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A005891 Centered pentagonal numbers: (5n^2+5n+2)/2; crystal ball sequence for 3.3.3.4.4. planar net.

%I A005891 M4112 #127 Mar 26 2025 08:32:35
%S A005891 1,6,16,31,51,76,106,141,181,226,276,331,391,456,526,601,681,766,856,
%T A005891 951,1051,1156,1266,1381,1501,1626,1756,1891,2031,2176,2326,2481,2641,
%U A005891 2806,2976,3151,3331,3516,3706,3901,4101,4306,4516,4731,4951,5176,5406
%N A005891 Centered pentagonal numbers: (5n^2+5n+2)/2; crystal ball sequence for 3.3.3.4.4. planar net.
%C A005891 Equals the triangular numbers convolved with [1, 3, 1, 0, 0, 0, ...]. - _Gary W. Adamson_ and _Alexander R. Povolotsky_, May 29 2009
%C A005891 From _Ant King_, Jun 15 2012: (Start)
%C A005891 a(n) == 1 (mod 5) for all n.
%C A005891 The digital roots of the a(n) form a purely periodic palindromic 9-cycle 1, 6, 7, 4, 6, 4, 7, 6, 1.
%C A005891 The units' digits of the a(n) form a purely periodic palindromic 4-cycle 1, 6, 6, 1.
%C A005891 (End)
%C A005891 Binomial transform of (1, 5, 5, 0, 0, 0, ...) and second partial sum of (1, 4, 5, 5, 5, ...). - _Gary W. Adamson_, Sep 09 2015
%C A005891 a(n) = a(-1-n) for all n in Z. - _Michael Somos_, Jan 25 2019
%C A005891 On the plane start with a single regular pentagon, and repeat the following procedure, "For each edge of any pentagon not already connected to an existing pentagon create a mirror image such that the mirror image does not overlap with an existing pentagon." a(n) is the number of pentagons occupying the plane after n repetitions. - _Torlach Rush_, Sep 14 2022
%D A005891 N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%D A005891 B. K. Teo and N. J. A. Sloane, Magic numbers in polygonal and polyhedral clusters, Inorgan. Chem. 24 (1985), 4545-4558.
%H A005891 T. D. Noe, <a href="/A005891/b005891.txt">Table of n, a(n) for n = 0..1000</a>
%H A005891 Paul Barry, <a href="https://arxiv.org/abs/2104.01644">Centered polygon numbers, heptagons and nonagons, and the Robbins numbers</a>, arXiv:2104.01644 [math.CO], 2021.
%H A005891 Simon Plouffe, <a href="https://arxiv.org/abs/0911.4975">Approximations de séries génératrices et quelques conjectures</a>, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
%H A005891 Simon Plouffe, <a href="/A000051/a000051_2.pdf">1031 Generating Functions</a>, Appendix to Thesis, Montreal, 1992
%H A005891 Cliff Reiter, <a href="http://hdl.handle.net/10385/2864">Polygonal Numbers and Fifty One Stars</a>, Lafayette College, Easton, PA (2019).
%H A005891 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/CenteredPentagonalNumber.html">Centered Pentagonal Number.</a>
%H A005891 <a href="/index/Ce#CENTRALCUBE">Index entries for sequences related to centered polygonal numbers</a>
%H A005891 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%H A005891 <a href="/index/Cor#crystal_ball">Index entries for crystal ball sequences</a>
%F A005891 G.f.: (1 + 3*x + x^2)/(1 - x)^3. _Simon Plouffe_ in his 1992 dissertation
%F A005891 Narayana transform (A001263) of [1, 5, 0, 0, 0, ...]. - _Gary W. Adamson_, Dec 29 2007
%F A005891 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), a(0)=1, a(1)=6, a(2)=16. - _Jaume Oliver Lafont_, Dec 02 2008
%F A005891 a(n) = 5*A000217(n) + 1 = 5*T(n) + 1, for n = 0, 1, 2, 3, ... and where T(n) = n*(n+1)/2 = n-th triangular number. - _Thomas M. Green_, Nov 25 2009
%F A005891 a(n) = a(n-1) + 5*n, with a(0)=1. - _Vincenzo Librandi_, Nov 18 2010
%F A005891 a(n) = A028895(n) + 1. - _Omar E. Pol_, Oct 03 2011
%F A005891 a(n) = 2*a(n-1) - a(n-2) + 5. - _Ant King_, Jun 12 2012
%F A005891 Sum_{n>=0} 1/a(n) = 2*Pi /sqrt(15) *tanh(Pi/2*sqrt(3/5)) = 1.360613169863... - _Ant King_, Jun 15 2012
%F A005891 a(n) = A101321(5,n). - _R. J. Mathar_, Jul 28 2016
%F A005891 E.g.f.: (2 + 10*x + 5*x^2)*exp(x)/2. - _Ilya Gutkovskiy_, Jul 28 2016
%F A005891 From _Amiram Eldar_, Jun 20 2020: (Start)
%F A005891 Sum_{n>=0} a(n)/n! = 17*e/2.
%F A005891 Sum_{n>=0} (-1)^(n+1)*a(n)/n! = 3/(2*e). (End)
%e A005891 a(2)= 5*T(2) + 1 = 5*3 + 1 = 16, a(4) = 5*T(4) + 1 = 5*10 + 1 = 51. - _Thomas M. Green_, Nov 16 2009
%p A005891 A005891 := proc(n)
%p A005891     1+5*n*(1+n)/2 ;
%p A005891 end proc:
%p A005891 seq(A005891(n),n=0..40) ; # _R. J. Mathar_, Oct 07 2021
%t A005891 FoldList[#1 + #2 &, 1, 5 Range@ 40] (* _Robert G. Wilson v_, Feb 02 2011 *)
%t A005891 LinearRecurrence[{3,-3,1},{1,6,16},50] (* _Harvey P. Dale_, Sep 08 2018 *)
%t A005891 Table[ j! Coefficient[Series[Exp[x]*(1 + 5 x^2/2)-1, {x, 0, 20}], x, j], {j, 0, 20}] (* _Nikolaos Pantelidis_, Feb 07 2023 *)
%o A005891 (PARI) a(n)=5*n*(n+1)/2+1 \\ _Charles R Greathouse IV_, Mar 22 2016
%o A005891 (Magma) [5*n*(n+1)/2 + 1: n in [0..50]]; // _G. C. Greubel_, Nov 04 2017
%o A005891 (Python)
%o A005891 def A005891(n): return (5*n*(n+1)>>1)+1 # _Chai Wah Wu_, Mar 25 2025
%Y A005891 Cf. A028895, A001844, A003215, A004068 (partial sums), A006322, A001263.
%Y A005891 Partial sums of A008706.
%Y A005891 Equals second row of A167546 divided by 2.
%K A005891 nonn,easy
%O A005891 0,2
%A A005891 _N. J. A. Sloane_
%E A005891 Formula corrected and relocated by _Johannes W. Meijer_, Nov 07 2009