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 A001652 a(n) = 6*a(n-1) - a(n-2) + 2 with a(0) = 0, a(1) = 3. (Formerly M3074 N1247) 153
 0, 3, 20, 119, 696, 4059, 23660, 137903, 803760, 4684659, 27304196, 159140519, 927538920, 5406093003, 31509019100, 183648021599, 1070379110496, 6238626641379, 36361380737780, 211929657785303, 1235216565974040 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS Consider all Pythagorean triples (X, X+1, Z) ordered by increasing Z; sequence gives X values. Numbers n such that triangular number t(n) (see A000217) = n(n+1)/2 is a product of two consecutive integers (cf. A097571). 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). The index of all triangular numbers T(a(n)) for which 4T(n)+1 is a perfect square. 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 All Pythagorean triples {X(n), Y(n)=X(n)+1, Z(n)} with X 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 Let b(n) = A053141 then a(n)*b(n+1) = b(n)*a(n+1) + b(n). - Kenneth J Ramsey, Sep 22 2007 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 lim_{n -> infinity} a(n)/a(n-1) = 3+2*sqrt(2). - Klaus Brockhaus, Feb 17 2009 The remainder of the division of a(n) by 5 is: 0, 1, 3 or 4. - Mohamed Bouhamida, Aug 26 2009 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. - Mohamed Bouhamida, Sep 01 2009 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 p0,   T(a(n)+2k*A001653(n+1)) = 2*T(A053141(n-1)+k*A002315(n))+k^2 and   T(a(n)+(2k+1)*A001653(n+1)) = (A001109(n+1)+k*A002315(n))^2+k(k+1).   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 For n>0, A143608(n) divides a(n). - Kenneth J Ramsey, Jun 28 2012 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 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 Conjecture: For n>1 a(n) is the index of the first occurrence of n in sequence A123737. - Vaclav Kotesovec, Jun 02 2015 Numbers n such that Product_{k=1..n} (4*k^4+1) is a square (see A274307). - Chai Wah Wu, Jun 21 2016 Numbers n such that n^2+(n+1)^2 is a square. - César Aguilera, Aug 14 2017 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 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 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 REFERENCES A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 122-125, 1964. N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence). N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence). LINKS T. D. Noe, Table of n, a(n) for n = 0..200 I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), 181-193. 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. 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. Thibault Gauthier, Deep Reinforcement Learning for Synthesizing Functions in Higher-Order Logic, arXiv:1910.11797 [cs.AI], 2019. See also EPiC Series in Computing, (2020) Vol. 73, 230-248. L. J. Gerstein, Pythagorean triples and inner products, Math. Mag., 78 (2005), 205-213. Ron Knott, Pythagorean Triples and Online Calculators A. Martin, Table of prime rational right-angled triangles, The Mathematical Magazine, 2 (1910), 297-324. A. Martin, Table of prime rational right-angled triangles (annotated scans of a few pages) A. Martin, On rational right-angled triangles, Proceedings of the Fifth International Congress of Mathematicians (Cambridge, 22-28 August 1912). S. P. Mohanty, Which triangular numbers are products of three consecutive integers, Acta Math. Hungar., 58 (1991), 31-36. Vladimir Pletser, Recurrent Relations for Multiple of Triangular Numbers being Triangular Numbers, arXiv:2101.00998 [math.NT], 2021. Vladimir Pletser, Closed Form Equations for Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2102.12392 [math.GM], 2021. Vladimir Pletser, Triangular Numbers Multiple of Triangular Numbers and Solutions of Pell Equations, arXiv:2102.13494 [math.NT], 2021. Vladimir Pletser, Congruence Properties of Indices of Triangular Numbers Multiple of Other Triangular Numbers, arXiv:2103.03019 [math.GM], 2021. Vladimir Pletser, Searching for multiple of triangular numbers being triangular numbers, 2021. Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992. Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992 Burkard Polster, Nice merging together, Mathologer video (2015). B. Polster and M. Ross, Marching in squares, arXiv:1503.04658 [math.HO], 2015. Zhang Zaiming, Problem #502, Pell's Equation - Once Again, The College Mathematics Journal, 25 (1994), 241-243. Index entries for linear recurrences with constant coefficients, signature (7,-7,1). FORMULA G.f.: x *(3 - x) / ((1 - 6*x + x^2) * (1 - x)). - Simon Plouffe in his 1992 dissertation 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 a(n) = a(n-2) + 4*sqrt(2*(a(n-1)^2)+2*a(n-1)+1). - Pierre CAMI, Mar 30 2005 a(n) = (sinh((2*n+1)*log(1+sqrt(2)))-1)/2 = (sqrt(1+8*A029549)-1)/2. - Bill Gosper, Feb 07 2010 Binomial(a(n)+1,2) = 2*binomial(A053141(n)+1,2) = A029549(n). See A053141. - Bill Gosper, Feb 07 2010 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 a(n)*a(n+1) + A046090(n)*A046090(n+1) = A001542(n+1)^2 = A084703(n+1). - Charlie Marion, Jul 01 2003 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 Sum_{k=0...n} (2*k+1)*a(n-k) = A001109(n+1) - A000217(n+1). - Charlie Marion, Jul 18 2003 a(n) = A055997(n) - 1 + sqrt(2*A055997(n)*A001108(n)). - Charlie Marion, Jul 21 2003 a(n) = {A002315(n) - 1}/2. - Lekraj Beedassy, Nov 25 2003 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 a(n) = (A001653(n+1) -3*A001653(n) - 2)/4. - Lekraj Beedassy, Jul 13 2004 a(n) = {2*A084159(n) - 1 + (-1)^(n+1)}/2. - Lekraj Beedassy, Jul 21 2004 a(n) = 5*(a(n-1) + a(n-2)) - a(n-3) + 4. - Mohamed Bouhamida, Sep 02 2006 a(n+1) = 3*a(n) + sqrt(8*a(n)^2 +8*a(n) +4) + 1, a(1)=0. - Richard Choulet, Sep 18 2007 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 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 From Charlie Marion, Jan 04 2010: (Start) a(n) = ( (1 + sqrt(2))^(2*n+1) + (1-sqrt(2))^(2*n+1) - 2)/4 = (A001333(2n+1) - 1)/2. 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); 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) a(n) = A048739(2*n-1) for n > 0. - Richard R. Forberg, Aug 31 2013 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 a(n)^2 + (a(n)+1)^2 = A001653(n+1)^2. - Pierre CAMI, Mar 30 2005; clarified by Hermann Stamm-Wilbrandt, Aug 31 2014 a(n+1) = 3*A001541(n) + 10*A001109(n) + A001108(n). - Hermann Stamm-Wilbrandt, Sep 09 2014 For n>0, a(n) = Sum_{k=1..2*n} A000129(k). - Charlie Marion, Nov 07 2015 a(n) = 3*A053142(n) - A053142(n-1). - R. J. Mathar, Feb 05 2016 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 a(n) = A001108(n) + 2*sqrt(A000217(A001108(n))). - Dimitri Papadopoulos, Jul 06 2017 a(A000217(n-1)) = ((A001653(n)+1)/2) * ((A001653(n)-1)/2), n > 1. - Ezhilarasu Velayutham, Mar 10 2019 a(n) = ((a(n-1)+1)*(a(n-1)-3))/a(n-2) for n > 2. - Vladimir Pletser, Apr 08 2020 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 EXAMPLE The first few triples are (0,1,1), (3,4,5), (20,21,29), (119,120,169), ... MAPLE A001652 := proc(n)     option remember;     if n <= 1 then         op(n+1, [0, 3]) ;     else         6*procname(n-1)-procname(n-2)+2 ;     end if; end proc: # R. J. Mathar, Feb 05 2016 MATHEMATICA LinearRecurrence[{7, -7, 1}, {0, 3, 20}, 30] (* Harvey P. Dale, Aug 19 2011 *) With[{c=3+2*Sqrt}, NestList[Floor[c*#]+3&, 3, 30]] (* Harvey P. Dale, Oct 22 2012 *) CoefficientList[Series[x (3 - x)/((1 - 6 x + x^2) (1 - x)), {x, 0, 30}], x] (* Vincenzo Librandi, Oct 21 2014 *) Table[(LucasL[2*n + 1, 2] - 2)/4, {n, 0, 30}] (* G. C. Greubel, Jul 15 2018 *) PROG (PARI) {a(n) = subst( poltchebi(n+1) - poltchebi(n) - 2, x, 3) / 4}; /* Michael Somos, Aug 11 2006 */ (PARI) concat(0, Vec(x*(3-x)/((1-6*x+x^2)*(1-x)) + O(x^50))) \\ Altug Alkan, Nov 08 2015 (PARI) {a=1+sqrt(2); b=1-sqrt(2); Q(n) = a^n + b^n}; for(n=0, 30, print1(round((Q(2*n+1) - 2)/4), ", ")) \\ G. C. Greubel, Jul 15 2018 (MAGMA) Z:=PolynomialRing(Integers()); N:=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 (MAGMA) m:=30; R:=PowerSeriesRing(Integers(), m);  cat Coefficients(R!(x*(3-x)/((1-6*x+x^2)*(1-x)))); // G. C. Greubel, Jul 15 2018 (Haskell) a001652 n = a001652_list !! n a001652_list = 0 : 3 : map (+ 2) (zipWith (-) (map (* 6) (tail a001652_list)) a001652_list) -- Reinhard Zumkeller, Jan 10 2012 (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 (Sage) (x*(3-x)/((1-6*x+x^2)*(1-x))).series(x, 30).coefficients(x, sparse=False) # G. C. Greubel, Mar 08 2019 CROSSREFS Cf. A046090(n) = -a(-1-n). Cf. A001108, A143608, A089950 (partial sums). Cf. A156035 (decimal expansion of 3+2*sqrt(2)). - Klaus Brockhaus, Feb 17 2009 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 Cf. A001541, A001109, A274307. Cf. A002315, A001653 (solutions of x^2 - 2*y^2 = -1). Sequence in context: A005096 A275796 A164535 * A128910 A037788 A037669 Adjacent sequences:  A001649 A001650 A001651 * A001653 A001654 A001655 KEYWORD nonn,easy,nice AUTHOR EXTENSIONS Additional comments from Wolfdieter Lang, Feb 10 2000 STATUS approved

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Last modified October 19 19:44 EDT 2021. Contains 348091 sequences. (Running on oeis4.)