|
| |
|
|
A046729
|
|
Expansion of 4*x/((1+x)*(1-6*x+x^2)).
|
|
15
|
|
|
|
0, 4, 20, 120, 696, 4060, 23660, 137904, 803760, 4684660, 27304196, 159140520, 927538920, 5406093004, 31509019100, 183648021600, 1070379110496, 6238626641380, 36361380737780, 211929657785304, 1235216565974040
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,2
|
|
|
COMMENTS
|
Related to Pythagorean triples: alternate terms of A001652 and A046090.
Even-valued legs of nearly isosceles right triangles: legs differ by 1. 0 is smaller leg of degenerate triangle with legs 0 and 1 and hypotenuse 1. - Charlie Marion, Nov 11 2003
The complete (nearly isosceles) primitive Pythagorean triple is given by {a(n), a(n)+(-1)^n, A001653(n)}. - Lekraj Beedassy, Feb 19 2004
Note also that A046092 is the even leg of this other class of nearly isosceles Pythagorean triangles {A005408(n), A046092(n), A001844(n)}, i.e., {2n+1, 2n(n+1), 2n(n+1)+1} where longer sides (viz. even leg and hypotenus) are consecutive. - Lekraj Beedassy, Apr 22 2004
|
|
|
REFERENCES
|
A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 122-125, 1964.
W. Sierpiński, Pythagorean triangles, Dover Publications, Inc., Mineola, NY, 2003, p. 17. MR2002669.
|
|
|
LINKS
|
|
|
|
FORMULA
|
a(n) = ((1+sqrt(2))^(2n+1) + (1-sqrt(2))^(2n+1) + 2*(-1)^(n+1))/4.
a(n) = ceiling((sqrt(2)+1)^(2*n+1) - (sqrt(2)-1)^(2*n+1) - 2*(-1)^n)/4. - Lambert Klasen (Lambert.Klasen(AT)gmx.net), Nov 12 2004
a(n) is the k-th entry amongst the complete near-isosceles primitive Pythagorean triple A114336(n), where k={3*(2n-1) - (-1)^n}/2, i.e., a(n)=A114336(A047235(n)), for positive n. - Lekraj Beedassy, Jun 04 2006
2*a(n)*(a(n) + (-1)^n) + 1 = (A000129(2*n+1))^2;
n > 0, 2*a(n)*(a(n) + (-1)^n) + 1 = ((a(n+1) - a(n-1))/4)^2, a perfect square.
a(n+1) = (3*a(n) + 2*(-1)^n) + 2*sqrt(2*a(n)*(a(n) + (-1)^n)+ 1).
a(n-1) = (3*a(n) + 2*(-1)^n) - 2*sqrt(2*a(n)*(a(n) + (-1)^n)+ 1).
a(n+1) = 6*a(n) - a(n-1) + 4*(-1)^n.
a(n+1) = 5*a(n) + 5*a(n-1) - a(n-2).
a(n+1) *a(n-1) = a(n)*(a(n) + 4*(-1)^n).
a(n) = (sqrt(1 + 8*A029549(n)) - (-1)^n)/2.
Limit_{n->infinity} a(n)/a(n-1) = 3 + 2*sqrt(2).
Limit_{n->infinity} a(n)/a(n-2) = 17 + 12*sqrt(2).
Limit_{n->infinity} a(n)/a(n-r) = (3 + 2*sqrt(2))^r.
Limit_{n->infinity} a(n-r)/a(n) = (3 - 2*sqrt(2))^r. (End)
a(n) = (A001333(2*n+1) - 2*(-1)^n)/4.
|
|
|
EXAMPLE
|
[1,0,1]*[1,2,2; 2,1,2; 2,2,3]^0 gives (degenerate) primitive Pythagorean triple [1, 0, 1], so a(0) = 0. [1,0,1]*[1,2,2; 2,1,2; 2,2,3]^7 gives primitive Pythagorean triple [137903, 137904, 195025] so a(7) = 137904.
G.f. = 4*x + 20*x^2 + 120*x^3 + 696*x^4 + 4060*x^5 + 23660*x^6 + ...
|
|
|
MATHEMATICA
|
|
|
|
PROG
|
(PARI) a(n)=n%2+(real((1+quadgen(8))^(2*n+1))-1)/2
(PARI) a(n)=if(n<0, -a(-1-n), polcoeff(4*x/(1+x)/(1-6*x+x^2)+x*O(x^n), n))
(Magma) [4*Floor(((Sqrt(2)+1)^(2*n+1)-(Sqrt(2)-1)^(2*n+1)-2*(-1)^n) / 16): n in [0..30]]; // Vincenzo Librandi, Jul 29 2019
(SageMath) [(lucas_number2(2*n+1, 2, -1) -2*(-1)^n)/4 for n in range(41)] # G. C. Greubel, Feb 11 2023
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,easy
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
