Let $n=15r$ where $r>5$ is an odd prime number. If $r\!\!\! \mod 15 \equiv w$ then is it true that $\Phi_{n}(x)$ is not flat whenever $2<w<13$ ? In other words, are the flat ones necessarily among those satisfying $r\equiv \pm 1 \!\!\! \mod 15$ and $r\equiv \pm 2 \!\!\! \mod 15$. I am looking for a Yes or No answer, possibly with links please.
Yes
More precisely, the flat ones are exactly those where $r$ is $\pm 1$ modulo $15$, all others (including those $\pm 2$) are not flat.
The following result is due to Kaplan (Theorems 2 and 3 in the paper linked below):
Let $p<q<r$ be primes, and let $s > q$ be prime such that $s$ is $r$ or $r$ modulo $pq$. Then the heights of the cyclotomic polynomials of order $pqr$ and $pqs$ are equal.
Using this result for $p=3$ and $q=5$, the problem in the question is reduced to checking it for a single prime in the relevant classes modulo $15$.
Using WolframAlpha get
for $r= 17$ it is not flat:
x^128+x^127+x^126x^123x^122x^121+x^113+x^112x^110x^109x^108x^107+x^105+x^104+x^98+x^97x^95x^94x^93x^92+x^90+x^89+x^83+x^82x^80x^79x^78+x^76+2 x^75+x^74x^72x^71x^70+x^68+x^67x^65x^64x^63+x^61+x^60x^58x^57x^56+x^54+2 x^53+x^52x^50x^49x^48+x^46+x^45+x^39+x^38x^36x^35x^34x^33+x^31+x^30+x^24+x^23x^21x^20x^19x^18+x^16+x^15x^7x^6x^5+x^2+x+1
for $r=19$ it is not flat:
x^144+x^143+x^142x^139x^138x^137+x^129+x^128+x^127x^1252 x^1242 x^123x^122+x^120+x^119+x^118+x^114+x^113+x^112x^1102 x^1092 x^108x^107+x^105+x^104+x^103+x^99+x^98+x^97x^952 x^942 x^93x^92+x^90+x^89+x^88+x^87+x^86+x^85+x^84+x^83x^812 x^802 x^792 x^78x^77+x^75+x^74+x^73+x^72+x^71+x^70+x^69x^672 x^662 x^652 x^64x^63+x^61+x^60+x^59+x^58+x^57+x^56+x^55+x^54x^522 x^512 x^50x^49+x^47+x^46+x^45+x^41+x^40+x^39x^372 x^362 x^35x^34+x^32+x^31+x^30+x^26+x^25+x^24x^222 x^212 x^20x^19+x^17+x^16+x^15x^7x^6x^5+x^2+x+1
for $r=31$ it is flat:
x^240+x^239+x^238x^235x^234x^233+x^225+x^224+x^223x^220x^219x^218+x^210x^207x^205+x^202+x^195x^192x^190+x^187+x^180x^177x^175+x^172+x^165x^162x^160+x^157+x^150+x^146x^141x^140+x^135+x^131x^126x^125+x^120x^115x^114+x^109+x^105x^100x^99+x^94+x^90+x^83x^80x^78+x^75+x^68x^65x^63+x^60+x^53x^50x^48+x^45+x^38x^35x^33+x^30x^22x^21x^20+x^17+x^16+x^15x^7x^6x^5+x^2+x+1
for $r=37$ it is not flat:
x^288+x^287+x^286x^283x^282x^281+x^273+x^272+x^271x^268x^267x^266+x^258+x^257+x^256x^253x^2522 x^251x^250x^249+x^246+x^245+x^244+x^243+x^242+x^241x^238x^2372 x^236x^235x^234+x^231+x^230+x^229+x^228+x^227+x^226x^223x^2222 x^221x^220x^219+x^216+x^215+x^214+x^213+x^212+x^211x^208x^2072 x^206x^205x^204+x^201+x^200+x^199+x^198+x^197+x^196x^193x^1922 x^191x^190x^189+x^186+x^185+x^184+x^183+x^182+x^181x^178x^176x^174x^172+x^169+x^168+x^167+x^166x^163x^161x^159x^157+x^154+x^153+x^152+x^151x^148x^146x^144x^142x^140+x^137+x^136+x^135+x^134x^131x^129x^127x^125+x^122+x^121+x^120+x^119x^116x^114x^112x^110+x^107+x^106+x^105+x^104+x^103+x^102x^99x^982 x^97x^96x^95+x^92+x^91+x^90+x^89+x^88+x^87x^84x^832 x^82x^81x^80+x^77+x^76+x^75+x^74+x^73+x^72x^69x^682 x^67x^66x^65+x^62+x^61+x^60+x^59+x^58+x^57x^54x^532 x^52x^51x^50+x^47+x^46+x^45+x^44+x^43+x^42x^39x^382 x^37x^36x^35+x^32+x^31+x^30x^22x^21x^20+x^17+x^16+x^15x^7x^6x^5+x^2+x+1
The $17$ shows it is not flat for $r$ congruent $\pm 2$, the $19$ shows it for $\pm 4$, the $37$ for $\pm 7$. While the $31$ shows flatness for $\pm 1$. This are all prime classes modulo $15$ and we are done.
The result for $\pm 1$ and $\pm 2$ could also be obtained as special cases of results of Kaplan and Elder, respectively. See section 8 of Elder's preprint.
Nathan Kaplan, Flat cyclotomic polynomials of order three, J. Number Theory 127 (2007), no. 1, 118126.

$\begingroup$ Come to think of it I could have used smaller primes. But I will leave it, at least for now. $\endgroup$ – user9072 Jan 23 '16 at 18:51

1$\begingroup$ Not a problem. It answers my question. Thank you very much. $\endgroup$ – user204463 Jan 23 '16 at 19:45

$\begingroup$ Is $(3,5)$ the only pair where the flat ones are squeezed in the margin of $\pm 1$? For example the "next" pair $(3,7)$ has a "gap" for $w=10,11$. I mean the flat ones occur at margin of $\pm 1,\pm 2$ as well as $\pm 10$ using the theorem you mentioned. $\endgroup$ – user204463 Jan 23 '16 at 20:08

$\begingroup$ The pair $(3,7)$ is different in that $7$ is $1$ modulo $3$ and thus $\pm 2$ is flat. This is true for any pair $p<q$. I have no explanation for $\pm 10$. I have no good intuition on this problem. One could do some numerical tests. $\endgroup$ – user9072 Jan 23 '16 at 20:13

1$\begingroup$ If my calculations are correct then $(5,7)$ is another pair like that. I mean the flat ones occur only within the margin of $\pm 1$. My wild guess is the gap $qp=2$ is responsible for that. $\endgroup$ – user204463 Jan 24 '16 at 0:06