A064322 - OEIS

%I #21 Dec 26 2024 15:20:11

%S 0,1,21,231,1540,7260,26796,82621,222111,536130,1186570,2445366,

%T 4747821,8763391,15487395,26357430,43398586,69401871,108140571,

%U 164629585,245433090,359026206,516216646,730632651,1019283825,1403201800,1908167976,2565535896,3413156131

%N Triply triangular numbers.

%H Harry J. Smith, <a href="/A064322/b064322.txt">Table of n, a(n) for n = 0..400</a>

%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (9,-36,84,-126,126,-84,36,-9,1).

%F a(n) = A000217(A000217(A000217(n))) = n*(n+1)*(n^2 + n + 2)*(n^4 + 2n^3 + 3n^2 + 2n + 8)/128 = A002817(n)*(A002817(n) + 1)/2.

%F G.f.: x*(1 + 12*x + 78*x^2 + 133*x^3 + 78*x^4 + 12*x^5 + x^6)/(1 - x)^9. - _Colin Barker_, Apr 19 2012

%e a(4) = 1540 because 4th triangular number is 10, 10th triangular number is 55 and 55th triangular number is 1540.

%p a:= n-> ((k-> binomial(k+1,2))@@3)(n):

%p seq(a(n), n=0..30);  # _Alois P. Heinz_, Apr 19 2012

%t f[n_] := n(n + 1)/2; Table[ Nest[f, n, 3], {n, 0, 25}] (* _Robert G. Wilson v_, Jun 30 2004 *)

%o (PARI) a(n) = { my(Tri(m)= m*(m + 1)/2); Tri(Tri(Tri(n))) } \\ _Harry J. Smith_, Sep 11 2009

%Y Cf. A000217, A002817, A066370.

%K nonn,easy,changed

%O 0,3

%A _Henry Bottomley_, Oct 15 2001