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A362202
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Lexicographic earliest sequence of distinct 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, 62, 46, 11, 68, 60, 18, 42, 7, 3, 8, 790, 470, 36, 87, 44, 17, 76, 64, 20, 48, 2, 9, 5, 14, 7905, 179, 35, 28, 25, 85, 61, 15, 29, 21, 50, 460, 684, 69, 762, 621, 444, 39, 80, 465, 1110, 41, 288, 256, 65, 117, 32, 84, 4609, 23, 26, 89, 53, 110, 52, 643, 7622, 83, 175, 24, 1780, 49, 13
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OFFSET
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1,1
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COMMENTS
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We conjecture that this is a permutation of the positive integers, but a proof seems without reach. Can it be disproved?
The sequence greedily extends to infinity, i.e., without backtracking, always choosing a(n) as the smallest possible term compatible with the digits given so far, and not leaving a 0 as next digit to be the initial digit of a future term a(n') or 2^a(n').
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LINKS
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EXAMPLE
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The first term a(1) must start with the same digits as 2^a(1), the smallest solution is a(1) = 6 with 2^a(1) = 64.
Then the next digit must be 4, and we can indeed choose a(2) = 4 with 2^a(2) = 16.
Then the next digit must be 1, and we can indeed choose a(3) = 1 with 2^a(3) = 2.
Then the next digit must be 6 (last digit of 2^a(2)), but since 6 = a(1) is already used, we have to consider a(4) with at least two digits, the second of which must be 2 from 2^a(3). We can indeed choose a(4) = 62 with 2^a(4) = 4611686018427387904.
Then the next digits must be 4 and 6 from 2^a(4). Since 4 = a(2) is already used, we must choose a(5) = 46 with 2^a(5) = 70368744177664.
After a(13), the next digits must be 7, 9, and 0. Although 79 is not used earlier, we can't take a(14) = 79, since this would require the next term to start with a digit 0, which is impossible. Therefore, a(14) = 790.
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PROG
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(PARI) {upto(N, d=[], i=1, j=1, U=[])=vector(N, n, my(L=#d, dk, dz, N, F, k);
while(k++, setsearch(U, k) && next;
dk = if(k, digits(k), [0]); dz = digits(2^k);
for( ii = 0, min(L-i, #dk-1), d[ i+ii ] == dk[ 1+ii ] || next(2));
if ( L >= i + #dk && ! d[i + #dk] && setsearch(U, 0), k = k*10-1; next);
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; U=setunion(U, [k]); k)/*+print(d)*/}
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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