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A362189
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Lexicographically earliest sequence of positive integers having the same concatenation of digits as the sequence 2^a(n).
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1
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6, 4, 1, 6, 2, 6, 4, 4, 6, 4, 1, 6, 1, 6, 6, 4, 1, 6, 2, 6, 4, 2, 6, 4, 6, 4, 1, 6, 2, 6, 4, 4, 6, 4, 1, 6, 4, 6, 4, 1, 6, 6, 4, 1, 6, 2, 6, 4, 4, 6, 4, 1, 6, 1, 6, 6, 4, 1, 6, 2, 6, 4, 1, 6, 6, 4, 1, 6, 2, 6, 4, 6, 4, 1, 6, 2, 6, 4, 4, 6, 4, 1, 6, 1, 6, 6, 4, 1, 6, 2, 6, 4, 2
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OFFSET
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0,1
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COMMENTS
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The sequence consists of an infinite number of each of the elements of {1, 2, 4, 6}, and appears to be non-periodic. Can this be proved?
The next to minimal solution starts (10, 2, 4, 4, 1, 6, 1, 6, 2, 6, 4, 2, 6, 4, 4, ...), and appears to have the same property as this one, except for the initial term. See A362191 for the variant where all terms must be distinct.
Is it true that the sequence is cubefree, i.e., with no substring of the form xxx where x is any substring of length >= 1? Can this be proved?
Fixed point of the morphism 1 -> 2, 2 -> 4, 4 -> 16, 6 -> 64 starting from 6. - Rémy Sigrist, Apr 15 2023
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LINKS
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EXAMPLE
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The digits of the terms of the sequence (a(0) = 6, a(1) = 4, a(2) = 1, ...) and those of the sequence b(k) = 2^a(k): (2^6 = 64, 2^4 = 16, 2^1 = 2, ...), are as follows:
n : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, ...
a(n) : 6, 4, 1, 6, 2, 6, 4, 4, 6, 4, 1, 6, 1, 6, 6, 4, 1, 6, ...
b(k) : 64, 16, 2, 64, 4, 64, 16, 16, 64, 16, ...
k : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, ...
Explanations:
The initial term a(0) must start with the same digits as 2^a(0). The smallest solution is a(0) = 6 with 2^a(0) = 64, the second digit of which must equal the first digit of the next term. (10 would also be a solution for the initial term, with 2^10 = 1024 starting with the same digits, '1' and then '0'.)
So the term a(1) must start with a digit 4. The smallest possible choice is a(1) = 4 with 2^a(1) = 16. These digits determine the next terms:
The first digit of the next term must now be 1, and we can indeed choose a(2) = 1 with 2^a(2) = 2, dictating the digit that must come after the '6' from the preceding 16.
The next term a(3) must again start with a digit 6, from the second digit of 2^a(1) = 16. Again the smallest possible choice is a(3) = 6 (since we allow for repeated terms, otherwise the smallest possible choice would be 62, cf. A362191), with 2^a(3) = 64. And so on.
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PROG
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(PARI) /* The sequence could be produced more efficiently; the code below is rather for illustrative purpose and generalizations */
{upto(N, d = [], i = 1, j = 1)=vector(N, n, my(L = #d, dk, dz, N, F, k);
while(k++, dk = digits(k); dz = digits(2^k);
for( ii = 0, min(L-i, #dk-1), d[ i+ii ] == dk[ 1+ii ] || next(2));
for( ii = 0, min(L-j, #dz-1), d[ j+ii ] == dz[ 1+ii ] || next(2));
(N = max ( i + #dk, j + #dz)-1) > #d && d = Vec(d, N);
F = i + #dk > j + #dz; for ( ii = L+1, N,
d[ ii ] = if ( F, dk[ ii-i+1 ], dz[ ii-j+1 ] ));
if ( F, for ( jj = L+1, j+#dz-1, d[ jj ] == dz[ jj-j+1 ] || next(2)),
for ( jj = L+1, i+#dk-1, d[ jj ] == dk[ jj-i+1 ] || next(2))); break);
i += #dk; j += #dz; k)/*+print(d)*/}
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CROSSREFS
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KEYWORD
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nonn,easy,base
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AUTHOR
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STATUS
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approved
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