# Sequence of integer square roots

Let's define a sequence of integer square roots. First, a(1) = 1. Then, a(n) is the smallest positive integer not seen before such that

sqrt(a(n) + sqrt(a(n-1) + sqrt(... + sqrt(a(1)))))


is an integer. Some examples:

a(2) is 3 because it's the smallest integer such that sqrt(a(2) + sqrt(a(1))) = sqrt(a(2) + 1) is integer, and 3 hasn't occured in the sequence before.

a(3) is 2 because it's the smallest integer such that sqrt(a(3) + sqrt(a(2) + sqrt(a(1)))) = sqrt(a(3) + 2) is integer, and 2 hasn't occured in the sequence before.

a(4) is 7 because sqrt(a(4) + 2) is integer. We couldn't have a(4) = 2 because 2 already occured in our sequence.

Write a program or function that given a parameter n returns a sequence of numbers a(1) to a(n).

The sequence starts 1,3,2,7,6,13,5, ....

Source of this sequence is from this Math.SE question.

A plot of the first 1000 elements in the sequence:

• :'-( Oct 12 '17 at 13:57
• @Mr.Xcoder That just makes it interesting!
– orlp
Oct 12 '17 at 13:58
• @Mr.Xcoder Yeah I agree it's so bad you can't just copy-paste the formula... Oct 12 '17 at 14:03
• @EriktheOutgolfer No. When you get n as input you should return or print a list of a(1) to a(n). In other words, the first n numbers in the sequence. There is no 'indexing'.
– orlp
Oct 12 '17 at 14:12
• Are errors caused by floating point inaccuracies acceptable for very large inputs? Oct 12 '17 at 14:12

# Python 2, 80 bytes

s=[]
exec'x=q=1\nwhile(x in s)+q%1:x+=1;q=(v+x)**.5\nv=q;s+=x,;'*input()
print s


Try it online!

Horribly inefficient, but does not rely on floating point arithmetic. Here a(x) = sqrt(f(x)+a(x-1)) is a helper sequence, that simplifies the computation.

a 0=0
a x=[k|k<-[1..],m<-[k^2-a(x-1)],m>0,notElem m$f<$>[1..x-1]]!!0
f x=(a x)^2-a(x-1)


Try it online!

# Python 2, 87 bytes

t,=s=1,
for n in~-input()*s:
while(n in s)+(t+n)**.5%1:n+=1
s+=n,;t=(t+n)**.5
print s


Try it online!

-3 thanks to Mr. Xcoder.
-5 thanks to ovs.

• 92 bytes -> while n in s or(t+n)**.5%1>0 -> while(n in s)+(t+n)**.5%1 Oct 12 '17 at 14:28
• 87 bytes
– ovs
Oct 12 '17 at 14:30
• @ovs clever one Oct 12 '17 at 16:43

# MATL, 30 27 bytes

lXHiq:"@ymH@+X^1\+}8MXHx@h


Try it online! Or see a graphical display (takes a while; times out for inputs exceeding approximately 60).

### Explanation

l          % Push 1. This is the array that holds the sequence, initialized to
% a single term. Will be extended with subsequent terms
XH         % Copy into clipboard H, which holds the latest result of the
% "accumulated" square root
iq:"       % Input n. Do the following n-1 times
%   Do...while
@      %     Push interaton index k, starting at 1. This is the candidate
%     to being the next term of the sequence
y      %     Push copy of array of terms found so far
m      %     Ismbmer? True if k is in the array
H      %     Push accumulated root
X^     %     Square root
1\     %     Modulo 1. This gives 0 if k gives an integer square root
+      %     Add. Gives nonzero if k is in the array or doesn't give an
%     integer square root; that is, if k is invalid.
%   The body of the do...while loop ends here. If the top of the
%   stack is nonzero a new iteration will be run. If it is zero that
%   means that the current k is a new term of the sequence
}        %   Finally: this is executed after the last iteration, right before
%   the loop is exited
8M     %     Push latest result of the square root
XH     %     Copy in clipboard K
x      %     Delete
@      %     Push current k
h      %     Append to the array
% End do...while (implicit)
% Display (implicit)


# Mathematica, 104 bytes

(s=f={i=1};Do[t=1;While[!IntegerQ[d=Sqrt[t+s[[i]]]]||!f~FreeQ~t,t++];f~(A=AppendTo)~t;s~A~d;i++,#-1];f)&


The sequence of the square roots is also very interesting...
and outputs a similar pattern

1,2,2,3,3,4,3,5,3,6,4,4,5,4,6,5,5,6,6,7,4,7,5,7,6,8,4,8,5,8,6,9,5,9,6,10,5,10,6,11,5,11,6,12,6,13,6,14,7,7,8,7,9,7,10,7,11,7,12,7,13,7,14,8,8,9,8,10...

also here are the differences of the main sequence

# Python 2, 11711511210299 87 bytes

t,=r=1,;exec"x=1\nwhile(t+x)**.5%1or x in r:x+=1\nr+=x,;t=(t+x)**.5;"*~-input();print r


Try it online!

Used the t=(t+x)**.5 logic from Erik's answer

• Oct 12 '17 at 14:20
• @JonathanFrech Thanks :) Oct 12 '17 at 14:20

# JavaScript (ES7), 898277 76 bytes

i=>(g=k=>(s=(++n+k)**.5)%1||u[n]?g(k):i--?[u[n]=n,...g(s,n=0)]:[])(n=0,u=[])


### Demo

let f =

i=>(g=k=>(s=(++n+k)**.5)%1||u[n]?g(k):i--?[u[n]=n,...g(s,n=0)]:[])(n=0,u=[])

console.log(JSON.stringify(f(10)))

### Formatted and commented

i => (                             // given i = number of terms to compute
u = [],                          // u = array of encountered values
g = p =>                         // g = recursive function taking p = previous square root
(s = (++n + p) ** .5) % 1      // increment n; if n + p is not a perfect square,
|| u[n] ?                      // or n was already used:
g(p)                         //   do a recursive call with p unchanged
:                              // else:
i-- ?                        //   if there are other terms to compute:
[u[n] = n, ...g(s, n = 0)] //     append n, set u[n] and call g() with p = s, n = 0
:                            //   else:
[]                         //     stop recursion
)(n = 0)                         // initial call to g() with n = p = 0


# R, 138105 99 bytes

function(n){for(i in 1:n){j=1
while(Reduce(function(x,y)(y+x)^.5,g<-c(T,j))%%1|j%in%T)j=j+1
T=g}
T}


Try it online!

-33 bytes using Tfeld's clever sqrt()%%1 trick in the while loop

-6 bytes using T instead of F

function(n,l={}){g=function(L)Reduce(function(x,y)(y+x)^.5,L,0)
for(i in 1:n){T=1
while(g(c(l,T))!=g(c(l,T))%/%1|T%in%l)T=T+1
l=c(l,T)}
l}


Try it online!

# Husk, 21 bytes

!¡oḟȯΛ±sFo√+Som:-N;1


Try it online!

### How?

!¡oḟȯΛ±sFo√+Som:-N;1    Function that generates a list of prefixes of the sequence and indexes into it
;1    The literal list [1]
¡                       Iterate the following function, collecting values in a list
oḟȯΛ±sFo√+Som:-N        This function takes a prefix of the sequence, l, and returns the next prefix.
-N      Get all the natural numbers that are not in l.
Som:         Append l in front each of these numbers, generates all possible prefixes.
ȯΛ±sFo√+               This predicate tests if sqrt(a(n) + sqrt(a(n-1) + sqrt(... + sqrt(a(1))))) is an integer.
F                Fold from the left
o√+             the composition of square root and plus
s                 Convert to string
ȯΛ±                  Are all the characters digits, (no '.')
oḟ                     Find the first list in the list of possible prefixes that satisfies the above predicate
!                        Index into the list