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A299412
Pentagonal pyramidal numbers divisible by 3.
2
0, 6, 18, 75, 126, 288, 405, 726, 936, 1470, 1800, 2601, 3078, 4200, 4851, 6348, 7200, 9126, 10206, 12615, 13950, 16896, 18513, 22050, 23976, 28158, 30420, 35301, 37926, 43560, 46575, 53016, 56448, 63750, 67626, 75843, 80190, 89376, 94221, 104430, 109800, 121086, 127008, 139425, 145926, 159528, 166635
OFFSET
0,2
FORMULA
a(n) = A007494(n)*A117748(n).
a(n) = (3*n/2)^2*(3*n/2+1)/2 if n even.
a(n) = ((3*n+1)/2)^2*((3*n+1)/2+1)/2 if n odd.
From Omar E. Pol, Feb 21 2018: (Start)
a(n) = 3*A001318(n)*A007494(n).
a(n) = A001318(n)*abs(A269416(n-1)), n >= 1. (End)
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
EXAMPLE
The first 6 pentagonal pyramidal numbers are 0, 1, 6, 18, 40, 75; of these, 0, 6, 18, 75 are divisible by 3.
MAPLE
f:= proc(n) if n::even then (3*n/2)^2*(3*n/2+1)/2 else
((3*n+1)/2)^2*((3*n+1)/2+1)/2 fi end proc:
map(f, [$0..100]); # Robert Israel, Feb 28 2018
MATHEMATICA
Array[((3 #1 + #2)/2)^2*((3 #1 + #2)/2 + 1)/2 & @@ {#, Boole@ OddQ@ #} &, 47, 0] (* Michael De Vlieger, Feb 21 2018 *)
LinearRecurrence[{1, 3, -3, -3, 3, 1, -1}, {0, 6, 18, 75, 126, 288, 405}, 50] (* Harvey P. Dale, Jul 16 2021 *)
PROG
(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
(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
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Justin Gaetano, Feb 20 2018
STATUS
approved