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A160451 a(n) = (4/3)*u*(u^3+6*u^2+8*u-3) where u=floor((3*n+5)/2).

%I A160451 #23 Feb 16 2025 08:33:10
%S A160451 1008,2080,6440,10208,22360,31416,57408,75208,122816,153680,232408,
%T A160451 281520,402600,476008,652400,757016,1003408,1147008,1479816,1671040,
%U A160451 2108408,2356760,2918560,3234408,3942240,4336816,5214008,5699408,6771016,7360200,8653008,9359800
%N A160451 a(n) = (4/3)*u*(u^3+6*u^2+8*u-3) where u=floor((3*n+5)/2).
%C A160451 It appears that the 4-tuple (3, (u^2-1)/3, (floor((3*n+11)/2)^2-1)/3, a(n)) has the Diophantus' property that the product of any two distinct terms plus one is a square.
%H A160451 Lenny Jones, <a href="http://www.jstor.org/stable/2691315">A polynomial Approach to a Diophantine Problem</a>, Math. Mag. 72 (1999) 52-55.
%H A160451 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/DiophantusProperty.html">Diophantus Property.</a>
%H A160451 <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (1,4,-4,-6,6,4,-4,-1,1).
%F A160451 From _R. J. Mathar_, May 15 2009: (Start)
%F A160451 a(n) = a(n-1)+4*a(n-2)-4*a(n-3)-6*a(n-4)+6*a(n-5)+4*a(n-6)-4*a(n-7)-a(n-8)+a(n-9).
%F A160451 G.f.: -8*x*(126+134*x+41*x^2-65*x^3+95*x^4+52*x^5-61*x^6-13*x^7+15*x^8)/((1+x)^4*(x-1)^5). (End)
%e A160451 For n=1 we get the 4-tuple (3,5,16,1008), and 3*5+1=16=4^2, 3*16+1=49=7^2, 3*1008+1=3025=55^2, 5*16+1=81=9^2, 5*1008+1=5041=71^2, 16*1008+1=16129=127^2.
%t A160451 Table[u=Floor[(3n+5)/2];4/3 u(u^3+6u^2+8u-3),{n,30}] (* or *) LinearRecurrence[{1,4,-4,-6,6,4,-4,-1,1},{1008,2080,6440,10208,22360,31416,57408,75208,122816},30] (* _Harvey P. Dale_, Nov 19 2013 *)
%Y A160451 Cf. A086302, A160372.
%K A160451 nonn,easy
%O A160451 1,1
%A A160451 _John W. Layman_, May 14 2009