Search: id:A061278
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%I A061278
%S A061278 0,1,5,20,76,285,1065,3976,14840,55385,206701,771420,2878980,10744501,
%T A061278 40099025,149651600,558507376,2084377905,7779004245,29031639076,
%U A061278 108347552060,404358569165,1509086724601,5631988329240,21018866592360
%N A061278 a(n) = 5a(n-1)-5a(n-2)+a(n-3) with a(1) = 1 and a(k) = 0 if k< = 0.
%C A061278 Indices m of triangular numbers T(m) which are one-third of another triangular 
               number: 3*T(m)=T(k); the k's are given by A001571. - Bruce Corrigan 
               (scentman(AT)myfamily.com), Oct 31 2002
%C A061278 Also numbers n such that the n-th centered 24-gonal number 12n(n+1)+1 
               is a perfect square A001834(n)^2, where A001834(n) is defined by 
               the recursion: a(0) = 1, a(1) = 5, a(n) = 4a(n-1) - a(n-2). - Alexander 
               Adamchuk (alex(AT)kolmogorov.com), Apr 21 2007
%C A061278 Also numbers n such that RootMeanSquare(1,5,...,6*n-1) is an integer. 
               [From Ctibor O. Zizka (c.zizka(AT)email.cz), Dec 17 2008]
%C A061278 Also numbers n such that n*(n+1) = sum(n+1,n+2,n+3,...,n+x) for some 
               x. (This does not apply to the first term.) [From Gil Broussard (gilbroussard(AT)bellsouth.net), 
               Dec 23 2008]
%H A061278 Eric Weisstein, Link to a section of The World of Mathematics, <a href="http:/
               /mathworld.wolfram.com/CenteredPolygonalNumber.html">Centered Polygonal 
               Number</a>.
%F A061278 a(n) = 4a(n-1)-a(n-2)+1 = A001075(n)-a(n-1)-1 = (A001835(n+1)-1)/2 = 
               (A001353(n+1)-A001353(n)-1)/2 = a(n-1)+A001353(n), i.e. partial sum 
               of A001353.
%F A061278 From the recursion: a(n+2)=4a(n+1)-a(n)+1 a(0)=0, a(1)=1 g.f: A(x)= x/
               ((1-x)*(1-4x+x^2)) closed form: a(n)=(1/12)*(-6+(3-sqrt(3))*(2-sqrt(3))^n+(3+sqrt(3))*(2+sqrt(3))^n) 
               - Bruce Corrigan (scentman(AT)myfamily.com), Oct 31 2002
%F A061278 a(n)=(1/12)(A003500(n)+A003500(n+1)-6). - Mario Catalani (mario.catalani(AT)unito.it), 
               Apr 11 2003
%F A061278 a(n+1)=sum{k=0..n, U(k, 2)}=sum{k=0..n, S(k, 4)} - Paul Barry (pbarry(AT)wit.ie), 
               Nov 14 2003
%F A061278 G.f.: x/((1-x)*(1-4*x+x^2)) = x/(1-5*x+5*x^2-x^3).
%e A061278 a(2)=5 and T(5)=15 which is 1/3 of 45=T(9)
%o A061278 (PARI) M = [ 1, 1, 0; 1, 3, 1; 0, 1, 1]; for(i=1,30,print1(([1,0,0]*M^i)[3],
               ",")) - from Lambert Klasen (Lambert.Klasen(AT)gmx.net), Jan 25 2005
%Y A061278 Cf. A001571.
%Y A061278 Cf. A001834.
%Y A061278 Sequence in context: A030191 A093131 A000344 this_sequence A000758 A005283 
               A057552
%Y A061278 Adjacent sequences: A061275 A061276 A061277 this_sequence A061279 A061280 
               A061281
%K A061278 nonn
%O A061278 0,3
%A A061278 Henry Bottomley (se16(AT)btinternet.com), Jun 04 2001
%E A061278 More terms from Lambert Klasen (Lambert.Klasen(AT)gmx.net), Jan 25 2005

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