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%I M3154 #74 Mar 16 2023 08:14:54
%S 0,3,48,675,9408,131043,1825200,25421763,354079488,4931691075,
%T 68689595568,956722646883,13325427460800,185599261804323,
%U 2585064237799728,36005300067391875,501489136705686528,6984842613812219523,97286307456665386800,1355023461779503195683
%N Numbers k such that the standard deviation of 1,...,k is an integer.
%C Gives solutions k to the Diophantine equation m^2 = k*(k+1)/3. - Anton Lorenz Vrba (anton(AT)a-l-v.net), Jun 28 2005
%C If x=a(n), y=a(n+1), z=a(n+2) are three consecutive terms, then x^2 - 16*y*x + 14*x*z + 16*y^2 - 16*z*y + z^2 = 144. The formula is symmetric in x and z, so it is also valid for x=a(n+2), y=a(n+1), z=a(n). - _Alexander Samokrutov_, Jul 02 2015
%C From _Bernard Schott_, Apr 09 2021 (Start):
%C Corresponding solutions m (of first comment) are in A011944.
%C Equivalently, numbers k such that k/3 and k+1 are both perfect squares. (End)
%D Guy Alarcon and Yves Duval, TS: Préparation au Concours Général, RMS, Collection Excellence, Paris, 2010, chapitre 13, Questions proposées aux élèves de Terminale S, Exercice 1, p. 220, p. 223.
%D D. A. Benaron, personal communication.
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H T. D. Noe, <a href="/A007654/b007654.txt">Table of n, a(n) for n = 1..100</a>
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H 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 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (15,-15,1).
%F a(n) = 3*A098301(n-1).
%F a(m) = 14*a(m-1) - a(m-2) + 6.
%F G.f.: -3*x^2*(1+x)/(-1+x)/(1-14*x+x^2) = -3 + (1/2)/(-1+x) + (1/2)*(-97*x+7)/(1-14*x+x^2). - _R. J. Mathar_, Nov 20 2007
%F a(n) = (-2 + (7-4*sqrt(3))^n*(7+4*sqrt(3)) + (7-4*sqrt(3))*(7+4*sqrt(3))^n)/4. - _Colin Barker_, Mar 05 2016
%F From _Bernard Schott_, Apr 09 2021: (Start)
%F a(n) = 3 * A001353(n-1)^2.
%F a(n) = A055793(n+1) - 1 = A001075(n-1)^2 - 1. (End)
%F 2*a(n) = A011943(n)-1. - _R. J. Mathar_, Mar 16 2023
%t RecurrenceTable[{a[m] == 14 a[m - 1] - a[m - 2] + 6, a[1] == 0, a[2] == 3}, a, {m, 1, 17}] (* _Michael De Vlieger_, Jul 02 2015 *)
%t CoefficientList[Series[-3 x^2*(1 + x)/(-1 + x)/(1 - 14 x + x^2), {x, 0, 17}], x] (* _Michael De Vlieger_, Feb 02 2016 *)
%o (PARI) concat(0,3*Vec((1+x)/(1-x)/(1-14*x+x^2)+O(x^98))) \\ _Charles R Greathouse IV_, May 14 2013
%o (Magma) I:=[0,3]; [n le 2 select I[n] else 14*Self(n-1)-Self(n-2)+6: n in [1..20]]; // _Vincenzo Librandi_, Mar 05 2016
%Y Cf. A001075, A001353, A007655, A011944, A055793, A098301.
%K easy,nonn
%O 1,2
%A _N. J. A. Sloane_
%E Corrected by Keith Lloyd, Mar 15 1996
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