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Numbers k where the root mean square (RMS) of k and 7 is an integer, i.e., sqrt((k^2 + 7^2)/2) is an integer.
4

%I #30 Sep 17 2020 06:17:44

%S 1,7,17,23,49,103,137,287,601,799,1673,3503,4657,9751,20417,27143,

%T 56833,118999,158201,331247,693577,922063,1930649,4042463,5374177,

%U 11252647,23561201,31322999,65585233,137324743,182563817,382258751,800387257,1064059903

%N Numbers k where the root mean square (RMS) of k and 7 is an integer, i.e., sqrt((k^2 + 7^2)/2) is an integer.

%H Vincenzo Librandi, <a href="/A076293/b076293.txt">Table of n, a(n) for n = 0..2000</a>

%H Jeremiah Bartz, Bruce Dearden, and Joel Iiams, <a href="https://arxiv.org/abs/1810.07895">Classes of Gap Balancing Numbers</a>, arXiv:1810.07895 [math.NT], 2018.

%H Jeremiah Bartz, Bruce Dearden, and Joel Iiams, <a href="https://ajc.maths.uq.edu.au/pdf/77/ajc_v77_p318.pdf">Counting families of generalized balancing numbers</a>, The Australasian Journal of Combinatorics (2020) Vol. 77, Part 3, 318-325.

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,6,0,0,-1).

%F a(n) = 6a(n-3) - a(n-6) = sqrt(2*A076294(n)^2 - 49) = A076295(n) + A076296(n).

%F a(3n+1) = 7*A002315(n).

%F G.f.: (x+1)*(x^2+3*x+1)^2 / (x^6-6*x^3+1). - _Colin Barker_, Sep 14 2014

%e 17 is in the sequence since sqrt((17^2 + 7^2)/2) = 13 is an integer.

%t Column[LinearRecurrence[{0, 0, 6, 0, 0, -1}, {1, 7, 17, 23, 49, 103}, 35] ] (* _Vincenzo Librandi_, Jul 30 2017 *)

%o (PARI) Vec((x+1)*(x^2+3*x+1)^2/(x^6-6*x^3+1) + O(x^100)) \\ _Colin Barker_, Sep 14 2014

%Y Cf. A002315, A076294, A076295, A076296.

%K nonn,easy

%O 0,2

%A _Henry Bottomley_, Oct 05 2002

%E More terms from _Colin Barker_, Sep 14 2014