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%I #26 Jul 08 2023 18:42:03
%S 7,7887,9101399,10503006367,12120460245927,13987000620793199,
%T 16140986595935105527,18626684544708490984767,
%U 21495177823607002661315399,24805416581757936362666985487,28625429240170834955515039936407,33033720537740561780727993419627999
%N Numbers n such that 3n^2-n = 6k^2-2k for some integer k>0.
%C Also indices of pentagonal numbers which are twice some other pentagonal number.
%C Note that A000326(n) = 2 A000326(k) <=> n(3n-1)=2k(3k-1), which is easily solved by standard Pell-type techniques (cf. link to D. Alpern's quadratic solver). Here we consider only positive solutions.
%C Inspired by a recent comment on A000326 by R. J. Mathar.
%H Colin Barker, <a href="/A137693/b137693.txt">Table of n, a(n) for n = 1..250</a>
%H Dario Alpern, <a href="https://www.alpertron.com.ar/QUAD.HTM">Quadratic two integer variable equation solver</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1155,-1155,1).
%F a(n) = f^{2n-2}(5,7)[2], where f(x,y) = (577x + 408y - 164, 816x + 577y - 232)
%F a(n) = (7,7,9,7,7,9,...) mod 10
%F G.f. x*(-7+198*x+x^2) / ( (x-1)*(x^2-1154*x+1) ). - _R. J. Mathar_, Apr 17 2011
%F a(0)=0, a(1)=7, a(2)=7887, a(3)=9101399, a(n)=1155*a(n-1)-1155*a(n-2)+ a(n-3). - _Harvey P. Dale_, Jun 21 2011
%t CoefficientList[Series[x (-7+198x+x^2)/((x-1)(x^2-1154x+1)),{x,0,20}],x] (* or *) Join[{0},LinearRecurrence[{1155,-1155,1},{7,7887,9101399}, 20]] (* _Harvey P. Dale_, Jun 21 2011 *)
%o (PARI) vector(20,i, (v=if(i>1,[577,408; 816,577]*v-[164;232], [5;7]))[2,1])
%Y Cf. A000326, A136112-A136118, A135768-A135769, A135771, A137694.
%K easy,nonn
%O 1,1
%A _M. F. Hasler_, Feb 08 2008
%E More terms from _Harvey P. Dale_, Jun 21 2011
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