A119474 -id:A119474 - cp's OEIS frontend

A119475 Inverse of permutation sequence A119474.

1, 2, 3, 5, 4, 9, 6, 17, 7, 10, 33, 8, 11, 18, 13, 12, 65, 19, 34, 14, 21, 20, 15, 129, 35, 25, 16, 22, 66, 37, 23, 36, 257, 26, 24, 67, 41, 38, 27, 130, 69, 29, 28, 39, 49, 68, 42, 40, 30, 513, 131, 73, 31, 43, 70, 258, 45, 32, 44, 50, 133, 71, 81, 132, 46, 51, 74, 72, 47, 1025

Offset: 1

Leroy Quet, May 22 2006

Sequence is a permutation of the positive integers.

Rémy Sigrist, Table of n, a(n) for n = 1..10000
Rémy Sigrist, PARI program
Index entries for sequences that are permutations of the natural numbers

Cf. A119474.

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More terms from Joshua Zucker, Jul 04 2006

A101267 a(1) = 1; a(n) = a(2^ceiling(log_2(n)) + 1 - n)th smallest positive integer not yet in the sequence.

Original entry on oeis.org

1, 2, 4, 3, 7, 9, 6, 5, 13, 15, 19, 17, 11, 14, 10, 8, 24, 27, 32, 29, 37, 40, 35, 33, 21, 23, 30, 26, 18, 22, 16, 12, 44, 49, 56, 52, 62, 67, 59, 57, 73, 76, 82, 79, 69, 74, 66, 63, 39, 43, 50, 46, 58, 64, 54, 51, 34, 38, 47, 42, 28, 36, 25, 20, 84, 90, 102, 94, 110, 116, 106

Offset: 1

Leroy Quet, Dec 18 2004

Sequence is a permutation of the positive integers. 2^ceiling(log_2(n)) + 1 - n is sequence A080079 with a change of offset.

			Since 2^ceiling(log_2(n)) +1 -n = 3 at n = 6, a(6) = the a(3)th (the 4th) smallest positive integer not among the first 5 terms of the sequence. The positive integers not among the first 5 terms are 5,6,8,9,10,... The 4th of these is 9, which is a(6).

Ivan Neretin, Table of n, a(n) for n = 1..8192

Cf. A101283, A119474.

a[1] = 1; a[n_] := a[n] = Complement[ Range[100], Table[ a[i], {i, n - 1}]] [[ a[2^Ceiling[ Log[2, n]] + 1 - n]]]; Table[ a[n], {n, 71}] (* Robert G. Wilson v, Jan 13 2005 *)

More terms from Robert G. Wilson v, Jan 13 2005

A119804 a(0) = 0. For m >= 0 and 0 <= k <= 2^m -1, a(2^m +k) = number of earlier terms of the sequence which equal k.

Original entry on oeis.org

0, 1, 1, 2, 1, 3, 1, 1, 1, 6, 1, 1, 0, 0, 1, 0, 4, 9, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 13, 14, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 36, 21, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0

Offset: 0

Leroy Quet, May 24 2006

			8 = 2^3 + 0; so for a(8) we want the number of terms among terms a(1), a(2),... a(7) which equal 0. So a(8) = 1.

John Tyler Rascoe, Table of n, a(n) for n = 0..8192

Cf. A000079, A119474, A119802, A119805.

A119804(mmax)= { local(a,ncopr); a=[0]; for(m=0,mmax, for(k=0,2^m-1, ncopr=0; for(i=1,2^m+k, if( a[i]==k, ncopr++; ); ); a=concat(a,ncopr); ); ); return(a); } { print(A119804(6)); } \\ R. J. Mathar, May 30 2006

More terms from R. J. Mathar, May 30 2006

cp's OEIS Frontend

A119475 Inverse of permutation sequence A119474.

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A101267 a(1) = 1; a(n) = a(2^ceiling(log_2(n)) + 1 - n)th smallest positive integer not yet in the sequence.

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A119804 a(0) = 0. For m >= 0 and 0 <= k <= 2^m -1, a(2^m +k) = number of earlier terms of the sequence which equal k.

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