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A046090 Consider all Pythagorean triples (X,X+1,Z) ordered by increasing Z; sequence gives X+1 values. 34
1, 4, 21, 120, 697, 4060, 23661, 137904, 803761, 4684660, 27304197, 159140520, 927538921, 5406093004, 31509019101, 183648021600, 1070379110497, 6238626641380, 36361380737781, 211929657785304, 1235216565974041 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

COMMENTS

Solution to a(a-1) = 2b(b-1) in natural numbers: a = a(n), b = b(n) = A011900(n).

n such that n^2 = (1/2)*(n+floor(sqrt(2)*n*floor(sqrt(2)*n))). - Benoit Cloitre, Apr 15 2003

a(n) = A001109(n+1) - A053141(n). - Manuel Valdivia, Apr 03 2010

a(n) balls in an urn; b(n) = A011900(n) are red; draw 2 balls without replacement; 2*Probability(2 red balls) = Probability(2 balls); this is equivalent to the Pell equation A(n)^2-2*B(n)^2 = -1 with a(n) = (A(n)+1)/2; b(n) = (B(n)+1)/2; and the fundamental solution (7;5) and the solution (3;2) for the unit form. - Paul Weisenhorn, Aug 03 2010

With b(n) = A001109, a(n)*(a(n)-1)/2 = b(n)*b(n+1) and b(n) + b(n+1) = 2*a(n) - 1. - Kenneth J Ramsey, Apr 24 2011

Positive integers n such that n^2 is a centered square number (A001844). - Colin Barker, Feb 12 2015

REFERENCES

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

S. Northshield, An Analogue of Stern's Sequence for Z[sqrt(2)], Journal of Integer Sequences, 18 (2015), #15.11.6.

LINKS

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000

T. W. Forget and T. A. Larkin, Pythagorean triads of the form X, X+1, Z described by recurrence sequences, Fib. Quart., 6 (No. 3, 1968), 94-104.

L. J. Gerstein, Pythagorean triples and inner products, Math. Mag., 78 (2005), 205-213.

Ron Knott, Pythagorean Triples and Online Calculators

Eric Weisstein's World of Mathematics, Pythagorean Triple

Index entries for two-way infinite sequences

Index entries for linear recurrences with constant coefficients, signature (7,-7,1).

FORMULA

a(n) = (1+sqrt(1+8*b(n)*(b(n)+1)))/2 with b(n) = A011900(n).

a(n) = 6*a(n-1)-a(n-2)-2, n >= 2, a(0) = 1, a(1) = 4. a(n) = (A(n+1)-3*A(n)+2)/4 with A(n) = A001653(n).

G.f.: (1-3*x)/((1-6*x+x^2)*(1-x)). a(n) = partial sums of A001541(n). - Barry Williams, May 03 2000

A001652(n)*A001652(n+1) + a(n)*a(n+1) = A001542(n+1)^2 = A084703(n+1). - Charlie Marion, Jul 01 2003

a(n) = 1/2 + ((1-2^{1/2})/4)*(3 - 2^{3/2})^n + ((1+2^{1/2})/4)*(3 + 2^{3/2})^n. - Antonio Alberto Olivares, Oct 13 2003

Let a(n) = A001652(n), b(n) = this sequence and c(n) = A001653(n). Then for k>j, c(i)*(c(k) - c(j)) = a(k+i)+...+a(i+j+1) + a(k-i-1)+...+a(j-i) + k - j. For n<0, a(n) = -b(-n-1). Also a(n)*a(n+2k+1) + b(n)*b(n+2k+1) + c(n)*c(n+2k+1) = (a(n+k+1) - a(n+k))^2; a(n)*a(n+2k) + b(n)*b(n+2k) + c(n)*c(n+2k) = 2*c(n+k)^2. - Charlie Marion, Jul 01 2003

2*a(n) = 2*A084159(n) + 1 + (-1)^(n+1)=2*A046729(n) + 1 - (-1)^(n+1). - Lekraj Beedassy, Jul 16 2004

From Paul Weisenhorn, Aug 03 2010: (Start)

a(n+1) = round((1+(7+5*sqrt(2))*(3+2*sqrt(2))^n)/2);

b(n+1) = round((2+(10+7*sqrt(2))*(3+2*sqrt(2))^n)/4) = A011900(n+1).

(End)

Let T(n) be the n-th triangular number; then T(a(n)) = A011900(n)^2 + A001109(n). See also A001653. - Charlie Marion, Apr 25 2011

a(0)=1, a(1)=4, a(2)=21, a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3). - Harvey P. Dale, Apr 13 2012

Lim_{n->infinity} a(n+1)/a(n) = 3 + 2*sqrt(2) = A156035. - Ilya Gutkovskiy, Jul 10 2016

a(n) = A001652(n)+1. - Dimitri Papadopoulos, Jul 06 2017

EXAMPLE

For n=4: a(4)=697; b(4)=493; 2*binomial(493,2)=485112=binomial(697,2). - Paul Weisenhorn, Aug 03 2010

MAPLE

Digits:=100: seq(round((1+(7+5*sqrt(2))*(3+2*sqrt(2))^(n-1))/2)/2, n=0..20); # Paul Weisenhorn, Aug 03 2010

MATHEMATICA

Join[{1}, #+1&/@With[{c=3+2Sqrt[2]}, NestList[Floor[c #]+3&, 3, 20]]] (* Harvey P. Dale, Aug 19 2011 *)

LinearRecurrence[{7, -7, 1}, {1, 4, 21}, 25] (* Harvey P. Dale, Apr 13 2012 *)

a[n_] := (2-ChebyshevT[n, 3]+ChebyshevT[n+1, 3])/4; Array[a, 21, 0] (* Jean-Fran├žois Alcover, Jul 10 2016, adapted from PARI *)

PROG

(PARI) a(n)=(2-subst(poltchebi(abs(n))-poltchebi(abs(n+1)), x, 3))/4

(Haskell)

a046090 n = a046090_list !! n

a046090_list = 1 : 4 : map (subtract 2)

   (zipWith (-) (map (* 6) (tail a046090_list)) a046090_list)

-- Reinhard Zumkeller, Jan 10 2012

CROSSREFS

Other 2 sides are A001652 and A001653.

Cf. A011900, A001541. A001652(n) = -a(-1-n).

Sequence in context: A212419 A020048 A093426 * A182435 A045721 A101810

Adjacent sequences:  A046087 A046088 A046089 * A046091 A046092 A046093

KEYWORD

nonn,easy,nice

AUTHOR

Eric W. Weisstein

EXTENSIONS

Additional comments from Wolfdieter Lang

STATUS

approved

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Last modified September 26 01:25 EDT 2017. Contains 292500 sequences.