A267707 a(n) = A000217(A000217(n)+1).

1, 3, 10, 28, 66, 136, 253, 435, 703, 1081, 1596, 2278, 3160, 4278, 5671, 7381, 9453, 11935, 14878, 18336, 22366, 27028, 32385, 38503, 45451, 53301, 62128, 72010, 83028, 95266, 108811, 123753, 140185, 158203, 177906, 199396, 222778, 248160, 275653, 305371

Offset: 0

Author

Waldemar Puszkarz, Jan 19 2016

Keywords

• nonn
• easy

Comments

It is the sequence of triangular numbers (A000217) with progressive gaps that grow as 0,1,2,3, ... (consecutive numbers), by which I mean that the 0,1,2,3, ... consecutive triangular numbers are removed from A000217 to form this sequence. For instance, (1), 6, a triangular number, is missing between 3 and 10, which is the gap with 1 triangular number removed, (2), 15 and 21 (two consecutive triangular numbers) are missing between 10 and 28, which is the gap with 2 triangular numbers removed, and so on.

The differences between the consecutive terms of this sequence can be expressed through the sum of cubes of two numbers separated by 2 as (n^3+(n+2)^3)/4, which is the same as A229183, except for the first term in there.

The same pattern when applied to squares, A000290(A000290(n)+1), gives A082044(n). Triangular numbers are also linked in a similar manner to A027927(n) = A000217(A000217(n)+2)/3.

Examples

			For n=0, a(0)=1*2/2=1. For n=2, a(2)=4*5/2=10.
		

Links

• Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Crossrefs

Cf. A000217 (triangular numbers), A229183 (consecutive terms differences), A082044 (related sequence for squares), A027927 (related sequence for triangular numbers).

Programs

Magma

I:=[1,3,10,28,66]; [n le 5 select I[n] else 5*Self(n-1)-10*Self(n-2)+10*Self(n-3)-5*Self(n-4)+Self(n-5): n in [1..50]]; // Vincenzo Librandi, Jan 22 2016

Mathematica

S[n_] :=n*(n+1)/2; Table[S[S[n]+1], {n, 0, 50}]
Table[(n*(n+1)/2+1)(n*(n+1)/2+2)/2, {n, 0, 50}]
Table[(n^4+2*n^3+7*n^2+6*n+8)/8, {n, 0, 50}]
CoefficientList[Series[(1 - 2 x + 5 x^2 - 2 x^3 + x^4) / (1 - x)^5, {x, 0, 33}], x] (* or *) LinearRecurrence[{5, -10, 10, -5, 1}, {1, 3, 10, 28, 66}, 50] (* Vincenzo Librandi, Jan 22 2016 *)

PARI

for(n=0,50,print1((n^4+2*n^3+7*n^2+6*n+8)/8 ", "))

Formula

a(n) = A000217(A000217(n)+1) = (n*(n+1)/2+1)(n*(n+1)/2+2)/2.

a(n) = (n^4+2n^3+7n^2+6n+8)/8 = (n^2+n+2)(n^2+n+4)/8.

G.f.: (1-2*x+5*x^2-2*x^3+x^4)/(1-x)^5. - Vincenzo Librandi, Jan 22 2016

a(n) = 5*a(n-1)-10*a(n-2)+10*a(n-3)-5*a(n-4)+a(n-5). - Vincenzo Librandi, Jan 22 2016