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%I #39 Dec 30 2023 23:46:37
%S 1,165,31977,6203341,1203416145,233456528757,45289363162681,
%T 8785902997031325,1704419892060914337,330648673156820350021,
%U 64144138172531086989705,12443632156797874055652717,2414000494280615035709637361,468303652258282519053613995285
%N Indices of pentagonal numbers that are also hexagonal.
%C The reason we obtain the same Diophantine equation with various parameters is the following: the number that is written 361 in base 4*A046179(n)-2 is the square of 6*A046178(n)-1. That is, 361 in base 110770 is 3*110770^2 + 6*110770 + 1 = 36810643321, i.e., the square of 191861 if we consider the third terms of A046179 and A046178, which are 27693 and 31977, respectively. - _Richard Choulet_, Oct 03 2007
%C As n increases, this sequence is approximately geometric with common ratio r = lim_{n->infinity} a(n)/a(n-1)) = (2 + sqrt(3))^4 = 97 + 56*sqrt(3). - _Ant King_, Dec 14 2011
%H Colin Barker, <a href="/A046178/b046178.txt">Table of n, a(n) for n = 1..438</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/HexagonalPentagonalNumber.html">Hexagonal Pentagonal Number.</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (195,-195,1)
%F From _Warut Roonguthai_ Jan 08 2001: (Start)
%F a(n) = 194*a(n-1) - a(n-2) - 32.
%F G.f.: x*(1-30*x-3*x^2)/((1-x)*(1-194*x+x^2)). (End)
%F a(n+1) = 97*a(n) - 16 + 28*sqrt(12*a(n)^2 - 4*a(n) + 1). - _Richard Choulet_, Oct 09 2007
%F From _Ant King_, Dec 14 2011: (Start)
%F a(n) = 195*a(n-1) - 195*a(n-2) + a(n-3).
%F a(n) = (1/12)*((sqrt(3)-1)*(2+sqrt(3))^(4n-2) - (sqrt(3)+1)* (2-sqrt(3))^(4n-2) + 2).
%F a(n) = ceiling((1/12)*(sqrt(3)-1)*(2+sqrt(3))^(4n-2)).
%F (End)
%t LinearRecurrence[{195, -195, 1}, {1, 165, 31977}, 11] (* _Ant King_, Dec 14 2011 *)
%o (PARI) Vec(x*(3*x^2+30*x-1)/((x-1)*(x^2-194*x+1)) + O(x^20)) \\ _Colin Barker_, Jun 21 2015
%Y Cf. A046179, A046180.
%K nonn,easy
%O 1,2
%A _Eric W. Weisstein_
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