A030289
a(n+1) is the next larger cube with no digits in common with a(n), a(0) = 0.
Original entry on oeis.org
0, 1, 8, 27, 64, 125, 343, 512, 4096, 5832, 64000, 91125, 300763, 941192, 3375000, 8489664, 13312053, 86444696669696, 100175333300307, 488224224494488, 510657175657000, 2233398984434344, 5177717000000000, 6393843393228864
Offset: 0
Squares whose digits do not appear in previous term:
A030288.
Primes whose digits do not appear in previous term:
A030284.
Cf.
A030290: cube roots of the terms.
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bb={0}; idi1=IntegerDigits[0]; Do[idi=IntegerDigits[r=i^3]; If[Intersection[idi, idi1]=={}, bb={bb, r}; idi1=idi], {i, 1, 100000}]; fla=Flatten[bb] (* Zak Seidov, Feb 17 2005 *)
Nest[Append[#, Block[{k = Last@ # + 1, m = IntegerDigits[Last[#]^3]}, While[IntersectingQ[IntegerDigits[k^3], m], k++]; k]] &, {0}, 23]^3 (* Michael De Vlieger, Nov 13 2017 *)
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next_A030289(n,D(n)=Set(digits(n)),S=D(n))={if(n>6e15,S[1]&&return(1000^(logint(n,1000)+1));n=n*20\9); for(k=sqrtnint(n,3)+1,oo,#setintersect(D(k^3),S)||return(k^3))} \\ M. F. Hasler, Nov 12 2017
A100373
Lexicographically earliest increasing sequence of composite numbers such that the digits of a(n) do not appear in a(n-1).
Original entry on oeis.org
4, 6, 8, 9, 10, 22, 30, 42, 50, 62, 70, 81, 90, 111, 200, 314, 500, 611, 700, 812, 900, 1111, 2000, 3111, 4000, 5111, 6000, 7111, 8000, 9111, 20000, 31111, 40000, 51111, 60000, 71111, 80000, 91111, 200000, 311113, 400000, 511112, 600000, 711111
Offset: 1
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f:= proc(x) local L,S,carry,m,nL,b,d0,Lz,z,i,d;
L:= convert(x,base,10);
nL:= nops(L);
S:= sort(convert({$0..9} minus convert(L,set),list));
b:= nops(S);
d0:= min(select(`>`,S,L[-1]));
if d0 = infinity then
if S[1] = 0 then Lz:= Vector([0$nL, S[2]])
else Lz:= Vector([S[1]$(nL+1)])
fi
else
Lz:= Vector([S[1]$(nL-1),d0])
fi;
d:= LinearAlgebra:-Dimension(Lz);
do
z:= add(Lz[i]*10^(i-1),i=1..d);
if not isprime(z) then return z fi;
carry:= true;
for i from 1 to d while carry do
if Lz[i] = S[-1] then Lz[i]:= S[1]
else
carry:= false; if member(Lz[i],S,'m') then Lz[i]:= S[m+1] fi
fi
od;
if carry then d:= d+1; if S[1] = 0 then Lz(d):= S[2] else Lz(d) := S[1] fi fi
od;
end proc:
R:= 4: r:= 4:
for i from 2 to 100 do
r:= f(r);
R:= R,r
od:
R; # Robert Israel, Feb 27 2025
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ta={1};Do[s1=IntegerDigits[Part[ta, Length[ta]]]; s2=IntegerDigits[n];If[Equal[Intersection[s1, s2], {}] &&!PrimeQ[n], Print[{Last[ta], n}];ta=Append[ta, n]], {n, 1, 1000000}];ta=Delete[ta, 1]
A294662
Least k > a(n-1) such that k^3 has no digit in common with a(n-1) and a(n+1), a(0)=0.
Original entry on oeis.org
0, 1, 2, 4, 5, 6, 7, 8, 9, 10, 29, 55, 88, 90, 111, 200, 211, 400, 518, 654, 888, 889, 1111, 2825, 3131, 4244, 11111, 28222, 31535, 42449, 53355, 90000, 111181, 590000, 618181, 900000, 1111115, 9000000, 11111115, 60660090, 114144155
Offset: 0
a(3) cannot be 3 because 3^3 = 27 would have the digit '2' in common with a(2) = 2, therefore a(3) = 4, which does not violate this condition.
After a(9) = 10, none of the numbers { 11, ..., 19 } can follow, because they have the digit '1' in common with a(9)^3 = 1000. Numbers { 20, ..., 28 } are excluded because their cube would have a digit '0' or '1' in common with a(9). Therefore, a(10) = 29 which hasn't a digit in common with a(9)^3, nor has 29^3 = 24389 a digit in common with a(9).
a(38) = 11111115 with 11111115^3 = 1371743552812575445875 using all digits except for 0, 6 and 9. So a(39) = 60660090 is possible, with a(39)^3 = 223207688999086038729000 having all digits except for 1, 4 and 5.
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nxt(a,L=oo,D(a)=Set(digits(a)),S=D(a),T=D(a^3))={for(k=a+1,L, #setintersect(D(k),T)||#setintersect(D(k^3),S)||return(k))}
A294662=List(); a=0; until(!a=nxt(a,1e7),write("/tmp/A294662.txt",#A294662," ",a);listput(A294662,a))
Showing 1-3 of 3 results.
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