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

Author

Eric Angelini and Carole Dubois, Feb 20 2021

Keywords

• base
• nonn

Comments

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

Examples

			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.
		

Links

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

Crossrefs

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