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%I #64 Sep 08 2022 08:45:50
%S 0,1,15,54,130,255,441,700,1044,1485,2035,2706,3510,4459,5565,6840,
%T 8296,9945,11799,13870,16170,18711,21505,24564,27900,31525,35451,
%U 39690,44254,49155,54405,60016,66000,72369,79135,86310,93906,101935,110409,119340
%N a(n) = (4*n^3 + n^2 - 3*n)/2.
%C 14-gonal (or tetradecagonal) pyramidal numbers generated by the formula n*(n+1)*(2*d*n-(2*d-3))/6 for d=6.
%C 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
%C Except for the initial 0, this is the principal diagonal of the convolution array A213761. - _Clark Kimberling_, Jul 04 2012
%C Starting (1, 15, 54, ...), this is the binomial transform of (1, 14, 25, 12, 0, 0, 0, ...). - _Gary W. Adamson_, Jul 29 2015
%D E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 93. - _Bruno Berselli_, Feb 13 2014
%H Vincenzo Librandi, <a href="/A172073/b172073.txt">Table of n, a(n) for n = 0..1000</a>
%H Bruno 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), 2008.
%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) = n*(n+1)*(4*n-3)/2.
%F G.f.: x*(1+11*x)/(1-x)^4. - _Bruno Berselli_, Dec 15 2010
%F a(n) = Sum_{i=0..n} A051866(i). - _Bruno Berselli_, Dec 15 2010
%F a(0)=0, a(1)=1, a(2)=15, a(3)=54; for n > 3, a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - _Harvey P. Dale_, Jan 29 2013
%F a(n) = Sum_{i=0..n-1} (n-i)*(12*i+1), with a(0)=0. - _Bruno Berselli_, Feb 10 2014
%F From _Amiram Eldar_, Jan 10 2022: (Start)
%F Sum_{n>=1} 1/a(n) = 4*Pi/21 + 8*log(2)/7 - 2/7.
%F 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)
%p seq(n*(n+1)*(4*n-3)/2, n=0..40); # _G. C. Greubel_, Aug 30 2019
%t f[n_]:= n(n+1)(4n-3)/2; Array[f, 40, 0]
%t LinearRecurrence[{4,-6,4,-1},{0,1,15,54},40] (* _Harvey P. Dale_, Jan 29 2013 *)
%t CoefficientList[Series[x (1+11x)/(1-x)^4, {x, 0, 40}], x] (* _Vincenzo Librandi_, Jan 01 2014 *)
%o (Magma) [(4*n^3+n^2-3*n)/2: n in [0..50]]; // _Vincenzo Librandi_, Jan 01 2014
%o (PARI) a(n)=(4*n^3+n^2-3*n)/2 \\ _Charles R Greathouse IV_, Oct 07 2015
%o (Sage) [n*(n+1)*(4*n-3)/2 for n in (0..40)] # _G. C. Greubel_, Aug 30 2019
%o (GAP) List([0..40], n-> n*(n+1)*(4*n-3)/2); # _G. C. Greubel_, Aug 30 2019
%Y Cf. similar sequences listed in A237616.
%Y Cf. A051866, A194454.
%K nonn,easy
%O 0,3
%A _Vincenzo Librandi_, Jan 25 2010
%E Edited by _Bruno Berselli_, Dec 14 2010
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