A046090 Consider all Pythagorean triples (X,X+1,Z) ordered by increasing Z; sequence gives X+1 values.

1, 4, 21, 120, 697, 4060, 23661, 137904, 803761, 4684660, 27304197, 159140520, 927538921, 5406093004, 31509019101, 183648021600, 1070379110497, 6238626641380, 36361380737781, 211929657785304, 1235216565974041, 7199369738058940, 41961001862379597, 244566641436218640

Offset: 0

Author

Eric W. Weisstein

Keywords

• nonn
• easy
• nice

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

Place a(n) balls in an urn, of which 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

Find base x in which repdigit yy has a square that is repdigit zzzz, corresponding to Diophantine equation zzzz_x = (yy_x)^2; then, solution z = a(n) with x = A002315(n) and y = A001653(n+1) for n >= 1 (see Maurice Protat reference). - Bernard Schott, Dec 21 2022

Examples

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

References

• A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 122-125, 1964.
• Maurice Protat, Des Olympiades à l'Agrégation, De zzzz_x = (yy_x)^2 à Pell-Fermat, Problème 23, pp. 52-54, Ellipses, Paris, 1997.

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.
• H. J. Hindin, Stars, hexes, triangular numbers and Pythagorean triples, J. Rec. Math., 16 (1983/1984), 191-193. (Annotated scanned copy)
• Ron Knott, Pythagorean Triples and Online Calculators
• S. Northshield, An Analogue of Stern's Sequence for Z[sqrt(2)], Journal of Integer Sequences, 18 (2015), #15.11.6.
• 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).

Crossrefs

Other 2 sides are A001652 and A001653.

Cf. A001009, A001541, A001542, A011900, A046729, A053141, A084159, A084703, A156035.

See comments in A301383.

Cf. A001109, A001653, A002024, A002315, A029549.

Programs

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

Magma

m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((1-3*x)/((1-6*x+x^2)*(1-x)))); // G. C. Greubel, Jul 15 2018

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-François Alcover, Jul 10 2016, adapted from PARI *)

PARI

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

PARI

x='x+O('x^30); Vec((1-3*x)/((1-6*x+x^2)*(1-x))) \\ G. C. Greubel, Jul 15 2018

Formula

a(n) = (-1+sqrt(1+8*b(n)*(b(n)+1)))/2 with b(n) = A011900(n). [corrected by Michel Marcus, Dec 23 2022]

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).

A001652(n) = -a(-1-n).

From Barry E. Williams, May 03 2000: (Start)

G.f.: (1-3*x)/((1-6*x+x^2)*(1-x)).

a(n) = partial sums of A001541(n). (End)

From Charlie Marion, Jul 01 2003: (Start)

A001652(n)*A001652(n+1) + a(n)*a(n+1) = A001542(n+1)^2 = A084703(n+1).

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. (End)

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

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

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

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)

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

T(a(n)) = A011900(n)^2 + A001109(n), where T(n) is the n-th triangular number. 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

Limit_{n->oo} 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

a(n) = (A002315(n) + 1)/2. - Bernard Schott, Dec 21 2022

E.g.f.: (exp(x) + exp(3*x)*(cosh(2*sqrt(2)*x) + sqrt(2)*sinh(2*sqrt(2)*x)))/2. - Stefano Spezia, Mar 16 2024

a(n) = A002024(A029549(n))+1. - Pontus von Brömssen, Sep 11 2024

Extensions

Additional comments from Wolfdieter Lang

Comment moved to A001653 by Claude Morin, Sep 22 2023