36mod3=0 so we need n so that the congruency is true, i.e., 2n+3+1=3m or 2n+1=3m' where m' is an integer, making 2n=3m'-1. For congruency, 3m'-1 must be an even number so m' must be odd. Let m'=1, then n=1; if m'=3 then n=4; if m'=5 then n=7; so in general n=1+3k where k is an integer.
2n+4=2(1+3k)+4=6k+6=6(k+1). This must be congruent to 36mod3. Since 6 is a multiple of 3 the congruency is preserved and n=1+3k. A sequence for n would be 1, 4, 7, 10, ... in other words, nmod3=1.
