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a(n) = n*12^n - 1.
5

%I #26 Dec 23 2024 17:47:56

%S 11,287,5183,82943,1244159,17915903,250822655,3439853567,46438023167,

%T 619173642239,8173092077567,106993205379071,1390911669927935,

%U 17974858503684095,231105323618795519,2958148142320582655,37716388814587428863,479219999055934390271,6070119988041835610111,76675199848949502443519

%N a(n) = n*12^n - 1.

%H Harry J. Smith, <a href="/A064758/b064758.txt">Table of n, a(n) for n = 1..150</a>

%H Paul Leyland, <a href="http://www.leyland.vispa.com/numth/factorization/cullen_woodall/cw.htm">Factors of Cullen and Woodall numbers</a>.

%H Paul Leyland, <a href="http://www.leyland.vispa.com/numth/factorization/cullen_woodall/gcw.htm">Generalized Cullen and Woodall numbers</a>.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (25,-168,144).

%F G.f.: x*(11 + 12*x - 144*x^2)/((1 - 12*x)^2*(1 - x)). - _Vincenzo Librandi_, Jun 21 2018

%F From _Elmo R. Oliveira_, Sep 07 2024: (Start)

%F E.g.f.: 1 + exp(x)*(12*x*exp(11*x) - 1).

%F a(n) = 25*a(n-1) - 168*a(n-2) + 144*a(n-3) for n > 3.

%F a(n) = A064750(n) - 2. (End)

%t CoefficientList[Series[(11 + 12 x - 144 x^2) / ((1 - 12 x)^2 (1 - x)), {x, 0, 33}], x] (* _Vincenzo Librandi_, Jun 21 2018 *)

%o (PARI) a(n) = { n*12^n - 1 } \\ _Harry J. Smith_, Sep 24 2009

%o (Magma) [n*12^n - 1: n in [1..30]]; // _Vincenzo Librandi_, Jun 21 2018

%Y Cf. for a(n) = n*k^n - 1: -A000012(k=0), A001477(k=1), A003261 (k=2), A060352 (k=3), A060416 (k=4), A064751 (k=5), A064752 (k=6), A064753 (k=7), A064754 (k=8), A064755 (k=9), A064756 (k=10), A064757 (k=11), this sequence (k=12).

%Y Cf. A064750.

%K nonn,easy

%O 1,1

%A _N. J. A. Sloane_, Oct 19 2001