elementary number theory - What's the pattern to $\frac12(m+1)(2n+3m)$ for positive $n$, $m$? - Mathematics Stack Exchange

this derived $s_a=\sum_{i=0}^m(n+ai)=\frac12(m+1)(2n+am)$, $(n,m)\ge1$ when $a=3$

a number works number $n,m$ plugged $s_a$ result in number. $[n,m]=[1,2]$ plugged $s_1=6$, 6 works $s_1$

note: if $a=1$, $s_1$ can every number not power of $2$, , never power of $2$. if $a=2$, $s_2$ can every composite number never prime.

i found same reasons $s_1$ can never power of $2$, $s_3$ can never power of $2$

i found $s_3$ can every odd number greater $4$($5$ being smallest number $s_3$ can ever be) because $(n)+n+3=2n+3$, if n positive, generate every odd number greater $4$.

i found numbers generated $s_3=\{5+2x\}\cup\{12+3x\}\cup\{22+4x\}\cup\{35+5x\}\cup\cdots\cup\{\frac12(3k^2-k)+kx\}\cup\cdots$, $x=\{0,1,2,3,\cdots\}$

i used find numbers didn't work: https://www.desmos.com/calculator/ebklyzijdy $s_3\ne1,2,3,4,6,8,10,14,16,20,28,32,44,52,64,68,76,88,104,128,136,152,184,208,232,248,256,272,296,304,328,344,368,464,496,512,592,656,688,736,752,848,928,944,976,992,1024,1072,1136,1168,1184,1264,1312,1328,1376,1424,\cdots$

i see pattern though. primes times powers of $2$ don't work, instance $496=2^4\cdot31$, $1376=2^5\cdot43$.

so that's in finding pattern. think may have every number $2$ odd factors working don't know.

first let's define $s_a$:$$s_a=\left \{ x \mid (\exists n,m \in \mathbb{n}) \left [x = \frac12(m+1)(2n+am) \right] \right \}$$

we interested in set $s_3$. separate numbers 2 classes based on parity of $m$.

1. say $m$ even. let's write $m = 2k$. can write our expression $\frac12(2k+1)(2n+6k) = (2k+1)(3k+n)$.

2. say $m$ odd. let's write $m = 2k+1$. can write our expression $\frac12(2k+2)(2n+6k+3) = (k+1)(2n+6k+3)$.

what can conclude this? in case 1 $2k+1$ factor, , in case 2 $2n+6k+3$ factor, both of odd. therefore can conclude in cases $x \in s3$ must contain odd factor. in other words, powers of 2 can't occur, , observation of yours correct.

when $k = 0$, case 1 impossible (because we'd have $m = 0$), see case 2 becomes $2n + 3$. every odd number $\geq 5$ in $s_3$.

the category of numbers left explore of form $2^s \cdot d$ $d$ odd. in fact, factor further , write $x = 2^s \cdot p \cdot r$ $p$ smallest odd factor of $x$, , $r$ odd (and may $1$).

1. $(2k+1)(3k+n) = 2^s\cdot p\cdot r$. maximize amount of numbers valid find $2k+1 = p$ , $3k+n = 2^s\cdot r$. possible when $3k < 2^s\cdot r$, , $p < \frac{2}{3}\cdot 2^s\cdot r + 1$ criterion case 1 allows numbers.

2. $(k+1)(2n+6k+3) = 2^s\cdot p\cdot r$. here permissive strategy set $k+1 = 2^s$, , $2n + 6k + 3 = p\cdot r$. possible if $6k + 3 < p\cdot r$, simplifies $\frac{6\cdot 2^s - 3}{r} < p$.

combining both cases find number where

$$p < \frac{2}{3} \cdot 2^s \cdot r + 1 \quad \vee\quad \frac{6\cdot 2^s - 3}{r} < p $$

is in $s_3$ (in addition odd number $\geq 5$). using demorgan's law can invert find criterion numbers not in $s_3$:

$$\frac{2}{3} \cdot 2^s \cdot r + 1 \leq p \leq \frac{6\cdot 2^s - 3}{r}$$

but if $r \neq 1$ inequality impossible, can conclude $p$ only prime factor of $d$. numbers bigger $5$ not in $s_3$ (in addition powers of two) of form $p \cdot 2^s$ $p$ prime , $s \geq 1$ and

$$\frac{2}{3} \cdot 2^s + 1 \leq p \leq 6\cdot 2^s - 3$$