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Search: id:A099776
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| A099776 |
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Length of the hypotenuse of an integer right triangle with the hypotenuse being one more than the longer side. The shorter sides are just consecutive odd numbers 3, 5, 7, ... |
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+0
2
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| 5, 13, 25, 41, 61, 85, 113, 145, 181, 221, 265, 313, 365, 421, 481, 545, 613, 685, 761, 841, 925, 1013, 1105, 1201, 1301, 1405, 1513, 1625, 1741, 1861, 1985, 2113, 2245, 2381, 2521, 2665, 2813, 2965, 3121, 3281, 3445, 3613, 3785, 3961, 4141, 4325, 4513
(list; graph; listen)
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OFFSET
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1,1
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COMMENT
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Largest hypotenuse of primitive Pythagorean triangles with inradius n. (For smallest hypotenuse of PPT with inradius n, see A087484) Essentially the same as A001844. - Lekraj Beedassy (blekraj(AT)yahoo.com), May 08 2006
The complete triple {X(n), Y(n), Z(n)=Y(n)+1}, with X<Y<Z, {X(n)=A005408(n);Y(n)=A046092(n), Z(n)=A001844(n)} 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 [1 -2 2 / 2 -1 2 / 2 -2 3 ]. Such triples correspond to successive number pair Pythagorean generators(p,q=p+1) yielding {X=p+q,Y=2p*q,Z=p^2 + q^2} - Lekraj Beedassy (blekraj(AT)yahoo.com), Jun 04 2006
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FORMULA
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a(n) = ((2*n+1)^2-1)/2 + 1
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PROGRAM
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(C) #include "stdio.h" int main(int argc, char* argv[]){ unsigned int i; for (i=1; i<50; i++) printf ("%u, ", (((2*i+1)*(2*i+1)-1)/2)+1); return 0; }
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CROSSREFS
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Adjacent sequences: A099773 A099774 A099775 this_sequence A099777 A099778 A099779
Sequence in context: A081961 A096891 A001844 this_sequence A133322 A098483 A064276
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KEYWORD
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easy,nonn
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AUTHOR
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Nick Robins (nrobins(AT)hackettfreedman.com), Nov 12 2004
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