# [seqfan] Re: Three Sequences

Tom Duff eigenvectors at gmail.com
Sat Aug 6 18:23:48 CEST 2022

```BTW, if we let the prime be >= a(n) rather than strictly greater, 2 is
another term in the sequence, b/c 2+2 is a power of 2.

On Sat, Aug 6, 2022 at 12:15 PM Tom Duff <eigenvectors at gmail.com> wrote:

> 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/
>>
>

```