Search: id:A099776
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%I A099776
%S A099776 5,13,25,41,61,85,113,145,181,221,265,313,365,421,481,545,613,685,761,
%T A099776 841,925,1013,1105,1201,1301,1405,1513,1625,1741,1861,1985,2113,2245,
%U A099776 2381,2521,2665,2813,2965,3121,3281,3445,3613,3785,3961,4141,4325,4513
%N A099776 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, ...
%C A099776 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
%C A099776 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
%C A099776 sum of two consecutive squares: 1^4=5,4+9=13,9+16=25,16+25=41,.. [From 
               Vladimir Orlovsky (4vladimir(AT)gmail.com), Sep 25 2009]
%F A099776 a(n) = ((2*n+1)^2-1)/2 + 1
%F A099776 a(n)=4*n+a(n-1), (with a(1)=5) [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), 
               Oct 24 2009]
%e A099776 For n=2, a(2)=4*2+5=13; n=3, a(3)=4*3+13=25; n=4, a(4)=4*4+25=41 [From 
               Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Oct 24 2009]
%t A099776 lst={};Do[a=(n^2+(n+1)^2);AppendTo[lst,a],{n,5!}];lst [From Vladimir 
               Orlovsky (4vladimir(AT)gmail.com), Sep 25 2009]
%o A099776 (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; }
%Y A099776 Sequence in context: A081961 A096891 A001844 this_sequence A133322 A146590 
               A098483
%Y A099776 Adjacent sequences: A099773 A099774 A099775 this_sequence A099777 A099778 
               A099779
%K A099776 easy,nonn
%O A099776 1,1
%A A099776 Nick Robins (nrobins(AT)hackettfreedman.com), Nov 12 2004

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