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 A160451 (4/3)u(u^3+6*u^2+8u-3) where u=Floor[{3n+5)/2]. 0
 1008, 2080, 6440, 10208, 22360, 31416, 57408, 75208, 122816, 153680, 232408, 281520, 402600, 476008, 652400, 757016, 1003408, 1147008, 1479816, 1671040, 2108408, 2356760, 2918560, 3234408, 3942240, 4336816, 5214008, 5699408 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS 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. LINKS Lenny Jones, A polynomial Approach to a Diophantine Problem, Math. Mag. 72 (1999) 52-55. Eric Weisstein's World of Mathematics, Diophantus Property. Index entries for linear recurrences with constant coefficients, signature (1, 4, -4, -6, 6, 4, -4, -1, 1). FORMULA 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] EXAMPLE 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. MATHEMATICA 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 *) CROSSREFS Sequence in context: A067918 A163557 A241932 * A254973 A092924 A331770 Adjacent sequences: A160448 A160449 A160450 * A160452 A160453 A160454 KEYWORD nonn AUTHOR John W. Layman, May 14 2009 STATUS approved

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Last modified December 3 09:50 EST 2022. Contains 358517 sequences. (Running on oeis4.)