You are given a natural number N which represents sequence [1,2...N]. We have to determine the number of pairs (x,y) from this sequence that satisfies the given conditions.
[1,2,3..x-1]) = sum of numbers from x+1 to y (i.e sum of [x+1...y])Example:-
If N = 3 there is only 1 pair (x=1,y=1) for which (sum of x-1 numbers) = 0 = (sum of x+1 to y)
any other pairs like (1,2),(1,3) or (2,3) does not satisfy the properties. so the answer is 1 as there is only one pair.
Another Example:-
If N=10, for pair (6,8) we can see sum of x-1 numbers i.e [1,2,3,4,5] = 15 = sum of numbers from x+1 to y i.e [7,8], Also another such pair would be (1,1). No other such pair exists so the answer, in this case, would be 2.
How can we approach and solve such problems to find the number of pairs in such a sequence?
Other things I have been able to deduce so far:-
| Condition | Answer | Pairs |
|---|---|---|
| If 1<=N<=7 | 1 | {(1,1)} |
| If 8<=N<=48 | 2 | {(1,1),(6,8)} |
| If 49<=N<=287 | 3 | {(1,1),(6,8),(35,49)} |
| If 288<=N<=1680 | 4 | - |
I tried but am unable to find any pattern or any such thing in these numbers.
Also, 1<=N<=10^16
--edit--
Courtesy of OEIS (link in comments): you can find the k'th value of y using this formula: ( (0.25) * (3.0+2.0*(2**0.5))**k ).floor
This gives us the k'th value in O(log k). First few results:
1
8
49
288
1681
9800
57121
332928
1940449
11309768
65918161
384199200
2239277041
13051463048
76069501249
443365544448
2584123765441
15061377048200
87784138523761
511643454094368
2982076586042447
17380816062160312
101302819786919424
590436102659356160
3441313796169217536
20057446674355949568
116903366249966469120
681362750825442836480
3971273138702690287616
23146276081390697054208
134906383349641499377664
786292024016458181771264
4582845760749107960348672
26710782540478185822224384
155681849482119992477483008
907380314352241764747706368
Notice that the ratio of successive numbers quickly approaches 5.828427124746. Given a value of n, take the log of n base 5.828427124746. The answer will be an integer close to this log.
E.g., say n = 1,000,000,000. Then log(n, 5.8284271247461) = 11.8. The answer is probably 12, but we can check the neighbors to be sure.
11: 65,918,161
12: 384,199,200
13: 2,239,277,041
Confirmed.
-- end edit --
Here's some ruby code to do this. Idea is to have two pointers and increment the pointer for x or y as appropriate. I'm using s(n) to calculate the sums, though this could be done without multiplication by just keeping a running total.
def s(n)
return n*(n+1)/2
end
def f(n)
count = 0
x = 1
y = 1
while y <= n do
if s(x-1) == s(y) - s(x)
count += 1
puts "(#{x}, #{y})"
end
if s(x-1) <= s(y) - s(x)
x += 1
else
y += 1
end
end
end
Here are the first few pairs:
(1, 1)
(6, 8)
(35, 49)
(204, 288)
(1189, 1681)
(6930, 9800)
(40391, 57121)
(235416, 332928)
(1372105, 1940449)
(7997214, 11309768)
(46611179, 65918161)