OFFSET
0,2
COMMENTS
Consider a natural number r such that r has 19 proper divisors and 6 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 13. Where n is a positive integer, d(r^n)=(4*n+9)*n^2.
The integers that satisfy the proper-divisor-prime-factor requirement are those of A179644.
LINKS
Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
FORMULA
G.f.: x*(13+16*x-5*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
EXAMPLE
The smallest integer that satisfies this is 240: It has 19 proper divisors (1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 30, 40, 48, 60, 80, 120) and 6 prime factors (2, 2, 2, 2, 3, 5), so d(240)=13. The square of 240, 57600, we would expect to have a difference of 68 between the number of its proper divisors and prime factors, and with respectively 80 and 12, d(57600)=68 indeed. Checking this with further integer powers of 240 will continue to generate terms in this sequence.
MATHEMATICA
Table[(4 n + 9) n^2, {n, 0, 40}] (* Vincenzo Librandi, Jun 06 2015 *)
LinearRecurrence[{4, -6, 4, -1}, {0, 13, 68, 189}, 40] (* Harvey P. Dale, Sep 12 2020 *)
PROG
(Magma) [(4*n+9)*n^2: n in [0..40]]; // Vincenzo Librandi, Jun 06 2015
(PARI) vector(50, n, n--; (4*n+9)*n^2) \\ Derek Orr, Jun 21 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Garrett Frandson, Jun 05 2015
EXTENSIONS
More terms from Vincenzo Librandi, Jun 06 2015
STATUS
approved