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A011943
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Numbers k such that any group of k consecutive integers has integral standard deviation (viz. A011944(k)).
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25
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1, 7, 97, 1351, 18817, 262087, 3650401, 50843527, 708158977, 9863382151, 137379191137, 1913445293767, 26650854921601, 371198523608647, 5170128475599457, 72010600134783751, 1002978273411373057, 13969685227624439047, 194572614913330773601, 2710046923559006391367
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OFFSET
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1,2
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COMMENTS
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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
a(n) = sqrt(12*A011944(n)^2 + 1).
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
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REFERENCES
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P.-F. Teilhet, Reply to Query 2094, L'Intermédiaire des Mathématiciens, 10 (1903), 235-238. - N. J. A. Sloane, Mar 03 2022
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LINKS
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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
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (14,-1).
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FORMULA
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a(n) = 14*a(n-1) - a(n-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
a(n) = (-1)^(n+1)*hypergeom([n-1, -n+1], [1/2], 4). - Peter Luschny, Jul 26 2020
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MAPLE
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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
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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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
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KEYWORD
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nonn,easy
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AUTHOR
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E. K. Lloyd
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EXTENSIONS
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Better description from Lekraj Beedassy, Jun 27 2002
Chebyshev comments from Wolfdieter Lang, Nov 08 2002
More terms from Vincenzo Librandi, Apr 19 2015
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STATUS
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approved
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