A295867 Numbers of the form A000217(n)*A007494(n) that are divisible by 3.
0, 9, 30, 60, 120, 189, 432, 630, 825, 1122, 1404, 2205, 2760, 3264, 3978, 4617, 6300, 7392, 8349, 9660, 10800, 13689, 15498, 17052, 19140, 20925, 25344, 28050, 30345, 33390, 35964, 42237, 46020, 49200, 53382, 56889, 65340, 70380, 74589, 80088, 84672, 95625, 102102, 107484, 114480, 120285, 134064, 142158
Offset: 0
Examples
For n = 1, let m = floor((6*1+4)/5) = 2, a(1) = (3*2/2)*binomial(2+1,2) = 3*3 = 9. For n = 2, let m = floor((6*2+4)/5) = 3, a(2) = ((3*3+1)/2)*binomial(3+1,2) = 5*6 = 30.
Links
- Colin Barker, Table of n, a(n) for n = 0..1000
- L. Euler and J. Hewlett, Elements of Algebra, Chapter V, Of Figurate or Polygonal Numbers, Springer-Verlag. 1984. See pages 139-145.
- Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,3,-3,0,0,0,-3,3,0,0,0,1,-1).
Programs
-
Mathematica
Select[Array[Floor[(3 # + 1)/2] (# + 1) #/2 &, 58, 0], Divisible[#, 3] &] (* Michael De Vlieger, Feb 17 2018 *) 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 *)
-
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
-
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
Formula
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).
Conjectures from Colin Barker, Feb 15 2018: (Start)
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).
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.
(End)
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