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%I #20 Jun 17 2017 02:52:58
%S 0,0,0,0,19,32,48,67,89,114,142,173,207,244,284,327,373,422,474,529,
%T 587,648,712,779,849,922,998,1077,1159,1244,1332,1423,1517,1614,1714,
%U 1817,1923,2032,2144,2259,2377,2498,2622,2749
%N a(n) = largest k such that A000326(n+1) = A000326(n) + (A000326(n) mod k), or 0 if no such k exists.
%C From the definition, a(n) = A000326(n) - A016777(n) if A000326(n) - A016777(n) > A016777(n), 0 otherwise, where A000326 are the pentagonal numbers and A016777 are the gaps between pentagonal numbers: 3n + 1.
%H Rémi Eismann, <a href="/A184750/b184750.txt">Table of n, a(n) for n = 1..1000</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F a(n) = (3n^2-7n-2)/2 for n >= 5 and a(n) = 0 for n <= 4.
%F a(n) = 3*a(n-1)-3*a(n-2)+a(n-3) for n>7. G.f.: x^5*(9*x^2-25*x+19) / (1-x)^3. - _Colin Barker_, Apr 05 2014
%F a(n) = A000326(n) - A016777(n), n>=5, (see a comment above). - _Wolfdieter Lang_, Apr 19 2014
%e For n = 3 we have A000326(3) = 12, A000326(4) = 22; there is no k such that 22 - 12 = 10 = (12 mod k), hence a(3) = 0.
%e For n = 5 we have A000326(5) = 35, A000326(6) = 51; 19 is the largest k such that 51 - 35 = 16 = (35 mod k), hence a(5) = 19; a(5) = (75-35-2)/2 = 19.
%e For n = 25 we have A000326(25) = 925, A000326(26) = 1001; 849 is the largest k such that 1001 - 925 = 76 = (925 mod k), hence a(25) = 849; a(25) = (1875-175-2)/2 = 849.
%p A184750:=n->(3*n^2 - 7*n - 2)*signum(floor(n/5))/2; seq(A184750(n), n=1..50); # _Wesley Ivan Hurt_, Apr 05 2014
%t Table[(3 n^2 - 7 n - 2) Sign[Floor[n/5]]/2, {n, 50}] (* _Wesley Ivan Hurt_, Apr 05 2014 *)
%o (PARI) concat([0,0,0,0], Vec(-x^5*(9*x^2-25*x+19)/(x-1)^3 + O(x^100))) \\ _Colin Barker_, Apr 05 2014
%Y Cf. A000326, A016777, A133151, A184751, A117078, A117563, A001223, A118534.
%K nonn,easy
%O 1,5
%A _Rémi Eismann_, Jan 21 2011
%E Edited - _Wolfdieter Lang_, Apr 19 2014
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