A007585 10-gonal (or decagonal) pyramidal numbers: a(n) = n*(n + 1)*(8*n - 5)/6.

0, 1, 11, 38, 90, 175, 301, 476, 708, 1005, 1375, 1826, 2366, 3003, 3745, 4600, 5576, 6681, 7923, 9310, 10850, 12551, 14421, 16468, 18700, 21125, 23751, 26586, 29638, 32915, 36425, 40176, 44176, 48433, 52955, 57750

Offset: 0

Author

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

Keywords

• nonn
• easy
• nice

Comments

Binomial transform of [1, 10, 17, 8, 0, 0, 0, ...] = (1, 11, 38, 90, ...). - Gary W. Adamson, Mar 18 2009

This sequence is related to A000384 by a(n) = n*A000384(n) - Sum_{i=0..n-1} A000384(i) and this is the case d=4 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, Apr 21 2010

For n>0, (a(n)) is the principal diagonal of the convolution array A213750. - Clark Kimberling, Jun 20 2012

From Ant King, Oct 30 2012: (Start)

The partial sums of the figurate decagonal numbers A001107.

For n>1, the digital roots of this sequence A010888(A007585(n)) form the purely periodic 27-cycle {1,2,2,9,4,4,8,6,6,7,8,8,6,1,1,5,3,3,4,5,5,3,7,7,2,9,9}.

For n>1, the units’ digits of this sequence A010879(A007585(n)) form the purely periodic 20-cycle {1,1,8,0,5,1,6,8,5,5,6,6,3,5,0,6,1,3,0,0}.

(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

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

Crossrefs

Cf. A000384.

Cf. A093565 ((8, 1) Pascal, column m=3). Partial sums of A001107.

Cf. similar sequences listed in A237616.

Programs

GAP

List([0..40], n-> n*(n+1)*(8*n-5)/6); # G. C. Greubel, Aug 30 2019

Magma

[n*(n+1)*(8*n-5)/6: n in [0..40]]; // G. C. Greubel, Aug 30 2019

Maple

seq(n*(n+1)*(8*n-5)/6, n=0..40); # G. C. Greubel, Aug 30 2019

Mathematica

Table[n(n+1)(8n-5)/6, {n,0,40}] (* Vladimir Joseph Stephan Orlovsky, Apr 18 2011 *)
LinearRecurrence[{4,-6,4,-1},{0,1,11,38},40] (* Harvey P. Dale, Dec 20 2011 *)

PARI

a(n)=(8*n^3+3*n^2-5*n)/6 \\ Charles R Greathouse IV, Sep 24 2015

Sage

[n*(n+1)*(8*n-5)/6 for n in (0..40)] # G. C. Greubel, Aug 30 2019

Formula

a(n) = (8*n-5)*binomial(n+1, 2)/3.

G.f.: x*(1+7*x)/(1-x)^4.

a(n) = (8*n^3 + 3*n^2 - 5*n)/6. - Vincenzo Librandi, Aug 01 2010

a(0)=0, a(1)=1, a(2)=11, a(3)=38, a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Harvey P. Dale, Dec 20 2011

From Ant King, Oct 30 2012: (Start)

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

a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 8.

a(n) = (n+1)*(2*A001107(n) + n)/6.

a(n) = A000292(n) + 7*A000292(n-1).

a(n) = A007584(n) + A000292(n-1).

a(n) = A000217(n) + 8*A000292(n-1).

a(n) = binomial(n+2,3) + 7*binomial(n+1,3).

Sum_{n>=1} 1/a(n) = 6*(4*pi*(sqrt(2)-1) + 4*(8-sqrt(2))*log(2) + 8*sqrt(2)*log(2-sqrt(2))-5)/65 = 1.145932345...

(End)

a(n) = Sum_{i=0..n-1} (n-i)*(8*i+1), with a(0)=0. - Bruno Berselli, Feb 10 2014

E.g.f.: x*(6 + 27*x + 8*x^2)*exp(x)/6. - Ilya Gutkovskiy, May 12 2017