1,2

Consider all Pythagorean triples (X,X+1,Z) ordered by increasing Z; sequence gives Z values.

The defining equation is X^2 + (X+1)^2 = Z^2, which when doubled gives 2Z^2 = (2X+1)^2 + 1. So the sequence gives Z's such that 2Z^2 = odd square + 1.

(x,y)=(a(n),a(n+1)) are the solutions with x<y of x/(yz)+y/(xz)+z/(xy)=3 with z=2. - Floor van Lamoen (fvlamoen(AT)hotmail.com), Nov 29 2001

Consequently the sum n^2*(2n^2 - 1) of the first n odd cubes (A002593) is also a square. - Lekraj Beedassy (blekraj(AT)yahoo.com), Jun 05 2002

Numbers n such that 2*n^2=ceil(sqrt(2)*n*floor(sqrt(2)*n)) Benoit Cloitre (benoit7848c(AT)orange.fr), May 10 2003

Also, number of domino tilings in S_5 X P_2n. - R. Stephan, Mar 30 2004. Here S_5 is the star graph on 5 vertices with the edges {1,2}, {1,3}, {1,4}, {1,5}.

If x is in the sequence then so is x*(8*x^2-3). - James Buddenhagen (jbuddenh(AT)gmail.com), Jan 13 2005

In general, sum{k=0..n, binomial(2n-k,k)j^(n-k)}=(-1)^n*U(2n,I*sqrt(j)/2), I=sqrt(-1). - Paul Barry (pbarry(AT)wit.ie), Mar 13 2005

a(n) = L(n,6), where L is defined as in A108299; see also A002315 for L(n,-6). - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Jun 01 2005

Define a T-circle to be a first-quadrant circle with integral radius that is tangent to the x- and y-axes. Such a circle has coordinates equal to its radius. Let C(0) be the T-circle with radius 1. Then for n>0, define C(n) to be the largest T-circle that intersects C(n-1). C(n) has radius a(n) and the coordinates of its points of intersection with C(n-1) are A001108(n) and A055997(n). Cf. A001109 - Charlie Marion (charliemath(AT)optonline.net), Sep 14 2005

Number of 01-avoiding words of length n on alphabet {0,1,2,3,4,5} which do not end in 0. - Tanya Khovanova (tanyakh(AT)yahoo.com), Jan 10 2007

The lower principal convergents to 2^(1/2), beginning with 1/1, 7/5, 41/29, 239/169, comprise a strictly increasing sequence; numerators=A002315 and denominators=A001653. - Clark Kimberling (ck6(AT)evansville.edu), Aug 26 2008

Apparently Ljunggren shows that 169 is the last square term.

The remainder of the division of a(n) by 5 is: 0, 1 or 4. [From Mohamed Bouhamida (bhmd95(AT)yahoo.fr), Aug 26 2009]

Number of units of a(n) belongs to a periodic sequence: 1, 5, 9, 9, 5, 1. The remainder of the division of a(n) by 5 belongs to a periodic sequence: 1, 0, 4, 4, 0, 1. [From 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 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]

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]

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).

I. Adler, Three Diophantine equations - Part II, Fib. Quart., 7 (1969), 181-193.

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

Fielder, Daniel C.; Special integer sequences controlled by three parameters. Fibonacci Quart 6 1968 64-70.

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.

W. Ljunggren, "Zur Theorie der Gleichung x^2+1=Dy^4", Avh. Norske Vid. Akad. Oslo I. 5, 27pp.

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

Problem 47, Amer. Math. Monthly, 4 (1897), 25-28.

Michael Z. Spivey and Laura L. Steil, The k-Binomial Transforms and the Hankel Transform, Journal of Integer Sequences, Vol. 9 (2006), Article 06.1.1.

David Wells, The Penguin Dictionary of Curious and Interesting Numbers (Rev. ed. 1997), p. 91.

T. D. Noe, Table of n, a(n) for n=1..201

J.-P. Ehrmann et al., POLYA003: Integers of the form a/(bc) + b/(ca) + c/(ab).

Tanya Khovanova, Recursive Sequences

Ron Knott, Pythagorean Triples and Online Calculators

INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 403

S. Plouffe, Approximations de S\'{e}ries G\'{e}n\'{e}ratrices et Quelques Conjectures, Dissertation, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.

S. Plouffe, 1031 Generating Functions and Conjectures, Universit\'{e} du Qu\'{e}bec \`{a} Montr\'{e}al, 1992.

Eric Weisstein's World of Mathematics, NSW Number

Index entries for two-way infinite sequences

Index entries for sequences related to linear recurrences with constant coefficients

Index entries for sequences related to Chebyshev polynomials.

G.f.: (1-x)/(1-6*x+x^2). a(n)=6a(n-1)-a(n-2) with a(0)=1, a(1)=5. a(-1-n)=a(n).

a(n) = S(n, 6)-S(n-1, 6) with S(n, 6) = A001109(n+1), S(-2, 6) := -1. S(n, x)=U(n, x/2) are Chebyshev's polynomials of the second kind. Cf. triangle A049310. a(n) = T(2*n+1, sqrt(2))/sqrt(2), with T(n, x) Chebyshev's polynomials of the first kind.

a(n) ~ 1/4*sqrt(2)*(sqrt(2) + 1)^(2*n+1) - Joe Keane (jgk(AT)jgk.org), May 15 2002

a(n) = [[(3 + 2*Sqrt(2))^(n+1) - (3 - 2*Sqrt(2))^(n+1)] - [(3 + 2*Sqrt(2))^n - (3 - 2*Sqrt(2))^n] ] / (4*Sqrt(2)). Lim. n->Inf. a(n)/a(n-1) = 3 + 2*Sqrt(2). - Gregory V. Richardson (omomom(AT)hotmail.com), Oct 12 2002

Let q(n, x)=sum(i=0, n, x^(n-i)*binomial(2*n-i, i)); then q(n, 4)=a(n) - Benoit Cloitre (benoit7848c(AT)orange.fr), Nov 10 2002

For n and j >= 1, sum_{k=0..j}a(k)*a(n)-sum_{k=0..j-1}a(k)*a(n-1) = A001109(j+1)*a(n)-A001109(j)*a(n-1) = a(n+j); e.g. (1+5+29)*5-(1+5)*1=169 - Charlie Marion (charliem(AT)bestweb.net), Jul 07 2003

For n>=k>=0, a(n)^2 = a(n+k)*a(n-k) - A084703(k)^2; e.g. 169^2 = 5741*5 - 144. - Charlie Marion (charliem(AT)bestweb.net), Jul 16 2003

For n > 0, a(n) ^2 - a(n-1)^2 = 4*Sum_{k=0...2n-1}a(k) = 4*A001109(2n); e.g. 985^2 - 169^2 = 4*(1 + 5 + 29 + ... + 195025) = 4*235416 - Charlie Marion (charliem(AT)bestweb.net), Jul 16 2003

Sum_{k=0...n}((-1)^(n-k)*a(k)) = A079291(n+1); e.g. -1 + 5 - 29 + 169 = 144 - Charlie Marion (charliem(AT)bestweb.net), Jul 16 2003

A001652(n) + A046090(n) - a(n) = A001542(n); e.g. 119 + 120 - 169 = 70 - Charlie Marion (charliem(AT)bestweb.net), Jul 16 2003

Sum_{k=0...n}((2k+1)*a(n-k))=A001333(n+1)^2 - (1 + (-1)^(n+1))/2; e.g. 1*169+3*29+5*5+7*1=288=17^2-1; 1*29+3*5+5*1=49=7^2 - Charlie Marion (charliem(AT)bestweb.net), Jul 18 2003

Sum_{k = 0...n}a(k)*a(n) = sum_{k = 0...n}a(2k) and Sum_{k = 0...n}a(k)*a(n+1) = sum_{k = 0...n}a(2k+1); e.g. (1+5+29)*29 = 1+29+985 and (1+5+29)*169 = 5+169+5741 - Charlie Marion (charliem(AT)bestweb.net), Sep 22 2003

For n >= 3, a_{n} = 7(a_{n-1} - a_{n-2}) + a_{n-3}, with a_1 = 1, a_2 = 5 and a_3 = 29. a(n) = ((-1+2^(1/2))/2^(3/2))*(3 - 2^(3/2))^n + ((1+2^(1/2))/2^(3/2))*(3 + 2^(3/2))^n . - Antonio Olivares (olivares14031(AT)yahoo.com), Oct 13, 2003

Let a(n) = A001652(n), b(n) = A046090(n) and c(n) = this sequence. 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

Let a(n) = A001652(n), b(n) = A046090(n) and c(n) = this sequence. Then for n>0, a(n)*b(n)*c(n)/(a(n)+b(n)+c(n))=sum_{k=0...n}c(2*k+1); e.g. 20*21*29/(20+21+29) = 5+169 = 174; a(n)*b(n)*c(n)/(a(n-1)+b(n-1)+c(n-1))=sum_{k=0...n}c(2*k); e.g. 119*120*169/(20+21+29)=1+29+985+33461=34476 - Charlie Marion (charliem(AT)bestweb.net), Dec 01 2003

Also solutions x>0 of the equation floor(x*r*floor(x/r))==floor(x/r*floor(x*r)) where r=1+sqrt(2) - Benoit Cloitre (benoit7848c(AT)orange.fr), Feb 15 2004

a(n)a(n+3) = 24 + a(n+1)a(n+2). - R. Stephan, May 29 2004

For n>=k, a(n)*a(n+2k+1)-a(n+k)*a(n+k+1)=a(k)^2-1; e.g. 29*195025-985*5741=840=29^2-1; 1*169-5*29=24=5^2-1; a(n)*a(n+2k)-a(n+k)^2=A001542(k)^2; e.g. 169*195025-5741^2=144=12^2; 1*29-5^2=4=2^2 - Charlie Marion (charliemath(AT)verizon.net), Jun 02 2004

For all k, a(n) is a factor of a((2n+1)*k+n). a((2n+1)k+n)=a(n)*(sum_{j=0...k-1}(-1)^j*(a((2n+1)(k-j))+a((2n+1)(k-j)-1))+(-1)^k); e.g. 195025=5*(33461+5741-169-29+1); 7645370045=169*(6625109+1136689-1) - Charlie Marion (charliemath(AT)verizon.net), Jun 04 2004

a(n)=sum{k=0..n, binomial(n+k, 2k)4^k} - Paul Barry (pbarry(AT)wit.ie), Aug 30 2004

a(n)=sum{k=0..n, binomial(2n+1, 2k+1)2^k}. - Paul Barry (pbarry(AT)wit.ie), Sep 30 2004

For n<k, a(n)*A001541(k)=A011900(n+k)+A053141(k-n-1); e.g. 5*99=495=493+2. For n>=k, a(n)*A001541(k)=A011900(n+k)+A053141(n-k); e.g. 29*3=87=85+2 - Charlie Marion (charliemath(AT)optonline.net), Oct 18 2004

a(n)=(-1)^n*U(2n, I*sqrt(4)/2)=(-1)^n*U(2n, I), U(n, x) Chebyshev polynomial of second kind, I=sqrt(-1); - Paul Barry (pbarry(AT)wit.ie), Mar 13 2005

a(n)=Pell(2n+1)=Pell(n)^2+Pell(n+1)^2; - Paul Barry (pbarry(AT)wit.ie), Jul 18 2005

a(n)*a(n+k)=Pell(2n+k+1)^2+Pell(k)^2;e.g., 5*29=12^2+1^2; 29*985=169^2+2^2 - Charlie Marion (charliemath(AT)optonline.net) Aug 02 2005

Let a(n)*a(n+k)=x. Then 2x^2-A001541(k)*x+A001109(k)^2=A001109(2n+k+1)^2; e.g., let x=29*985; then 2x^2-17x+6^2=40391^2; cf. A076218 - Charlie Marion (charliemath(AT)optonline.net) Aug 02 2005

a(n)*a(n+k)=A000129(k)^2+A000129(2n+k+1)^2;e.g., 29*5741=12^2+169^2; - Charlie Marion (charliemath(AT)optonline.net) Aug 03 2005

With a=3+2sqrt(2), b=3-2sqrt(2): a(n)=(a^((2n+1)/2)+b^((2n+1)/2))/(2sqrt(2)). a(n)=A001109(n+1)-A001109(n). - Mario Catalani (mario.catalani(AT)unito.it), Mar 31 2003

If k is in the sequence, then the next term is floor[k*(3+2*sqrt(2))]. - Lekraj Beedassy (blekraj(AT)yahoo.com), Jul 19 2005

a(n)=Jacobi_P(n,-1/2,1/2,3)/Jacobi_P(n,-1/2,1/2,1); - Paul Barry (pbarry(AT)wit.ie), Feb 03 2006

a(n)=sum{k=0..n, sum{j=0..n-k, C(n,j)C(n-j,k)Pell(n-j+1)}}, Pell(n)=A000129(n); - Paul Barry (pbarry(AT)wit.ie), May 19 2006

a(n)=round[sqrt({A002315(n)}^2/2)]. - Lekraj Beedassy (blekraj(AT)yahoo.com), Jul 15 2006

a(n) = A079291(n) + A079291(n+1). - Lekraj Beedassy (blekraj(AT)yahoo.com), Aug 14 2006

a(n+1)=3*a(n)+(8*a(n)^2-4)^0.5, a(1)=1. - Richard Choulet (richardchoulet(AT)yahoo.fr), Sep 18 2007

6*a(n)*a(n+1)=a(n)^2+a(n+1)^2+4; e.g., 6*5*29=29^2+5^2+4; 6*169*985=169^2+985^2+4 - Charlie Marion (charliemath(AT)optonline.net), Oct 07 2007 - Charlie Marion (charliemath(AT)optonline.net), Oct 07 2007

2*A001541(k)*a(n)*a(n+k)=a(n)^2+a(n+k)^2+A001542(k)^2; e.g., 2*3*5*29=5^2+29^2+2^2; 2*99*29*5741=2*99*29*5741=29^2+5741^2+70^2 - Charlie Marion (charliemath(AT)optonline.net), Oct 12 2007

[a(n), A001109(n)] = [1,4; 1,5]^n * [1,0]. - Gary W. Adamson (qntmpkt(AT)yahoo.com), Mar 21 2008

a(n)=6*a(n-1)-a(n-2), a(0)=1, a(1)=5. [From Ctibor O. Zizka (c.zizka(AT)email.cz), Mar 31 2009]

Contribution from Charlie Marion (charliemath(AT)optonline.net), Apr 10 2009: (Start)

In general, for n>=k, a(n+k)= 2*A001541(k)*a(n)-a(n-k);

e.g., a(n+0)=2*1*a(n)-a(n); a(n+1)=6*a(n)-a(n-1); a(6+0)=33461=2*33461-33461;

a(5+1)=33461=6*5741-985; a(4+2)=33461=34*985-29; a(3+3)=33461=198*169-1.

(End)

a[0]:=1: a[1]:=5: for n from 2 to 26 do a[n]:=6*a[n-1]-a[n-2] od: seq(a[n], n=0..20); - Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Jul 26 2006

A001653:=-(-1+5*z)/(z**2-6*z+1); [Conjectured (correctly) by S. Plouffe in his 1992 dissertation. Gives sequence except for one of the leading 1's.]

a[n_] := (MatrixPower[{{1, 2, 2}, {2, 1, 2}, {2, 2, 3}}, n].{{1}, {0}, {1}})[[3, 1]]; Table[ a[n], {n, 0, 20}] (from Robert G. Wilson v Jan 08 2005)

(PARI) a(n)=subst(poltchebi(abs(n+1))+poltchebi(abs(n)), x, 3)/4 (from Michael Somos)

(PARI) a(n)=([5, 2; 2, 1]^n)[1, 1] (from Lambert Klasen)

(PARI) a1=1; a2=5; print1(a1, ", ", a2); for(n=0, 10, bak=a1; a1=a2; a2=6*a1-bak; print1(a2, ", ")) - Ray Copalla (rhcp(AT)gmx.co.uk), Apr 25 2004

(Other) sage: [lucas_number1(n, 6, 1)-lucas_number1(n-1, 6, 1) for n in xrange(1, 22)]# [From Zerinvary Lajos (zerinvarylajos(AT)yahoo.com), Nov 10 2009]

Other two sides are A001652, A046090. a(n)=sqrt{(A002315(n)^2 +1)/2}.

Cf. A001519.

These numbers are the odd-indexed Pell numbers from A000129. The even-indexed Pell numbers are A001542. - Ira Gessel (gessel(AT)brandeis.edu), Sep 27 2002.

Row 6 of array A094954.

Cf. A001109.

Sequence in context: A146178 A015537 A141812 this_sequence A141814 A122370 A088349

Adjacent sequences: A001650 A001651 A001652 this_sequence A001654 A001655 A001656

nonn,easy,nice,new

N. J. A. Sloane (njas(AT)research.att.com).

Additional comments from Wolfdieter Lang (wolfdieter.lang(AT)physik.uni-karlsruhe.de), Feb 10 2000

Better description from Harvey P. Dale (hpd1(AT)nyu.edu), Jan 15 2002

Edited by N. J. A. Sloane (njas(AT)research.att.com) Nov 02 2002