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{(PARI) {upto(N, d=[], i=1, j=1, U=[])=vector(N, n, my(L=#d, dk, dz, N, F, k);

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We conjecture that the sequence greedily (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').)

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').

allocated for M. F. Hasler

Lexicographic earliest sequence of distinct positive integers having the same concatenation of digits as the sequence 2^a(n).

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

1,1

We conjecture that the sequence greedily (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').)

We conjecture that this is a permutation of the positive integers, but a proof seems without reach. Can it be disproved?

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.

{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)*/}

Cf. A000079 (2^n), A362191 (variant with nonnegative terms).

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M. F. Hasler, Apr 10 2023

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allocated for M. F. Hasler

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