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The number of homogeneous trisubstituted linear alkanes.
2

%I #14 Dec 01 2021 12:36:58

%S 2,6,16,36,70,122,196,296,426,590,792,1036,1326,1666,2060,2512,3026,

%T 3606,4256,4980,5782,6666,7636,8696,9850,11102,12456,13916,15486,

%U 17170,18972,20896,22946,25126,27440,29892,32486

%N The number of homogeneous trisubstituted linear alkanes.

%C See the paper by Valentin Vankov Iliev for details.

%C This sequence is related to A152947 by a(n) = (n-1)*A152947(n) + sum( A152947(i), i=1..n-1 ). - _Bruno Berselli_, Dec 19 2013

%H Valentin Vankov Iliev, <a href="https://doi.org/10.1007/s10910-009-9534-4">A mathematical characterization of the groups of substitution isomerism of the linear alkanes</a>, J. Math. Chem. 47 (2010), 52-61.

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

%F a(n) = (1/3)*(2*n^3 - 9*n^2 + 19*n - 12), where n is the number of carbons.

%F a(n) = 2*A081489(n-1) = (n-1)*(2*n^2-7*n+12)/3. - _R. J. Mathar_, Apr 28 2009

%F G.f.: 2*x^2*(1-x+2*x^2)/(1-x)^4. - _Colin Barker_, Aug 06 2012

%e The number of homogeneous trisubstituted linear alkane with ten carbon atoms is 426.

%Y Cf. A002522, A033816, A081489, A152947, A159940, A159941.

%K nonn,easy

%O 2,1

%A _Parthasarathy Nambi_, Apr 26 2009

%E More terms from _Colin Barker_, Aug 06 2012