cp's OEIS Frontend

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

%I A011943 #103 Feb 06 2025 08:21:05
%S A011943 1,7,97,1351,18817,262087,3650401,50843527,708158977,9863382151,
%T A011943 137379191137,1913445293767,26650854921601,371198523608647,
%U A011943 5170128475599457,72010600134783751,1002978273411373057,13969685227624439047,194572614913330773601,2710046923559006391367
%N A011943 Numbers k such that any group of k consecutive integers has integral standard deviation (viz. A011944(k)).
%C A011943 If k is in the sequence, then it has successor 7*k + 4*sqrt(3*(k^2 - 1)). - _Lekraj Beedassy_, 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 all solutions for x in Pell equation x^2 - 12*y^2 = 1. A011944 gives corresponding values for y. - _Herbert Kociemba_, Jun 05 2022
%C A011943 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
%C A011943 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
%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. - _Jonathan Vos Post_, Aug 05 2008
%C A011943 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
%D A011943 P.-F. Teilhet, Reply to Query 2094, L'Intermédiaire des Mathématiciens, 10 (1903), 235-238. - _N. J. A. Sloane_, Mar 03 2022
%H A011943 Robert Israel, <a href="/A011943/b011943.txt">Table of n, a(n) for n = 1..788</a>
%H A011943 Jim Delany, Roger Douglass, Mike Breen and Roger B. Eggleton, <a href="http://www.jstor.org/stable/2324410">Problem E 3302: Averaging to Integers</a>, The American Mathematical Monthly, Vol. 97, No. 5 (May, 1990), p. 432.
%H A011943 R. K. Guy, <a href="/A001075/a001075.pdf">Letter to N. J. A. Sloane concerning A001075, A011943, A094347</a> [Scanned and annotated letter, included with permission]
%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/abs/0808.0284">Uniqueness of certain polynomials constant on a hyperplane</a>, arXiv:0808.0284 [math.CV], 2008-2010.
%H A011943 E. Keith Lloyd, <a href="http://www.jstor.org/stable/3619201">The Standard Deviation of 1, 2,..., n: Pell's Equation and Rational Triangles</a>, Math. Gaz. vol 81 (1997), 231-243.
%H A011943 Ronald S. Tiberio, <a href="http://alum.wpi.edu/~tiberio/vita/AMM3302.pdf">Solution to Problem E 3302</a> [Broken link]
%H A011943 <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials</a>.
%H A011943 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (14,-1).
%F A011943 a(n) = 14*a(n-1) - a(n-2).
%F A011943 a(n) = sqrt(12*A011944(n)^2 + 1).
%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_, Apr 24 2008
%F A011943 a(n) = (-1)^(n+1)*hypergeom([n-1, -n+1], [1/2], 4). - _Peter Luschny_, Jul 26 2020
%F A011943 E.g.f.: exp(7*x)*cosh(4*sqrt(3)*x). - _Stefano Spezia_, Dec 12 2022
%p A011943 seq(orthopoly[T](n,7), n = 0..50); # _Robert Israel_, Jun 02 2015
%p A011943 a := n -> (-1)^(n+1)*hypergeom([n-1, -n+1], [1/2], 4):
%p A011943 seq(simplify(a(n)), n=1..20); # _Peter Luschny_, Jul 26 2020
%t A011943 LinearRecurrence[{14,-1},{1,7},30] (* _Harvey P. Dale_, Dec 16 2013 *)
%t A011943 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 *)
%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","))) \\ _Cino Hilliard_, Mar 05 2005
%o A011943 (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
%Y A011943 a(n) = A001075(2n).
%Y A011943 Row 2 of array A188644
%Y A011943 Cf. A000330, A007654, A007655, A011944, A102344.
%K A011943 nonn,easy
%O A011943 1,2
%A A011943 E. K. Lloyd
%E A011943 Better description from _Lekraj Beedassy_, Jun 27 2002
%E A011943 Chebyshev comments from _Wolfdieter Lang_, Nov 08 2002
%E A011943 More terms from _Vincenzo Librandi_, Apr 19 2015