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Numbers of the form 2*t^2-4 when t > 1 is a term in A001541.
2

%I #27 Mar 01 2024 14:53:01

%S 14,574,19598,665854,22619534,768398398,26102926094,886731088894,

%T 30122754096398,1023286908188734,34761632124320654,

%U 1180872205318713598,40114893348711941774,1362725501650887306814,46292552162781456489998,1572584048032918633353214,53421565080956452077519374

%N Numbers of the form 2*t^2-4 when t > 1 is a term in A001541.

%C Equivalently: integers k such that k$ / (k/2+1)! and k$ / (k/2+2)! are both squares when A000178 (k) = k$ = 1!*2!*...*k! is the superfactorial of k (see A348692 for further information).

%C The 3 subsequences of A349081 are A035008, A139098 and this one.

%H Rick Mabry and Laura McCormick, <a href="https://www.austms.org.au/wp-content/uploads/Gazette/2009/Nov09/TechPaperMabry.pdf">Square products of punctured sequences of factorials</a>, Gaz. Aust. Math. Soc., 2009, pages 346-352.

%H <a href="/index/O#Olympiads">Index to sequences related to Olympiads</a>.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (35, -35, 1).

%F a(n) = 2*(cosh(2*n*arcsinh(1)))^2 - 4.

%F a(n) = 16*A001110(n) - 2. - _Hugo Pfoertner_, Dec 04 2021

%e A001541(1) = 3, then for t = 3, 2*t^2-4 = 14; also for k = 14, 14$ / 8! = 1309248519599593818685440000000^2 and 14$ / 9! = 436416173199864606228480000000^2. Hence, 14 is a term.

%p with(orthopoly):

%p sequence = (2*T(n,3)^2-4, n=1..20);

%t (2*#^2 - 4) & /@ LinearRecurrence[{6, -1}, {3, 17}, 17] (* _Amiram Eldar_, Dec 04 2021 *)

%t LinearRecurrence[{35, -35, 1},{14, 574, 19598},17] (* _Ray Chandler_, Mar 01 2024 *)

%o (PARI) a(n) = my(t=subst(polchebyshev(n), 'x, 3)); 2*t^2-4; \\ _Michel Marcus_, Dec 04 2021

%Y Cf. A000178, A035008, A139098, A348692, A349079, A349080, A349081.

%Y Subsequence of A060626.

%K nonn

%O 1,1

%A _Bernard Schott_, Dec 04 2021