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

A294660 Least nonnegative integer not occurring earlier whose square has no digit in common with the square of the previous term, a(0) = 0.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 9, 8, 10, 15, 12, 16, 20, 11, 22, 13, 18, 14, 28, 19, 17, 21, 23, 26, 29, 24, 30, 25, 33, 58, 27, 34, 47, 38, 45, 31, 48, 41, 50, 37, 52, 44, 65, 40, 57, 76, 32, 63, 35, 60, 39, 62, 36, 88, 46, 67, 51, 183, 75, 43, 55, 42, 53, 56, 70, 61, 64, 85, 59, 77, 69, 73, 78, 89
Offset: 0

Views

Author

M. F. Hasler, Nov 08 2017

Keywords

Comments

This is not a permutation of the nonnegative integers, since numbers whose square has all digits '1' through '9' (cf. A294661, e.g., 11826 with 11826^2 = 139854276) can never appear - and these numbers have asymptotic density 1.
Will all integers whose square does not have all of the digits 1-9, eventually appear? Or might the sequence be finite? Since a(n)^2 has no digits in common with a(n-1)^2, it is sufficient for a(n+1) to exist, to find a number whose square has a subset of the digits of a(n-1)^2. Is this always possible? This problem sometimes has only "sporadic k-digital solutions", see, e.g., A058430, A030175, ... and the link to De Geest's page.

Examples

			Since a(7)^2 = 7^2 = 49, the subsequent term cannot be 8, since 8^2 = 64 has the digit 4 in common with 49. Therefore, a(8) = 9, with 9^2 = 81 having no common digit with 49.
a(1201) = 1037. So the square of the next term must not have any of the digits in {0, 1, 3, 5, 6, 7, 9}, only 2, 4, 8 are allowed. The least such number that has not been used before is a(1202) = 210912978, with a(1202)^2 = 210912978^2 = 44484284288828484. - _Alois P. Heinz_, Nov 09 2017

Links

Crossrefs

Cf. A030287 (strictly increasing), A067581 (do not take squares).

Programs

  • PARI
    {u=a=0; for(n=0, 99, print1(a", "); u+=1<
  • PARI
    {u=[a=0]; for(n=0, 99, print1(a", "); D=Set(if(a, digits(a^2))); for(k=u[1]+1, oo, setsearch(u, k)&&next; #setintersect(D, Set(digits(k^2)))&&next; u=setunion(u,[a=k]); break); while(#u>1&&u[2]==u[1]+1,u=u[^1])); a}

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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