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%I #30 Sep 08 2022 08:46:12
%S 0,10,52,144,304,550,900,1372,1984,2754,3700,4840,6192,7774,9604,
%T 11700,14080,16762,19764,23104,26800,30870,35332,40204,45504,51250,
%U 57460,64152,71344,79054,87300,96100,105472,115434,126004,137200,149040,161542,174724
%N a(n) = (3*n+7)*n^2.
%C 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.
%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*(10+12*x-4*x^2)/(1-x)^4. - _Vincenzo Librandi_, Apr 15 2015
%F a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>3. - _Vincenzo Librandi_, Apr 15 2015
%e 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.
%e The integers which satisfy the proper-divisor-prime-factor requirement are those of A189975.
%p A257042:=n->(3*n+7)*n^2: seq(A257042(n), n=0..50); # _Wesley Ivan Hurt_, Apr 16 2015
%t 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 *)
%o (Magma) [(3*n+7)*n^2: n in [0..65]] // _Vincenzo Librandi_, Apr 15 2015
%o (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
%Y Cf. A189975.
%K nonn,easy
%O 0,2
%A _Garrett Frandson_, Apr 14 2015
%E More terms from _Vincenzo Librandi_, Apr 15 2015
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