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a(n) = 13*(2^n - 1) - 3*n^2 - 9*n.
2

%I #60 Sep 08 2022 08:46:12

%S 1,9,37,111,283,657,1441,3051,6319,12909,26149,52695,105859,212265,

%T 425161,851043,1702903,3406725,6814477,13630095,27261451,54524289,

%U 109050097,218101851,436205503,872412957,1744828021,3489658311,6979319059,13958640729

%N a(n) = 13*(2^n - 1) - 3*n^2 - 9*n.

%C These numbers belong to a family of sequences obtained as follows:

%C . A000225: 1*(2^n-1);

%C . A050488: 3*(2^n-1) - 2*n;

%C . a(n): 13*(2^n-1) - 3*n^2 - 9*n;

%C . A257449: 75*(2^n-1) - 4*n^3 - 18*n^2 - 52*n;

%C . A257450: 541*(2^n-1) - 5*n^4 - 30*n^3 - 130*n^2 - 375*n,

%C where the sequence 1, 3, 13, 75, 541, ... is A000670 (after the first term), and A208744 gives the triangle of coefficients:

%C 2;

%C 3, 9;

%C 4, 18, 52;

%C 5, 30, 130, 375;

%C 6, 45, 260, 1125, 3246;

%C 7, 63, 455, 2625, 11361, 32781, etc.

%C Also, the antidiagonal sums in the array are given by the formula (6*n^2 + 6*k*n + (k-1)*k)*(k+n)!/((k+3)!*(n-1)!) for k = 0, 1, 2, 3, 4, ... (see Example field).

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (5, -9, 7, -2).

%F G.f.: x*(1+4*x+x^2)/((1-x)^3*(1-2*x)).

%F a(n) = 5*a(n-1) - 9*a(n-2) + 7*a(n-3) - 2*a(n-4) for n>4. - _Ray Chandler_, Jul 25 2015

%e By the second comment, the array begins:

%e k=0: 1, 8, 27, 64, 125, 216, ... A000578

%e k=1: 1, 9, 36, 100, 225, 441, ... A000537

%e k=2: 1, 10, 46, 146, 371, 812, ... A024166

%e k=3: 1, 11, 57, 203, 574, 1386, ... A101094

%e k=4: 1, 12, 69, 272, 846, 2232, ... A101097

%e k=5: 1, 13, 82, 354, 1200, 3432, ... A101102

%e k=6: 1, 14, 96, 450, 1650, 5082, ... A254469

%e ...

%e See also A254469 (Example field).

%t Table[13 (2^n - 1) - 3 n^2 - 9n, {n, 30}]

%t CoefficientList[Series[x (1 + 4 x + x^2)/((1 - x)^3*(1 - 2 x)), {x, 0, 30}], x] (* _Michael De Vlieger_, Nov 14 2016 *)

%o (Magma) [13*(2^n-1)-3*n^2-9*n: n in [1..30]]; // _Vincenzo Librandi_, Apr 24 2015

%Y Cf. A000225, A000670, A050488, A208744, A257449, A257450.

%K nonn,easy

%O 1,2

%A _Luciano Ancora_, Apr 23 2015

%E Edited by _Bruno Berselli_, Apr 28 2015