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The continued square root map applied to a sequence (x,y,z,...) is CSR(x,y,z,...) = sqrt(x + sqrt(y + sqrt(z + ...))); this is well defined if the logarithm of the terms is O(2^n).

Similar to A257582, but converging to e.

Also decimal expansion of the positive root of (x+1)^n - x^(2n). (x+1)^n - x^(2n) = 0 has only two real roots x1 = -(sqrt(5)-1)/2 and x2 = (sqrt(5)+1)/2 for all n > 0. - Cino Hilliard, May 27 2004

The golden ratio phi is the most irrational among irrational numbers; its successive continued fraction convergents F(n+1)/F(n) are the slowest to approximate to its actual value (I. Stewart, in "Nature's Numbers", Basic Books, 1997). - Lekraj Beedassy, Jan 21 2005

Let t=golden ratio. The lesser sqrt(5)-contraction rectangle has shape t-1, and the greater sqrt(5)-contraction rectangle has shape t. For definitions of shape and contraction rectangles, see A188739. - Clark Kimberling, Apr 16 2011

The golden ratio (often denoted by phi or tau) is the shape (i.e., length/width) of the golden rectangle, which has the special property that removal of a square from one end leaves a rectangle of the same shape as the original rectangle. Analogously, removals of certain isosceles triangles characterize side-golden and angle-golden triangles. Repeated removals in these configurations result in infinite partitions of golden rectangles and triangles into squares or isosceles triangles so as to match the continued fraction, [1,1,1,1,1,...] of tau. For the special shape of rectangle which partitions into golden rectangles so as to match the continued fraction [tau, tau, tau, ...], see A188635. For other rectangular shapes which depend on tau, see A189970, A190177, A190179, A180182. For triangular shapes which depend on tau, see A152149 and A188594; for tetrahedral, see A178988. - Clark Kimberling, May 06 2011

Given a pentagon ABCDE, 1/(phi)^2 <= (A*C^2 + C*E^2 + E*B^2 + B*D^2 + D*A^2) / (A*B^2 + B*C^2 + C*D^2 + D*E^2 + E*A^2) <= (phi)^2. - Seiichi Kirikami, Aug 18 2011

If a triangle has sides whose lengths form a geometric progression in the ratio of 1:r:r^2 then the triangle inequality condition requires that r be in the range 1/phi < r < phi. - Frank M Jackson, Oct 12 2011

The graphs of x-y=1 and x*y=1 meet at (tau,1/tau). - Clark Kimberling, Oct 19 2011

Also decimal expansion of the first root of x^sqrt(x+1) = sqrt(x+1)^x. - Michel Lagneau, Dec 02 2011

Also decimal expansion of the root of (1/x)^(1/sqrt(x+1)) = (1/sqrt(x+1))^(1/x). - Michel Lagneau, Apr 17 2012

This is the case n=5 of (Gamma(1/n)/Gamma(3/n))*(Gamma((n-1)/n)/Gamma((n-3)/n)): (1+sqrt(5))/2 = (Gamma(1/5)/Gamma(3/5))*(Gamma(4/5)/Gamma(2/5)). - Bruno Berselli, Dec 14 2012

Also decimal expansion of the only number x>1 such that (x^x)^(x^x) = (x^(x^x))^x = x^((x^x)^x). - Jaroslav Krizek, Feb 01 2014

For n >= 1, round(phi^prime(n)) == 1 (mod prime(n)) and, for n >= 3, round(phi^prime(n)) == 1 (mod 2*prime(n)). - Vladimir Shevelev, Mar 21 2014

The continuous radical sqrt(1+sqrt(1+sqrt(1+...))) tends to phi. - Giovanni Zedda, Jun 22 2019

Equals sqrt(2+sqrt(2-sqrt(2+sqrt(2-...)))). - Diego Rattaggi, Apr 17 2021

Given any complex p such that real(p) > -1, phi is the only real solution of the equation z^p+z^(p+1)=z^(p+2), and the only attractor of the complex mapping z->M(z,p), where M(z,p)=(z^p+z^(p+1))^(1/(p+2)), convergent from any complex plane point. - Stanislav Sykora, Oct 14 2021

The only positive number such that its decimal part, its integral part and the number itself (x-[x], [x] and x) form a geometric progression is phi, with respectively (phi -1, 1, phi) and a ratio = phi. This is the answer to the 4th problem of the 7th Canadian Mathematical Olympiad in 1975 (see IMO link and Doob reference). - Bernard Schott, Dec 08 2021

The golden ratio is the unique number x such that f(n*x)*c(n/x) - f(n/x)*c(n*x) = n for all n >= 1, where f = floor and c = ceiling. - Clark Kimberling, Jan 04 2022

In The Second Scientific American Book Of Mathematical Puzzles and Diversions, Martin Gardner wrote that, by 1910, Mark Barr (1871-1950) gave phi as a symbol for the golden ratio. - Bernard Schott, May 01 2022

Phi is the length of the equal legs of an isosceles triangle with side c = phi^2, and internal angles (A,B) = 36 degrees, C = 108 degrees. - Gary W. Adamson, Jun 20 2022

The positive solution to x^2 - x - 1 = 0. - Michal Paulovic, Jan 16 2023

The minimal polynomial of phi^n, for nonvanishing integer n, is P(n, x) = x^2 - L(n)*x + (-1)^n, with the Lucas numbers L = A000032, extended to negative arguments with L(n) = (-1)^n*L(n). P(0, x) = (x - 1)^2 is not minimal. - Wolfdieter Lang, Feb 20 2025

This is the largest real zero x of (x^4 + x^2 + 1)^2 = 2*(x^8 + x^4 + 1). - Thomas Ordowski, May 14 2025

a(n+1) is the least k > a(n) + 1 such that A000217(a(n)) + A000217(k) is a square. - David Wasserman, Jun 30 2005

Values of Fibonacci polynomial n^2 - n - 1 for n = 2, 3, 4, 5, ... - Artur Jasinski, Nov 19 2006

A127701 * [1, 2, 3, ...]. - Gary W. Adamson, Jan 24 2007

Row sums of triangle A135223. - Gary W. Adamson, Nov 23 2007

Equals row sums of triangle A143596. - Gary W. Adamson, Aug 26 2008

a(n-1) gives the number of n X k rectangles on an n X n chessboard (for k = 1, 2, 3, ..., n). - Aaron Dunigan AtLee, Feb 13 2009

sqrt(a(0) + sqrt(a(1) + sqrt(a(2) + sqrt(a(3) + ...)))) = sqrt(1 + sqrt(5 + sqrt(11 + sqrt(19 + ...)))) = 2. - Miklos Kristof, Dec 24 2009

When n + 1 is prime, a(n) gives the number of irreducible representations of any nonabelian group of order (n+1)^3. - Andrew Rupinski, Mar 17 2010

a(n) = A176271(n+1, n+1). - Reinhard Zumkeller, Apr 13 2010

The product of any 4 consecutive integers plus 1 is a square (see A062938); the terms of this sequence are the square roots. - Harvey P. Dale, Oct 19 2011

Or numbers not expressed in the form m + floor(sqrt(m)) with integer m. - Vladimir Shevelev, Apr 09 2012

Left edge of the triangle in A214604: a(n) = A214604(n+1,1). - Reinhard Zumkeller, Jul 25 2012

Another expression involving phi = (1 + sqrt(5))/2 is a(n) = (n + phi)(n + 1 - phi). Therefore the numbers in this sequence, even if they are prime in Z, are not prime in Z[phi]. - Alonso del Arte, Aug 03 2013

a(n-1) = n*(n+1) - 1, n>=0, with a(-1) = -1, gives the values for a*c of indefinite binary quadratic forms [a, b, c] of discriminant D = 5 for b = 2*n+1. In general D = b^2 - 4ac > 0 and the form [a, b, c] is a*x^2 + b*x*y + c*y^2. - Wolfdieter Lang, Aug 15 2013

a(n) has prime factors given by A038872. - Richard R. Forberg, Dec 10 2014

A253607(a(n)) = -1. - Reinhard Zumkeller, Jan 05 2015

An example of a quadratic sequence for which the continued square root map (see A257574) produces the number 2. There are infinitely many sequences with this property - another example is A028387. See Popular Computing link. - N. J. A. Sloane, May 03 2015

Left edge of the triangle in A260910: a(n) = A260910(n+2,1). - Reinhard Zumkeller, Aug 04 2015

Numbers m such that 4m+5 is a square. - Bruce J. Nicholson, Jul 19 2017

The numbers represented as 131 in base n: 131_4 = 29, 131_5 = 41, ... . If 'digits' larger than the base are allowed then 131_2 = 11 and 131_1 = 5 also. - Ron Knott, Nov 14 2017

From Klaus Purath, Mar 18 2019: (Start)

Let m be a(n) or a prime factor of a(n). Then, except for 1 and 5, there are, if m is a prime, exactly two squares y^2 such that the difference y^2 - m contains exactly one pair of factors {x,z} such that the following applies: x*z = y^2 - m, x + y = z with

x < y, where {x,y,z} are relatively prime numbers. {x,y,z} are the initial values of a sequence of the Fibonacci type. Thus each a(n) > 5, if it is a prime, and each prime factor p > 5 of an a(n) can be assigned to exactly two sequences of the Fibonacci type. a(0) = 1 belongs to the original Fibonacci sequence and a(1) = 5 to the Lucas sequence.

But also the reverse assignment applies. From any sequence (f(i)) of the Fibonacci type we get from its 3 initial values by f(i)^2 - f(i-1)*f(i+1) with f(i-1) < f(i) a term a(n) or a prime factor p of a(n). This relation is also valid for any i. In this case we get the absolute value |a(n)| or |p|. (End)

a(n-1) = 2*T(n) - 1, for n>=1, with T = A000217, is a proper subsequence of A089270, and the terms are 0,-1,+1 (mod 5). - Wolfdieter Lang, Jul 05 2019

a(n+1) is the number of wedged n-dimensional spheres in the homotopy of the neighborhood complex of Kneser graph KG_{2,n}. Here, KG_{2,n} is a graph whose vertex set is the collection of subsets of cardinality 2 of set {1,2,...,n+3,n+4} and two vertices are adjacent if and only if they are disjoint. - Anurag Singh, Mar 22 2021

Also the number of squares between (n+2)^2 and (n+2)^4. - Karl-Heinz Hofmann, Dec 07 2021

(x, y, z) = (A001105(n+1), -a(n-1), -a(n)) are solutions of the Diophantine equation x^3 + 4*y^3 + 4*z^3 = 8. - XU Pingya, Apr 25 2022

The least significant digit of terms of this sequence cycles through 1, 5, 1, 9, 9. - Torlach Rush, Jun 05 2024

Herschfeld calls this the Kasner number, after Edward Kasner. - Charles R Greathouse IV, Dec 30 2008

No closed-form expression is known for this constant.

"It was discovered by T. Vijayaraghavan that the infinite radical sqrt( a_1 + sqrt( a_2 + sqrt ( a_3 + sqrt( a_4 + ...)))), where a_n >= 0, will converge to a limit if and only if the limit of (log a_n)/2^n exists" - Clawson, p. 229. Obviously if a_n = n, the limit of (log a_n) / 2^n as n -> infinity is 0.

The continued fraction is A072450.

Clawson misstates Vijayaraghavan's theorem. Vijayaraghavan proved that for a_n > 0, the infinite radical sqrt(a_1 + sqrt(a_2 + sqrt(a_3 + ...))) converges if and only if limsup (log a_n)/2^n < infinity. (For example, suppose a_n = 1 if n is odd, and a_n = e^2^n if n is even. Then (log a_n)/2^n = 0, 1, 0, 1, 0, 1, ... for n >= 1, so the limit does not exist. However, limsup (log a_n)/2^n = 1 and the infinite radical converges.) - Jonathan Sondow, Mar 25 2014

sqrt(1 + sqrt(2 + sqrt(3 + sqrt(4 + ...)))) = 1.75793275661800...

"It was discovered by T. Vijayaraghavan that the infinite radical, sqrt( a_1 + sqrt( a_2 + sqrt ( a_3 + sqrt( a_4 + ... )))) where a_n >= 0, will converge to a limit if and only if the limit of log(a_n)/2^n exists." [Clawson, 229; cf. A072449].

We know the asymptotic limit of primes and hence that the Prime Nested Radical converges.

Clawson misstates Vijayaraghavan's theorem. Vijayaraghavan proved that for a_n > 0, the infinite radical sqrt(a_1 + sqrt(a_2 + sqrt(a_3 + ...))) converges if and only if limsup (log a_n)/2^n < infinity. (For example, suppose a_n = 1 if n is odd, and a_n = e^2^n if n is even. Then (log a_n)/2^n = 0, 1, 0, 1, 0, 1, ... for n >= 1, so the limit does not exist. However, limsup (log a_n)/2^n = 1 and the infinite radical converges.) - Jonathan Sondow, Mar 25 2014

The continued fraction expression of this is A105818. "It was discovered by T. Vijayaraghavan that the infinite radical, sqrt( a_1 + sqrt( a_2 + sqrt ( a_3 + sqrt( a_4 + ... where a_n => 0, will converge to a limit if and only if the limit of (ln a_n)/2^n exists." [Clawson, 229; Sloane]. We know the asymptotic limit of Fibonacci numbers is Phi^n (Binet expansion) and that Phi^n < 2^n and hence that the Fibonacci Nested Radical converges.

Clawson misstates Vijayaraghavan's theorem. Vijayaraghavan proved that for a_n > 0, the infinite radical sqrt(a_1 + sqrt(a_2 + sqrt(a_3 + ...))) converges if and only if limsup (log a_n)/2^n < infinity. (For example, suppose a_n = 1 if n is odd, and a_n = e^2^n if n is even. Then (log a_n)/2^n = 0, 1, 0, 1, 0, 1, ... for n >= 1, so the limit does not exist. However, limsup (log a_n)/2^n = 1 and the infinite radical converges.) - Jonathan Sondow, Mar 25 2014

An example of a quadratic sequence for which the continued square root map (see A257574) produces the number 2. There are infinitely many sequences with this property - another example is A028387.