cp's OEIS Frontend

A122770 Numbers k such that A056109(k) is a square.

%I A122770 #31 Jul 19 2024 08:22:18
%S A122770 0,6,88,1230,17136,238678,3324360,46302366,644908768,8982420390,
%T A122770 125108976696,1742543253358,24270496570320,338044408731126,
%U A122770 4708351225665448,65578872750585150,913395867282526656,12721963269204788038,177194089901584505880
%N A122770 Numbers k such that A056109(k) is a square.
%C A122770 All terms are even. Sequence is infinite. Corresponding squares are s^2 with s = 1, 11, 153, 2131, 29681, 413403, 5757961, 80198051, 1117014753, 15558008491, 216695104121, 3018173449203, 42037733184721, ... (see A122769).
%C A122770 Numbers m such that the distance from (0,0,-1) to (m,m,m) in R^3 is an integer. - _James R. Buddenhagen_, Jun 15 2013
%C A122770 Also n such that the sum of the pentagonal numbers P(n) and P(n+1) is equal to the sum of two consecutive triangular numbers. - _Colin Barker_, Dec 07 2014
%H A122770 Colin Barker, <a href="/A122770/b122770.txt">Table of n, a(n) for n = 0..874</a>
%H A122770 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (15,-15,1).
%F A122770 a(n) = ((b+1)*(7+4*b)^n - (b-1)*(7-4*b)^n - 2)/6, where b = sqrt(3).
%F A122770 a(n) = 14*a(n-1) - a(n-2) + 4, with a(0)=0, a(1)=6.
%F A122770 a(n) = 2*A011916(n) = (A001353(n+1)^2 - A001075(n)^2)/2. - _Richard R. Forberg_, Aug 26 2013
%F A122770 a(n) = 15*a(n-1)-15*a(n-2)+a(n-3). - _Colin Barker_, Dec 07 2014
%F A122770 G.f.: 2*x*(x-3) / ((x-1)*(x^2-14*x+1)). - _Colin Barker_, Dec 07 2014
%t A122770 LinearRecurrence[{15, -15, 1}, {0, 6, 88}, 25] (* _Paolo Xausa_, Jul 19 2024 *)
%o A122770 (PARI) concat(0, Vec(2*x*(x-3) / ((x-1)*(x^2-14*x+1)) + O(x^100))) \\ _Colin Barker_, Dec 07 2014
%Y A122770 Cf. A001353, A001075, A011916, A056109, A122769, A251730.
%K A122770 nonn,easy
%O A122770 0,2
%A A122770 _Zak Seidov_, Oct 21 2006
%E A122770 More terms from _Colin Barker_, Dec 07 2014