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%I #26 Dec 21 2023 11:57:34
%S 1,61,3781,234361,14526601,900414901,55811197261,3459393815281,
%T 214426605350161,13290990137894701,823826961944121301,
%U 51063980650397625961,3165142973362708688281,196187800367837541047461,12160478479832564836254301,753753477949251182306719201
%N a(0)=1, a(1)=61; for n>1, a(n) = 62*a(n-1) - a(n-2).
%C Centered hexagonal numbers (A003215) with index in A145607. Example: 35 is a member of A145607, therefore A003215(35) = 3781 is a term of this sequence.
%C Also, centered 10-gonal numbers (A062786) with index in A182432. Example: 28 is a member of A182432 and A062786(28) = 3781.
%H Colin Barker, <a href="/A302329/b302329.txt">Table of n, a(n) for n = 0..500</a>
%H Christian Aebi and Grant Cairns, <a href="https://arxiv.org/abs/2312.10866">Less than Equable Triangles on the Eisenstein lattice</a>, arXiv:2312.10866 [math.CO], 2023.
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (62,-1).
%F G.f.: (1 - x)/(1 - 62*x + x^2).
%F a(n) = a(-1-n).
%F a(n) = cosh((2*n + 1)*arccosh(4))/4.
%F a(n) = ((4 + sqrt(15))^(2*n + 1) + 1/(4 + sqrt(15))^(2*n + 1))/8.
%F a(n) = (1/4)*T(2*n+1, 4), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. - _Peter Bala_, Jul 08 2022
%t LinearRecurrence[{62, -1}, {1, 61}, 20]
%o (PARI) x='x+O('x^99); Vec((1-x)/(1-62*x+x^2)) \\ _Altug Alkan_, Apr 06 2018
%Y Cf. A003215, A145607; A062786, A182432.
%Y Fourth row of the array A188646.
%Y First bisection of A041449, A042859.
%Y Similar sequences of the type cosh((2*n+1)*arccosh(k))/k: A000012 (k=1), A001570 (k=2), A077420 (k=3), this sequence (k=4), A302330 (k=5), A302331 (k=6), A302332 (k=7), A253880 (k=8).
%K nonn,easy
%O 0,2
%A _Bruno Berselli_, Apr 05 2018
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