A050533 Thickened pyramidal numbers: a(n) = 2*(n+1)*n + Sum_{i=1..n} (4*i*(i-1) + 1).

0, 5, 22, 59, 124, 225, 370, 567, 824, 1149, 1550, 2035, 2612, 3289, 4074, 4975, 6000, 7157, 8454, 9899, 11500, 13265, 15202, 17319, 19624, 22125, 24830, 27747, 30884, 34249, 37850, 41695, 45792, 50149, 54774, 59675, 64860, 70337, 76114, 82199, 88600, 95325

Offset: 0

Author

Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Dec 29 1999

Keywords

• nonn
• easy
• nice

Comments

This sequence is the partial sums of A053755. - J. M. Bergot, May 31 2012

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. A000292, A000447, A000217, A050492, A053755.

Programs

Magma

I:=[0, 5, 22, 59]; [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..40]]; // Vincenzo Librandi, Apr 27 2012

Mathematica

CoefficientList[Series[x*(5+2*x+x^2)/(1-x)^4,{x,0,40}],x] (* Vincenzo Librandi, Apr 27 2012 *)
LinearRecurrence[{4,-6,4,-1},{0,5,22,59},40] (* Harvey P. Dale, May 08 2012 *)

PARI

a(n)=n*(4*n^2+6*n+5)/3 \\ Charles R Greathouse IV, Apr 16 2012

Formula

a(n) = (1/3)*n*(5 + 6*n + 4*n^2) = binomial(2*n+1, 3) + 2*(n+1)*n = A000447(n) + 4*A000217(n).

G.f.: x*(5+2*x+x^2)/(1-x)^4. - Colin Barker, Apr 16 2012

a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Vincenzo Librandi, Apr 27 2012

E.g.f.: exp(x)*x*(15 + 18*x + 4*x^2)/3. - Elmo R. Oliveira, Aug 08 2025