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a(n) = 4*n^3 - 6*n^2 + 3*n - 1.
3

%I #36 Sep 08 2022 08:46:15

%S 0,13,62,171,364,665,1098,1687,2456,3429,4630,6083,7812,9841,12194,

%T 14895,17968,21437,25326,29659,34460,39753,45562,51911,58824,66325,

%U 74438,83187,92596,102689,113490,125023,137312,150381,164254,178955,194508,210937,228266,246519,265720

%N a(n) = 4*n^3 - 6*n^2 + 3*n - 1.

%C Nonnegative numbers n such that 2*n+1 is a cube.

%C Or, (y^k-1)/2 for k odd. - _N. J. A. Sloane_, Mar 05 2022

%D H. Brocard, #2158, L'Intermédiaire des Mathématiciens, 10 (1903), 282-283

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4, -6, 4, -1).

%F G.f.: (13*x + 10*x^2 + x^3)/(-1 + x)^4. - _Michael De Vlieger_, Apr 16 2016

%F a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>4. - _Wesley Ivan Hurt_, Apr 17 2016

%e a(1) = 0 because 4*1^3 - 6*1^2 + 3*1 - 1 = 0.

%e a(2) = 13 because 4*2^3 - 6*2^2 + 3*2 - 1 = 13.

%p A268201:=n->4*n^3 - 6*n^2 + 3*n - 1: seq(A268201(n), n=1..80); # _Wesley Ivan Hurt_, Apr 17 2016

%t Table[((2 n - 1)^3 - 1)/2, {n, 41}] (* or *)

%t CoefficientList[Series[(13*x + 10*x^2 + x^3)/(-1 + x)^4, {x, 0, 40}],

%t x] (* _Michael De Vlieger_, Apr 16 2016 *)

%o (Magma) [((2*n-1)^3-1)/2: n in [0..41]];

%o (PARI) lista(nn) = for(n=1, nn, print1(4*n^3-6*n^2+3*n-1, ", ")); \\ _Altug Alkan_, Apr 17 2016

%Y Cf. nonnegative numbers n such that 2*n + k is a cube: A271828 (k=-3), A050492 (k=-1), this sequence (k=1).

%K nonn,easy

%O 1,2

%A _Juri-Stepan Gerasimov_, Apr 16 2016