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%I #31 Sep 08 2022 08:46:06
%S 1,12,39,82,141,216,307,414,537,676,831,1002,1189,1392,1611,1846,2097,
%T 2364,2647,2946,3261,3592,3939,4302,4681,5076,5487,5914,6357,6816,
%U 7291,7782,8289,8812,9351,9906,10477,11064,11667,12286,12921,13572,14239,14922
%N a(n) = 8n^2 + 3n + 1.
%C Positions a(n) of hexagonal numbers such that h(a(n)) = h(a(n)-1) + h(4*n+1), where h=A000384.
%C First bisection of A057029. The sequence contains infinitely many squares: 1, 676, 779689, 899760016, ... [_Bruno Berselli_, Jan 24 2014]
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F G.f.: -(6*x^2+9*x+1) / (x-1)^3. - _Colin Barker_, Jan 21 2014
%F a(n) = 3*a(n-1)-3*a(n-2)+a(n-3). - _Colin Barker_, Jan 21 2014
%e For n=5, A000384(a(5)) = 93096 = A000384(a(5)-1)+A000384(4*5+1) = 92235 + 861.
%t Table[8 n^2 + 3 n + 1, {n, 0, 50}] (* _Bruno Berselli_, Jan 24 2014 *)
%t LinearRecurrence[{3,-3,1},{1,12,39},50] (* _Harvey P. Dale_, May 26 2019 *)
%o (PARI) Vec(-(6*x^2+9*x+1)/(x-1)^3 + O(x^100)) \\ _Colin Barker_, Jan 21 2014
%o (Magma) [8*n^2+3*n+1: n in [0..50]]; // _Bruno Berselli_, Jan 24 2014
%Y Cf. A000384, A057029, A064225, A152948, A236257.
%K nonn,easy
%O 0,2
%A _Vladimir Shevelev_, Jan 21 2014
%E More terms from _Colin Barker_, Jan 21 2014
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