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A242604 a(n) = (n - 1)*(n^3 + 1) = n^4 - n^3 + n - 1, for n >= 1.

%I A242604 #20 Jul 23 2025 11:35:13
%S A242604 0,9,56,195,504,1085,2064,3591,5840,9009,13320,19019,26376,35685,
%T A242604 47264,61455,78624,99161,123480,152019,185240,223629,267696,317975,
%U A242604 375024,439425,511784,592731,682920,783029,893760,1015839,1150016,1297065
%N A242604 a(n) = (n - 1)*(n^3 + 1) = n^4 - n^3 + n - 1, for n >= 1.
%C A242604 1/a(n), for n >= 2, is in base n represented by 0.repeat(0,0,0,1,1,1). This is instance p = 3 of the general formula for 0.repeat(0^(q),1^(q)) (meaning here q zeros followed by q 1's) in base b >= 2 which is 1/a(q,b) with a(q,b) = (b - 1)*(b^p + 1), for p >= 1.
%H A242604 Harvey P. Dale, <a href="/A242604/b242604.txt">Table of n, a(n) for n = 1..1000</a>
%H A242604 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F A242604 a(n) = (n - 1)*(n^3 + 1) = n^4 - n^3 + n - 1, n >= 1.
%F A242604 O.g.f.: x^2*(9 + 11*x + 5*x^2 - x^3)/(1 - x)^5.
%F A242604 a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - _Wesley Ivan Hurt_, Jul 23 2025
%p A242604 A242604:=n->(n-1)*(n^3+1); seq(A242604(n), n=1..50); # _Wesley Ivan Hurt_, Jun 20 2014
%t A242604 Table[n^4-n^3+n-1,{n,40}] (* _Harvey P. Dale_, Jun 20 2019 *)
%o A242604 (Magma) [(n - 1)*(n^3 + 1) : n in [1..50]]; // _Wesley Ivan Hurt_, Jun 20 2014
%Y A242604 Cf. A005563 (case q=1), A062158 (case q=2).
%K A242604 nonn,easy
%O A242604 1,2
%A A242604 _Wolfdieter Lang_, Jun 20 2014