## 6.9.16

## 4.9.16

### A Couple Of Sequences

Two integer sequences that I recently submitted to the OEIS have been approved. The first sequence is the sequence of Collatz primes(A276260) , which I talked about here.

The mesmerizing thing about this sequence is the fact that there are only 9 known terms of it so far, and if a 10th term exists, it must be greater than 10^9.

The other sequence that I submitted was the one I talked about here, which is now sequence A276290 in OEIS.

The interesting thing to me is the fact that so far no counter example has been found to my conjecture that if n is the product of two odd primes p and q and p = 3 then neither p nor q is in the trajectory of n+1 under the Collatz 3x+1 map.

This makes me wonder if there are primes other than 3 for which this conjecture also appears to be true.

http://oeis.org/A276260 |

The mesmerizing thing about this sequence is the fact that there are only 9 known terms of it so far, and if a 10th term exists, it must be greater than 10^9.

The other sequence that I submitted was the one I talked about here, which is now sequence A276290 in OEIS.

The interesting thing to me is the fact that so far no counter example has been found to my conjecture that if n is the product of two odd primes p and q and p = 3 then neither p nor q is in the trajectory of n+1 under the Collatz 3x+1 map.

This makes me wonder if there are primes other than 3 for which this conjecture also appears to be true.

Labels:
Abstract Algebra
,
Collatz Conjecture
,
Discrete Mathematics
,
Mathematics
,
Number Theory

### A Square of A Square

In my Recurring Oddities entry I talked about a recurrence relation where each subsequent term was equal to the previous term plus an odd integer, which I calculated using the index of the previous term and I claimed that this recurrence relation generates the sequence formed by squaring integers consecutively:

1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225,... [A000290]

Proof: Although I never proved this when I first made this claim, the proof of this is very simple and it involves basic algebra.

Since a

n

= (n

= n~~2*n~~ + ~~2*n~~ - ~~2~~ + ~~2~~

= n

.'.

I used this recurrence relation to find the smallest square root of ((n+1)/2)

There are also another type of interesting integers, which I endearingly call

Example: Let n = 35, then k = 23 since 23

1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225,... [A000290]

**9 = 4 + (2*(3-1) + 1)****16 = 9 + (2*(4-1) + 1)****25 = 16 + (2*(5-1) + 1)****.****.****.**

**Claim**: If a_{2}= 4 then a_{i}= a_{i-1}+ (2*(i-1) + 1) where i > 2 generates the sequence formed by squaring integers consecutively.Proof: Although I never proved this when I first made this claim, the proof of this is very simple and it involves basic algebra.

Since a

_{i}= n^{2}, a_{i-1}= (n-1)^{2}, and (i-1) = (n-1), the a_{i}= a_{i-1}+ (2*(i-1) + 1) recurrence relation becomes:n

^{2}= (n-1)^{2}+ (2*(n-1) + 1)= (n

^{2}- 2*n + 1) + (2*n - 2 + 1)= n

^{2}-= n

^{2 }.'.

I used this recurrence relation to find the smallest square root of ((n+1)/2)

^{2}mod n, which I then used to find lonely squares, or integers k such that k^{2}= 1 mod n where n is the product of distinct primes because I claimed a lonely square is just 2 times the smallest square root of ((n+1)/2)^{2}mod n.There are also another type of interesting integers, which I endearingly call

*foursome squares*.**Definition:**A*foursome square*is an integer k such that k^{2}= 4 mod nExample: Let n = 35, then k = 23 since 23

^{2}= 4 mod 35 but also 12^{2}= 4 mod 35 and also 2^{2}= 4 mod 35 and 33^{2}= 4 mod 35

**Claim:**If n is the product of 2 distinct odd primes and k is a foursome square, then k*((n+1)/2) mod n is a lonely square.
Labels:
Abstract Algebra
,
Discrete Mathematics
,
Mathematics
,
skewy

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