[seqfan] Re: Three Sequences
Tom Duff
eigenvectors at gmail.com
Sat Aug 6 18:15:43 CEST 2022
For definiteness, you should probably say that sequence (1) lists the terms
in increasing order.
Is there a good reason why the terms have to be positive?
Without that restriction, the first few terms (up to 2^63-1) are
-1
0
3
15
61
255
2043
4093
32765
65535
262141
8388599
33554397
134217699
268435453
1073741821
17179869159
137438953463
274877906937
1099511627761
8796093022179
17592186044409
70368744177649
140737488355323
281474976710635
562949953421243
2251799813685227
4503599627370475
9007199254740987
36028797018963965
144115188075855863
1152921504606846943
The corresponding sequence of primes is
2
2
5
17
67
257
2053
4099
32771
65537
262147
8388617
33554467
134217757
268435459
1073741827
17179869209
137438953481
274877906951
1099511627791
8796093022237
17592186044423
70368744177679
140737488355333
281474976710677
562949953421381
2251799813685269
4503599627370517
9007199254740997
36028797018963971
144115188075855881
1152921504606847009
and, for completeness, the exponents of 2 are
0, 1, 3, 5, 7, 9, 12, 13, 16, 17, 19, 24, 26, 28, 29, 31, 35,
38, 39, 41, 44, 45, 47, 48, 49, 50, 52, 53, 54, 56, 58, 61
On Sat, Aug 6, 2022 at 9:41 AM Ali Sada via SeqFan <seqfan at list.seqfan.eu>
wrote:
> Hi everyone,
>
> Please see the three sequences below. If they fit OEIS, I would really
> appreciate help with definitions and terms.
>
> 1)
>
> a(n) is a positive integer such that a(n)+p equals to a power of two,
> where p is the least prime larger than a(n). a(1) = 3 (3+5 = 8)
> a(2) = (15+17 = 32)
> a(3) = 61 (61+67 = 128)etc.
>
> 2)
> a(1) = 1 2 is a multiple of the first term, so a(2) = 2/1 = 2
> Now, to get a multiple of a(2) we need to add 3 of the unused numbers
> (3+4+5 = 12). So, a(3) = 12/2 = 6.
> a(4) = 6/6 =1
> a(5)= 7/1= 7
> a(6)= (8+9+10+11+12+13)/7 = 9
> a(7) = (14+15+16)/9 = 5
> and so on.
>
> 3)
> We have a computer program that draws 4n equal line segments in each step
> n. The goal is to draw as many squares (from all sizes) as possible.
> Ideally, we get A111715(n) squares. However, at each step, directly after
> the 4n segments are drawn, another program erases n segments trying to
> terminate as many squares as possible. The sequence is the total number of
> squares in each step.
> Best,
>
> Ali
>
>
>
> --
> Seqfan Mailing list - http://list.seqfan.eu/
>
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