A328290 - OEIS

%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