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%I #11 Oct 14 2019 11:29:40
%S 1,4,1,0,0,1,0,1,15,5,9,0,2,3,2,0,4,1,1,0,2,1,2,0,0,1,1,0,0,0,0,0,0,0,
%T 2,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,1,64,42,
%U 21,9,0,14,8,4,7,0,6,4,3,6,0,3,2,5,2,0,4,5,3,2,0,0,2,1,2,0,2,1,2,1,0,1,2,1,2,0,1,3,1,1,0,1,1,2,1,0,0,0,0,2
%N Table T(b,n) = #{ k > 0 | nk has only distinct and nonzero digits in base b }, b >= 2, 1 <= n <= A051846(b).
%C The table could also be considered as an infinite square array with T(b,n) = 0 for n > A051846(b) = the largest pandigital number in base b.
%C Can anyone find a simple formula for the index of the last terms > 1 in each row b?
%F T(b,b) = 0, since any multiple of b has a trailing digit 0 in base b.
%F T(b,A051846(b)) = 1 and T(b,n) = 0 for n > A051846(b) = (b-1)(b-2)..21 in base b.
%e The table reads: (column n >= 2 corresponds to the base)
%e B \ n = 1 2 3 4 5 6 7 8 9 10 ...
%e 2 1 (0 ...)
%e 3 4 1 0 0 1 0 1 (0 ...)
%e 4 15 5 9 0 2 3 2 0 4 1 ...
%e 5 64 42 21 9 0 14 8 4 7 0 ...
%e 6 325 130 65 65 161 0 48 23 32 66 ...
%e 7 1956 651 1140 319 386 221 0 156 362 128 ...
%e 8 13699 5871 4506 1957 2748 1944 6277 0 1470 1189 ...
%e 9 109600 73588 27400 56930 21973 18397 15641 8305 0 14826 ...
%e 10 986409 438404 572175 219202 109601 255752 140515 109601 432645 0 ...
%e (...)
%e In base 2, 1 is the only number with distinct nonzero digits, so T(2,1) = 1, T(2,n) = 0 for n > 1.
%e In base 3, {1, 2, 12_3 = 5, 21_3 = 7} are the only numbers with distinct nonzero digits, so T(3,1) = 4, T(3,2) = T(3,7) = T(3,7) = 1, T(3,n) = 0 for n > 7.
%e In base 4, {1, 2, 3, 12_4 = 6, 13_4 = 7, 21_4 = 9, ..., 321_4 = 57} are the only numbers with distinct nonzero digits, so T(4,n) = 0 for n > 57.
%o (PARI) T(B,n)={my(S,T,U); for(L=1,B-1,T=vectorv(L,k,B^(k-1)); forperm(L,p, U=vecextract(T,p); forvec(D=vector(L,i,[1,B-1]),D*U%n||S++,2)));S}
%Y Cf. A328287 (row 10), A328288, A328277.
%Y Column 1 is A007526 (number of nonnull variations of n distinct objects).
%K nonn,base,tabf
%O 2,2
%A _M. F. Hasler_, Oct 11 2019
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