A155883 a(n) = 14*n^3 - 30*n^2 + 24*n - 7.

1, 33, 173, 505, 1113, 2081, 3493, 5433, 7985, 11233, 15261, 20153, 25993, 32865, 40853, 50041, 60513, 72353, 85645, 100473, 116921, 135073, 155013, 176825, 200593, 226401, 254333, 284473, 316905, 351713, 388981, 428793, 471233, 516385, 564333, 615161, 668953

Offset: 1

Author

David Z Crookes, Jan 29 2009

Keywords

• nonn
• easy
• changed

Comments

A three-dimensional version of the centered hexagonal numbers (A003215). Two examples: the third term 173 is built up as 19 + 37 + 61 + 37 + 19 and the fourth term 505 is built up as 37 + 61 + 91 + 127 + 91 + 61 + 37.

The sequence's digital roots are 1, 6, 2 (repeat).

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• David Z Crookes, De Pulchritudine Numerorum Figuratorum (On the Beauty of Figurate Numbers), Mathematics in School, Vol. 17, No. 3 (May, 1988), pp. 38-39.
• Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Crossrefs

Cf. A003215.

Programs

Magma

I:=[1, 33, 173, 505]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..50]];  // Vincenzo Librandi, Jun 30 2012

Mathematica

CoefficientList[Series[(1+29*x+47*x^2+7*x^3)/(1-x)^4,{x,0,40}],x] (* Vincenzo Librandi, Jun 30 2012 *)

Formula

From N. J. A. Sloane, Feb 16 2025: (Start)

Let h(i) denote the centered hexagonal number A003215(i). Then for n >= 1,

a(n) = h(2*n-2) + 2*Sum_{i=n-1..2*n-3} h(i).

E.g. a(3) = h(2) + h(3) + h(4) + h(3) + h(2), as in the COMMENTS.

This sequence should really have had offset 0, not 1, which would have given a simpler formula. (End)

G.f.: x*(1+29*x+47*x^2+7*x^3)/(1-x)^4. - Colin Barker, Jun 16 2012

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Vincenzo Librandi, Jun 30 2012

E.g.f.: 7 + exp(x)*(-7 + 8*x + 12*x^2 + 14*x^3). - Elmo R. Oliveira, Sep 02 2025

Extensions

More terms from Colin Barker, Jun 16 2012

New name using explicit formula from Joerg Arndt, Jan 15 2021

Edited by N. J. A. Sloane, Feb 16 2025