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A217490 Let t be the length of the shortest computation yielding a positive multiple of n! using addition, subtraction and multiplication. Then a(n) is the least k > 0 such that k*n! can be computed in t steps.

%I A217490 #20 May 02 2013 13:57:05
%S A217490 1,1,1,1,2,1,13,26,11830,1183,1,561,48048,3432,3718,3718,
%T A217490 956689100690500088178176,187,8983799529705,6061484504517072231744,
%U A217490 26002249020,1181920410,8006931102170352452004696490160949546032818169320135140000
%N A217490 Let t be the length of the shortest computation yielding a positive multiple of n! using addition, subtraction and multiplication. Then a(n) is the least k > 0 such that k*n! can be computed in t steps.
%C A217490 Least k > 0 such that A173419(k*n!) = A217031(n).
%C A217490 Related to the algebraic version of the P =? NP problem, see A173419 and A217031.
%C A217490 a(n) = 1 if and only if A217031(n) = A217032(n).
%F A217490 Trivial bound: 1 <= a(n) <= 2^(2^(A217031(n))/n! <= 2^(2^(2n-2))/n! . Can this be improved?
%e A217490 a(1) = 1 since A173419(1!) = 0.
%e A217490 a(2) = 1 since A173419(2!) = 1.
%e A217490 a(3) = 1 since A173419(3!) = 3.
%e A217490 a(4) = 1 since A173419(4!) = 4.
%e A217490 a(5) = 2 since A173419(2*5!) = 5.
%e A217490 a(6) = 1 since A173419(6!) = 6.
%e A217490 a(7) = 13 since A173419(13*7!) = 6.
%e A217490 a(8) = 26 since A173419(26*8!) = 7.
%e A217490 a(9) = 11830 since A173419(11830*9!) = 7.
%e A217490 a(10) = 1183 since A173419(1183*10!) = 7.
%e A217490 a(11) = 1 since A173419(11!) = 9.
%e A217490 a(12) = 561 since A173419(561*12!) = 9.
%e A217490 a(22) = 1181920410
%e A217490 Because of the following 12 step computation:
%e A217490 1, 2, 4, 16, 256, 18, 324, 104976, 104720, 10993086720, 120847955633440358400, 10992982000, 1328479401015208457964748800000
%e A217490 The last number is 1181920410 * 22!
%K A217490 nonn,hard,more
%O A217490 1,5
%A A217490 _Charles R Greathouse IV_, Oct 04 2012
%E A217490 Extended until a(23) doing full enumeration of all 12 step computations, from _Gil Dogon_, May 02 2013