The given sequence is:
1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, ... Here, the number nnn appears nnn times consecutively.

Step 1: Understand the pattern

• Number 1 appears 1 time (positions 1)
• Number 2 appears 2 times (positions 2–3)
• Number 3 appears 3 times (positions 4–6)
• Number 4 appears 4 times (positions 7–10)
• Number 5 appears 5 times (positions 11–15)
• Number nnn appears nnn times.

The position at which the number nnn ends is the sum of the first nnn natural numbers:

End position for n=n(n+1)2\text{End position for }n=\frac{n(n+1)}{2}End position for n=2n(n+1)​

Step 2: Find nnn such that the 2320th position lies within the block of

nnn

We want to find nnn such that:

n(n+1)2≥2320\frac{n(n+1)}{2}\geq 23202n(n+1)​≥2320

and

(n−1)n2<2320\frac{(n-1)n}{2}<23202(n−1)n​<2320

Multiply both sides by 2:

n(n+1)≥4640n(n+1)\geq 4640n(n+1)≥4640

Solve the quadratic inequality:

n2+n−4640≥0n^2+n-4640\geq 0n2+n−4640≥0

Using the quadratic formula for equality:

n=−1±1+4×46402=−1±185612n=\frac{-1\pm \sqrt{1+4\times 4640}}{2}=\frac{-1\pm \sqrt{18561}}{2}n=2−1±1+4×4640​​=2−1±18561​​

18561≈136.25\sqrt{18561}\approx 136.2518561​≈136.25, so

n≈−1+136.252=135.252=67.625n\approx \frac{-1+136.25}{2}=\frac{135.25}{2}=67.625n≈2−1+136.25​=2135.25​=67.625

Since nnn must be an integer, check n=67n=67n=67 and n=68n=68n=68:

• For n=67n=67n=67:

67×682=45562=2278<2320\frac{67\times 68}{2}=\frac{4556}{2}=2278<2320267×68​=24556​=2278<2320

• For n=68n=68n=68:

68×692=46922=2346≥2320\frac{68\times 69}{2}=\frac{4692}{2}=2346\geq 2320268×69​=24692​=2346≥2320

So the 2320th position is in the block where the number 68 appears.

Step 3: Confirm the exact position within the block

The block for 68 starts at position 227922792279 (one after 2278) and ends at position 234623462346. Since 232023202320 is between 227922792279 and 234623462346, the number at position 2320 is 68. Final answer: The number at the 2320th position in the sequence is 68.