified March 29 15:31 EDT 2017. Contains 284273 sequences.