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%I #14 Jan 03 2024 07:38:01
%S 1008,2080,6440,10208,22360,31416,57408,75208,122816,153680,232408,
%T 281520,402600,476008,652400,757016,1003408,1147008,1479816,1671040,
%U 2108408,2356760,2918560,3234408,3942240,4336816,5214008,5699408
%N (4/3)u(u^3+6*u^2+8u-3) where u=Floor[(3n+5)/2].
%C It appears that the 4-tuple (3, ((u^2-1)/3, (Floor[(3n+11)/2]^2-1)/3, a(n)}, where a(n)=(4/3)u(u^3+6*u^2+8u-3) with u=Floor[{3n+5)/2] has Diophantus' property that the product of any two distinct terms plus one is a square.
%H 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 Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/DiophantusProperty.html">Diophantus Property.</a>
%H <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 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). 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). [_R. J. Mathar_, May 15 2009]
%e 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 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 A086302, A160372
%K nonn
%O 1,1
%A _John W. Layman_, May 14 2009
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