A permutation based on Pascal's triangle

[Nick Hobson]

Hi Seqfans,

Is this sequence of interest?  Or is it artificially derivative?

4, 7, 8, 11, 13, 16, 19, 12, 22, 26, 29, 34, 37, 43, 46, 53, 17, 18, 56,  
64, 67, 76, 79, 89, 92, 103, 106, 118, 23, 25, 121, 134, 137, 151, 154,  
169, 172, 188, 191, 208, 24, 211, 229, 30, 33, 232, 251, 254, 274, 277,  
298, 301, 323, 326, 349, 352, 376, 379, 404, 38, 42, 407, 433, 436, 463,  
466, 494, 497, 526, ... .

The sequence is based upon A007318: Pascal's triangle read by rows (offset  
0): 1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 4, 6, 4, 1, ... .

In A007318, the number 2 appears in position 4, the number 3 appears in  
positions 7 and 8, and so on.  So the new sequence is a permutation of the  
integers n such that A007318(n) > 1.

Nick



I would describe it as:
Triangle, read by rows, in which row n (n >= 2) lists
the indices in A007318 where n appears.

Please send it in!
Neil