cp's OEIS Frontend

Table starts

.1........3............7................17...................41

.1.......17..........119...............866.................6328

.1.......99.........2008.............46105..............1010078

.1......577........33873...........2460824............162430981

.1.....3363.......571358.........131347807..........26135748494

.1....19601......9637322........7010554450........4205264399891

.1...114243....162555929......374176705928......676612123702617

.1...665857...2741882383....19970973695923...108863121975329470

.1..3880899..46248186229..1065911268786396.17515379016294357354

.1.22619537.780082534460.56890876705125765

The left column, A(n,0), is A000129(n), Pell Numbers.

The top row, A(0,k), is A077957(k) plus an initial 0, which is the inverse binomial transform of A000129.

These may be considered initializing values, or results, depending the perspective taken, since there are several ways to generate the array. See Formula section for details.

The columns of the array hold all values, in sequential order, of numbers m such that 2m^2 + 2^k or 2m^2 - 2^k are squares, and their corresponding square roots in the next column, which then form the "next round" of m values for k+1.

For example A(n,0) are numbers such that 2m^2 +- 1 are squares, the integer square roots of each are in A(n,1), which are then numbers m such that 2m^2 +- 2 are squares, with those square roots in A(n,2), etc.

A(n, k)/A(n,k-2) = 2; A(n,k)/A(n,k-1) converges to sqrt(2) for large n.

A(n,k)/A(n-1,k) converges to 1 + sqrt(2) for large n.

The other columns of this array hold current OEIS sequences as follows:

A(n,1) = A001333(n); A(n,2) = A163271(n); A(n,3) = A002203(n);

Bisections of these column-oriented sequences also appear in the OEIS, corresponding to the even and odd rows of the array, which in turn correspond to the two different recursive square root equations in the formula section below.

Farey fraction denominators interleave columns 0 and 1, and the corresponding numerators interleave columns 1 and 2, for approximating sqrt(2). See A002965 and A119016, respectively.

The other rows of this array hold current OEIS sequences as follows:

A(1,k) = A016116(k); A(2,k) = A029744(k) less the initial 1;

A(3,k) = A070875(k); A(4,k) = A091523(k) less the initial 8.

The Pell Numbers (A000219) are the only initializing set of numbers where the two recursive square root equations (see below) produce exclusively integer values, for all iterations of k. For any other initial values only even iterations (at k = 2, 4, ...) produce integers.

The numbers in this array, especially the first three columns, are also integer square roots of these expressions: floor(m^2/2), floor(m^2/2 + 1), floor (m^2/2 - 1). See A227972 for specific arrangements and relationships. Also: ceiling(m^2/2), ceiling(m^2/2 + 1), ceiling (m^2/2 -1), m^2+1, m^2-1, m^2*(m^2-1)/2, m^2*(m^2-1)/2, in various different arrangements. Many of these involve: A000129(2n)/2 = A001109(n).

A001109 also appears when multiplying adjacent columns: A(n,k) * A(n,k+1) = (k+1) * A001109(n), for all k.

Lim_{n -> inf} a(n)/a(n-2) = 3 + 2*sqrt(2) = R1*R2. Lim_{k -> inf} a(2*k-1)/a(2*k) = (9 + 4*sqrt(2))/7 = R1 = A156649 (ratio #1). Lim_{k -> inf} a(2*k)/a(2*k-1) = (11 + 6*sqrt(2))/7 = R2 (ratio #2).

Also gives solutions > 3 to the equation x^2-4 = floor(x*r*floor(x/r)) where r=sqrt(2). - Benoit Cloitre, Feb 14 2004

From Paul Curtz, Dec 15 2012: (Start)

a(n-1) and A006452(n) are companions. Like A000129 and A001333.

Reduced mod 10 this is a sequence of period 12: 3, 5, 3, 7, 5, 7, 7, 5, 7, 3, 5, 3.

(End)

The Pisano periods (periods of the sequence reducing a(n) modulo m) for m>=1 are 1, 1, 8, 4, 12, 8, 6, 4, 24, 12, 24, 8, 28, ... R. J. Mathar, Dec 15 2012

Positive values of x (or y) satisfying x^2 - 6xy + y^2 + 56 = 0. - Colin Barker, Feb 08 2014

From Wolfdieter Lang, Feb 05 2015: (Start)

a(n+1) gives for n >= 0 all positive x solutions of the (generalized) Pell equation x^2 - 2*y^2 = +7.

The corresponding y solutions are given in A077442(n), n >= 0. The, e.g., the Nagell reference for finding all solutions.

Because the primitive Pythagorean triangle (3,4,5) is the only one with the sum of legs equal to 7 all positive solutions (x(n),y(n)) = (a(n+1),A077442(n)) of the Pell equation x^2 - 2*y^2 = +7 satisfy x(n) - y(n) < y(n) if n >= 1; only the first solution (x(0),y(0)) = (3,2) satisfies 3-1 > 1. Proof: Primitive Pythagorean triangles are characterized by the positive integer pairs [u,v] with u+v odd, gcd(u,v) = 1 and u > v. See the Niven et al. reference, Theorem 5.5, p. 232. The leg sum is L = (u+v)^2 - 2*v^2. With L = 7, x = u+v and y = v, every solution (x(n),y(n)) with x(n)-y(n) = u(n) > v(n) = y(n) will correspond to a primitive Pythagorean triangle. Note that because of gcd(x,y) = 1 also gcd(u,v) = 1. But there is only one such triangle with L=7, namely the one with [u(0),v(0)] = [2,1]. All other solutions with n >= 1 must therefore satisfy x(n)-y(n) < y(n). (End)

For n > 0, a(n+1) is the n-th almost Lucas-cobalancing number of first type (see Tekcan and Erdem). - Stefano Spezia, Nov 25 2022

a(1)=1, a(n) is the smallest integer > a(n-1) such that sqrt(2)*a(n) is closer and > to an integer than sqrt(2)*a(n-1) (i.e., a(n) is the smallest integer > a(n-1) such that frac(sqrt(2)*a(n)) < frac(sqrt(2)*a(n-1))).

n such that floor(sqrt(2)*n^2) = n*floor(sqrt(2)*n).

The sequence 1,1,3,5,17,... has g.f. (1+x-3x^2-x^3)/(1-6x^2+x^4); a(n) = Sum_{k=0..floor(n/2)} C(n,2k)*2^(n-k-floor((n+1)/2)); a(n) = -(sqrt(2)-1)^n*((sqrt(2)/8-1/4)*(-1)^n - sqrt(2)/8 - 1/4) - (sqrt(2)+1)^n*((sqrt(2)/8-1/4)*(-1)^n - sqrt(2)/8 - 1/4); a(2n) = A001541(n) = A001333(2n); a(2n+1) = A001653(n) = A000129(2n+1). - Paul Barry, Jan 22 2005

The lower principal and intermediate convergents to 2^(1/2), beginning with 1/1, 4/3, 7/5, 24/17, 41/29, form a strictly increasing sequence; essentially, numerators=A143608 and denominators=A079496. - Clark Kimberling, Aug 27 2008

From Richard Choulet, May 09 2010: (Start)

This sequence is a particular case of the following situation: a(0)=1, a(1)=a, a(2)=b with the recurrence relation a(n+3)=(a(n+2)*a(n+1)+q)/a(n) where q is given in Z to have Q=(a*b^2+q*b+a+q)/(a*b) itself in Z.

The g.f is f: f(z)=(1+a*z+(b-Q)*z^2+(a*b+q-a*Q)*z^3)/(1-Q*z^2+z^4); so we have the linear recurrence: a(n+4)=Q*a(n+2)-a(n).

The general form of a(n) is given by:

a(2*m)=sum((-1)^p*binomial(m-p,p)*Q^(m-2*p),p=0..floor(m/2))+(b-Q)*sum((-1)^p*binomial(m-1-p,p)*Q^(m-1-2*p),p=0..floor((m-1)/2)) and

a(2*m+1)=a*sum((-1)^p*binomial(m-p,p)*Q^(m-2*p),p=0..floor(m/2))+(a*b+q-a*Q)*sum((-1)^p*binomial(m-1-p,p)*Q^(m-1-2*p),p=0..floor((m-1)/2)). (End)

The integer square roots of floor(n^2/2 + 1) or (A007590 + 1). - Richard R. Forberg, Aug 01 2013

With the exception of 0, a subsequence of A075114. - R. J. Mathar, Dec 15 2008

Consequently, A014105(k) is a square if and only if k = a(n). - Bruno Berselli, Oct 14 2011

From M. F. Hasler, Jan 17 2012: (Start)

Bisection of A079291. The squares 2*k+1 are given in A055792.

A204576 is this sequence written in binary. (End)

a(n+1), n >= 0, is the perimeter squared (x(n) + y(n) + z(n))^2 of the ordered primitive Pythagorean triple (x(n), y(n) = x(n) + 1, z(n)). The first two terms are (x(0)=0, y(0)=1, z(0)=1), a(1) = 2^2, and (x(1)=3, y(1)=4, z(1)=5), a(2) = 12^2. - George F. Johnson, Nov 02 2012

Inverse binomial transform of A077957.

The inverse of the g.f. is 3-x-2/(1+x) which generates 1, 1, -2, +2, -2, +2, ... (-2, +2 periodically continued). - Gary W. Adamson, Jan 10 2011

Pisano period lengths: 1, 1, 8, 4, 12, 8, 6, 4, 24, 12, 24, 8, 28, 6, 24, 8, 16, 24, 40, 12, ... - R. J. Mathar, Aug 10 2012

a(n) is the rational part of the Q(sqrt(2)) integer (sqrt(2) - 1)^n = a(n) + A077985(n-1)*sqrt(2), with A077985(-1) = 0. - Wolfdieter Lang, Dec 07 2014

3^n*a(n) = A251732(n) gives the rational part of the integer in Q(sqrt(2)) giving the length of a variant of Lévy's C-curve at iteration step n. - Wolfdieter Lang, Dec 07 2014

Define u(0) = 1/0, u(1) = -1/1, and u(n) = -(8 + 3*u(n-1)*u(n-2))/(3*u(n-1) + 2*u(n-2)) for n>1. Then u(n) = a(n)/A000219(n). - Michael Somos, Apr 19 2022

From Charlie Marion, Jan 07 2009: (Start)

In general, denominators, a(k,n) and numerators, b(k,n), of continued

fraction convergents to sqrt((k+1)/k) may be found as follows:

a(k,0) = 1, a(k,1) = 2k; for n>0, a(k,2n) = 2*a(k,2n-1)+a(k,2n-2)

and a(k,2n+1)=(2k)*a(k,2n)+a(k,2n-1);

b(k,0) = 1, b(k,1) = 2k+1; for n>0, b(k,2n) = 2*b(k,2n-1)+b(k,2n-2)

and b(k,2n+1)=(2k)*b(k,2n)+b(k,2n-1).

For example, the convergents to sqrt(3/2) start 1/1, 5/4, 11/9,

49/40, 109/89.

In general, if a(k,n) and b(k,n) are the denominators and numerators,

respectively, of continued fraction convergents to sqrt((k+1)/k)

as defined above, then

k*a(k,2n)^2-a(k,2n-1)*a(k,2n+1)=k=k*a(k,2n-2)*a(k,2n)-a(k,2n-1)^2 and

b(k,2n-1)*b(k,2n+1)-k*b(k,2n)^2=k+1=b(k,2n-1)^2-k*b(k,2n-2)*b(k,2n);

for example, if k=2 and n=3, then b(2,n)=a(n) and

2*a(2,6)^2-a(2,5)*a(2,7)=2*881^2-396*3920=2;

2*a(2,4)*a(2,6)-a(2,5)^2=2*89*881-396^2=2;

b(2,5)*b(2,7)-2*b(2,6)^2=485*4801-2*1079^2=3;

b(2,5)^2-2*b(2,4)*b(2,6)=485^2-2*109*1079=3.

Cf. A000129, A001333, A142239, A153315-A153318. (End)

Numbers n such that 2*triangular(n) + 1 is a triangular number. Equivalently, numbers n such that n^2 + n + 1 is a triangular number. - Alex Ratushnyak, Apr 18 2013

For n > 0, a(n) is the n-th almost cobalancing number of first type (see Tekcan and Erdem). - Stefano Spezia, Nov 25 2022

Table starts

...2....3......5.......8........13.........21...........34............55

...4....7.....19......40........97........217..........508..........1159

...8...17.....77.....216.......809.......2529.........8832.........28793

..16...41....313....1152......6737......28977.......152048........699833

..32...99...1277....6160.....56549.....333517......2644336......17124415

..64..239...5215...32928....475809....3837761.....46125216.....419022831

.128..577..21305..176032...4008817...44171841....806190208...10258304689

.256.1393..87049..941056..33795201..508425617..14105294112..251170142257

.512.3363.355685.5030848.284980061.5852202757.246929287360.6150224353031

The grid has m rows and n columns.