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A106330
Numbers k such that k^2 = 24*j^2 + 25.
1
5, 7, 11, 25, 59, 103, 245, 583, 1019, 2425, 5771, 10087, 24005, 57127, 99851, 237625, 565499, 988423, 2352245, 5597863, 9784379, 23284825, 55413131, 96855367, 230496005, 548533447, 958769291, 2281675225, 5429921339, 9490837543, 22586256245, 53750679943
OFFSET
1,1
COMMENTS
The ratio k(n) /(2*j(n)) tends to sqrt(6) as n increases.
k(n) = 2*b + 1, for n > 0, where b is a side of the Heronian triangle (5, b, b+1). - Andrés Ventas, Dec 13 2024
FORMULA
Recurrence: k(1)=5, k(2)=7, k(3)=11, k(4)=25, k(5)=10*k(2)-k(3), k(6)=10*k(3)-k(2) then k(n)=10*k(n-3)-k(n-6).
From Ralf Stephan, Nov 15 2010: (Start)
G.f.: (-7x^5-11x^4-25x^3+11x^2+7x+5)/(x^6-10x^3+1).
a(3n+1) = 5*A001079(n), a(3n+2) = A077409(n), a(3n+3) = A077250(n). (End)
PROG
(PARI) Vec(-x*(7*x^5+11*x^4+25*x^3-11*x^2-7*x-5)/(x^6-10*x^3+1) + O(x^100)) \\ Colin Barker, Apr 16 2014
CROSSREFS
Cf. A106331.
Sequence in context: A067289 A036491 A036490 * A057247 A157437 A213677
KEYWORD
nonn,easy
AUTHOR
Pierre CAMI, Apr 29 2005
EXTENSIONS
More terms from Ralf Stephan, Nov 15 2010
More terms from Colin Barker, Apr 16 2014
STATUS
approved