A007586 11-gonal (or hendecagonal) pyramidal numbers: a(n) = n*(n+1)*(3*n-2)/2.

0, 1, 12, 42, 100, 195, 336, 532, 792, 1125, 1540, 2046, 2652, 3367, 4200, 5160, 6256, 7497, 8892, 10450, 12180, 14091, 16192, 18492, 21000, 23725, 26676, 29862, 33292, 36975, 40920, 45136, 49632, 54417, 59500, 64890, 70596, 76627, 82992, 89700, 96760, 104181

Offset: 0

Author

N. J. A. Sloane, R. K. Guy

Keywords

• nonn
• easy
• nice

Comments

Starting with 1 equals binomial transform of [1, 11, 19, 9, 0, 0, 0, ...]. - Gary W. Adamson, Nov 02 2007

Examples

			From _Vincenzo Librandi_, Feb 12 2014: (Start)
After 0, the sequence is provided by the row sums of the triangle (see above, third formula):
  1;
  2, 10;
  3, 20, 19;
  4, 30, 38, 28;
  5, 40, 57, 56, 37;
  6, 50, 76, 84, 74, 46; etc. (End)
		

References

• A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 194.
• E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 93.
• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Crossrefs

Cf. A051682.

Cf. A093644 ((9,1) Pascal, column m=3).

Cf. similar sequences listed in A237616.

Programs

GAP

List([0..45], n-> n*(n+1)*(3*n-2)/2); # G. C. Greubel, Aug 30 2019

Magma

I:=[0,1,12,42]; [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, Feb 12 2014

Maple

seq(n*(n+1)*(3*n-2)/2, n=0..45); # G. C. Greubel, Aug 30 2019

Mathematica

Table[n(n+1)(3n-2)/2,{n,0,45}] (* or *) LinearRecurrence[{4,-6,4,-1},{0,1,12,42}, 45] (* Harvey P. Dale, Apr 09 2012 *)
CoefficientList[Series[x(1+8x)/(1-x)^4, {x, 0, 45}], x] (* Vincenzo Librandi, Feb 12 2014 *)

PARI

a(n)=n*(n+1)*(3*n-2)/2 \\ Charles R Greathouse IV, Oct 07 2015

Sage

[n*(n+1)*(3*n-2)/2 for n in (0..45)] # G. C. Greubel, Aug 30 2019

Formula

G.f.: x*(1+8*x)/(1-x)^4.

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>3, a(0)=0, a(1)=1, a(2)=12, a(3)=42. - Harvey P. Dale, Apr 09 2012

a(n) = Sum_{i=0..n-1} (n-i)*(9*i+1), with a(0)=0. - Bruno Berselli, Feb 10 2014

From Amiram Eldar, Jun 28 2020: (Start)

Sum_{n>=1} 1/a(n) = (9*log(3) + sqrt(3)*Pi - 4)/10.

Sum_{n>=1} (-1)^(n+1)/a(n) = (sqrt(3)*Pi + 2 - 4*log(2))/5. (End)

E.g.f.: exp(x)*x*(2 + 10*x + 3*x^2)/2. - Elmo R. Oliveira, Aug 03 2025

Extensions

More terms from Vincenzo Librandi, Feb 12 2014