|
%I #78 Sep 08 2022 08:46:20
%S 0,6,18,75,126,288,405,726,936,1470,1800,2601,3078,4200,4851,6348,
%T 7200,9126,10206,12615,13950,16896,18513,22050,23976,28158,30420,
%U 35301,37926,43560,46575,53016,56448,63750,67626,75843,80190,89376,94221,104430,109800,121086,127008,139425,145926,159528,166635
%N Pentagonal pyramidal numbers divisible by 3.
%H Justin Gaetano, <a href="/A299412/b299412.txt">Table of n, a(n) for n = 0..10000</a>
%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (1,3,-3,-3,3,1,-1)
%F a(n) = A007494(n)*A117748(n).
%F a(n) = (3*n/2)^2*(3*n/2+1)/2 if n even.
%F a(n) = ((3*n+1)/2)^2*((3*n+1)/2+1)/2 if n odd.
%F From _Omar E. Pol_, Feb 21 2018: (Start)
%F a(n) = 3*A001318(n)*A007494(n).
%F a(n) = A001318(n)*abs(A269416(n-1)), n >= 1. (End)
%F G.f.: 3*x*(3*x^4 + 5*x^3 + 13*x^2 + 4*x + 2)/((x-1)^4*(x+1)^3). - _Robert Israel_, Feb 28 2018
%e The first 6 pentagonal pyramidal numbers are 0, 1, 6, 18, 40, 75; of these, 0, 6, 18, 75 are divisible by 3.
%p f:= proc(n) if n::even then (3*n/2)^2*(3*n/2+1)/2 else
%p ((3*n+1)/2)^2*((3*n+1)/2+1)/2 fi end proc:
%p map(f, [$0..100]); # _Robert Israel_, Feb 28 2018
%t Array[((3 #1 + #2)/2)^2*((3 #1 + #2)/2 + 1)/2 & @@ {#, Boole@ OddQ@ #} &, 47, 0] (* _Michael De Vlieger_, Feb 21 2018 *)
%t LinearRecurrence[{1,3,-3,-3,3,1,-1},{0,6,18,75,126,288,405},50] (* _Harvey P. Dale_, Jul 16 2021 *)
%o (PARI) lista(nn) = {for (n=0, nn, if (!(n^2*(n+1)/2 % 3), print1(n^2*(n+1)/2, ", ")););} \\ _Michel Marcus_, Feb 21 2018
%o (PARI) x='x+O('x^99); concat(0, Vec(3*x*(3*x^4+5*x^3+13*x^2+4*x+2)/((x-1)^4*(x+1)^3))) \\ _Altug Alkan_, Mar 14 2018
%o (Magma) [IsEven(n) select (3*n/2)^2*(3*n/2+1)/2 else ((3*n+1)/2)^2*((3*n+1)/2+1)/2: n in [0..50] ]; // _Vincenzo Librandi_, Mar 14 2018
%Y Cf. A001318, A002411, A007494, A117748, A269416.
%K nonn
%O 0,2
%A _Justin Gaetano_, Feb 20 2018
|
