Python 2, 84 82 81 bytes

-2 bytes thanks to ElPedro.

n=input();n+=n/7;w=''
while-~n:w+=" vibgyor"[n%8];print' '*n+w+w[-1]+w[::-1];n-=1

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JavaScript (ES6), 100 bytes

Returns an array of strings.

f=(n,a=[i=' '])=>++i<n+n/7?f(n,[c=' vibgyor'[i&7],...a].map(s=>c+s+c)):a.map(s=>' '.repeat(--i)+s)

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05AB1E, 32 31 23 bytes

.•VvÈ©•¹∍¬„ v:Rηε¬ý}.c

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-1 thanks to Kevin Cruijssen and -8 thanks to Adnan

Explanation (Stack example w/ input of 3):

.•VvÈ©• # Push 'aibgyor' | ['aibgyor']
 ¹∍ # Extend to input length. | ['aib']
 ¬ # Push head. | ['aib','a']
 „ v: # Replace with ' v'. | [' vib']
 R # Reverse. | ['biv ']
 η # Prefixes. | ['b', 'bi', 'biv', 'biv ']
 ε } # For each.... | 
 ¬ý # Bifurcate, join by head. | ['b','b'] -> ['bbb']
 | ['bi','ib'] -> ['biiib']
 | ['biv','vib'] -> ['bivvvib']
 | ['biv ',' vib'] -> ['biv vib']
 .c # Center the result. | Expected output.

Canvas, 29 28 26 bytes

７÷Ｕ＋｛ <ibgyor＠¹×／ｎ｝⇵Ｋ２＊∔─↶

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Explanation:

7÷U+ ceil(input/7) + input
 for n in 1..the above
 <ibgyor@ in the string " <ibgyor", pick the nth character
 ¹× repeat n times
 / create a diagonal of that
 n and overlap the top 2 stack items (the 1st time around this does nothing, leaving an item for the next iterations)
 ⇵ reverse the result vertically
 K take off the last line (e.g. " <ibgyor <ib")
 2* repeat that vertically twice
 ∔ and append that back to the diagonals
 ─ palindromize vertically
 ↶ and rotate 90° anti-clockwise. This rotates "<" to "v"

25 24 22 bytes after fixing that ｍold should cycle if the wanted length is bigger than the inputs length and fixing ／ for like the 10th time

Haskell, 114 110 101 bytes

Thanks to [nimi][1] for -4 13 bytes!

f n=""#(n+1+div n 7)
w#0=
w#n|x<-cycle"r vibgyo"!!n=((' '<$[2..n])++reverse w++x:x:x:w):(x:w)#(n-1)

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Dyalog APL, 41 39 38 bytes

↑⌽(⌽,⊃,A↑⊢)⍵↑A⍴' vibgyor'¨-⍳A←⌈⎕×8÷7

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A similar approach to others: A←⌈⎕×8÷7 finds the height of the rainbow (also the width of the longest 'half row' to the left/right of the centre) and assigns it to A for later use, while ¨-⍳ iterates through the values 1..A, negating them to select on the correct side when used with ↑.

A⍴' vibgyor' generates a 'half row' and ⍵↑ selects the correct length substring. (⌽,⊃,A↑⊢) generates the full row in reverse (which takes fewer characters to do), starting with a reversed half row (⌽), then the centre character taken from the beginning of the half row string (⊃) and finally a right padded version of the half row (A↑⊢). The final ⌽ reverses the row into the correct orientation and ↑ turns the vector of rows into a 2D array.

Edit: -2 thanks to dzaima

Edit: -1 thanks to ngn

Python 2, 108 bytes

n=input()-1
n+=n/7+2
o=
for s in(' vibgyor'*n)[:n]:o=[s+l+s for l in[s]+o]
for l in o:print l.center(n-~n)

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Charcoal, 30 bytes

↶≔… vibgyor⁺²÷×⁸⊖Ｎ⁷θθ⸿Ｅθ✂θκ‖Ｏ←

Try it online! Link is to verbose version of code. Explanation:

↶

Change the drawing direction to upwards.

≔… vibgyor⁺²÷×⁸⊖Ｎ⁷θ

Calculate the height of the rainbow and repeat the literal string to that length.

θ⸿

Print the central line of the rainbow.

Ｅθ✂θκ

Print the right half of the rainbow by taking successive slices and printing each on its own "line".

‖Ｏ←

Reflect to complete the rainbow.

Jelly, 31 bytes

:7+‘µ“ vibgyor”ṁṚ,Ṛjṛ/ƲƤṭ"ḶṚ⁶ẋƲ

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Check out a test suite!

ಠ_ಠ This is overly complicated because Jelly doesn't have a centralize function...

Jelly, 28 bytes

:7+‘“ vibgyor”ṁµṫJZz⁶U;"⁸ŒBṚ

A monadic link accepting an integer which yields a list of lists of characters.

Try it online! (footer joins with newline characters)

Or see the test-suite.

How?

:7+‘“ vibgyor”ṁµṫJZz⁶U;"⁸ŒBṚ - Link: integer
:7 - integer divide by seven (number of full rainbows)
 ‘ - increment (the input integer)
 + - add (gets the number bands)
 “ vibgyor” - list of characters = " vibgyor"
 ṁ - mould like the result above (as a range)
 µ - start a new monadic chain
 J - range of length
 ṫ - tail (vectorises) (gets the suffixes)
 Z - transpose
 z⁶ - transpose with filler space character
 - (together these pad with spaces to the right)
 U - reverse each
 - (now we have the left side of the rainbow upside down)
 ⁸ - chain's left argument, as right argument of...
 " - zip with:
 ; - concatenation
 - (adds the central character)
 Ã…Â’B - bounce (vectorises at depth 1)
 - (reflects each row like [1,2,3,4] -> [1,2,3,4,3,2,1])
 Ṛ - reverse (turn the rainbow up the right way)

R, 130 bytes

function(n,k=n%/%7*8+1+n%%7,a=el(strsplit(' vibgyor'/k,'')))for(i in k:1)cat(d<-' '/(i-1),a[c(k:i,i,i:k)],d,sep='','
')
"/"=strrep

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• -6 bytes thanks to @JayCe

PowerShell 5.1, 108 98 89 Bytes

param($x)$a=" vibgyor"*($x+=$x/7-replace'..*');$x..0|%' '*$_+-join$a[$x..$_+$_+$_..$x]

This one feels pretty alright now. Banker's rounding is still the devil and I figured out how to make a non-dumb join. I tried monkeying with $ofs to not much success. Speaking of, the results without joins look pretty good, a bit melty:

 vvv
 v v
 v rrr v
 v r ooo r v
 v r o yyy o r v
 v r o y ggg y o r v
 v r o y g bbb g y o r v
 v r o y g b iii b g y o r v
 v r o y g b i vvv i b g y o r v
v r o y g b i v v i b g y o r v

Haskell, 106 113 bytes

I can't yet comment other posts (namely this) so I have to post the solution as a separate answer.

Golfed away 7 bytes by ovs

p x=reverse x++x!!0:x
u m|n<-m+div(m-1)7=[(' '<$[z..n])++p(drop(n-z)$take(n+1)$cycle" vibgyor")|z<-[0..n]]

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(Old version, 113 bytes)

Python 2, 132 131 bytes

def f(n):
 t=n+n/7;s=('vibgyor '*n)[:t];r=[s[~i:]+t*' 'for i in range(t)]
 for l in zip(*r+3*[' '+s]+r[::-1])[::-1]:print''.join(l)

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Saved:

• -1 byte, thanks to Jonathan Frech

Red, 153 bytes

func[n][r: take/last/part append/dup copy"""roygbiv "n l: 9 * n + 8 / 8
repeat i l[print rejoin[t: pad/left take/part copy r i l last t reverse copy t]]]

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Slightly more readable:

f: func[ n ] [
 r: copy ""
 append/dup r "roygbiv " n
 r: take/last/part r l: 9 * n + 8 / 8
 repeat i l [
 print rejoin [ t: pad/left take/part copy r i l
 last t 
 reverse copy t ]
 ]
]

Java (JDK 10), 184 bytes

n->int h=n+n/7,i=h+1,w=i*2+1,j,k=0;var o=new char[i][w];for(;i-->0;o[i][w/2]=o[i][w/2+1])for(j=w/2;j-->0;)o[i][j]=o[i][w+~j]=i<h?j<1?32:o[i+1][j-1]:" vibgyor".charAt(k++%8);return o;

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Prints an extra leading and trailing space for each multiple of 7.

Explanation

n-> // IntFunction
 int h=n+n/7, // Declare that height = n + n/7
 i=h+1, // that index = h + 1
 w=i*2+1, // that width = (h+1)*2+1
 j, // j
 k=0; // that k = 0
 var o=new char[i][w]; // Declare a 2D char array
 for(; // Loop
 i-->0; // Until i is 0
 o[i][w/2]=o[i][w/2+1] // After each run, copy the middle letter.
 )
 for(j=w/2; // Loop on j = w/2
 j-->0; // Until j = 0
 ) //
 o[i][j] // copy letters to the left side,
 =o[i][w+~j] // and the right side
 =i<h // if it's not the last line
 ?j<1 // if it's the first (and last) character
 ?32 // set it to a space.
 :o[i+1][j-1] // else set it to the previous character on the next line.
 :" vibgyor".charAt(k++%8); // else assign the next letter.
 return o; // return everything

Credits

• -2 bytes thanks to Kevin Cruijssen
  

Stax, 23 bytes

⌡G'5h!M╩EV[Ez ▼>≈<S⌡⌡0`

Run and debug it

Unpacked, ungolfed, and commented, it looks like this.

" vibgyor" string literal
,8*7/^ input * 8 / 7 + 1
:m repeat literal to that length
|] get all prefixes
Mr rectangularize, transpose array of arrays, then reverse
 this is the same as rotating counter-clockwise
m map over each row with the rest of the program, then implicitly output
 the stack starts with just the row itself
 _h push the first character of the row
 _r push the reversed row
 L wrap the entire stack in a single array

Run this one