A328290 Table T(b,n) = #{ k > 0 | nk has only distinct and nonzero digits in base b }, b >= 2, 1 <= n <= A051846(b).
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, 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, 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
Offset: 2
Examples
The table reads: (column n >= 2 corresponds to the base) B \ n = 1 2 3 4 5 6 7 8 9 10 ... 2 1 (0 ...) 3 4 1 0 0 1 0 1 (0 ...) 4 15 5 9 0 2 3 2 0 4 1 ... 5 64 42 21 9 0 14 8 4 7 0 ... 6 325 130 65 65 161 0 48 23 32 66 ... 7 1956 651 1140 319 386 221 0 156 362 128 ... 8 13699 5871 4506 1957 2748 1944 6277 0 1470 1189 ... 9 109600 73588 27400 56930 21973 18397 15641 8305 0 14826 ... 10 986409 438404 572175 219202 109601 255752 140515 109601 432645 0 ... (...) In base 2, 1 is the only number with distinct nonzero digits, so T(2,1) = 1, T(2,n) = 0 for n > 1. 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. 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.
Crossrefs
Programs
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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}
Comments