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A295867 Numbers of the form A000217(n)*A007494(n) that are divisible by 3. 1

%I #53 Mar 01 2018 11:59:40

%S 0,9,30,60,120,189,432,630,825,1122,1404,2205,2760,3264,3978,4617,

%T 6300,7392,8349,9660,10800,13689,15498,17052,19140,20925,25344,28050,

%U 30345,33390,35964,42237,46020,49200,53382,56889,65340,70380,74589,80088,84672,95625,102102,107484,114480,120285,134064,142158

%N Numbers of the form A000217(n)*A007494(n) that are divisible by 3.

%H Colin Barker, <a href="/A295867/b295867.txt">Table of n, a(n) for n = 0..1000</a>

%H L. Euler and J. Hewlett, <a href="https://archive.org/stream/elementsalgebra00lagrgoog#page/n173/mode/2up">Elements of Algebra, Chapter V, Of Figurate or Polygonal Numbers</a>, Springer-Verlag. 1984. See pages 139-145.

%H <a href="/index/Rec#order_16">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,0,3,-3,0,0,0,-3,3,0,0,0,1,-1).

%F Let m = floor((6*n+4)/5), a(n) = (3*m/2)*binomial(m+1,2) if m is even, otherwise ((3*m+1)/2)*binomial(m+1,2).

%F Conjectures from _Colin Barker_, Feb 15 2018: (Start)

%F G.f.: 3*x*(3 + 7*x + 10*x^2 + 20*x^3 + 23*x^4 + 72*x^5 + 45*x^6 + 35*x^7 + 39*x^8 + 25*x^9 + 33*x^10 + 8*x^11 + 3*x^12 + x^13) / ((1 - x)^4*(1 + x + x^2 + x^3 + x^4)^3).

%F a(n) = a(n-1) + 3*a(n-5) - 3*a(n-6) - 3*a(n-10) + 3*a(n-11) + a(n-15) - a(n-16) for n>15.

%F (End)

%F Colin Barker's conjecture is true. This is a cubic quasipolynomial of order 5: a(n) = 162/125*n^3 + 27/25*n^2 if n is 0 mod 5, 162/125*n^3 + 261/125*n^2 + 111/125*n + 12/125 if n is 4 mod 5, 162/125*n^3 + 297/125*n^2 + 144/125*n + 21/125 if n is 3 mod 5, 162/125*n^3 + 423/125*n^2 + 339/125*n + 84/125 if n is 2 mod 5, and 162/125*n^3 + 459/125*n^2 + 396/125*n + 108/125 if n is 1 mod 5. Generally a(n) = 162/125*n^3 + O(n^2). - _Charles R Greathouse IV_, Feb 20 2018

%e For n = 1, let m = floor((6*1+4)/5) = 2, a(1) = (3*2/2)*binomial(2+1,2) = 3*3 = 9.

%e For n = 2, let m = floor((6*2+4)/5) = 3, a(2) = ((3*3+1)/2)*binomial(3+1,2) = 5*6 = 30.

%t Select[Array[Floor[(3 # + 1)/2] (# + 1) #/2 &, 58, 0], Divisible[#, 3] &] (* _Michael De Vlieger_, Feb 17 2018 *)

%t LinearRecurrence[{1, 0, 0, 0, 3, -3, 0, 0, 0, -3, 3, 0, 0, 0, 1, -1}, {0, 9, 30, 60, 120, 189, 432, 630, 825, 1122, 1404, 2205, 2760, 3264, 3978, 4617, 6300}, 48] (* _Robert G. Wilson v_, Feb 19 2018 *)

%o (PARI) a(n) = my(k=n%5); 162/125*n^3 + if(k==0, 27/25*n^2, k==1, 459/125*n^2 + 396/125*n + 108/125, k==2, 423/125*n^2 + 339/125*n + 84/125, k==3, 297/125*n^2 + 144/125*n + 21/125, 261/125*n^2 + 111/125*n + 12/125) \\ _Charles R Greathouse IV_, Feb 20 2018

%o (PARI) concat(0, Vec(3*x*(3 + 7*x + 10*x^2 + 20*x^3 + 23*x^4 + 72*x^5 + 45*x^6 + 35*x^7 + 39*x^8 + 25*x^9 + 33*x^10 + 8*x^11 + 3*x^12 + x^13) / ((1 - x)^4*(1 + x + x^2 + x^3 + x^4)^3) + O(x^60))) \\ _Colin Barker_, Mar 01 2018

%Y Cf. A000217, A007494, A047248.

%K nonn,easy

%O 0,2

%A _Justin Gaetano_, Feb 13 2018

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