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
Keywords
Examples
The first 6 pentagonal pyramidal numbers are 0, 1, 6, 18, 40, 75; of these, 0, 6, 18, 75 are divisible by 3.
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)
Programs
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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
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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
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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 *) -
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
Formula
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)
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