Consider all Pythagorean triples (X, X+1, Z) ordered by increasing Z; sequence gives X values.
Numbers n such that triangular number t(n) (see
A000217) = n(n+1)/2 is a product of two consecutive integers (cf.
A097571).
Members of Diophantine pairs. Solution to a*(a+1) = 2*b*(b+1) in natural numbers including 0; a = a(n), b = b(n) =
A053141(n); The solution of a special case of a binomial problem of H. Finner and K. Strassburger (strass(AT)godot.dfi.uni-duesseldorf.de).
The index of all triangular numbers T(a(n)) for which 4T(n)+1 is a perfect square.
All Pythagorean triples {X(n), Y(n)=X(n)+1, Z(n)} with X<Y<Z, may be recursively generated through the mapping W(n) -> M*W(n), where W(n)=transpose of vector [X(n) Y(n) Z(n)] and M a 3 X 3 matrix given by [2 1 2 / 1 2 2 / 2 2 3]. -
Lekraj Beedassy, Aug 14 2006
In general, if b(n) =
A053141(n), then a(n)*b(n+k) = a(n+k)*b(n)+b(k); e.g., 3*84 = 119*2+14; 3*2870 = 4059*2+492; 20*2870 = 5741*14+84. -
Charlie Marion, Nov 19 2007
If (p,q) is a solution of the Diophantine equation: X^2 + (X+1)^2 = Y^2 then (p+q) or (p+q+1) are perfect squares. If (p,q) is a solution of the Diophantine equation: X^2 + (X+1)^2 = Y^2 then (p+q) or (p+q)/8 are perfect squares. If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation: X^2 + (X+1)^2 = Y^2 with p<r then s-r=p+q+1. -
Mohamed Bouhamida, Aug 29 2009
If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation: X^2 + (X + 1)^2 = y^2 with p<r then r=3p+2q+1 and s=4p+3q+2. -
Mohamed Bouhamida, Sep 02 2009
The numbers 3*
A001652 = (0, 9, 60, 357, 2088, 12177, 70980, ...) are all the nonnegative values of X such that X^2 + (X+3)^2 = Z^2 (Z is in
A075841). -
Bruno Berselli, Aug 26 2010
Let T(n) = n*(n+1)/2 (the n-th triangular number). For n > 0,
Set a(n)=p; a(n)+1=q; the generated triple x=p^2+pq; y=q^2+pq; k=p^2+q^2 satisfies x^2+y^2=k(x+y). -
Carmine Suriano, Dec 17 2013
The arms of the triangle are found with (b(n),c(n)) for 2*b(n)*c(n) and c(n)^2 - b(n)^2. Let b(1) = 1 and c(1) = 2, then b(n) = c(n-1) and c(n) = 2*c(n-1) + b(n-1). Alternatively, b(n) = c(n-1) and c(n) equals the nearest integer to b(n)*(1+sqrt(2)). -
J. M. Bergot, Oct 09 2014
Conjecture: For n>1 a(n) is the index of the first occurrence of n in sequence
A123737. -
Vaclav Kotesovec, Jun 02 2015
Numbers m such that Product_{k=1..m} (4*k^4+1) is a square (see
A274307). -
Chai Wah Wu, Jun 21 2016
Numbers m such that m^2+(m+1)^2 is a square. -
César Aguilera, Aug 14 2017
For integers a and d, let P(a,d,1) = a, P(a,d,2) = a+d, and, for n>2, P(a,d,n) = 2*P(a,d,n-1) + P(a,d,n-2). Further, let p(n) = Sum_{i=1..2n} P(a,d,i). Then p(n)^2 + (p(n)+d)^2 + a^2 = P(a,d,2n+1)^2 + d^2. When a = 1 and d = 1, p(n) = a(n) and P(a,d,n) =
A000129(n), the n-th Pell number. -
Charlie Marion, Dec 08 2018
The terms of this sequence satisfy the Diophantine equation k^2 + (k+1)^2 = m^2, which is equivalent to (2k+1)^2 - 2*m^2 = -1. Now, with x=2k+1 and y=m, we get the Pell-Fermat equation x^2 - 2*y^2 = -1. The solutions (x,y) of this equation are respectively in
A002315 and
A001653. The relation k = (x-1)/2 explains
Lekraj Beedassy's Nov 25 2003 formula. Thus, the corresponding numbers m = y, which express the length of the hypotenuse of these right triangles (k,k+1,m) are in
A001653. -
Bernard Schott, Mar 10 2019
Members of Diophantine pairs. Related to solutions of p^2 = 2q^2 + 2 in natural numbers; p = p(n) = 2*sqrt(4T(a(n))+1), q = q(n) = sqrt(8*T(a(n))+1). Note that this implies that 4*T(a(n))+1 is a perfect square (numbers of the form 8*T(n)+1 are perfect squares for all n); these T(a(n))'s are the only solutions to the given Diophantine equation. -
Steven Blasberg, Mar 04 2021
