Search: id:A001652
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%I A001652 M3074 N1247
%S A001652 0,3,20,119,696,4059,23660,137903,803760,4684659,27304196,159140519,
%T A001652 927538920,5406093003,31509019100,183648021599,1070379110496,
%U A001652 6238626641379,36361380737780,211929657785303,1235216565974040
%N A001652 a(n)=6a(n-1)-a(n-2)+2 with a(0) = 0, a(1) = 3.
%C A001652 Consider all Pythagorean triples (X,X+1,Z) ordered by increasing Z; sequence 
               gives X values.
%C A001652 Numbers n such that triangular number t(n) (see A000217) = n(n+1)/2 is 
               a product of two consecutive integers (cf. A097571).
%C A001652 Members of Diophantine pairs. Solution to a(a+1)=2b(b+1) in natural numbers 
               including 0; a=a(n), b=b(n)= A053141(n) - The solution of a special 
               case of a binomial problem of H. Finner and K. Strassburger (strass(AT)godot.dfi.uni-duesseldorf.de).
%C A001652 The three sequences x (A001652), y (A046090) and z (A001653) may be obtained 
               by setting u and v equal to the Pell numbers (A000129) in the formulae 
               x = 2uv, y = u^2 - v^2, z = u^2 + v^2. [Joseph Wiener and Donald 
               Skow]. - Antonio Alberto Olivares (olivares14031(AT)yahoo.com), Dec 
               22 2003
%C A001652 Define a(1)=0 a(2)=3 such that 2*(a(1)^2)+2*a(1)+1=j(1)^2=1^2 and 2*(a(2)^2)+2*a(2)+1=j(2)^2=5^2=25. 
               Then a(n)=a(n-2)+4*sqrt(2*(a(n-1)^2)+2*a(n-1)+1). Another definition: 
               a(n) such that 2*(a(n)^2)+2*a(n)+1 = j(n)^2. - Pierre CAMI (pierrecami(AT)tele2.fr), 
               Mar 30 2005
%C A001652 The complete Pythagorean triple {X(n),Y(n)=X(n)+1,Z(n)} with X<Y<Z,may 
               be recursively generated through the mapping W(n) -> M*W(n), where 
               W(n)=transpose of vector [X(n) Y(n) Z(n)] and M a 3 X 3 matrix given 
               by [2 1 2 / 1 2 2 / 2 2 3]. - Lekraj Beedassy (blekraj(AT)yahoo.com), 
               Aug 14 2006
%C A001652 Let b(n) = A053141 then a(n)*b(n+1) = b(n)*a(n+1) + b(n) - Kenneth J 
               Ramsey (Ramsey2879(AT)msn.com), Sep 22 2007
%C A001652 In general, if b(n)= A053141(n), then a(n)*b(n+k) = a(n+k)*b(n)+b(k); 
               e.g. 3*84 = 119*2+14; 3*2870 = 4059*2+492; 20*2870=5741*14+84 - Charlie 
               Marion (charliemath(AT)optonline.net), Nov 19 2007
%C A001652 lim_{n -> infinity} a(n)/a(n-1) = 3+2*sqrt(2). [From Klaus Brockhaus 
               (klaus-brockhaus(AT)t-online.de), Feb 17 2009]
%C A001652 The remainder of the division of a(n) by 5 is: 0, 1, 3 or 4. [From Mohamed 
               Bouhamida (bhmd95(AT)yahoo.fr), Aug 26 2009]
%C A001652 Number of units of a(n) belongs to a periodic sequence: 0, 3, 0, 9, 6, 
               9. The remainder of the division of a(n) by 5 belongs to a periodic 
               sequence: 0, 3, 0, 4, 1, 4. [From Mohamed Bouhamida (bhmd95(AT)yahoo.fr), 
               Sep 01 2009]
%C A001652 If (p,q) is a solution of the Diophantine equation: X^2 + (X+1)^2 = Y^2 
               then (p+q) or (p+q+1) are perfect squares. If (p,q) is a solution 
               of the Diophantine equation: X^2 + (X+1)^2 = Y^2 then (p+q) or (p+q)/
               8 are perfect squares. If (p,q) and (r,s) are two cosecutive solutions 
               of the Diophantine equation: X^2 + (X+1)^2 = Y^2 with p<r then s-r=p+q+1. 
               [From Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Aug 29 2009]
%C A001652 If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation: 
               X^2 + (X + 1)^2 = y^2 with p<r then r=3p+2q+1 and s=4p+3q+2. [From 
               Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Sep 02 2009]
%D A001652 N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, 
               Academic Press, 1995 (includes this sequence).
%D A001652 N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 
               (includes this sequence).
%D A001652 I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), 
               181-193.
%D A001652 A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, 
               pp. 122-125, 1964.
%D A001652 Martin V. Bonsangue, Gerald E. Gannon and Laura J. Pheifer, Misinterpretations 
               can sometimes be a good thing, Math. Teacher, vol. 95, No. 6 (2002) 
               pp. 446-449.
%D A001652 T. W. Forget and T. A. Larkin, Pythagorean triads of the form X, X+1, 
               Z described by recurrence sequences, Fib. Quart., 6 (No. 3, 1968), 
               94-104.
%D A001652 L. J. Gerstein, Pythagorean triples and inner products, Math. Mag., 78 
               (2005), 205-213.
%D A001652 A. Martin, Table of prime rational right-angled triangles, Math. Mag., 
               2 (1910), 297-324.
%D A001652 S. P. Mohanty, Which triangular numbers are products of three consecutive 
               integers, Acta Math. Hungar., 58 (1991), 31-36.
%D A001652 Zhang Zaiming, Problem #502, Pell's Equation - Once Again, Two-Year College 
               Math. Jnl., 25 (1994), 241-243.
%H A001652 T. D. Noe, <a href="b001652.txt">Table of n, a(n) for n=0..200</a>
%H A001652 <a href="Sindx_Tu.html#2wis">Index entries for two-way infinite sequences</
               a>
%H A001652 <a href="Sindx_Rea.html#recLCC">Index entries for sequences related to 
               linear recurrences with constant coefficients</a>
%H A001652 Ron Knott, <a href="http://www.mcs.surrey.ac.uk/Personal/R.Knott/Pythag/
               pythag.html">Pythagorean Triples and Online Calculators</a>
%H A001652 S. Plouffe, <a href="http://www.lacim.uqam.ca/%7Eplouffe/articles/MasterThesis.pdf">
               Approximations de S\'{e}ries G\'{e}n\'{e}ratrices et Quelques Conjectures</
               a>, Dissertation, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 
               1992.
%H A001652 S. Plouffe, <a href="http://www.lacim.uqam.ca/%7Eplouffe/articles/FonctionsGeneratrices.pdf">
               1031 Generating Functions and Conjectures</a>, Universit\'{e} du 
               Qu\'{e}bec \`{a} Montr\'{e}al, 1992.
%F A001652 a(n) = [ (1+ sqrt(2))^(2n+1) + (1-sqrt(2))^(2n+1) - 2 ]/4.
%F A001652 G.f.: x(3-x)/((1-6x+x^2)(1-x)). - Michael Somos, Apr 07 2003
%F A001652 a_{n} = 7(a_{n-1} - a_{n-2}) + a_{n-3}. a_{n} = -1/2 + ((1-2^{1/2})/4)*(3 
               - 2^{3/2})^n + ((1+2^{1/2})/4)*(3 + 2^{3/2})^n . - Antonio Olivares 
               (olivares14031(AT)yahoo.com), Oct 13, 2003
%F A001652 Let b(n) = A046090(n) and c(n) = A001653(n). Then for k>j, c(i)*(c(k) 
               - c(j)) = a(k+i)+...+a(i+j+1) + a(k-i-1)+...+a(j-i) + k - j. For 
               n<0, a(n) = -b(-n-1). Also a(n)*a(n+2k+1) + b(n)*b(n+2k+1) + c(n)*c(n+2k+1) 
               = (a(n+k+1) - a(n+k))^2; a(n)*a(n+2k) + b(n)*b(n+2k) + c(n)*c(n+2k) 
               = 2*c(n+k)^2. - Charlie Marion (charliem(AT)bestweb.net), Jul 01 
               2003
%F A001652 a(n)*a(n+1) + A046090(n)*A046090(n+1) = A001542(n+1)^2 = A084703(n+1). 
               - Charlie Marion (charliem(AT)bestweb.net), Jul 01 2003
%F A001652 For n and j >= 1, sum_{k=0..j}A001653(k)*a(n) - sum_{k=0...j-1}A001653(k)*a(n-1) 
               + A053141(j) = A001109(j+1)*a(n) - A001109(j)*a(n-1) + A053141(j)= 
               a(n+j); e.g. (1+5+29)*119-(1+5)*20+14=4059 - Charlie Marion (charliem(AT)bestweb.net), 
               Jul 07 2003
%F A001652 Sum_{k=0...n}((2k+1)*a(n-k))=A001109(n+1)-A000217(n+1); e.g. 1*119+3*20+5*3+7*0=194=204-10 
               - Charlie Marion (charliem(AT)bestweb.net), Jul 18 2003
%F A001652 a(n)=A055997(n)-1+(2*A055997(n)*A001108(n))^.5; e.g. 119=50-1+(2*50*49)^.5 
               - Charlie Marion (charliem(AT)bestweb.net), Jul 21 2003
%F A001652 a(n)={A002315(n) - 1}/2. - Lekraj Beedassy (blekraj(AT)yahoo.com), Nov 
               25 2003
%F A001652 Sum_{k=0...n}a(k)=A089950(n); e.g., 0+3+20+119=142 - Charlie Marion (charliemath(AT)verizon.net), 
               Jan 21 2004
%F A001652 a(2n+k)+a(k)+1=A001541(n)*A002315(n+k); e.g. 23660+696+1=3*8119; for 
               k>0, a(2n+k)-a(k-1)=A001541(n+k)*A002315(n);e.g. 803760-119=19601*41 
               - Charlie Marion (charliemath(AT)verizon.net), Mar 17 2003
%F A001652 a(n)=(A001653(n+1) - 3*A001653(n) - 2)/4. - Lekraj Beedassy (blekraj(AT)yahoo.com), 
               Jul 13 2004
%F A001652 a(n)={2*A084159(n) - 1 + (-1)^(n+1)}/2. - Lekraj Beedassy (blekraj(AT)yahoo.com), 
               Jul 21 2004
%F A001652 a(n)=5*(a(n-1)+a(n-2))-a(n-3)+4; a(n)=7*(a(n-1)-a(n-2))+a(n-3). - Mohamed 
               Bouhamida (bhmd95(AT)yahoo.fr), Sep 02 2006
%F A001652 a(n+1)=3*a(n)+(8*a(n)^2+8*a(n)+4)^0.5+1, a(1)=0. - Richard Choulet (richardchoulet(AT)yahoo.fr), 
               Sep 18 2007
%F A001652 As noted (Sep 20 2006), a(n) = 5*(a(n-1)+a(n-2))-a(n-3)+4. In general, 
               for n>2k, a(n)=A001653(k)*(a(n-k)+a(n-k-1)+1)-a(n-2k-1)-1; e.g. 696=5*(119+20+1)-3-1; 
               137903=29(4059+696+1)-20-1. Also a(n) = 7*(a(n-1)-a(n-2))+a(n-3). 
               In general, for n>2k, A002378(k)*(a(n-k)-a(n-k-1))+a(n-2k-1); e.g. 
               696=7(119-20)+3; 137903=41(4059-696)+20 - Charlie Marion (charliemath(AT)optonline.net), 
               Dec 26 2007
%F A001652 In general, for n>=k>0, a(n)= (A001653(n+k)-A001541(k)*A001653(n)-2*A001109(k-1))/
               (4*A001109(k-1)); e.g. 4059=(33461-3*5741-2*1)/(4*1); 4059=(195025-17*5741-2*6)/
               (4*6) - Charlie Marion (charliemath(AT)optonline.net), Jan 21 2008
%e A001652 The first few triples are (0,1,1), (3,4,5), (20,21,29), (119,120,169), 
               ...
%p A001652 A001652:=z*(-3+z)/(z-1)/(z**2-6*z+1); [S. Plouffe in his 1992 dissertation.]
%t A001652 pythList[ n_Integer?Positive ] := With[ {p = 3 + 2 Sqrt[ 2 ]}, NestList[ 
               Floor[ p # ] + 3 &, 3, n - 1 ] ]
%o A001652 (PARI) {a(n)=subst(poltchebi(n+1)-poltchebi(n)-2, x, 3)/4} /* Michael 
               Somos Aug 11 2006 */
%o A001652 (MAGMA) Z<x>:=PolynomialRing(Integers()); N<r2>:=NumberField(x^2-2); 
               S:=[ (-2+(r2+1)*(3+2*r2)^n-(r2-1)*(3-2*r2)^n)/4: n in [1..20] ]; 
               [ Integers()!S[j]: j in [1..#S] ]; [From Klaus Brockhaus (klaus-brockhaus(AT)t-online.de), 
               Feb 17 2009]
%Y A001652 Cf. A001653. A046090(n)=-a(-1-n).
%Y A001652 Cf. A001108.
%Y A001652 A156035 (decimal expansion of 3+2*sqrt(2)). [From Klaus Brockhaus (klaus-brockhaus(AT)t-online.de), 
               Feb 17 2009]
%Y A001652 Sequence in context: A108911 A005096 A164535 this_sequence A128910 A037788 
               A037669
%Y A001652 Adjacent sequences: A001649 A001650 A001651 this_sequence A001653 A001654 
               A001655
%K A001652 nonn,easy,nice
%O A001652 0,2
%A A001652 N. J. A. Sloane (njas(AT)research.att.com).
%E A001652 Mathematica formula from Harvey P. Dale (hpd1(AT)is2.nyu.edu).
%E A001652 Additional comments from Wolfdieter Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de), 
               Feb 10 2000
%E A001652 Removed attribute "conjectured" from Plouffe g.f R. J. Mathar (mathar(AT)strw.leidenuniv.nl), 
               Mar 11 2009

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