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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 (list; graph; refs; listen; history; text; internal format)
 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 5 first pentagonal pyramidal numbers are 0, 1, 6, 18, 40, 75; among them 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 *) 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: A121156 A057051 A318069 * A218080 A277609 A188119 Adjacent sequences:  A299409 A299410 A299411 * A299413 A299414 A299415 KEYWORD nonn AUTHOR Justin Gaetano, Feb 20 2018 STATUS approved

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Last modified December 12 03:27 EST 2018. Contains 318052 sequences. (Running on oeis4.)