A299412 Pentagonal pyramidal numbers divisible by 3.

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

Links

Justin Gaetano, Table of n, a(n) for n = 0..10000

Index entries for linear recurrences with constant coefficients, signature (1,3,-3,-3,3,1,-1)

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

Cf. A001318, A002411, A007494, A117748, A269416.

Sequence in context: A057051 A318069 A332939 * A218080 A277609 A188119

Adjacent sequences: A299409 A299410 A299411 * A299413 A299414 A299415

Keyword

nonn

Author

Justin Gaetano, Feb 20 2018

Status

approved