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10-gonal (or decagonal) pyramidal numbers: a(n) = n*(n + 1)*(8*n - 5)/6.
(Formerly M4791)
16

%I M4791 #64 Sep 08 2022 08:44:35

%S 0,1,11,38,90,175,301,476,708,1005,1375,1826,2366,3003,3745,4600,5576,

%T 6681,7923,9310,10850,12551,14421,16468,18700,21125,23751,26586,29638,

%U 32915,36425,40176,44176,48433,52955,57750

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

%C Binomial transform of [1, 10, 17, 8, 0, 0, 0, ...] = (1, 11, 38, 90, ...). - _Gary W. Adamson_, Mar 18 2009

%C 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

%C For n>0, (a(n)) is the principal diagonal of the convolution array A213750. - _Clark Kimberling_, Jun 20 2012

%C From _Ant King_, Oct 30 2012: (Start)

%C The partial sums of the figurate decagonal numbers A001107.

%C 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}.

%C 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}.

%C (End)

%D A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 194.

%D E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 93.

%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

%H Harvey P. Dale, <a href="/A007585/b007585.txt">Table of n, a(n) for n = 0..1000</a>

%H B. Berselli, A description of the recursive method in Comments lines: website <a href="http://www.lanostra-matematica.org/2008/12/sequenze-numeriche-e-procedimenti.html">Matem@ticamente</a> (in Italian).

%H <a href="/index/Ps#pyramidal_numbers">Index to sequences related to pyramidal numbers</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1)

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

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

%F a(n) = (8*n^3 + 3*n^2 - 5*n)/6. - _Vincenzo Librandi_, Aug 01 2010

%F 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

%F From _Ant King_, Oct 30 2012: (Start)

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

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

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

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

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

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

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

%F 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...

%F (End)

%F a(n) = Sum_{i=0..n-1} (n-i)*(8*i+1), with a(0)=0. - _Bruno Berselli_, Feb 10 2014

%F E.g.f.: x*(6 + 27*x + 8*x^2)*exp(x)/6. - _Ilya Gutkovskiy_, May 12 2017

%p seq(n*(n+1)*(8*n-5)/6, n=0..40); # _G. C. Greubel_, Aug 30 2019

%t Table[n(n+1)(8n-5)/6, {n,0,40}] (* _Vladimir Joseph Stephan Orlovsky_, Apr 18 2011 *)

%t LinearRecurrence[{4,-6,4,-1},{0,1,11,38},40] (* _Harvey P. Dale_, Dec 20 2011 *)

%o (PARI) a(n)=(8*n^3+3*n^2-5*n)/6 \\ _Charles R Greathouse IV_, Sep 24 2015

%o (Magma) [n*(n+1)*(8*n-5)/6: n in [0..40]]; // _G. C. Greubel_, Aug 30 2019

%o (Sage) [n*(n+1)*(8*n-5)/6 for n in (0..40)] # _G. C. Greubel_, Aug 30 2019

%o (GAP) List([0..40], n-> n*(n+1)*(8*n-5)/6); # _G. C. Greubel_, Aug 30 2019

%Y Cf. A000384.

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

%Y Cf. similar sequences listed in A237616.

%K nonn,easy,nice

%O 0,3

%A _N. J. A. Sloane_, _R. K. Guy_