[seqfan] Re: A175388(?)

[Hans Havermann]

Zak:

> A175388 = A176762, so A175388 may be killed.

I'm hoping that, with a proper index adjustment, my arguments about  
extending A175388 still apply (to A176762). Here is a copy of my email  
(this morning) to MathFun that details a possible extension to A176762:


If it is true that, from a(18) on, a(n) is the sum-of-squares-of- 
digits of a(n+1), I have an empirically derived table that would allow  
one to derive a(n+1) from a(n) mod 81...

mod  x   y

00   0   0
01   0   1
02   2   688
03   2   1688
04   0   2
05   0   12
06   2   2688
07   3   37888
08   1   58
09   0   3
10   0   13
11   2   3688
12   1   258
13   3   8888
14   3   18888
15   2   788
16   0   4
17   1   77
18   1   177
19   1   68
20   1   168
21   1   277
22   3   38888
23   1   268
24   2   3788
25   0   5
26   0   15
27   2   5688
28   1   368
29   0   25
30   2   888
31   2   1888
32   1   78
33   1   178
34   2   2888
35   1   468
36   0   6
37   0   16
38   3   58888
39   2   3888
40   0   26
41   1   378
42   1   577
43   5   8888888
44   1   568
45   0   36
46   2   4888
47   1   88
48   1   188
49   0   7
50   0   17
51   1   288
52   0   46
53   0   27
54   0   127
55   2   5888
56   1   388
57   1   578
58   0   37
59   0   137
60   4   888888
61   0   56
62   3   78888
63   1   488
64   0   8
65   0   18
66   2   6888
67   2   16888
68   0   28
69   0   128
70   2   26888
71   3   378888
72   1   588
73   0   38
74   0   57
75   0   157
76   1   2588
77   3   88888
78   3   188888
79   2   7888
80   0   48

More specifically, I am suggesting that if a(n) mod 81 = z -> {x,y},  
then a(n+1) = (y+1)*10^((a(n)-z-81*x)/81)-1. So...

a(19) mod 81 = 42 -> a(20) =  578*10^196-1
a(20) mod 81 = 55 -> a(21) = 5889*10^((578*10^196-218)/81)-1
a(21) mod 81 =  2 -> a(22) =  689*10^((5889*10^((578*10^196-218)/ 
81)-165)/81)-1
a(22) mod 81 = 31 -> a(23) =  
1889*10^((689*10^((5889*10^((578*10^196-218)/81)-165)/81)-194)/81)-1