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%I #21 Jan 09 2024 16:30:55
%S 1,0,2,7,27,114,523,2589,13744,77821,467767,2972432,19895813,
%T 139824045,1028804338,7905124379,63287544055,526827208698,
%U 4551453462543,40740750631417,377254241891064,3608700264369193,35613444194346451,362161573323083920,3790824599495473121
%N A000110 / (1,2,3,...): (convolved with (1,2,3,...) = Bell numbers).
%C This is the sequence which must be convolved with (1,2,3,...), offset 0, to generate the Bell numbers starting (1, 2, 5, 15, 52, ...) offset 1;
%C equivalent to row sums of triangle A154109 = (1, 2, 5, 15, 52, ...).
%C A variant of A011965. - _R. J. Mathar_, Jan 07 2009
%F A000110 / (1,2,3,...); where A000110 (the Bell numbers) begins with offset 1: (1, 2, 5, 15, 52, 203, 877, ...).
%F G.f.: (A000110(x)-1)*(x-1)^2, where A000110(x) is the g.f. of the Bell numbers. - _R. J. Mathar_, Nov 27 2018
%e A000110(5) = 52 = (1, 0, 2, 7, 27) convolved with (1, 2, 3, 4, 5) = (5 + 0 + 6 + 14 + 27).
%t nmax = 30; Table[a[j]/.SolveAlways[Table[Sum[a[k]*(n-k), {k, 0, n}]==BellB[n], {n, 1, nmax+1}], a][[1]], {j, 0, nmax}] (* _Vaclav Kotesovec_, Jul 26 2021 *)
%Y Cf. A000110, A154109.
%K nonn
%O 1,3
%A _Gary W. Adamson_, Jan 04 2009
%E More terms from _Vaclav Kotesovec_, Jul 26 2021
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