cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A030288 a(n+1) is smallest square > a(n) having no digits in common with a(n), with a(0) = 0.

Original entry on oeis.org

0, 1, 4, 9, 16, 25, 36, 49, 81, 225, 361, 400, 529, 676, 841, 900, 1156, 2209, 3136, 4225, 6889, 7225, 8100, 24336, 58081, 69696, 70225, 84681, 90000, 111556, 200704, 316969, 407044, 511225, 608400, 923521, 4000000, 5112121, 6036849
Offset: 0

Views

Author

Patrick De Geest

Keywords

Comments

It appears that from a(102) on, there is a 4-periodic pattern: a(4k) ~ 3*10^(k-3) a(4k+1) ~ 6.1111...*10^(k-3), a(4k+2) ~ 7*10^(k-3), a(4k+3) ~ 8.1111...*10^(k-3), where ~ means the next larger square which has only digits {0, 3, 4, 5, 7} for even-indexed terms, or {1, 2, 6, 8, 9} for odd-indexed terms. - M. F. Hasler, Nov 12 2017

Links

Crossrefs

Cf. A000290, A019544, A030098, A030287, A030289.

Programs

  • Mathematica
    FromDigits /@ NestList[Block[{k = Sqrt@ FromDigits@ # + 1, m}, While[ContainsAny[#, Set[m, IntegerDigits[k^2]]], k++]; m] &, {0}, 38] (* Michael De Vlieger, Nov 02 2017 *)
ssga[a_]:=Module[{k=Floor[Sqrt[a]]+1},While[Length[Intersection[IntegerDigits[k^2],IntegerDigits[ a]]]> 0,k++];k^2]; NestList[ssga,0,40] (* Harvey P. Dale, Sep 10 2024 *)
  • PARI
    next_A030288(n, D(n)=Set(digits(n)), S=D(n))={for(k=sqrtint(n)+1, oo, #setintersect(D(k^2), S)||return(k^2))} \\ Could be made more efficient by implementing the observed patterns, in particular for n >= 104. - M. F. Hasler, Nov 12 2017

Formula

a(n) = A030287(n)^2. - Michel Marcus, Nov 03 2017

A030290 a(n) is the smallest k > a(n-1) such that k^3 has no digit in common with a(n-1)^3.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 7, 8, 16, 18, 40, 45, 67, 98, 150, 204, 237, 44216, 46443, 78742, 79930, 130714, 173000, 185604, 1000000, 1304963, 10000000, 13049563, 100000000, 130495593, 1000000000, 1304955895, 10000000000, 13049558812, 100000000000, 130495588186, 1000000000000, 1304955880707
Offset: 0

Views

Author

Patrick De Geest

Keywords

Comments

From a(24) = 10^6 on, we have a(2k) = 10^(k-6) and a(2k+1) ~ c*a(2k) with c = (20/9)^(1/3) = 1.30495588... Indeed, a(2k)^3 = 1000^(k-6) has then only digits 0 and 1, and the next term must have a cube >= 2.2222...*1000^(k-6), so a(2k+1) will be the cube root of the next larger cube with no digit 0 and 1. - M. F. Hasler, Nov 12 2017

Examples

			a(5) = 5 and 5^3 = 125 has no digit in common with the cube of a(4) = 4, 4^3 = 64.
But a(6) cannot be equal to 6, because 6^3 = 216 has digits '1' and '2' in common with 5^3 = 125.

Links

Crossrefs

Cf. A030289.

Programs

  • PARI
    next_A030290(n, S=Set(digits(n^3)))={if(n>18e4,S[1]&&return(10^logint(n<<3,10));n\=sqrtn(.45,3));for(k=n+1,oo, #setintersect(Set(digits(k^3)), S)||return(k))} \\ M. F. Hasler, Nov 12 2017
print1(a=0); for(i=1, 99, print1(", "a=next_A030290(a))) \\ M. F. Hasler, Nov 08 2017

Formula

a(n) = A030289(n)^(1/3). - David W. Wilson, Nov 08 2017
For k >= 12, a(2k) = 10^(k-6), and a(2k+1) > c*a(2k) with approximate equality, where c = (20/9)^(1/3) = 1.30495588... - 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

Views

Author

Labos Elemer, Dec 01 2004

Keywords

Links

Crossrefs

Cf. A002808, A030283, A030284, A030285, A030286, A030287, A030288, A030289, A030290.

Programs

  • Maple
    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
  • Mathematica
    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]
Showing 1-3 of 3 results.

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