

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...


2



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
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OFFSET

1,1


COMMENTS

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 nth term = the (positive) n4th term of A229525.


LINKS

Colin Barker, Table of n, a(n) for n = 1..1000
Russell Walsmith, CLChemy III: HyperQuadratics


FORMULA

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(n2)3*a(n4)+a(n6). G.f.: x*(x^5+x^44*x^314*x^2+x+5) / ((x1)^3*(x+1)^3).  Colin Barker, Nov 02 2014
a(n) = (5+3*(1)^n)*(42*n+n^2)/8.  Colin Barker, Nov 03 2014


EXAMPLE

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


PROG

(PARI) Vec(x*(x^5+x^44*x^314*x^2+x+5)/((x1)^3*(x+1)^3) + O(x^100)) \\ Colin Barker, Nov 02 2014


CROSSREFS

The a coefficients are A168077, b coefficients are A171621, the sum of a, b and c coefficients is A229525.
Sequence in context: A140210 A010130 A206773 * A204007 A242404 A145295
Adjacent sequences: A229523 A229524 A229525 * A229527 A229528 A229529


KEYWORD

sign,easy


AUTHOR

Russell Walsmith, Sep 27 2013


EXTENSIONS

More terms from Colin Barker, Nov 02 2014


STATUS

approved



