A067581 -id:A067581 - cp's OEIS frontend

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

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

M. F. Hasler, Nov 08 2017

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.

			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

M. F. Hasler, Table of n, a(n) for n = 0..4829
P. De Geest, Squares containing at most three distinct digits, Index entries for related sequences

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

{u=a=0; for(n=0, 99, print1(a", "); u+=1<

{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}

A298482 a(n) = smallest integer not yet in the sequence with no digits in common with a(n-1), a(n-2), and a(n-3); a(0)=0, a(1)=1, a(2)=2.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 22, 33, 44, 11, 20, 35, 46, 17, 28, 30, 45, 16, 27, 38, 40, 15, 26, 37, 48, 19, 25, 36, 47, 18, 29, 50, 34, 61, 72, 55, 39, 14, 60, 52, 73, 41, 66, 58, 23, 49, 67, 51, 32, 80, 64, 57, 12, 83, 69, 54, 21, 70, 63, 59, 24, 71, 68

Offset: 0

Enrique Navarrete, Feb 28 2018

The first differences of this sequence are symmetrically distributed around 0, in a distribution that has a larger kurtosis than the Normal distribution.

Cf. A067581, A276633.

A284146 a(n+1) is the smallest prime not already in the sequence which shares no digit with a(n).

Original entry on oeis.org

2, 3, 5, 7, 11, 23, 17, 29, 13, 47, 19, 37, 41, 53, 61, 43, 59, 31, 67, 83, 71, 89, 73, 101, 79, 103, 97, 113, 227, 109, 223, 107, 229, 131, 257, 139, 277, 149, 233, 151, 239, 157, 263, 179, 283, 167, 293, 181, 269, 137, 409, 127, 349

Offset: 1

Zak Seidov, Mar 20 2017

nonn

The sequence is not a permutation of prime numbers.
E.g., after calculating 2001 terms of the sequence, the first absent primes  are 1973,3719,3917,7193,9137,9173,9371. It's evident that these numbers will never appear in the sequence because any last  term of the sequence should use at least one of digits 1,3,7,9.
The first nine terms {2, 3, 5, 7, 11, 23, 17, 29, 13} coincide with A068863(1..9).
The only fixed points are a(n) for n={1, 2, 3, 4, 5, 7, 12, 13, 17, 19} are {2, 3, 5, 7, 11, 17, 37, 41, 59, 67} that is for these n's a(n)=prime(n)=A000040(n).
a (100*k) for k = 1,20: {443, 1193, 1741, 1621, 4567, 6047, 5851, 6491, 7151, 7559, 9349, 10601, 11119, 11699, 13001, 11839, 14107, 16111, 15073, 16487}.

Giovanni Resta, Table of n, a(n) for n = 1..10000
Zak Seidov, Graph of first 2000 terms.

Cf. A000040, A068863, A067581, A107801.

a = {2}; While[ Length[a] < 100, d = IntegerDigits@ Last@ a; p = 2; While[ Intersection[ IntegerDigits@p, d] != {} || MemberQ[a, p], p = NextPrime@ p]; AppendTo[a, p]]; a (* Giovanni Resta, Mar 21 2017 *)

A301801 a(n) = smallest integer not yet in the sequence with no digits in common with a(n-1), a(n-2), a(n-3), and a(n-4); a(0)=0, a(1)=1, a(2)=2, a(3)=3.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 22, 33, 44, 55, 11, 20, 36, 47, 58, 19, 200, 63, 74, 85, 91, 202, 66, 34, 57, 18, 29, 60, 43, 75, 81, 92, 600, 333, 45, 17, 28, 69, 30, 54, 71, 82, 96, 300, 444, 15, 27, 68, 39, 40, 51, 72, 86, 93, 400, 111, 25, 67, 38, 49, 100

Offset: 0

Enrique Navarrete, Mar 26 2018

The first differences of this sequence are symmetrically distributed in a distribution that has a larger kurtosis than the Normal distribution.
It seems that appart from the initial terms, 39 and 40 are the only consecutive terms.
Unlike A298482, 3-digit terms appear as early as a(22)=200.

Rémy Sigrist, PARI program for A301801

Cf. A067581, A276633, A298482.

Nest[Append[#, Block[{k = 4, d}, While[Nand[FreeQ[#, k], ! IntersectingQ[Union@ Apply[Join, Take[#[[All, -1]], -4] ], Set[d, IntegerDigits[k]]] ], k++]; {k, d}]] &, Transpose@ {#, IntegerDigits@ #} &@ Range[0, 3], 62][[All, 1]] (* Michael De Vlieger, Apr 12 2018 *)

PARI
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See Links section.
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A343927 a(1) = 1, a(2) = 2; for n>2, a(n) is the smallest positive integer not yet in the sequence which shares a digit with a(n-1) but not with a(n-2), and where a(n) contains at least one digit not in a(n-1).

Original entry on oeis.org

1, 2, 20, 10, 13, 23, 24, 14, 15, 25, 26, 16, 17, 27, 28, 18, 19, 29, 32, 30, 40, 41, 12, 52, 35, 31, 21, 42, 34, 36, 56, 45, 43, 37, 57, 50, 60, 46, 47, 70, 80, 38, 39, 49, 48, 58, 51, 61, 62, 72, 71, 81, 68, 63, 53, 54, 64, 67, 73, 83, 82, 92, 59, 65, 76, 74, 84, 85, 75, 79, 69, 86, 78, 97, 90

Offset: 1

Scott R. Shannon, May 17 2021

This is the digit sequence equivalent of the Enots Wolley sequence A336957. Like that sequence to avoid the sequence halting rapidly an additional rule is placed on a(n) - it must have as least one digit not in a(n-1). This implies a(n) cannot be a repdigit as otherwise a(n+1) would not exist. If this rule is removed then the sequence terminates after five terms: 1, 2, 20, 10, 11. The next term then does not exist as it must both contain and not contain the digit 1.

The sequence is probably infinite as any a(n) must contain at least two distinct digits, thus a(n+2) can have at most eight distinct digits. This implies that a(n+3) can always be created using a digit in a(n+2) and a digit not in a(n+2). However the behavior of the sequence as n gets very large is unknown.

			a(3) = 20 as this is the smallest unused positive integer that contains a digit in a(2) = 2 while not containing any digit in a(1) = 1.
a(4) = 10 as this is the smallest unused positive integer that contains a digit in a(3) = 20 while not containing any digit in a(2) = 2.
a(5) = 13 as this is the smallest unused positive integer that contains a digit in a(3) = 10, contains a digit not in a(3), while not containing any digit in a(3) = 20.

Scott R. Shannon, Image of the first 500000 terms. The green line is a(n) = n.

Cf. A336957, A010785, A184992, A067581.

cp's OEIS Frontend

A294660 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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A298482 a(n) = smallest integer not yet in the sequence with no digits in common with a(n-1), a(n-2), and a(n-3); a(0)=0, a(1)=1, a(2)=2.

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A284146 a(n+1) is the smallest prime not already in the sequence which shares no digit with a(n).

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A301801 a(n) = smallest integer not yet in the sequence with no digits in common with a(n-1), a(n-2), a(n-3), and a(n-4); a(0)=0, a(1)=1, a(2)=2, a(3)=3.

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PARI

A343927 a(1) = 1, a(2) = 2; for n>2, a(n) is the smallest positive integer not yet in the sequence which shares a digit with a(n-1) but not with a(n-2), and where a(n) contains at least one digit not in a(n-1).

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