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A046729 Expansion of 4x/((1+x)(1-6x+x^2)). 14
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

Union of even entries of A001652 and A046090. Sum of legs of primitive Pythagorean triangles is A002315(n) = 2*a(n) + (-1)^n. - Lekraj Beedassy, Apr 30 2004

REFERENCES

A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 122-125, 1964.

W. SierpiƄski, Pythagorean triangles, Dover Publications, Inc., Mineola, NY, 2003, p. 17. MR2002669.

LINKS

Table of n, a(n) for n=0..20.

Index entries for linear recurrences with constant coefficients, signature (5,5,-1).

FORMULA

a(n) = ((1+sqrt(2))^(2n+1) + (1-sqrt(2))^(2n+1) + 2*(-1)^(n+1))/4; a(n) = A089499(n)*A089499(n+1); cf. A084159.

a(n) = 4*A084158(n). - Lekraj Beedassy, Jul 16 2004

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

a(n) = A046727(n) - (-1)^n = 2*A114620(n). - Lekraj Beedassy, Aug 14 2006

From George F. Johnson, Aug 29 2012: (Start)

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.

a(n) = A002315(n) - A084159(n) = A084159(n) - (-1)^n.

a(n)  = A001652(n) + (1 - (-1)^n)/2 = A046090(n) - (1 + (-1)^n)/2.

Lim_{n->infinity} a(n)/a(n-1) =  3 + 2*sqrt(2).

Lim_{n->infinity} a(n)/a(n-2) = 17 + 12*sqrt(2).

Lim_{n->infinity} a(n)/a(n-r) = (3 + 2*sqrt(2))^r.

Lim_{n->infinity} a(n-r)/a(n) = (3 - 2*sqrt(2))^r.

(End)

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

LinearRecurrence[{5, 5, -1}, {0, 4, 20}, 25] (* Vincenzo Librandi, Jul 29 2019 *)

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

CROSSREFS

Cf. A046727, A084159, A084158, A001652, A046090, A029549.

Sequence in context: A013197 A319788 A089498 * A277920 A093123 A092055

Adjacent sequences:  A046726 A046727 A046728 * A046730 A046731 A046732

KEYWORD

nonn,easy

AUTHOR

N. J. A. Sloane

STATUS

approved

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Last modified November 12 09:29 EST 2019. Contains 329054 sequences. (Running on oeis4.)