Search: id:A117950
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%I A117950
%S A117950 3,4,7,12,19,28,39,52,67,84,103,124,147,172,199,228,259,292,327,364,403,
%T A117950 444,487,532,579,628,679,732,787,844,903,964,1027,1092,1159,1228,1299,
%U A117950 1372,1447,1524,1603,1684,1767,1852,1939,2028,2119,2212,2307,2404,2503
%N A117950 a(n) = n^2 + 3.
%C A117950 Sequence allows us to find the solutions of the equation: X^3 - (X + 
               3)^2 + X + 6 = Y^2. To prove that X = n^2 + 3: Y^2 = X^3 - (X + 3)^2 
               + X + 6 = X^3 - X^2 - 5X - 3 = (X - 3)(X^2 + 2X + 1) = (X - 3)*(X 
               + 1)^2 it means: (X - 3) must be a perfect square, so X = n^2 + 3 
               and Y = n(n^2 + 4). - Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Nov 
               12 2007
%C A117950 An equivalent technique of integer factorization would work for example 
               for the equation X^3-3*X^2-9*X-5=(X-5)(X+1)^2=Y^2, looking for perfect 
               squares of the form X-5=n^2. - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), 
               Nov 20 2007
%C A117950 Take a square array of (n+1)-by-(n+1) dots (which correspond to the vertices 
               of a grid of n-by-n squares). Connect the dots with vertical and 
               horizontal line-segments of any length so that each dot is connected 
               to each of its orthogonal neighbors, and so that no line-segment 
               crosses any previously drawn line-segment. Then the minimum number 
               of line-segments is a(n), for n >= 1. [From Leroy Quet (q1qq2qqq3qqqq(AT)yahoo.com), 
               Apr 12 2009]
%H A117950 Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/
               Near-SquarePrime.html">Near-Square Prime</a>
%F A117950 G.f.: -(3-5*x+4*x^2)/(-1+x)^3 = -2/(-1+x)^3-3/(-1+x)^2-4/(-1+x) . - R. 
               J. Mathar (mathar(AT)strw.leidenuniv.nl), Nov 20 2007
%F A117950 a(n) = ((n-3)^2 + 3*(n+1)^2)/4. [From Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), 
               Feb 13 2009]
%F A117950 a(n) = A132111(n-1,2) for n>1. - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), 
               Aug 10 2007
%F A117950 a(n)=2*n+a(n-1)-3 (with a(1)=3) [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), 
               Nov 07 2009]
%e A117950 For n=2, a(2)=2*2+3-3=4; n=3, a(3)=2*3+4-3=7; n=4, a(4)=2*4+7-3=12 [From 
               Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Nov 07 2009]
%t A117950 a[n_]:=n^2+3; [From Vladimir Orlovsky (4vladimir(AT)gmail.com), Dec 15 
               2008]
%o A117950 (Other) sage: [lucas_number1(3,n,-3) for n in xrange(0, 51)] # [From 
               Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), May 16 2009]
%Y A117950 a(n) = A000290(n)+3. [From Omar E. Pol (info(AT)polprimos.com), Dec 20 
               2008]
%Y A117950 Cf. A028560, A005563.
%Y A117950 For primes in this sequence see A049422.
%Y A117950 Sequence in context: A130324 A020677 A158237 this_sequence A025047 A050342 
               A108700
%Y A117950 Adjacent sequences: A117947 A117948 A117949 this_sequence A117951 A117952 
               A117953
%K A117950 nonn,easy
%O A117950 0,1
%A A117950 Eric Weisstein (eric(AT)weisstein.com), Apr 04, 2006
%E A117950 Edited by N. J. A. Sloane Apr 15 2009 at the suggestion of Leroy Quet

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