OFFSET
1,1
COMMENTS
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.
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
LINKS
Robert Israel, Table of n, a(n) for n = 1..13
FORMULA
a(1)=2; for n >= 2: a(n) = a(n-1) + a(n-1)^2/2 = A046092(a(n-1)/2).
a(n) = A053630(n) - 1. - Robert G. Wilson v, Jul 29 2014
a(n) = 2*A007018(n-1). - Ivan Neretin, Jul 26 2015
EXAMPLE
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
MAPLE
a[1]:= 2:
for n from 2 to 10 do a[n]:= a[n-1] + a[n-1]^2/2 od:
seq(a[i], i=1..10); # Robert Israel, Jul 08 2015
MATHEMATICA
NestList[# + #^2/2 &, 2, 9] (* Robert G. Wilson v, Dec 12 2012 *)
PROG
(Maxima)
a[1]:2$
a[n]:=a[n-1] + (a[n-1]^2)/2$
A053631(n):=a[n]$
makelist(A053631(n), n, 1, 10); /* Martin Ettl, Nov 08 2012 */
(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 */
CROSSREFS
KEYWORD
nonn
AUTHOR
Henry Bottomley, Mar 21 2000
EXTENSIONS
Corrected and extended by James A. Sellers, Mar 22 2000
a(1) = 2 added by Zak Seidov, Apr 10 2007
STATUS
approved