[seqfan] Re: Two thingies with digits

[Alois Heinz]

Dear Eric,

S starts:

1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 13, 14, 15, 17, 16, 18, 12, 21, 19, 25,
22, 23, 26, 24, 27, 28, 29, 31, 33, 32, 34, 35, 36, 38, 39, 41, 42,
43, 37, 44, 45, 46, 47, 48, 51, 149, 53, 49, 52, 54, 55, 56, 57, 59,
58, 63, 64, 65, 66, 67, 68, 62, 69, 72, 73, 75, 85, 77, 74, 76, 78,
82, 79, 83, 84, 86, 88, 89, 111, 91, 113, 87, 115, 93, 92, 94, 114,
108, 116, 106, 117, 105, 118, 104, 119, 81, 139, 61, 141, 95, 131, 71,
133, 109, 135, 107, 137, 143, 134, 136, 138, 154, 144, 145, 147, 146,
151, 96, 161, 97, 171, 129, 174, 126, 177, 127, 176, 124, 179, 103,
159, 148, 152, 155, 165, 142, 158, 164, 156, 166, 157, 167, 168, 162,
172, 128, 175, 125, 178, 122, 181, 169, 184, 121, 182, 185, 188, 112,
191, 186, 189, 211, 192, 194, 311, 153, 251, 193, 212, 187, 206, 205,
208, 203, 238, 207, 209, 202, 229, 215, 218, 216, 217, 183, 221, 219,
225, 222, 223, 226, 227, 173, 231, 213, 195, 252, 196, 261, 197, 271

Note that S is NOT a permutation of N, missing numbers are:
10,20,30,40,50,60,70,80,90, ... this is because when you add 10, ...
the digit at the rightmost position of the previous number will be shared.

Best, Alois

Am 15.08.2011 10:49, schrieb Eric Angelini:
>   Hello SeqFans,
>   First, could someone please compute a hundred or so terms of S
>   where two consecutive terms of S share no digit with their sum?
>   This is http://oeis.org/A129562 dropping the "increasing" cons-
>
>   traint. I guess S starts:
>   S = 1,2,3,4,5,6,7,8,9,11,13,14,15,17,16,18,12,21...
>   Note that S is now a permutation of N. (Hope I didn't overlook S
>   in the OEIS.)