A341817 If a(n) is odd, add to a(n) its odd digits and iterate; if a(n) is even, add to a(n) its even digits and iterate; if an iteration reproduces a term already in the sequence, don't do this iteration and extend the sequence with the smallest integer not yet present in the sequence.

1, 2, 4, 8, 16, 22, 26, 34, 38, 46, 56, 62, 70, 3, 6, 12, 14, 18, 5, 10, 7, 9, 11, 13, 17, 25, 30, 15, 21, 19, 29, 20, 23, 24, 27, 28, 31, 35, 43, 32, 33, 39, 51, 57, 69, 78, 86, 100, 36, 42, 48, 60, 66, 37, 47, 54, 58, 40, 44, 52, 41, 45, 50, 49, 53, 61, 55, 65, 59, 73, 83, 63, 64, 74, 67, 68, 82, 92

Offset

1,2

Comments

This sequence is, by definition, a permutation of the positive integers.

Links

Carole Dubois, Table of n, a(n) for n = 1..5000

Example

a(1) = 1, odd, a(2) is thus 1 + 1 = 2;
a(2) = 2, even, thus a(3) = 2 + 2 = 4;
a(3) = 4, even, thus a(4) = 4 + 4 = 8;
a(4) = 8, even, thus a(5) = 8 + 8 = 16;
a(5) = 16, even, thus a(6) = 16 + 6 = 22;
a(6) = 22, even, thus a(7) = 22 + 2 + 2 = 26;
...
a(13) = 70, even, thus a(14) = 70 + 0 = 70 (already in the sequence, thus a(14) = 3 instead, the smallest integer not yet present in the sequence);
a(14) = 3, odd, thus a(15) = 3 + 3 = 6;
a(15)  = 6, even, thus a(16) = 6 + 6 = 12;
a(16) = 12, even, thus a(17) =  12 + 2 = 14;
a(17) = 14, even, thus a(18) = 14 + 4 = 18;
a(18) =  18, even, thus a(19) = 18 + 6 = 26 (already in the sequence, thus a(19) = 5 instead, the smallest integer not yet present in the sequence);
a(19) = 5, odd, thus a(20) = 5 + 5 = 10; etc.

Crossrefs

Cf. A004207 (sum of digits of all previous terms).

Sequence in context: A102039 A045844 A254062 * A230102 A063108 A161140

Adjacent sequences: A341814 A341815 A341816 * A341818 A341819 A341820

Keyword

base,nonn

Author

Eric Angelini and Carole Dubois, Feb 20 2021

Status

approved