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%I #10 Jun 22 2022 20:24:05
%S 2,3,4,4,5,6,6,6,7,7,8,8,8,9,9,10,10,10,10,10,11,11,12,12,12,12,12,12,
%T 13,13,13,14,14,14,14,15,15,15,15,16,16,16,16,16,16,16,16,17,17,17,18,
%U 18,18,18,18,18,18,18,18,19,19,19,19,20,20,20,20,20,20,21,21,21,21,21,22,22,22,22,22,22,22,22,22,22,22,23
%N Lazy cutter's sequence (see Comments).
%C From the infinite sequence G of fractions that may be used to demonstrate the countability of rational numbers, where a(n) = A092542(n)/A092543(n), form a new sequence H by taking only those terms of G that are proper fractions unequal to a fraction that appears earlier in H (making H the list of all proper fractions without repetitions). Let b/c be the n-th term of H and b be the number of congruent pizzas that have to be equally divided between c people by means of radial cuts. a(n) is the minimum number of cuts to achieve such a division.
%C H can be directly calculated as its n-th term equals A182972(n)/A182973(n). H starts with 1/2, 1/3, 1/4, 2/3, 1/5, 1/6, 2/5, 3/4, 3/5, 1/7, 1/8, 2/7, 4/5, 3/7, 1/9.
%C As a(n) is equal for all proper fractions b/c such that b + c = n, counting the number of equal consecutive terms of this sequence gives A023022 from its third term onwards (see _Geoffrey Critzer_'s and _Reinhard Zumkeller_'s comments at A023022).
%F h(n) = A182972(n)/A182973(n) = b/c, c = x*b + r and a(n) = (x+1)*b + r - 1.
%e To equally divide 4 pizzas between 7 people we can divide each pizza into 7 equal parts with 7 radial cuts making the total number of cuts 28 (far from minimal). Ancient Egyptians, representing 4/7 as 1/2 + 1/14, would cut all pizzas into halves (8 cuts) and one of the halves into 7 equal pieces (6 additional cuts), making the total number of cuts 8 + 6 = 14. We can do even better by cutting each pizza into two pieces (3/7 and 4/7), for a total of 8 cuts, and dividing one 3/7 piece to 3 equal pieces (2 additional cuts), minimizing the total number of cuts to 8 + 2 = 10. Since the 19th term of H sequence is 4/7, a(19) = 10.
%t a092542=Flatten[Table[Join[Range[2n-1],Reverse@Range[2n-2]],{n,12}]];
%t a092543=Take[Cases[Import["https://oeis.org/A092543/b092543.txt","Table"],{_,_}][[All,2]],276];g=a092542/a092543; h=DeleteDuplicates[Select[g,#<1&]];
%t a[n_]:=Module[{x=Floor[Denominator[n]/Numerator[n]],r=Mod[Denominator[n],
%t Numerator[n]]},(x+1)*Numerator[n]+r-1];a/@h
%Y Cf. A023022, A092542, A092543, A182972, A182973.
%K nonn
%O 1,1
%A _Ivan N. Ianakiev_, May 31 2022
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