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%I #22 Sep 08 2022 08:46:12
%S 0,13,68,189,400,725,1188,1813,2624,3645,4900,6413,8208,10309,12740,
%T 15525,18688,22253,26244,30685,35600,41013,46948,53429,60480,68125,
%U 76388,85293,94864,105125,116100,127813,140288,153549,167620,182525,198288,214933
%N a(n) = (4*n+9)*n^2.
%C 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.
%C The integers that satisfy the proper-divisor-prime-factor requirement are those of A179644.
%H Harvey P. Dale, <a href="/A258618/b258618.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F G.f.: x*(13+16*x-5*x^2)/(1-x)^4. - _Vincenzo Librandi_, Jun 06 2015
%F a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4). - _Vincenzo Librandi_, Jun 06 2015
%e 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.
%t Table[(4 n + 9) n^2, {n, 0, 40}] (* _Vincenzo Librandi_, Jun 06 2015 *)
%t LinearRecurrence[{4,-6,4,-1},{0,13,68,189},40] (* _Harvey P. Dale_, Sep 12 2020 *)
%o (Magma) [(4*n+9)*n^2: n in [0..40]]; // _Vincenzo Librandi_, Jun 06 2015
%o (PARI) vector(50,n,n--;(4*n+9)*n^2) \\ _Derek Orr_, Jun 21 2015
%Y Cf. A179644.
%K nonn,easy
%O 0,2
%A _Garrett Frandson_, Jun 05 2015
%E More terms from _Vincenzo Librandi_, Jun 06 2015
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