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.

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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

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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
		

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}
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