A258402 a(n) = (n^2 + 4*n + 6) * n^2.

0, 11, 72, 243, 608, 1275, 2376, 4067, 6528, 9963, 14600, 20691, 28512, 38363, 50568, 65475, 83456, 104907, 130248, 159923, 194400, 234171, 279752, 331683, 390528, 456875, 531336, 614547, 707168, 809883, 923400, 1048451, 1185792, 1336203, 1500488, 1679475

Offset: 0

Author

Garrett Frandson, Jun 05 2015

Keywords

• nonn
• easy

Comments

Consider a natural number r such that r has 15 proper divisors and 4 prime factors. (Note that these prime factors do not have to be distinct). The difference between these two values, say d(r), is in this case 11. Where n is a positive integer, d(r^n) = (n^2 + 4*n + 6) * n^2.

The integers which satisfy the proper-divisor-prime-factor requirement are those of A033993.

Examples

			The smallest integer which satisfies this is 210: It has 15 proper divisors (1, 2, 3, 5, 6, 7, 10, 14, 15, 21, 30, 35, 42, 70, 105) and 4 prime factors (2, 3, 5, 7), so d(210) = 11. The square of 210, 44100, we would expect to have a difference of 72 between the number of its proper divisors and prime factors, and with respectively 80 and 8, d(44100) = 72 indeed. Checking this with further integer powers of 210 will continue to generate terms in this sequence.
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Crossrefs

Cf. A033993.

Programs

Magma

[(n^2+4*n+6)*n^2: n in [0..40]]; // Vincenzo Librandi, Jun 06 2015

Magma

I:=[0,11, 72,243,608]; [n le 5 select I[n] else 5*Self(n-1)-10*Self(n-2)+10*Self(n-3)-5*Self(n-4)+Self(n-5): n in [1..40]]; // Vincenzo Librandi, Jun 06 2015

Mathematica

Table[(n^2 + 4n + 6) * n^2, {n, 0, 39}] (* Alonso del Arte, Jun 06 2015 *)
CoefficientList[Series[x (11 + 17 x - 7 x^2 + 3 x^3)/(1 - x)^5, {x, 0, 40}], x] (* Vincenzo Librandi, Jun 06 2015 *)
LinearRecurrence[{5,-10,10,-5,1},{0,11,72,243,608},40] (* Harvey P. Dale, May 05 2018 *)

PARI

a(n)=(n^2+4*n+6)*n^2 \\ Charles R Greathouse IV, Jun 15 2015

Formula

From Vincenzo Librandi, Jun 06 2015: (Start)

G.f.: x*(11 + 17*x - 7*x^2 + 3*x^3)/(1 - x)^5.

a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). (End)

E.g.f.: exp(x)*x*(11 + 25*x + 10*x^2 + x^3). - Stefano Spezia, Oct 28 2023

Extensions

More terms from Alonso del Arte, Jun 06 2015