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A294660
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Least nonnegative integer not occurring earlier whose square has no digit in common with the square of the previous term, a(0) = 0.
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2
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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
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OFFSET
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0,3
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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
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PROG
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(PARI) {u=a=0; for(n=0, 99, print1(a", "); u+=1<<a; D=Set(if(a, digits(a^2))); for(k=1, oo, bittest(u, k)&&next; #setintersect(D, Set(digits(k^2)))&&next; a=k; break)); a}
(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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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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