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A053631
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Pythagorean spiral: a(n-1)+1, a(n) and a(n)+1 are the sides of a right triangle (a primitive Pythagorean triangle).
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6
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2, 4, 12, 84, 3612, 6526884, 21300113901612, 226847426110843688722000884, 25729877366557343481074291996721923093306518970391612, 331013294649039928396936390888878360035026305412754995683702777533071737279144813617823976263475290370884
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OFFSET
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1,1
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COMMENTS
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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
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LINKS
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FORMULA
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a(1)=2; for n >= 2: a(n) = a(n-1) + a(n-1)^2/2 = A046092(a(n-1)/2).
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EXAMPLE
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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
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MAPLE
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a[1]:= 2:
for n from 2 to 10 do a[n]:= a[n-1] + a[n-1]^2/2 od:
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MATHEMATICA
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PROG
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(Maxima)
a[1]:2$
a[n]:=a[n-1] + (a[n-1]^2)/2$
(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 */
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CROSSREFS
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Apart from the initial term, the sequence is the same as A127690.
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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