A174162 - cp's OEIS frontend

A174162 a(1) = 2. Let k >= 1 be the minimal integer such that 2ka(n-1) + 1 has at least one prime divisor which is not already in the sequence. Then a(n) is the smallest such divisor.

n	a(n)
1	2
2	5
3	11
4	23
5	47
6	19
7	3
8	7
9	29
10	59
11	17
12	103
13	619
14	2477
15	991
16	661
17	3967
18	2267
19	907
20	191
21	383
22	13
23	53
24	107
25	43
26	173
27	347
28	139
29	31
30	83
31	167
32	67
33	269
34	359
35	719
36	1439
37	2879
38	443
39	887
40	71
41	61
42	41
43	137
44	823
45	37
46	149
47	199
48	797
49	1063
50	709
51	2837
52	227
53	101
54	607
55	3643
56	21859

[2, 5, 11, 23, 47, 19, 3, 7, 29, 59, 17, 103, 619, 2477, 991, 661, 3967, 2267, 907, 191, 383, 13, 53, 107, 43, 173, 347, 139, 31, 83, 167, 67, 269, 359, 719, 1439, 2879, 443, 887, 71, 61, 41, 137, 823, 37, 149, 199, 797, 1063, 709, 2837, 227, 101, 607, 3643, 21859]

cp's OEIS Frontend

A174162 a(1) = 2. Let k >= 1 be the minimal integer such that 2ka(n-1) + 1 has at least one prime divisor which is not already in the sequence. Then a(n) is the smallest such divisor.

Table of values

List of values

cp's OEIS Frontend

A174162 a(1) = 2. Let k >= 1 be the minimal integer such that 2*k*a(n-1) + 1 has at least one prime divisor which is not already in the sequence. Then a(n) is the smallest such divisor.

Table of values

List of values

A174162 a(1) = 2. Let k >= 1 be the minimal integer such that 2ka(n-1) + 1 has at least one prime divisor which is not already in the sequence. Then a(n) is the smallest such divisor.