## Longest Consecutive Sequence Of Squares

### December 11, 2015

This is similar to our previous exercise to find the longest consecutive sequence of odd numbers that sums to a given number, and is solved the same way: Initialize `lo`, `hi`, and `sum` to 1. If `sum` is less than the target, increase `hi` and add its square to `sum`. If `sum` is greater than the target, increase `lo` and subtract its square from `sum`. If `sum` is equal to the target, you’re done. If the square of the `hi` value exceeds the target before you have found a solution, then there is no solution.

```(define (squares n)
(define (square x) (* x x))
(let loop ((lo 1) (hi 1) (sum 1))
(cond ((< n (* hi hi)) (list))
((< sum n) (loop lo (+ hi 1) (+ sum (square (+ hi 1)))))
((< n sum) (loop (+ lo 1) hi (- sum (square lo))))
(else (list lo hi)))))```

Here is the opposite function that takes the two bounds and returns the sum of the squares within those bounds; it uses the function for the sum of the squares of the first n integers:

```(define (sum-squares lo hi)
(define (sum-squares n) (* n (+ n 1) (+ n n 1) 1/6))
(- (sum-squares hi) (sum-squares (- lo 1))))```

Here are some examples:

```> (squares 595) (6 12) > (sum-squares 6 12) 595```

And here is A034705, which shows all the numbers that can be written as the sum of consecutive squares:

```> (map (lambda (xs) (sum-squares (car xs) (cadr xs)))
(filter pair? (map squares (range 300)))))
(1 4 5 9 13 14 16 25 29 30 36 41 49 50 54 55 61 64 77 81 85
86 90 91 100 110 113 121 126 135 139 140 144 145 149 169 174
181 190 194 196 199 203 204 221 225 230 245 255 256 265 271
280 284 285 289 294)```

You can run the program at http://ideone.com/121NjK.

Pages: 1 2

### 13 Responses to “Longest Consecutive Sequence Of Squares”

1. Rutger said

in Python

```def longest_consequtive_squares(n):
results = []
for i in range(1,int(n**0.5+1)):
for j in range(i, int(n**0.5+2)):
if sum([x**2 for x in range(i,j)]) == n:
results.append(range(i,j))
if results:
results.sort(key=len, reverse=True)
return results
else:
return None

for n in range(1,1000):
print n, longest_consequtive_squares(n)
```
2. Paul said

In Python.

```def longest_conseq_sq(target):
lo = hi = hisq = sumsq = 1
while hisq <= target:
if sumsq < target:
hi += 1
hisq = hi ** 2
sumsq += hisq
elif sumsq > target:
sumsq -= lo ** 2
lo += 1
else:
return lo, hi
return None
```
3. FA said
```package programmingpraxis

// https://programmingpraxis.com/2015/12/11/longest-consecutive-sequence-of-squares/
// Some numbers can be represented as the sum of a sequence of consecutive squares;
// for instance, 595 = 6^2 + 7^2 + 8^2 + 9^2 + 10^2 + 11^2 + 12^2. Other numbers are not so representable.
// Your task is to write a program that finds the longest consecutive sequence of squares that sums to a given number, or determines that no such sequence exists.
object LongestConsecutiveSequenceOfSquares {

def squares(start: Int = 0): Stream[Int] = Stream.cons(start * start, squares(start + 1))
//> squares: (start: Int)Stream[Int]

def squaresSum(start: Int = 0, sum: Int = 0): Stream[Int] = Stream.cons(sum + start * start, squaresSum(start + 1, sum + start * start))
//> squaresSum: (start: Int, sum: Int)Stream[Int]
def sum(from: Int, to: Int): Int = squaresSum(from, 0).take(to - from + 1).lastOption.getOrElse(-1)
//> sum: (from: Int, to: Int)Int

def find(n: Int): Set[String] = {
def find(n: Int, from: Int, to: Int): Set[String] = {
val s = sum(from, to)
if (s == n) Set(""+from+"^2 - "+ to+"^2")
else if (s < n) find(n, from + 1, to+1) ++ find(n, from, to + 1)
else Set()
}
find(n, 2, 2)
}                                               //> find: (n: Int)Set[String]

find(595)                                 //> res6: Set[String] = Set(6^2 - 12^2)
find(2)                                   //> res8: Set[String] = Set()

(1 to 1000).map(n=>(n,find(n))).filter(!_._2.isEmpty).map(p=>p._1+": "+p._2.mkString(" or ")).mkString("\r\n")
//> res9: String = 4: 2^2 - 2^2
//| 9: 3^2 - 3^2
//| 13: 2^2 - 3^2
//| 16: 4^2 - 4^2
//| 25: 5^2 - 5^2 or 3^2 - 4^2
//| 29: 2^2 - 4^2  ...
}
```
4. matthew said

A Python solution using second order differences, so no multiplication:

```def squaresum(n):
d0, n0 = -1, 0; d1, n1 = 1, 1; t = 1
while t != 0:
if t < n:   d1 += 2; n1 += d1; t += n1
elif t > n: d0 += 2; n0 += d0; t -= n0
else: return (n, (d0+3)//2, (d1+1)//2)

for res in filter(None,map(squaresum,range(1,301))):
print res
```
5. Ernie said

Is it known whether any integer can be represented as the sum of consecutive squares in more than one way?

6. programmingpraxis said

@Ernie: Yes; 25 can be represented by the sequences (3 4) or (5). Of these, the longest is (3 4), so it is the desired sequence. The suggested solution will always find the longest sequence because it starts from 1 and increases.

7. matthew said

@Ernie: good question. Many numbers have 2 representations, as @praxis says, 25 is the smallest, 365 is the smallest that isn’t a perfect square.

Here’s a modified program that will show us numbers with 3 or more representations:

```def squaresums(n):
d0, n0 = -1, 0; d1, n1 = 1, 1; t = 1; s = []
while t != 0:
if t == n: s.append((n, (d0+3)//2, (d1+1)//2))
if t <= n:   d1 += 2; n1 += d1; t += n1
else: d0 += 2; n0 += d0; t -= n0
return s

n = 1
while True:
s = squaresums(n)
if len(s) > 2: print s
n += 1
```

Output so far:

```[(20449, 7, 39), (20449, 38, 48), (20449, 143, 143)]
[(147441, 18, 76), (147441, 29, 77), (147441, 85, 101)]
[(910805, 35, 140), (910805, 144, 178), (910805, 550, 552)]
[(1026745, 1, 145), (1026745, 51, 147), (1026745, 716, 717)]
[(2403800, 298, 322), (2403800, 368, 384), (2403800, 583, 589)]
[(2513434, 66, 198), (2513434, 286, 313), (2513434, 473, 483)]
[(3198550, 1, 212), (3198550, 127, 226), (3198550, 225, 275)]
```

Another question: do numbers with a sum of consecutive squares representation become scarcer as the numbers grow? Experimentally, they seem to, so there is a possibility there is a only a finite number (apart from the perfect squares of course).

8. matthew said

To answer my own question: of course there are an infinite number as we can generate new ones as differences of sums of squares. Still not sure about the density. Something like N^(2/3) might be about about right (there are approximately N^(1/3) sums of squares less than N, and we can choose 2 of them to take the difference).

9. Paul said

Here is a quadruple representation: 5545037505 (480, 1210), (3570, 3612), (3613, 3654), (7442, 7451)

10. matthew said

@Paul: Good stuff! How did you find that one? I’m guessing by generating all differences-of-sums-of-squares directly rather than checking each number for a representation.

Did some investigation of density, N^(2/3)/N seems about right, but I suspect my reasons for coming up with it in the first were bogus.

11. matthew said

@Paul: BTW, I think you meant 554503705. Good stuff anyway.

12. Paul said

@Matthew. Sorry for the typo. I first generate the cumulative sum of squares. Then simply loop over the start and the end indeces and fill a dictionary. The main problem is that the memory is filled up quickly. Therefore I limit the range for the dict.

```def all_longest_conseq_sq(minimum, maximum):
squares = [i**2 for i in range(int(math.sqrt(maximum)))]
acc_squares = list(accumulate(squares))
D = defaultdict(list)
for i, aci in enumerate(acc_squares):
for j, acj in enumerate(acc_squares[i+1:], i+1):
if minimum < acj - aci <= maximum:
D[acj - aci].append((i+1, j))
return D
```
13. Yuri said

Java Script

```//task is to write a program that finds the longest consecutive sequence of squares that sums to a given number, or determines that no such sequence exists.
var x = 190; // enter the number for checking here
document.write("Number " + x);
document.write(" is a sum of consecutive squares of following numbers (if any): "+ conseq(x));

function conseq(x){
var max = Math.sqrt(x);
for (var i=1; i<= max; i++ ){
var solution =[];
var sum=0;
var j=i;
while (sum <= x){
sum += Math.pow(j, 2);
solution.push(j);
j += 1;
if (sum == x) {return solution };
}; //end of while cicle
};
return "no solution  :("
}; //end of function

```