A229525 -id:A229525 - cp's OEIS frontend

A229526 The c coefficients of the transform ax^2 + (4a/k - b)x + 4a/k^2 + 2b/k + c= 0 for a,b,c = 1,-1,-1, k = 1,2,3...

5, 1, 1, -1, -11, -5, -31, -11, -59, -19, -95, -29, -139, -41, -191, -55, -251, -71, -319, -89, -395, -109, -479, -131, -571, -155, -671, -181, -779, -209, -895, -239, -1019, -271, -1151, -305, -1291, -341, -1439, -379, -1595, -419, -1759, -461, -1931, -505

Offset: 1

Russell Walsmith, Sep 27 2013

The positive/negative roots of ax^2 + bx + c = 0 combine with the negative/positive roots of (ck^2 - bk + c)x^2 +(2ck - b)x + c = 0 to define a point on the hyperbola kxy + x + y = 0. To shift such points (roots) to the hyperbola’s other line, put the coefficients of these equations into the formula ax^2 + (4a/k - b)x + 4a/k^2 + 2b/k + c = 0. Let a,b,c = 1,-1,-1 and k = 1,2,3... Then the coefficients given by this last equation are the sequence 1,5,5; 1,3,1; 1,7/3,1/9... Clearing fractions, the c coefficients are the sequence above.

The n-th term = the (positive) n-4th term of A229525.

			For k = 5, the coefficients are 1, 9/5, -11/25. Clearing fractions gives 25, 45, -11 and -11 = a[5].

Colin Barker, Table of n, a(n) for n = 1..1000
Russell Walsmith, CL-Chemy III: Hyper-Quadratics

The a coefficients are A168077, b coefficients are A171621, the sum of a, b and c coefficients is A229525.

Vec(-x*(x^5+x^4-4*x^3-14*x^2+x+5)/((x-1)^3*(x+1)^3) + O(x^100)) \\ Colin Barker, Nov 02 2014

ax^2 + (4a/k - b)x + 4a/k^2 + 2b/k + c; a,b,c = 1,-1,-1, k = 1,2,3..n.
a(n) = 3*a(n-2)-3*a(n-4)+a(n-6). G.f.: -x*(x^5+x^4-4*x^3-14*x^2+x+5) / ((x-1)^3*(x+1)^3). - Colin Barker, Nov 02 2014
a(n) = (-5+3*(-1)^n)*(-4-2*n+n^2)/8. - Colin Barker, Nov 03 2014

More terms from Colin Barker, Nov 02 2014

A363347 Denominator of the continued fraction 1/(2-3/(3-4/(4-5/(...(n-1)-n/(-4))))).

Original entry on oeis.org

11, 5, 31, 11, 59, 19, 19, 29, 139, 41, 191, 1, 251, 71, 29, 89, 79, 109, 479, 131, 571, 31, 61, 181, 41, 1, 179, 239, 1019, 271, 1151, 61, 1291, 1, 1439, 379, 1, 419, 1759, 461, 1931, 101, 2111, 1, 1, 599, 499, 59, 2699, 701, 71, 151, 101, 811

Offset: 3

Mohammed Bouras, May 28 2023

nonn

Conjecture 1: Every term of this sequence is either a prime or 1.
Conjecture 2: The sequence contains all prime numbers which end with a 1 or 9.
Conjecture 3: Except for 5, the primes all appear exactly twice.
Conjecture: The sequence of record values is A028877. - Bill McEachen, May 20 2024

			For n=3, 1/(2 - 3/(-4)) = 4/11, so a(3) = 11.
For n=4, 1/(2 - 3/(3 - 4/(-4))) = 4/5, so a(4) = 5.
For n=5, 1/(2 - 3/(3 - 4/(4 -5/(-4)))) = 47/31, so a(5) = 31.
a(3) = a(6) = 3 + 6 + 2 = 11.
a(5) = a(24) = 5 + 24 + 2 = 31.
a(7) = a(50) = 7 + 50 + 2 = 59.

Mohammed Bouras, The Distribution Of Prime Numbers And Continued Fractions, (ppt) (2022)

Cf. A006530, A051403, A229525, A356247, A028877.

a(n) = (n^2 + 2*n - 4)/gcd(n^2 + 2*n - 4, 4*A051403(n-3) + n*A051403(n-4)).
a(n) = gpf(n^2 + 2*n - 4) if gpf(n^2 + 2*n - 4) > n, otherwise a(n) = 1 (where gpf(n) denotes the greatest prime factor of n).
If n != m and a(n) = a(m) != 1, then we have:
a(n) = n + m + 2.
a(n) = gcd(n^2 + 2*n - 4, m^2 + 2*m - 4).

cp's OEIS Frontend

A229526 The c coefficients of the transform ax^2 + (4a/k - b)x + 4a/k^2 + 2b/k + c= 0 for a,b,c = 1,-1,-1, k = 1,2,3...

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

Formula

Extensions