A258617 a(n) = (4*n+8)*n^2.

0, 12, 64, 180, 384, 700, 1152, 1764, 2560, 3564, 4800, 6292, 8064, 10140, 12544, 15300, 18432, 21964, 25920, 30324, 35200, 40572, 46464, 52900, 59904, 67500, 75712, 84564, 94080, 104284, 115200, 126852, 139264, 152460, 166464, 181300, 196992, 213564

Offset: 0

Author

Garrett Frandson, Jun 05 2015

Keywords

• nonn
• easy

Comments

Let r be a natural number such that r has 17 proper divisors and 5 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 12. Where n is a positive integer, d(r^n)=(4*n+8)*n^2.

The integers that satisfy the proper-divisor-prime-factor requirement are those of A179643.

Examples

			The smallest integer that satisfies the (17, 5) requirement is 180: it has 17 proper divisors (1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90) and 5 prime factors (2, 2, 3, 3, 5), so d(120)=12=a(1).
The square of 180, 32400, we would expect to have a difference of 64 between the number of its proper divisors and prime factors, and with respectively 74 and 10, d(32400)=64=a(2) indeed. Checking this with further integer powers of 180 will continue to generate terms in this sequence.
		

Links

• Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Crossrefs

Cf. A152618, A179643.

Programs

Magma

[(4*n+8)*n^2: n in [0..50]]; // Vincenzo Librandi, Jun 06 2015

Magma

I:=[0, 12, 64, 180]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..40]]; // Vincenzo Librandi, Jun 06 2015

Mathematica

Table[(4 n + 8) n^2, {n, 0, 40}] (* or *) CoefficientList[Series[4 x (3 + 4 x - x^2)/(1 - x)^4, {x, 0, 40}], x] (* Vincenzo Librandi, Jun 06 2015 *)

PARI

vector(50,n,n--;(4*n+8)*n^2) \\ Derek Orr, Jun 21 2015

Formula

a(n) = 4*A152618(n+1).

G.f.: 4*x*(3+4*x-x^2)/(1-x)^4. - Vincenzo Librandi, Jun 06 2015

a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4). - Vincenzo Librandi, Jun 06 2015

For any m, let x=A179643(m), then a(n) = A000005(x^n) - A001222(x^n). - Michel Marcus, Jul 09 2015

Extensions

More terms from Vincenzo Librandi, Jun 06 2015