It's not binary!

$\text{It's actually a proof question, so it wasn't meant to be posed the way I did it.}\\ Firstly,\quad N\quad =\quad 101010101...101.\\ Consider\quad the\quad case\quad of\quad 1010101.\quad It\quad has\\ an\quad even\quad number\quad of\quad ones\quad and\quad it\quad can\quad be\\ written\quad as\quad 1010000+101.\quad Similarly,\quad one\\ can\quad extend\quad it\quad to\quad any\quad N\quad with\quad an\quad even\quad number\quad \\ of\quad ones.\quad So\quad it\quad cannot\quad be\quad composite.\quad Try\quad a\quad few\\ to\quad convince\quad yourself.\\ \\ Now,\quad when\quad the\quad number\quad of\quad ones\quad is\quad odd,\quad as\quad in\quad the\quad \\ case\quad of\quad 10101,\quad you\quad can\quad see\quad that\quad it\quad can\quad be\quad written\\ as\quad \sum _{ i=0 }^{ 2n } 10^{ 2i },\quad where\quad n\quad is\quad any\quad natural\quad number.\quad This\quad \\ number\quad has\quad 2n+1\quad ones.\quad The\quad number\quad can\quad be\quad expressed\\ as\quad \frac { { 10 }^{ 4n+2 }-1 }{ 99 } .\\ { 10 }^{ 4n+2 }-1=({ 10 }^{ 2n+1 }-1)({ 10 }^{ 2n+1 }+1).\\ Now,\quad ({ 10 }^{ 2n+1 }-1)/9=\quad 1111....111\quad with\quad 2n+1\quad ones\quad and\quad \\ ({ 10 }^{ 2n+1 }+1)\quad is\quad divisible\quad by\quad 11.\quad So,\quad N\quad is\quad divisible\quad by\\ 1111...2n+1\quad ones.\quad Hence\quad it\quad is\quad not\quad prime.\\ \therefore \quad 101\quad is\quad the\quad only\quad prime\quad that\quad can\quad be\quad written\quad as\quad \\ an\quad alternating \quad sequence\quad of\quad 1s\quad and\quad 0s\quad that\quad starts\quad and\quad ends\quad \\with\quad 1.\quad \quad \\ \\$