A011943 Numbers k such that any group of k consecutive integers has integral standard deviation (viz. A011944(k)).

1, 7, 97, 1351, 18817, 262087, 3650401, 50843527, 708158977, 9863382151, 137379191137, 1913445293767, 26650854921601, 371198523608647, 5170128475599457, 72010600134783751, 1002978273411373057, 13969685227624439047, 194572614913330773601, 2710046923559006391367

Offset: 1

Author

E. K. Lloyd

Keywords

• nonn
• easy

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 all solutions for x in Pell equation x^2 - 12*y^2 = 1. A011944 gives corresponding values for y. - Herbert Kociemba, Jun 05 2022

Also numbers x of the form 3j+1 such that x^2 = 3m^2+1. Also solutions of x in x^2 - 3*y^2 = 1 in A001075 if x = 3j+1, j=1,2,... - Cino Hilliard, Mar 05 2005

In addition to having integral standard deviation, these k 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, 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

In a (square pyramidal) pile of cannonballs, paint the ball at the top and the balls on 2 opposite sides of the base. The sequence, after its 1st term, gives the numbers of painted balls, k, in piles where the total number of balls is twice a square multiple of k. - Peter Munn, Feb 06 2025

References

• P.-F. Teilhet, Reply to Query 2094, L'Intermédiaire des Mathématiciens, 10 (1903), 235-238. - N. J. A. Sloane, Mar 03 2022

Links

• Robert Israel, Table of n, a(n) for n = 1..788
• Jim Delany, Roger Douglass, Mike Breen and Roger B. Eggleton, Problem E 3302: Averaging to Integers, The American Mathematical Monthly, Vol. 97, No. 5 (May, 1990), p. 432.
• 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.
• Ronald S. Tiberio, Solution to Problem E 3302 [Broken link]
• Index entries for sequences related to Chebyshev polynomials.
• Index entries for linear recurrences with constant coefficients, signature (14,-1).

Crossrefs

a(n) = A001075(2n).

Row 2 of array A188644

Cf. A000330, A007654, A007655, A011944, A102344.

Programs

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

Maple

seq(orthopoly[T](n,7), n = 0..50); # Robert Israel, Jun 02 2015
a := n -> (-1)^(n+1)*hypergeom([n-1, -n+1], [1/2], 4):
seq(simplify(a(n)), n=1..20); # Peter Luschny, Jul 26 2020

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 *)

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

Formula

a(n) = 14*a(n-1) - a(n-2).

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

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

a(n) = (-1)^(n+1)*hypergeom([n-1, -n+1], [1/2], 4). - Peter Luschny, Jul 26 2020

E.g.f.: exp(7*x)*cosh(4*sqrt(3)*x). - Stefano Spezia, Dec 12 2022

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