OFFSET
0,2
COMMENTS
Consider a natural number r such that r has 15 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 10. Where n is a positive integer, d(r^n)=(3*n+7)*n^2.
LINKS
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).
FORMULA
From Vincenzo Librandi, Apr 15 2015: (Start)
G.f.: x*(10+12*x-4*x^2)/(1-x)^4.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>3. (End)
From Amiram Eldar, Jul 30 2024: (Start)
Sum_{n>=1} 1/a(n) = sqrt(3)*Pi/98 + Pi^2/42 + 9*log(3)/98 - 351/1372.
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/84 - sqrt(3)*Pi/49 - 6*log(2)/49 + 225/1372. (End)
EXAMPLE
The smallest integer that satisfies this is 120: it has 15 proper divisors (1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60) and 5 prime factors (2, 2, 2, 3, 5), so d(120)=10. The square of 120, 14400, we would expect to have a difference of 52 between the number of its proper divisors and prime factors, and with respectively 62 and 10, d(120)=52 indeed. Checking this with further integer powers of 120 will continue to generate terms in this sequence.
The integers which satisfy the proper-divisor-prime-factor requirement are those of A189975.
MAPLE
MATHEMATICA
Table[(3 n + 7) n^2, {n, 40}] (* or *) CoefficientList[Series[(10 + 12 x - 4 x^2) / (1 - x)^4, {x, 0, 40}], x] (* Vincenzo Librandi, Apr 15 2015 *)
PROG
(Magma) [(3*n+7)*n^2: n in [0..65]]; // Vincenzo Librandi, Apr 15 2015
(PARI) lista(nn) = {v = 1; while(!((numdiv(v)-1 == 15) && (bigomega(v) == 5)), v++); for (n=0, nn, vn = v^n; nb = numdiv(vn)-1-bigomega(vn); print1(nb, ", "); ); } \\ Michel Marcus, Apr 16 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Garrett Frandson, Apr 14 2015
EXTENSIONS
More terms from Vincenzo Librandi, Apr 15 2015
STATUS
approved