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A156702 Numbers k such that k^2 - 1 == 0 (mod 24^2).

%I A156702 #40 Oct 13 2024 19:02:52
%S A156702 1,127,161,287,289,415,449,575,577,703,737,863,865,991,1025,1151,1153,
%T A156702 1279,1313,1439,1441,1567,1601,1727,1729,1855,1889,2015,2017,2143,
%U A156702 2177,2303,2305,2431,2465,2591,2593,2719,2753,2879,2881,3007,3041,3167,3169
%N A156702 Numbers k such that k^2 - 1 == 0 (mod 24^2).
%C A156702 Numbers k that are == +-1 (mod 9) and == +-1 (mod 32). - _Charles R Greathouse IV_, Dec 23 2011
%H A156702 Vincenzo Librandi, <a href="/A156702/b156702.txt">Table of n, a(n) for n = 1..1000</a>
%H A156702 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,1,-1).
%F A156702 G.f.: (-x^4 + 2*x^3 + 126*x^2 + 34*x + 127)/(x^5 - x^4 - x + 1). - _Alexander R. Povolotsky_, Feb 15 2009
%F A156702 a(n) = -36 + 27*(-1)^n + (4-4*i)*(-i)^n + (4+4*i)*i^n + 72*n. - _Harvey P. Dale_, Apr 25 2012
%F A156702 Sum_{n>=1} (-1)^(n+1)/a(n) = (cot(Pi/288) - tan(17*Pi/288))*Pi/288. - _Amiram Eldar_, Feb 26 2023
%F A156702 E.g.f.: 1 + 8*cos(x) + 9*(8*x - 1)*cosh(x) - 8*sin(x) + 9*(8*x - 7)*sinh(x). - _Stefano Spezia_, Oct 13 2024
%t A156702 LinearRecurrence[{1,0,0,1,-1},{1,127,161,287,289},50] (* _Vincenzo Librandi_, Feb 08 2012 *)
%t A156702 With[{c=24^2},Select[Range[3200],Divisible[#^2-1,c]&]] (* _Harvey P. Dale_, Apr 25 2012 *)
%o A156702 (PARI) a(n)=n\4*288+[-1,1,127,161][n%4+1]
%K A156702 nonn,easy
%O A156702 1,2
%A A156702 _Vincenzo Librandi_, Feb 13 2009
%E A156702 Corrected and edited by _Vinay Vaishampayan_, Jun 23 2010