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%I #27 Sep 08 2022 08:45:49
%S 0,1,32896,21526641,2147516416,76294140625,1410555793536,
%T 16616468167201,140737496743936,926510115949281,5000000050000000,
%U 22974865038965521,92442129662509056,332708304999455281,1088976669642580096
%N a(n) = n^8*(n^8 + 1)/2.
%C Number of unoriented rows of length 16 using up to n colors. For a(0)=0, there are no rows using no colors. For a(1)=1, there is one row using that one color for all positions. For a(2)=32896, there are 2^16=65536 oriented arrangements of two colors. Of these, 2^8=256 are achiral. That leaves (65536-256)/2=32640 chiral pairs. Adding achiral and chiral, we get 32896. - _Robert A. Russell_, Nov 13 2018
%H Vincenzo Librandi, <a href="/A170780/b170780.txt">Table of n, a(n) for n = 0..10000</a>
%H <a href="/index/Rec#order_17">Index entries for linear recurrences with constant coefficients</a>, signature (17, -136, 680, -2380, 6188, -12376, 19448, -24310, 24310, -19448, 12376, -6188, 2380, -680, 136, -17, 1).
%F G.f.: (x + 32879*x^2 + 20967545*x^3 + 1786036695*x^4 + 42691617829* x^5 + 391057805899*x^6 + 1603741496717*x^7 + 3191399514435*x^8 + 3191399514435*x^9 + 1603741496717*x^10 + 391057805899*x^11 + 42691617829*x^12 + 1786036695*x^13 + 20967545*x^14 + 32879*x^15 + x^16) /(1-x)^17. - _G. C. Greubel_, Dec 05 2017
%F From _Robert A. Russell_, Nov 13 2018: (Start)
%F a(n) = (A010804(n) + A001016(n)) / 2 = (n^16 + n^8) / 2.
%F G.f.: (Sum_{j=1..16} S2(16,j)*j!*x^j/(1-x)^(j+1) + Sum_{j=1..8} S2(8,j)*j!*x^j/(1-x)^(j+1)) / 2, where S2 is the Stirling subset number A008277.
%F G.f.: x*Sum_{k=0..15} A145882(16,k) * x^k / (1-x)^17.
%F E.g.f.: (Sum_{k=1..16} S2(16,k)*x^k + Sum_{k=1..8} S2(8,k)*x^k) * exp(x) / 2, where S2 is the Stirling subset number A008277.
%F For n>16, a(n) = Sum_{j=1..17} -binomial(j-18,j) * a(n-j). (End)
%t Table[n^8*(n^8+1)/2, {n, 0, 30}] (* _G. C. Greubel_, Dec 05 2017 *)
%o (Magma) [n^8*(n^8+1)/2: n in [0..30]]; // _Vincenzo Librandi_, Aug 26 2011
%o (PARI) for(n=0, 30, print1(n^8*(n^8+1)/2, ", ")) \\ _G. C. Greubel_, Dec 05 2017
%o (Sage) [n^8*(n^8+1)/2 for n in range(30)] # _G. C. Greubel_, Nov 15 2018
%o (GAP) List([0..30], n -> n^8*(n^8+1)/2); # _G. C. Greubel_, Nov 15 2018
%o (Python) for n in range(0,20): print(int(n**8*(n**8 + 1)/2), end=', ') # _Stefano Spezia_, Nov 15 2018
%Y Row 16 of A277504.
%Y Cf. A010804 (oriented), A001016 (achiral).
%K nonn,easy
%O 0,3
%A _N. J. A. Sloane_, Dec 11 2009
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