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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)
124
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 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, Dec 22 2003

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

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 (bhmd95(AT)yahoo.fr), 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 (bhmd95(AT)yahoo.fr), 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 p<r then s-r=p+q+1. - Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Aug 29 2009

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 (bhmd95(AT)yahoo.fr), Sep 02 2009

a(n+k) = A001541(k)*a(n)+A001542(k)*A001653(n+1)+A001108(k). - Charlie Marion, Dec 10 2010

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

Let T(n) = n*(n+1)/2 (the n-th triangular number).  For n>0,

  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 triplet 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

REFERENCES

A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 122-125, 1964.

A. Martin, Table of prime rational right-angled triangles, Math. Mag., 2 (1910), 297-324.

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.

L. J. Gerstein, Pythagorean triples and inner products, Math. Mag., 78 (2005), 205-213.

Ron Knott, Pythagorean Triples and Online Calculators

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.

Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992.

Simon Plouffe, 1031 Generating Functions and Conjectures, Université du Québec à Montréal, 1992.

Zhang Zaiming, Problem #502, Pell's Equation - Once Again, The College Mathematics Journal, 25 (1994), 241-243.

Index entries for two-way infinite sequences

Index entries for sequences related to linear recurrences with constant coefficients, signature (7,-7,1)

FORMULA

G.f.: x *(3 - x) / ((1 - 6*x + x^2) * (1 - x)). - Michael Somos, Apr 07 2003

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(A001652+1,2) = 2*binomial(A053141+1,2) = A029549. 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+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, 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); e.g., (1+5+29)*119-(1+5)*20+14=4059. - Charlie Marion, Jul 07 2003

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, Jul 18 2003

a(n) = A055997(n)-1+(2*A055997(n)*A001108(n))^.5; e.g., 119=50-1+(2*50*49)^.5. - Charlie Marion, Jul 21 2003

a(n) = {A002315(n) - 1}/2. - Lekraj Beedassy, Nov 25 2003

Sum_{k=0...n}a(k)=A089950(n); e.g., 0+3+20+119=142. - Charlie Marion, Jan 21 2004

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, 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 (bhmd95(AT)yahoo.fr), Sep 02 2006

a(n+1) = 3*a(n)+(8*a(n)^2+8*a(n)+4)^0.5+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>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, 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

a(n) = ( (1 + sqrt(2))^(2n+1) + (1-sqrt(2))^(2n+1) - 2)/4 = (A001333(2n+1) - 1)/2.

a(2n+k-1) = Pell(2n-1)*Pell(2n+2k) + Pell(2n-2)*Pell(2n+2k+1) + A001108(k+1);

a(2n+k) = Pell(2n)*Pell(2n+2k+1) + Pell(2n-1)*Pell(2n+2k+2) - A055997(k+2).

  - Charlie Marion, Jan 04 2010

a(n) = A048739(2n-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

EXAMPLE

The first few triples are (0,1,1), (3,4,5), (20,21,29), (119,120,169), ...

MAPLE

A001652:=z*(-3+z)/(z-1)/(z**2-6*z+1); # Simon Plouffe in his 1992 dissertation

MATHEMATICA

LinearRecurrence[{7, -7, 1}, {0, 3, 20}, 30] (* Harvey P. Dale, Aug 19 2011 *)

With[{c=3+2*Sqrt[2]}, NestList[Floor[c*#]+3&, 3, 30]] (* Harvey P. Dale, Oct 22 2012 *)

PROG

(PARI) {a(n) = subst( poltchebi(n+1) - poltchebi(n) - 2, x, 3) / 4}; /* Michael Somos, Aug 11 2006 */

(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

(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

CROSSREFS

Cf. A001653. A046090(n) = -a(-1-n).

Cf. A001108, A143608.

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.

Sequence in context: A108911 A005096 A164535 * A128910 A037788 A037669

Adjacent sequences:  A001649 A001650 A001651 * A001653 A001654 A001655

KEYWORD

nonn,easy,nice,changed

AUTHOR

N. J. A. Sloane

EXTENSIONS

Additional comments from Wolfdieter Lang, Feb 10 2000

STATUS

approved

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Last modified September 17 03:34 EDT 2014. Contains 246833 sequences.