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A184220 a(n) = largest k such that A000290(n+1) = A000290(n) + (A000290(n) mod k), or 0 if no such k exists.

%I A184220 #13 Mar 15 2016 11:41:16
%S A184220 0,0,0,0,14,23,34,47,62,79,98,119,142,167,194,223,254,287,322,359,398,
%T A184220 439,482,527,574,623,674,727,782,839,898,959,1022,1087,1154,1223,1294,
%U A184220 1367,1442,1519,1598,1679,1762
%N A184220 a(n) = largest k such that A000290(n+1) = A000290(n) + (A000290(n) mod k), or 0 if no such k exists.
%C A184220 From the definition, a(n) = A000290(n) - A005408(n) if A000217(n) - A005408(n) > A005408(n), 0 otherwise, where A000290 are the squares and A005408 are the gaps between squares: 2n + 1.
%H A184220 Rémi Eismann, <a href="/A184220/b184220.txt">Table of n, a(n) for n = 1..10000</a>
%H A184220 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3, -3, 1).
%F A184220 a(n) = (n-1)^2-2 = A008865(n-1) for n >= 5 and a(n) = 0 for n <= 4.
%e A184220 For n = 3 we have A000290(3) = 9, A000290(4) = 16; there is no k such that 16 - 9 = 7 = (9 mod k), hence a(3) = 0.
%e A184220 For n = 5 we have A000290(5) = 25, A000290(6) = 36; 14 is the largest k such that 36 - 25 = 11 = (25 mod k), hence a(5) = 14; a(5) = A000290(5) - A005408(5) = 25 - 11 = 14.
%e A184220 For n = 25 we have A000217(25) = 625, A000217(26) = 676; 574 is the largest k such that 676 - 625 = 51 = (625 mod k), hence a(25) = 574; a(25) = A000290(25) - A005408(25) = 574.
%Y A184220 Cf. essentially the same as A008865, A000290, A005408, A133150, A184221, A118534, A117078, A117563, A001223.
%K A184220 nonn,easy
%O A184220 1,5
%A A184220 _Rémi Eismann_, Jan 10 2011