%I #54 Mar 25 2024 08:02:08
%S 2,4,12,84,3612,6526884,21300113901612,226847426110843688722000884,
%T 25729877366557343481074291996721923093306518970391612,
%U 331013294649039928396936390888878360035026305412754995683702777533071737279144813617823976263475290370884
%N Pythagorean spiral: a(n-1)+1, a(n) and a(n)+1 are the sides of a right triangle (a primitive Pythagorean triangle).
%C To derive a list of Pythagorean triples from this sequence, we note that the difference between the second and the third terms in the Pythagorean triple is 1 and that the last term of the previous triple gives us the first term in the next triple. Therefore the sequence is completely determined by the initial triple.
%C A053631 gives us a list of Pythagorean triples beginning with (3,4,5), since a(1)=2. Using any initial value h>1, (2h-1,2h^2-2h,2h^2-2h+1) forms a Pythagorean triple; we can use b(1)=2h-1 and the recursive formula b(n)=b(n-1)^2-b(n-1)+1 for n>1, we can create infinitely many of spirals of this type. - _Haoqi Chen_, _Teena Carroll_
%H Robert Israel, <a href="/A053631/b053631.txt">Table of n, a(n) for n = 1..13</a>
%F a(1)=2; for n >= 2: a(n) = a(n-1) + a(n-1)^2/2 = A046092(a(n-1)/2).
%F a(n) = A053630(n) - 1. - _Robert G. Wilson v_, Jul 29 2014
%F a(n) = 2*A007018(n-1). - _Ivan Neretin_, Jul 26 2015
%e For n=3, a(n-1) = 4, so we want a right triangle with sides 4 + 1 = 5, a(n), and a(n)+1. Solving (x+1)^2 = x^2 + 5^2 gives x = 12, so a(3) = 12. - _Michael B. Porter_, Jul 19 2016
%p a[1]:= 2:
%p for n from 2 to 10 do a[n]:= a[n-1] + a[n-1]^2/2 od:
%p seq(a[i],i=1..10); # _Robert Israel_, Jul 08 2015
%t NestList[# + #^2/2 &, 2, 9] (* _Robert G. Wilson v_, Dec 12 2012 *)
%o (Maxima)
%o a[1]:2$
%o a[n]:=a[n-1] + (a[n-1]^2)/2$
%o A053631(n):=a[n]$
%o makelist(A053631(n),n,1,10); /* _Martin Ettl_, Nov 08 2012 */
%o (PARI) main(size)={v=vector(size); v[1]=2;for(n=2,size,v[n]=v[n-1]+v[n-1]^2/2);return(v)} /* _Anders Hellström_, Jul 08 2015 */
%Y Apart from the initial term, the sequence is the same as A127690.
%Y Cf. A046092
%K nonn
%O 1,1
%A _Henry Bottomley_, Mar 21 2000
%E Corrected and extended by _James A. Sellers_, Mar 22 2000
%E a(1) = 2 added by _Zak Seidov_, Apr 10 2007