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a(n) = 8^(n-1)*(2^n - 1).
4

%I #19 Aug 01 2024 02:20:15

%S 1,24,448,7680,126976,2064384,33292288,534773760,8573157376,

%T 137304735744,2197949513728,35175782154240,562881233944576,

%U 9006649498927104,144110790029344768,2305807824841605120

%N a(n) = 8^(n-1)*(2^n - 1).

%H Harry J. Smith, <a href="/A060195/b060195.txt">Table of n, a(n) for n = 1..100</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (24,-128).

%F a(1)=1, a(2)=24, a(n) = 24*a(n-1) - 128*a(n-2). - _Harvey P. Dale_, Oct 20 2014

%F From _G. C. Greubel_, Aug 01 2024: (Start)

%F a(n) = A000225(n)*A001018(n-1).

%F G.f.: x/((1 - 8*x)*(1 - 16*x)).

%F E.g.f.: (1/4)*exp(12*x)*sinh(4*x). (End)

%t Table[8^(n-1) (2^n-1),{n,20}] (* or *) LinearRecurrence[{24,-128},{1,24},20] (* _Harvey P. Dale_, Oct 20 2014 *)

%o (PARI) { for (n=1, 100, write("b060195.txt", n, " ", 8^(n - 1)*(2^n - 1)); ) } \\ _Harry J. Smith_, Jul 02 2009

%o (Magma) [8^(n-1)*(2^n-1): n in [1..40]]; // _G. C. Greubel_, Aug 01 2024

%o (SageMath) [8^(n-1)*(2^n-1) for n in range(1,41)] # _G. C. Greubel_, Aug 01 2024

%Y Cf. A000225, A001018.

%K nonn,easy

%O 1,2

%A Manish Kumar Gupta (M.Gupta(AT)math.canterbury.ac.nz), Mar 21 2001

%E More terms from _Jason Earls_ and Larry Reeves (larryr(AT)acm.org), Mar 21 2001