|
%I M3074 N1247 #404 Mar 01 2024 05:22:31
%S 0,3,20,119,696,4059,23660,137903,803760,4684659,27304196,159140519,
%T 927538920,5406093003,31509019100,183648021599,1070379110496,
%U 6238626641379,36361380737780,211929657785303,1235216565974040
%N a(n) = 6*a(n-1) - a(n-2) + 2 with a(0) = 0, a(1) = 3.
%C Consider all Pythagorean triples (X, X+1, Z) ordered by increasing Z; sequence gives X values.
%C Numbers n such that triangular number t(n) (see A000217) = n(n+1)/2 is a product of two consecutive integers (cf. A097571).
%C 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 The index of all triangular numbers T(a(n)) for which 4T(n)+1 is a perfect square.
%C 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 formulas x = 2uv, y = u^2 - v^2, z = u^2 + v^2 [Joseph Wiener and Donald Skow]. - _Antonio Alberto Olivares_, Dec 22 2003
%C All Pythagorean triples {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_, Aug 14 2006
%C Let b(n) = A053141 then a(n)*b(n+1) = b(n)*a(n+1) + b(n). - _Kenneth J Ramsey_, Sep 22 2007
%C 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_, Nov 19 2007
%C Limit_{n -> oo} a(n)/a(n-1) = 3+2*sqrt(2) = A156035. - _Klaus Brockhaus_, Feb 17 2009
%C 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 consecutive solutions of the Diophantine equation: X^2 + (X+1)^2 = Y^2 with p<r then s-r=p+q+1. - _Mohamed Bouhamida_, Aug 29 2009
%C 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. - _Mohamed Bouhamida_, Sep 02 2009
%C a(n+k) = A001541(k)*a(n) + A001542(k)*A001653(n+1) + A001108(k). - _Charlie Marion_, Dec 10 2010
%C The numbers 3*A001652 = (0, 9, 60, 357, 2088, 12177, 70980, ...) are all the nonnegative values of X such that X^2 + (X+3)^2 = Z^2 (Z is in A075841). - _Bruno Berselli_, Aug 26 2010
%C Let T(n) = n*(n+1)/2 (the n-th triangular number). For n > 0,
%C T(a(n) + 2*k*A001653(n+1)) = 2*T(A053141(n-1) + k*A002315(n)) + k^2 and
%C T(a(n) + (2*k+1)*A001653(n+1)) = (A001109(n+1) + k*A002315(n))^2 + k*(k+1).
%C Also (a(n) + k*A001653(n))^2 + (a(n) + k*A001653(n) + 1)^2 = (A001653(n+1) + k*A002315(n))^2 + k^2. - _Charlie Marion_, Dec 09 2010
%C For n>0, A143608(n) divides a(n). - _Kenneth J Ramsey_, Jun 28 2012
%C Set a(n)=p; a(n)+1=q; the generated triple x=p^2+pq; y=q^2+pq; k=p^2+q^2 satisfies x^2+y^2=k(x+y). - _Carmine Suriano_, Dec 17 2013
%C The arms of the triangle are found with (b(n),c(n)) for 2*b(n)*c(n) and c(n)^2 - b(n)^2. Let b(1)=1 and c(1)=2, then b(n)=c(n-1) and c(n)= 2*c(n-1) + b(n-1). Alternatively, b(n)=c(n-1) and c(n) equals the nearest integer to b(n)*(1+sqrt(2)). - _J. M. Bergot_, Oct 09 2014
%C Conjecture: For n>1 a(n) is the index of the first occurrence of n in sequence A123737. - _Vaclav Kotesovec_, Jun 02 2015
%C Numbers m such that Product_{k=1..m} (4*k^4+1) is a square (see A274307). - _Chai Wah Wu_, Jun 21 2016
%C Numbers m such that m^2+(m+1)^2 is a square. - _César Aguilera_, Aug 14 2017
%C For integers a and d, let P(a,d,1) = a, P(a,d,2) = a+d, and, for n>2, P(a,d,n) = 2*P(a,d,n-1) + P(a,d,n-2). Further, let p(n) = Sum_{i=1..2n} P(a,d,i). Then p(n)^2 + (p(n)+d)^2 + a^2 = P(a,d,2n+1)^2 + d^2. When a = 1 and d = 1, p(n) = a(n) and P(a,d,n) = A000129(n), the n-th Pell number. - _Charlie Marion_, Dec 08 2018
%C The terms of this sequence satisfy the Diophantine equation k^2 + (k+1)^2 = m^2, which is equivalent to (2k+1)^2 - 2*m^2 = -1. Now, with x=2k+1 and y=m, we get the Pell-Fermat equation x^2 - 2*y^2 = -1. The solutions (x,y) of this equation are respectively in A002315 and A001653. The relation k = (x-1)/2 explains _Lekraj Beedassy_'s Nov 25 2003 formula. Thus, the corresponding numbers m = y, which express the length of the hypotenuse of these right triangles (k,k+1,m) are in A001653. - _Bernard Schott_, Mar 10 2019
%C Members of Diophantine pairs. Related to solutions of p^2 = 2q^2 + 2 in natural numbers; p = p(n) = 2*sqrt(4T(a(n))+1), q = q(n) = sqrt(8*T(a(n))+1). Note that this implies that 4*T(a(n))+1 is a perfect square (numbers of the form 8*T(n)+1 are perfect squares for all n); these T(a(n))'s are the only solutions to the given Diophantine equation. - _Steven Blasberg_, Mar 04 2021
%D A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 122-125, 1964.
%D N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H T. D. Noe, <a href="/A001652/b001652.txt">Table of n, a(n) for n = 0..200</a>
%H I. Adler, <a href="http://www.fq.math.ca/Scanned/7-2/adler.pdf">Three Diophantine equations - Part II</a>, Fib. Quart., 7 (1969), 181-193.
%H Christian Aebi and Grant Cairns, <a href="http://math.colgate.edu/~integers/x48/x48.pdf">Lattice equable quadrilaterals III: tangential and extangential cases</a>, Integers (2023) Vol. 23, #A48.
%H Martin V. Bonsangue, Gerald E. Gannon and Laura J. Pheifer, <a href="http://www.jstor.org/stable/20871097">Misinterpretations can sometimes be a good thing</a>, Math. Teacher, vol. 95, No. 6 (2002) pp. 446-449.
%H T. W. Forget and T. A. Larkin, <a href="http://www.fq.math.ca/Scanned/6-3/forget.pdf">Pythagorean triads of the form X, X+1, Z described by recurrence sequences</a>, Fib. Quart., 6 (No. 3, 1968), 94-104.
%H Thibault Gauthier, <a href="https://arxiv.org/abs/1910.11797">Deep Reinforcement Learning for Synthesizing Functions in Higher-Order Logic</a>, arXiv:1910.11797 [cs.AI], 2019. See also <a href="https://doi.org/10.29007/7jmg">EPiC Series in Computing</a>, (2020) Vol. 73, 230-248.
%H L. J. Gerstein, <a href="http://www.jstor.org/stable/30044157">Pythagorean triples and inner products</a>, Math. Mag., 78 (2005), 205-213.
%H Ron Knott, <a href="http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Pythag/pythag.html">Pythagorean Triples and Online Calculators</a>
%H A. Martin, <a href="http://www.mathunion.org/ICM/ICM1912.2/Main/icm1912.2.0040.0058.ocr.pdf">Table of prime rational right-angled triangles</a>, The Mathematical Magazine, 2 (1910), 297-324.
%H A. Martin, <a href="/A001652/a001652.pdf">Table of prime rational right-angled triangles</a> (annotated scans of a few pages)
%H A. Martin, <a href="https://web.archive.org/web/20171111031012/http://www.mathunion.org/ICM/ICM1912.2/Main/icm1912.2.0040.0058.ocr.pdf">On rational right-angled triangles</a>, Proceedings of the Fifth International Congress of Mathematicians (Cambridge, 22-28 August 1912).
%H S. P. Mohanty, <a href="http://dx.doi.org/10.1007/BF01903544">Which triangular numbers are products of three consecutive integers</a>, Acta Math. Hungar., 58 (1991), 31-36.
%H Vladimir Pletser, <a href="https://arxiv.org/abs/2101.00998">Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers</a>, arXiv:2101.00998 [math.NT], 2021.
%H Vladimir Pletser, <a href="https://arxiv.org/abs/2102.12392">Closed Form Equations for Triangular Numbers Multiple of Other Triangular Numbers</a>, arXiv:2102.12392 [math.GM], 2021.
%H Vladimir Pletser, <a href="https://arxiv.org/abs/2102.13494">Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations</a>, arXiv:2102.13494 [math.NT], 2021.
%H Vladimir Pletser, <a href="https://arxiv.org/abs/2103.03019">Congruence Properties of Indices of Triangular Numbers Multiple of Other Triangular Numbers</a>, arXiv:2103.03019 [math.GM], 2021.
%H Vladimir Pletser, <a href="https://doi.org/10.13140/RG.2.2.35428.91527">Searching for multiple of triangular numbers being triangular numbers</a>, 2021.
%H Vladimir Pletser, <a href="https://www.researchgate.net/profile/Vladimir-Pletser/publication/359808848_USING_PELL_EQUATION_SOLUTIONS_TO_FIND_ALL_TRIANGULAR_NUMBERS_MULTIPLE_OF_OTHER_TRIANGULAR_NUMBERS/">Using Pell equation solutions to find all triangular numbers multiple of other triangular numbers</a>, 2021.
%H Simon Plouffe, <a href="https://arxiv.org/abs/0911.4975">Approximations de séries génératrices et quelques conjectures</a>, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
%H Simon Plouffe, <a href="/A000051/a000051_2.pdf">1031 Generating Functions</a>, Appendix to Thesis, Montreal, 1992
%H Burkard Polster, <a href="http://youtu.be/rjHFkx6eTL8">Nice merging together</a>, Mathologer video (2015).
%H B. Polster and M. Ross, <a href="http://arxiv.org/abs/1503.04658">Marching in squares</a>, arXiv:1503.04658 [math.HO], 2015.
%H Zhang Zaiming, <a href="http://www.jstor.org/stable/2687658">Problem #502, Pell's Equation - Once Again</a>, The College Mathematics Journal, 25 (1994), 241-243.
%H <a href="/index/Tu#2wis">Index entries for two-way infinite sequences</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (7,-7,1).
%F G.f.: x *(3 - x) / ((1 - 6*x + x^2) * (1 - x)). - _Simon Plouffe_ in his 1992 dissertation
%F a(n) = 7*a(n-1) - 7*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 Alberto Olivares_, Oct 13 2003
%F a(n) = a(n-2) + 4*sqrt(2*(a(n-1)^2)+2*a(n-1)+1). - _Pierre CAMI_, Mar 30 2005
%F a(n) = (sinh((2*n+1)*log(1+sqrt(2)))-1)/2 = (sqrt(1+8*A029549)-1)/2. - _Bill Gosper_, Feb 07 2010
%F Binomial(a(n)+1,2) = 2*binomial(A053141(n)+1,2) = A029549(n). See A053141. - _Bill Gosper_, Feb 07 2010
%F 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+2*k+1) + b(n)*b(n+2*k+1) + c(n)*c(n+2*k+1) = (a(n+k+1) - a(n+k))^2; a(n)*a(n+2*k) + b(n)*b(n+2*k) + c(n)*c(n+2*k) = 2*c(n+k)^2. - _Charlie Marion_, Jul 01 2003
%F a(n)*a(n+1) + A046090(n)*A046090(n+1) = A001542(n+1)^2 = A084703(n+1). - _Charlie Marion_, Jul 01 2003
%F 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). - _Charlie Marion_, Jul 07 2003
%F Sum_{k=0...n} (2*k+1)*a(n-k) = A001109(n+1) - A000217(n+1). - _Charlie Marion_, Jul 18 2003
%F a(n) = A055997(n) - 1 + sqrt(2*A055997(n)*A001108(n)). - _Charlie Marion_, Jul 21 2003
%F a(n) = {A002315(n) - 1}/2. - _Lekraj Beedassy_, Nov 25 2003
%F a(2*n+k) + a(k) + 1 = A001541(n)*A002315(n+k). For k>0, a(2*n+k) - a(k-1) = A001541(n+k)*A002315(n); e.g., 803760-119 = 19601*41. - _Charlie Marion_, Mar 17 2003
%F a(n) = (A001653(n+1) - 3*A001653(n) - 2)/4. - _Lekraj Beedassy_, Jul 13 2004
%F a(n) = {2*A084159(n) - 1 + (-1)^(n+1)}/2. - _Lekraj Beedassy_, Jul 21 2004
%F a(n+1) = 3*a(n) + sqrt(8*a(n)^2 + 8*a(n) +4) + 1, a(1)=0. - _Richard Choulet_, Sep 18 2007
%F As noted (Sep 20 2006), a(n) = 5*(a(n-1) + a(n-2)) - a(n-3) + 4. In general, for n > 2*k, a(n) = A001653(k)*(a(n-k) + a(n-k-1) + 1) - a(n-2*k-1) - 1. Also a(n) = 7*(a(n-1) - a(n-2)) + a(n-3). In general, for n > 2*k, A002378(k)*(a(n-k)-a(n-k-1)) + a(n-2*k-1). - _Charlie Marion_, Dec 26 2007
%F 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_, Jan 21 2008
%F From _Charlie Marion_, Jan 04 2010: (Start)
%F a(n) = ( (1 + sqrt(2))^(2*n+1) + (1-sqrt(2))^(2*n+1) - 2)/4 = (A001333(2n+1) - 1)/2.
%F a(2*n+k-1) = Pell(2*n-1)*Pell(2*n+2*k) + Pell(2*n-2)*Pell(2*n+2*k+1) + A001108(k+1);
%F a(2*n+k) = Pell(2*n)*Pell(2*n+2*k+1) + Pell(2*n-1)*Pell(2*n+2*k+2) - A055997(k+2). (End)
%F a(n) = A048739(2*n-1) for n > 0. - _Richard R. Forberg_, Aug 31 2013
%F a(n+1) = 3*a(n) + 2*A001653(n) + 1 [Mohamed Bouhamida's 2009 (p,q)(r,s) comment above rewritten]. - _Hermann Stamm-Wilbrandt_, Jul 27 2014
%F a(n)^2 + (a(n)+1)^2 = A001653(n+1)^2. - _Pierre CAMI_, Mar 30 2005; clarified by _Hermann Stamm-Wilbrandt_, Aug 31 2014
%F a(n+1) = 3*A001541(n) + 10*A001109(n) + A001108(n). - _Hermann Stamm-Wilbrandt_, Sep 09 2014
%F For n>0, a(n) = Sum_{k=1..2*n} A000129(k). - _Charlie Marion_, Nov 07 2015
%F a(n) = 3*A053142(n) - A053142(n-1). - _R. J. Mathar_, Feb 05 2016
%F E.g.f.: (1/4)*(-2*exp(x) - (sqrt(2) - 1)*exp((3-2*sqrt(2))*x) + (1 + sqrt(2))*exp((3+2*sqrt(2))*x)). - _Ilya Gutkovskiy_, Apr 11 2016
%F a(n) = A001108(n) + 2*sqrt(A000217(A001108(n))). - _Dimitri Papadopoulos_, Jul 06 2017
%F a(A000217(n-1)) = ((A001653(n)+1)/2) * ((A001653(n)-1)/2), n > 1. - _Ezhilarasu Velayutham_, Mar 10 2019
%F a(n) = ((a(n-1)+1)*(a(n-1)-3))/a(n-2) for n > 2. - _Vladimir Pletser_, Apr 08 2020
%F In general, for each k >= 0, a(n) = ((a(n-k)+a(k-1)+1)*(a(n-k)-a(k)))/a(n-2*k) for n > 2*k. - _Charlie Marion_, Dec 27 2020
%F A generalization of the identity a(n)^2 + A046090(n)^2 = A001653(n+1)^2 follows. Let P(k,n) be the n-th k-gonal number. Then P(k,a(n)) + P(k,A046090(n)) = P(k,A001653(n+1)) + (4-k)*A001109(n). - _Charlie Marion_, Dec 07 2021
%e The first few triples are (0,1,1), (3,4,5), (20,21,29), (119,120,169), ...
%p A001652 := proc(n)
%p option remember;
%p if n <= 1 then
%p op(n+1,[0,3]) ;
%p else
%p 6*procname(n-1)-procname(n-2)+2 ;
%p end if;
%p end proc: # _R. J. Mathar_, Feb 05 2016
%t LinearRecurrence[{7,-7,1}, {0,3,20}, 30] (* _Harvey P. Dale_, Aug 19 2011 *)
%t With[{c=3+2*Sqrt[2]},NestList[Floor[c*#]+3&,3,30]] (* _Harvey P. Dale_, Oct 22 2012 *)
%t CoefficientList[Series[x (3 - x)/((1 - 6 x + x^2) (1 - x)), {x, 0, 30}], x] (* _Vincenzo Librandi_, Oct 21 2014 *)
%t Table[(LucasL[2*n + 1, 2] - 2)/4, {n, 0, 30}] (* _G. C. Greubel_, Jul 15 2018 *)
%o (PARI) {a(n) = subst( poltchebi(n+1) - poltchebi(n) - 2, x, 3) / 4}; /* _Michael Somos_, Aug 11 2006 */
%o (PARI) concat(0, Vec(x*(3-x)/((1-6*x+x^2)*(1-x)) + O(x^50))) \\ _Altug Alkan_, Nov 08 2015
%o (PARI) {a=1+sqrt(2); b=1-sqrt(2); Q(n) = a^n + b^n};
%o for(n=0, 30, print1(round((Q(2*n+1) - 2)/4), ", ")) \\ _G. C. Greubel_, Jul 15 2018
%o (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] ]; // _Klaus Brockhaus_, Feb 17 2009
%o (Magma) m:=30; R<x>:=PowerSeriesRing(Integers(), m); [0] cat Coefficients(R!(x*(3-x)/((1-6*x+x^2)*(1-x)))); // _G. C. Greubel_, Jul 15 2018
%o (Haskell)
%o a001652 n = a001652_list !! n
%o a001652_list = 0 : 3 : map (+ 2)
%o (zipWith (-) (map (* 6) (tail a001652_list)) a001652_list)
%o -- _Reinhard Zumkeller_, Jan 10 2012
%o (GAP) a:=[0,3];; for n in [3..25] do a[n]:=6*a[n-1]-a[n-2]+2; od; a; # _Muniru A Asiru_, Dec 08 2018
%o (Sage) (x*(3-x)/((1-6*x+x^2)*(1-x))).series(x, 30).coefficients(x, sparse=False) # _G. C. Greubel_, Mar 08 2019
%Y Cf. A046090(n) = -a(-1-n).
%Y Cf. A001108, A143608, A089950 (partial sums), A156035.
%Y Cf. numbers m such that k*A000217(m)+1 is a square: A006451 for k=1; m=0 for k=2; A233450 for k=3; this sequence for k=4; A129556 for k=5; A001921 for k=6. - _Bruno Berselli_, Dec 16 2013
%Y Cf. A053141, A001541, A001109, A274307.
%Y Cf. A002315, A001653 (solutions of x^2 - 2*y^2 = -1).
%K nonn,easy,nice
%O 0,2
%A _N. J. A. Sloane_
%E Additional comments from _Wolfdieter Lang_, Feb 10 2000
| 