0,3

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

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

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]

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]

Eric Weisstein, Link to a section of The World of Mathematics, Centered Polygonal Number.

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.

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

a(n)=(1/12)(A003500(n)+A003500(n+1)-6). - Mario Catalani (mario.catalani(AT)unito.it), Apr 11 2003

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

G.f.: x/((1-x)*(1-4*x+x^2)) = x/(1-5*x+5*x^2-x^3).

a(2)=5 and T(5)=15 which is 1/3 of 45=T(9)

(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

Cf. A001571.

Cf. A001834.

Sequence in context: A030191 A093131 A000344 this_sequence A000758 A005283 A057552

Adjacent sequences: A061275 A061276 A061277 this_sequence A061279 A061280 A061281

nonn

Henry Bottomley (se16(AT)btinternet.com), Jun 04 2001

More terms from Lambert Klasen (Lambert.Klasen(AT)gmx.net), Jan 25 2005