A172073 a(n) = (4*n^3 + n^2 - 3*n)/2.

0, 1, 15, 54, 130, 255, 441, 700, 1044, 1485, 2035, 2706, 3510, 4459, 5565, 6840, 8296, 9945, 11799, 13870, 16170, 18711, 21505, 24564, 27900, 31525, 35451, 39690, 44254, 49155, 54405, 60016, 66000, 72369, 79135, 86310, 93906, 101935, 110409, 119340, 128740

Offset: 0

Author

Vincenzo Librandi, Jan 25 2010

Keywords

• nonn
• easy

Comments

14-gonal (or tetradecagonal) pyramidal numbers generated by the formula n*(n+1)*(2*d*n-(2*d-3))/6 for d=6.

In fact, the sequence is related to A000567 by a(n) = n*A000567(n) - Sum_{i=0..n-1} A000567(i) and this is the case d=6 in the identity n*(n*(d*n-d+2)/2) - Sum_{k=0..n-1} k*(d*k-d+2)/2 = n*(n+1)*(2*d*n-2*d+3)/6. - Bruno Berselli, Nov 29 2010

Except for the initial 0, this is the principal diagonal of the convolution array A213761. - Clark Kimberling, Jul 04 2012

Starting (1, 15, 54, ...), this is the binomial transform of (1, 14, 25, 12, 0, 0, 0, ...). - Gary W. Adamson, Jul 29 2015

References

• E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 93. - Bruno Berselli, Feb 13 2014

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• Bruno Berselli, A description of the recursive method in Comments lines: website Matem@ticamente (in Italian), 2008.
• Index to sequences related to pyramidal numbers.
• Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Crossrefs

Cf. similar sequences listed in A237616.

Cf. A000567, A051866, A194454.

Programs

GAP

List([0..40], n-> n*(n+1)*(4*n-3)/2); # G. C. Greubel, Aug 30 2019

Magma

[(4*n^3+n^2-3*n)/2: n in [0..50]]; // Vincenzo Librandi, Jan 01 2014

Maple

seq(n*(n+1)*(4*n-3)/2, n=0..40); # G. C. Greubel, Aug 30 2019

Mathematica

f[n_]:= n(n+1)(4n-3)/2; Array[f, 40, 0]
LinearRecurrence[{4,-6,4,-1},{0,1,15,54},40] (* Harvey P. Dale, Jan 29 2013 *)
CoefficientList[Series[x (1+11x)/(1-x)^4, {x, 0, 40}], x] (* Vincenzo Librandi, Jan 01 2014 *)

PARI

a(n)=(4*n^3+n^2-3*n)/2 \\ Charles R Greathouse IV, Oct 07 2015

Sage

[n*(n+1)*(4*n-3)/2 for n in (0..40)] # G. C. Greubel, Aug 30 2019

Formula

a(n) = n*(n+1)*(4*n-3)/2.

From Bruno Berselli, Dec 15 2010: (Start)

G.f.: x*(1+11*x)/(1-x)^4.

a(n) = Sum_{i=0..n} A051866(i). (End)

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)=15, a(3)=54. - Harvey P. Dale, Jan 29 2013

a(n) = Sum_{i=0..n-1} (n-i)*(12*i+1), with a(0)=0. - Bruno Berselli, Feb 10 2014

From Amiram Eldar, Jan 10 2022: (Start)

Sum_{n>=1} 1/a(n) = 4*Pi/21 + 8*log(2)/7 - 2/7.

Sum_{n>=1} (-1)^(n+1)/a(n) = 4*sqrt(2)*Pi/21 + 8*sqrt(2)*log(sqrt(2)+2)/21 - (20 + 4*sqrt(2))*log(2)/21 + 2/7. (End)

E.g.f.: exp(x)*x*(2 + 13*x + 4*x^2)/2. - Elmo R. Oliveira, Aug 04 2025

Extensions

Edited by Bruno Berselli, Dec 14 2010