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A001653 Numbers n such that 2*n^2 - 1 is a square.
(Formerly M3955 N1630)
158
1, 5, 29, 169, 985, 5741, 33461, 195025, 1136689, 6625109, 38613965, 225058681, 1311738121, 7645370045, 44560482149, 259717522849, 1513744654945, 8822750406821, 51422757785981, 299713796309065, 1746860020068409, 10181446324101389, 59341817924539925 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,2

COMMENTS

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, Jun 05 2002

These numbers are the odd-indexed Pell numbers from A000129. The even-indexed Pell numbers are A001542. - Ira M. Gessel, Sep 27 2002

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

Also, number of domino tilings in S_5 X P_2n. - Ralf 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 R. Buddenhagen, 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, Mar 13 2005

a(n) = L(n,6), where L is defined as in A108299; see also A002315 for L(n,-6). - Reinhard Zumkeller, 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, 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, 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, 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. - 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. - 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

Equals INVERT transform of A005054: (1, 4, 20, 100, 500, 2500,...) and INVERTi transform of A122074: (1, 6, 40, 268, 1796,...). - Gary W. Adamson, Jul 22 2010

a(n) is the number of compositions of n when there are 5 types of 1 and 4 types of other natural numbers. - Milan Janjic, Aug 13 2010

The remainder after division of a(n) by a(k) appears to belong to a periodic sequence: 1, 5,..., a(k-1), 0, a(k)-a(k-1),..., a(k)-1, a(k)-1,..., a(k)-a(k-1), 0, a(k-1),..., 5, 1. See Bouhamida’s Sep 01 2009 comment. - Charlie Marion, May 02 2011

Apart from initial 1: subsequence of A198389, see also A198385. - Reinhard Zumkeller, Oct 25 2011

(a(n+1),2*b(n+1)) and (a(n+2),2*b(n+1)), n>=0, with b(n):= A001109(n), give the (u(2*n),v(2*n)) and (u(2*n+1),v(2*n+1)) sequences, respectively, for Pythagorean triples (x,y,z), where x=|u^2-v^2|, y=2*u*v and z=u^2+v^2, with u odd and v even, which are generated from (u(0)=1,v(0)=2) by the substitution rule (u,v) -> (2*v+u,v) if u<v and (u,v) -> (u,2*u+v) if u>v. This leads to primitive triples because gcd(u,v) = 1 is respected. This corresponds to (primitive) Pythagorean triangles with |x-y|=1 (the catheti differ by one length unit). This (u,v) sequence starts with (1,2), (5,2), (5,12), (29,12), (29,70) ... - Wolfdieter Lang, 06 Mar 2012

a(n)^2 = 2*A182435(n)*(A182435(n)-1)+1. - Bruno Berselli, Oct 23 2012

Area of the Fibonacci snowflake of order n. - José Luis Ramírez Ramírez, Dec 13 2012

Area of the 3-generalized Fibonacci snowflake of order n, n>=3. - José Luis Ramírez Ramírez, Dec 13 2012

For the o.g.f. given by Johannes W. Meijer, Aug 01 2010, in the formula section see a comment under A077445. - Wolfdieter Lang, Jan 18 2013

Positive values of x (or y) satisfying x^2 - 6xy + y^2 + 4 = 0. - Colin Barker, Feb 04 2014

Length of period of the continued fraction expansion of a(n)*sqrt(2) is 1, the corresponding repeating value is A077444(n). - Ralf Stephan, Feb 20 2014

Positive values of x (or y) satisfying x^2 - 34xy + y^2 + 144 = 0. - Colin Barker, Mar 04 2014

REFERENCES

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

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.

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).

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

LINKS

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

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

A. Blondin-Massé, S. Brlek, S. Labbé, and M. Mendès France, Fibonacci snowflakes, Special Issue dedicated to Paulo Ribenboim, Annales des Sciences Mathématiques du Québec 35, No 2 (2011).

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

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

Daniel C. Fielder, Errata:Special integer sequences controlled by three parameters, Fibonacci Quarterly 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.

M. A. Gruber, Artemas Martin, A. H. Bell, J. H. Drummond, A. H. Holmes and H. C. Wilkes, Problem 47, Amer. Math. Monthly, 4 (1897), 25-28.

INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 403

Tanya Khovanova, Recursive Sequences

Ron Knott, Pythagorean Triples and Online Calculators

J.-C. Novelli, J.-Y. Thibon, Hopf Algebras of m-permutations,(m+1)-ary trees, and m-parking functions, arXiv preprint arXiv:1403.5962, 2014

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.

José L. Ramírez, Gustavo N. Rubiano, and Rodrigo de Castro, A Generalization of the Fibonacci Word Fractal and the Fibonacci Snowflake, arXiv preprint arXiv:1212.1368, 2012

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.

Eric Weisstein's World of Mathematics, NSW Number

Index entries for two-way infinite sequences

Index entries for sequences related to Chebyshev polynomials.

Index to sequences with linear recurrences with constant coefficients, signature (6,-1).

FORMULA

G.f.: x*(1-x)/(1-6*x+x^2).

a(n) = 6*a(n-1)-a(n-2) with a(1)=1, a(2)=5.

4*a(n) = A077445(n).

Can be extended backwards by a(-n+1) = a(n).

a(n) = sqrt((A002315(n)^2 +1)/2). [Inserted by N. J. A. Sloane, May 08 2000]

a(n+1) = S(n, 6)-S(n-1, 6), n>=0, 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+1) = T(2*n+1, sqrt(2))/sqrt(2),n>=0, with T(n, x) Chebyshev's polynomials of the first kind. [Offset corrected by Wolfdieter Lang, Mar 06 2012]

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, 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, 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, Jul 07 2003

From Charlie Marion, Jul 16 2003: (Start)

For n>=k>=0, a(n)^2 = a(n+k)*a(n-k) - A084703(k)^2; e.g., 169^2 = 5741*5 - 144.

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.

Sum_{k=0...n}((-1)^(n-k)*a(k)) = A079291(n+1); e.g., -1 + 5 - 29 + 169 = 144.

A001652(n) + A046090(n) - a(n) = A001542(n); e.g. 119 + 120 - 169 = 70.

(End)

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, 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, 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 Alberto Olivares, 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, 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, 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, Feb 15 2004

a(n)a(n+3) = 24 + a(n+1)a(n+2). - Ralf 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 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, Jun 04 2004

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

a(n) = sum{k=0..n, binomial(2n+1, 2k+1)2^k}. - Paul Barry, 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, 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, Mar 13 2005

a(n) = Pell(2n+1) = Pell(n)^2+Pell(n+1)^2. - Paul Barry, Jul 18 2005

a(n)*a(n+k) = A000129(k)^2+A000129(2n+k+1)^2; e.g., 29*5741=12^2+169^2. - Charlie Marion, 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, Aug 02 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, Jul 19 2005

a(n)=Jacobi_P(n,-1/2,1/2,3)/Jacobi_P(n,-1/2,1/2,1). - Paul Barry, 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, May 19 2006

a(n) = round(sqrt(A002315(n)^2/2)). - Lekraj Beedassy, Jul 15 2006

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

a(n+1) = 3*a(n)+(8*a(n)^2-4)^0.5, a(1)=1. - Richard Choulet, 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, 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, Oct 12 2007

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

From Charlie Marion, 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)

G.f.: sqrt(x)*tan(4*arctan(sqrt(x)))/4. - Johannes W. Meijer, Aug 01 2010

Given k=(sqrt(2)+1)^2=3+2*sqrt(2); and a(0)=1; then a(n)=a(n-1)*k-((k-1)/(k^n)). - Charles L. Hohn, Mar 06 2011

Given k=(sqrt(2)+1)^2=3+2*sqrt(2); and a(0)=1; then a(n)=(k^n)+(k^(-n))-a(n-1) (which = a(n)=A003499(n)-a(n-1)) - Charles L. Hohn, Apr 04 2011

Let T(n) be the n-th triangular number; then, for n>0,

  T(a(n))+A001109(n-1)=A046090(n)^2.  See also A046090. - Charlie Marion, Apr 25 2011

For k>0, a(n+2k-1) - a(n) = 4*A001109(n+k-1)*A002315(k-1); a(n+2k) - a(n) = 4*A001109(k)*A002315(n+k-1). - Charlie Marion, Jan 06 2012

a(k+j+1) = (A001541(k)*A001541(j)+A002315(k)*A002315(j))/2. - Charlie Marion, Jun 25 2012

a(n) = A143608(n-1)*A143608(n) + 1. - Charlie Marion, Dec 11 2012

G.f.: G(0)*(1-x)/(2-6*x), where G(k)= 1 + 1/(1 - x*(8*k-9)/( x*(8*k-1) - 3/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 12 2013

a(n+1) = 4*A001652(n) + 3*a(n) + 2 [Mohamed Bouhamida's 2009 (p,q)(r,s) comment above rewritten]. - Hermann Stamm-Wilbrandt, Jul 27 2014

a(n)^2 = A001652(n-1)^2 + (A001652(n-1)+1)^2. - Hermann Stamm-Wilbrandt, Aug 31 2014

MAPLE

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, Jul 26 2006

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

MATHEMATICA

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

LinearRecurrence[{6, -1}, {1, 5}, 40] (* Harvey P. Dale, Jul 12 2011 *)

PROG

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

(PARI) a(n)=([5, 2; 2, 1]^n)[1, 1] \\ Lambert Klasen (lambert.klasen(AT)gmx.de)

(Haskell)

a001653 n = a001653_list !! n

a001653_list = 1 : 5 : zipWith (-) (map (* 6) $ tail a001653_list) a001653_list

-- Reinhard Zumkeller, May 07 2013

(MAGMA) I:=[1, 5]; [n le 2 select I[n] else 6*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 22 2014

CROSSREFS

Other two sides are A001652, A046090.

Cf. A001519, A001109, A005054, A122074, A056220.

Row 6 of array A094954.

Row 1 of array A188647.

Cf. similar sequences listed in A238379.

Sequence in context: A182017 A141812 A227206 * A141814 A175883 A122370

Adjacent sequences:  A001650 A001651 A001652 * A001654 A001655 A001656

KEYWORD

nonn,easy,nice

AUTHOR

N. J. A. Sloane

EXTENSIONS

Additional comments from Wolfdieter Lang, Feb 10 2000

Better description from Harvey P. Dale, Jan 15 2002

Edited by N. J. A. Sloane, Nov 02 2002

STATUS

approved

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Last modified October 25 17:33 EDT 2014. Contains 248557 sequences.