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Irregular sequence derived from the b-numbers

If the n’th b-number is b(n), then b(n) == 3 (mod 8).

Therefore, b(n)+5 is divisible by 8, and we can form:

b'(n) := ( b(n) + 5)/8. For n = 1, 2, 3, …

we have b'(n) as copied below, an irregular sequence:

1
2
3
4
5
6
7
9
10
11
14
17
18
19
22
26
33
34
35
38
42
50
65
66
67
70
74
82
97
99
102
106
129
130
131
134
138
146
161
163
166
170
193
195
198
202
227
234
257
258
259
262
266
274
289
291
294
298
321
323
326
330
355
362
386
402
417
422
451
458
483
513
514
515
518
522
530
545
547
550
554
577
579
582
586
611
618
642
658
673
678
707
714
739
769
771
774
778
803
810
833
838
874
934
970
995
1025
1026
1027
1030
1034
1042
1057
1059
1062
1066
1089
1091
1094
1098
1123
1130
1154
1170
1185
1190
1219
1226
1251
1281
1283
1286
1290
1315
1322
1345
1350
1386
1446
1482
1507
1538
1554
1569
1574
1603
1610
1635
1666
1682
1697
1731
1795
1802
1827
1862
1958
1994

etc.

 

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Written by meditationatae

September 18, 2017 at 10:23 pm

Posted in History

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