Search: id:A005995 Results 1-1 of 1 results found. %I A005995 M2916 %S A005995 1,3,12,28,66,126,236,396,651,1001,1512,2184,3108,4284, %T A005995 5832,7752,10197,13167,16852,21252,26598,32890,40404,49140, %U A005995 59423,71253,85008,100688,118728,139128,162384,188496,218025 %N A005995 Alkane (or paraffin) numbers l(8,n). %C A005995 Contribution from M. F. Hasler (MHasler(AT)univ-ag.fr), May 01 2009: (Start) %C A005995 Also, number of 5-element subsets of {1,...,n+5} whose elements sum to an odd integer, i.e. column 5 of A159916. %C A005995 A linear recurrent sequence with constant coefficients and characteristic polynomial x^9 - 3*x^8 + 8*x^6 - 6*x^5 - 6*x^4 + 8*x^3 - 3*x + 1. (End) %D A005995 N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence). %D A005995 S. M. Losanitsch, Die Isomerie-Arten bei den Homologen der Paraffin-Reihe, Chem. Ber. 30 (1897), 1917-1926. %D A005995 L. Smith, Polynomial invariants of finite groups, Bull. Am. Math. Soc. 34 (1997), 211-250. %D A005995 Winston C. Yang (paper in preparation). %H A005995 N. J. A. Sloane, Classic Sequences %F A005995 G.f.: (1+3*x^2)/((1-x)^3*(1-x^2)^3). %F A005995 a(2n-1)=n(n+1)(n+2)(2n+1)(2n+3)/30, a(2n)=(n+1)(n+2)(n+3)(4n^2+6n+5)/ 30. [From M. F. Hasler (MHasler(AT)univ-ag.fr), May 01 2009] %p A005995 (Maple) a := n -> (Matrix([[1, 0$6, -3, -9]]).Matrix(9, (i,j)-> if (i=j-1) then 1 elif j=1 then [3, 0, -8, 6, 6, -8, 0, 3, -1][i] else 0 fi)^n)[1, 1]; seq (a(n), n=0..32); [From Alois P. Heinz (heinz(AT)hs-heilbronn.de), Jul 31 2008] %o A005995 (PARI) a(k)= if(k%2,(k+1)*(k+3)*(k+5),(k+6)*(k^2+3*k+5))*(k+2)*(k+4)/ 240 [From M. F. Hasler (MHasler(AT)univ-ag.fr), May 01 2009] %Y A005995 Sequence in context: A066643 A140065 A115549 this_sequence A034503 A026557 A124052 %Y A005995 Adjacent sequences: A005992 A005993 A005994 this_sequence A005996 A005997 A005998 %K A005995 nonn %O A005995 0,2 %A A005995 N. J. A. Sloane (njas(AT)research.att.com), Winston C. Yang (yang(AT)math.wisc.edu) Search completed in 0.001 seconds