A007585 10-gonal (or decagonal) pyramidal numbers: a(n) = n*(n + 1)*(8*n - 5)/6. (Formerly M4791)

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

0,3

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)

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

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 *)

Prog

(PARI) a(n)=(8*n^3+3*n^2-5*n)/6 \\ Charles R Greathouse IV, Sep 24 2015
(Magma) [n*(n+1)*(8*n-5)/6: n in [0..40]]; // G. C. Greubel, Aug 30 2019
(Sage) [n*(n+1)*(8*n-5)/6 for n in (0..40)] # G. C. Greubel, Aug 30 2019
(GAP) List([0..40], n-> n*(n+1)*(8*n-5)/6); # G. C. Greubel, Aug 30 2019

Crossrefs

Cf. A000384.

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

Cf. similar sequences listed in A237616.

Sequence in context: A139276 A010002 A143109 * A024202 A213775 A133258

Adjacent sequences: A007582 A007583 A007584 * A007586 A007587 A007588

Keyword

nonn,easy,nice

Author

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

Status

approved