cp's OEIS Frontend

From Hieronymus Fischer, Jan 02 2009: (Start)

Set c:=1+sqrt(2). Then the fractional part of c^n equals 1/c^n, if n odd. For even n, the fractional part of c^n is equal to 1-(1/c^n).

c:=1+sqrt(2) satisfies c-c^(-1)=floor(c)=2, hence c^n + (-c)^(-n) = round(c^n) for n>0, which follows from the general formula of A001622.

1/c = sqrt(2)-1.

See A001622 for a general formula concerning the fractional parts of powers of numbers x>1, which satisfy x-x^(-1)=floor(x).

Other examples of constants x satisfying the relation x-x^(-1)=floor(x) include A001622 (the golden ratio: where floor(x)=1) and A098316 (the "bronze" ratio: where floor(x)=3). (End)

In terms of continued fractions the constant c can be described by c=[2;2,2,2,...]. - Hieronymus Fischer, Oct 20 2010

Side length of smallest square containing five circles of diameter 1. - Charles R Greathouse IV, Apr 05 2011

Largest radius of four circles tangent to a circle of radius 1. - Charles R Greathouse IV, Jan 14 2013

An analog of Fermat theorem: for prime p, round(c^p) == 2 (mod p). - Vladimir Shevelev, Mar 02 2013

n*(1+sqrt(2)) is the perimeter of a 45-45-90 triangle with hypotenuse n. - Wesley Ivan Hurt, Apr 09 2016

This algebraic integer of degree 2, with minimal polynomial x^2 - 2*x - 1, is also the length ratio diagonal/side of the second largest diagonal in the regular octagon (not counting the side). The other two diagonal/side ratios are A179260 and A121601. - Wolfdieter Lang, Oct 28 2020

c^n = A001333(n) + A000129(n) * sqrt(2). - Gary W. Adamson, Apr 26 2023

c^n = c * A000129(n) + A000129(n-1), where c = 1 + sqrt(2). - Gary W. Adamson, Aug 30 2023

This is the case n=8 of the ratio Gamma(1/n)*Gamma((n-1)/n)/(Gamma(2/n)*Gamma((n-2)/n)). - Bruno Berselli, Dec 13 2012

An algebraic integer of degree 4: largest root of x^4 - 4x^2 + 2. - Charles R Greathouse IV, Nov 05 2014

This number is also the length ratio of the shortest diagonal (not counting the side) of the octagon and the side. This ratio is A121601 for the longest diagonal. - Wolfdieter Lang, May 11 2017 [corrected Oct 28 2020]

From Wolfdieter Lang, Apr 29 2018: (Start)

This constant appears in a historic problem posed by Adriaan van Roomen (Adrianus Romanus) in his Ideae mathematicae from 1593, solved by Viète. See the Havil reference, problem 3, pp. 69-74. See also the comments in A302711 with the Romanus link and his Exemplum tertium.

This problem is equivalent to R(45, 2*sin(Pi/120)) = 2*sin(3*Pi/8) with a special case of monic Chebyshev polynomials of the first kind, named R, given in A127672. For the constant 2*sin(Pi/120) see A302715. (End)

1 + csc(Pi/9) is the radius of the smallest circle into which 11 unit circles can be packed ("r=3.923+ Proved by Melissen in 1994.", according to the Friedman link, which has a diagram). csc(Pi/9) [=1/A019829] is the distance between the center of the larger circle and the center of each unit circle that touches the larger circle.

The 1st electrum rectangle is introduced here as a rectangle whose length L and width W satisfy L/W=(1+sqrt(3))/2. The name of this shape refers to the alloy of gold and silver known as electrum, in view of the existing names "golden rectangle" and "silver rectangle" and these continued fractions:

golden ratio: L/W=[1,1,1,1,1,1,1,1,1,1,1,...]

silver ratio: L/W=[2,2,2,2,2,2,2,2,2,2,2,...]

1st electrum ratio: L/W=[1,2,1,2,1,2,1,2,...]

2nd electrum ratio: L/W=[2,1,2,1,2,1,2,1,...].

Recall that removal of 1 square from a golden rectangle leaves a golden rectangle, and that removal of 2 squares from a silver rectangle leaves a silver rectangle. Removal of a square from a 1st electrum rectangle leaves a silver rectangle; removal of 2 squares from a 2nd electrum rectangle leaves a golden rectangle.

The 2nd electrum rectangle is introduced here as a rectangle whose length L and width W satisfy L/W=1+sqrt(3). The name of this shape refers to the alloy of gold and silver known as electrum, in view of the existing names "golden rectangle" and "silver rectangle" and these continued fractions:

golden ratio: L/W=[1,1,1,1,1,1,1,1,1,1,1,...]

silver ratio: L/W=[2,2,2,2,2,2,2,2,2,2,2,...]

1st electrum ratio: L/W=[1,2,1,2,1,2,1,2,...]

2nd electrum ratio: L/W=[2,1,2,1,2,1,2,1,...].

Recall that removal of 1 square from a golden rectangle leaves a golden rectangle, and that removal of 2 squares from a silver rectangle leaves a silver rectangle. Removal of a square from a 1st electrum rectangle leaves a silver rectangle; removal of 2 squares from a 2nd electrum rectangle leaves a golden rectangle.

Corresponding k values are in A121604. For these n, the centers of k unit circles can form a regular k-gon with sides of length 2 centered at the center of the larger circle. From the diagrams in the link it appears likely that 13,18,19 are the next three terms.

From the diagrams in the link it appears likely that 10,12,12 are the next three terms.

The length of row n is given in A338431(n), for n >= 1.

The length of the sublists t(n, k) of the power basis coefficients of DSR(n, k), for k = 1, 2, ..., floor(n/2), is 1 if n = 1, for n >= 2 it is k except for n = n(j) = A111774(j) for which the final A219839(n) sublists have fewer than k members.

Trailing vanishing coefficients of the delta(n) = A055034(n) power base elements <1 = rho(n)^0, rho(n)^1, ..., rho(n)^{delta(n)-1}> are not recorded. The coefficients of the minimal polynomial C(n, x) of rho(n) = 2*cos(Pi/n) of degree delta(n) are given in A187360. C(n, rho(n)) = 0 is used to eliminate all powers of rho with exponent >= delta(n).

The length ratios DSR(n, k) := diagonal(n, k)/side(n) of regular n-gons, for n >= 2, and k = 1, 2, ..., floor(n/2) (distinct diagonals, starting with the side for k = 1, in increasing order) are given by DSR(n, k) = S(k-1, rho(n)), with the Chebyshev S polynomials (A049310). See the W. Lang link.

For n = 2, the degenerate case, diagonal/side = side/side = 1 for k = 1. For n = 1 (a point) diagonal/side is undetermined, and T(1, 1) is set to 1.

For the power basis sublists t(n, k), for k = 1, 2, ..., delta(n), only the k coefficients of S(k-1, x) are present (trailing vanishing coefficients are not recorded). For k = delta(n)+1, ..., floor(n/2) less than k coefficients appear due to elimination via C(n, rho(n)) = 0. E.g., for n = 6 with delta(6) = 2 the only coefficient for k = 3 is 2 (coefficient of rho^0). This appears for n = n(j) = A111774(j), because then floor(n/2) - delta(n) = A219839(n) > 0.

Because A219839(n) = 0 means that n is from A174090, i.e., a prime or a power of 2 (complement of A111774), these rows n have all the sublists t(n, k) with the k coefficients of S(k-1, x), hence they are identical (but the basis differs). See especially the table for the pairs of consecutive numbers n with identical coefficients, like (2, 3), (4, 5), (16, 17), (256, 267), (65536, 65537), ?... (cf. Fermat primes A019434).