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A065071 Minimum number of identical bricks of length 1 which, when stacked without mortar in the naive way, form a stack of length >=n.

%I A065071 #44 Feb 16 2025 08:32:45
%S A065071 1,5,32,228,1675,12368,91381,675215,4989192,36865413,272400601,
%T A065071 2012783316,14872568832,109894245430,812014744423,6000022499694,
%U A065071 44334502845081,327590128640501,2420581837980562,17885814992891027
%N A065071 Minimum number of identical bricks of length 1 which, when stacked without mortar in the naive way, form a stack of length >=n.
%C A065071 Note that one can do "better" in terms of projections if one groups the bricks asymmetrically into lozenges with holes. See the Ainsley and Drummond references. Ainsley considers only the case of four bricks, but achieves an overhang of (15 - 4*sqrt(2))/8, compared with 25/24 for the harmonic pile. - D. G. Rogers, Aug 31 2005
%C A065071 Lim_{n -> inf} a(n)/a(n-1) = exp(2). - _Robert G. Wilson v_, Jan 26 2017
%D A065071 N. J. A. Sloane, Illustration for sequence M4299 (=A007340) in The Encyclopedia of Integer Sequences (with Simon Plouffe), Academic Press, 1995.
%H A065071 Robert G. Wilson v, <a href="/A065071/b065071.txt">Table of n, a(n) for n = 1..1000</a>
%H A065071 S. Ainley, <a href="http://www.jstor.org/stable/3618049">Finely Balanced</a>, Math. Gaz., 63 (1979), 272.
%H A065071 R. Dickau, <a href="http://www.robertdickau.com/BookStacking.html">Harmonic numbers and the book-stacking problem</a>
%H A065071 J. E. Drummond, <a href="http://www.jstor.org/stable/3617937">On stacking bricks to achieve a large overhang</a>, Math. Gaz., 65 (1981), 40-42.
%H A065071 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/BookStackingProblem.html">Book Stacking Problem</a>
%F A065071 a(n) = A002387(2n) + 1 = A014537(n) + 1.
%e A065071 Obviously a(1)=1. If the center of gravity of one brick is placed at the end of a second brick, the length of the stack of 2 bricks is 1.5. If the c.g. of that stack is placed at the end of a third brick, the length of the stack is 1.75. Continuing, we get a stack of length 1.916666... for 4 bricks and a stack of length 2.0416666... for 5 bricks. Thus a(2)=5.
%t A065071 A002387[n_] := Floor[ Exp[n - EulerGamma] + 1/2]; a[n_] := A002387[2n - 2] + 1; a[1] = 1; Table[a[n], {n, 1, 20}] (* _Jean-François Alcover_, Dec 13 2011, after _Charles R Greathouse IV_ *)
%t A065071 f[n_] := k /. FindRoot[HarmonicNumber[k -1] == 2n, {k, Exp[ 2n]}, WorkingPrecision -> 100] // Ceiling; Array[f, 21, 0] (* _Robert G. Wilson v_, Jan 26 2017 after _Jean-François Alcover_ in A014537 *) (* note that the index is off by one *)
%Y A065071 Cf. harmonic numbers H(n) = A001008/A002805, A002387, A004080.
%K A065071 nonn,easy
%O A065071 1,2
%A A065071 _John W. Layman_, Nov 08 2001
%E A065071 More terms from _Vladeta Jovovic_, Nov 14 2001