MHB What is the value of $a_{2013}$ in the sequence challenge II?

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The sequence is defined such that the integer $k$ appears $k$ times consecutively. The pattern starts with 1 appearing once, 2 appearing twice, 3 appearing three times, and so forth. To find $a_{2013}$, one must determine which integer corresponds to the position 2013 in this cumulative sequence. The cumulative count of terms can be calculated using the formula for the sum of the first $n$ integers, which is $n(n + 1)/2$. By solving for $n$ where this sum exceeds 2013, it is determined that $a_{2013} = 63$.
lfdahl
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Let $a_1 = 1$, $a_2 = a_3 = 2$, $a_4 = a_5 = a_6 = 3$, $a_7 = a_8 = a_9 = a_{10} = 4$, and so on. That is,
$a_n ∶ 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, . . . . $ What is $a_{2013}$?
 
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lfdahl said:
Let $a_1 = 1$, $a_2 = a_3 = 2$, $a_4 = a_5 = a_6 = 3$, $a_7 = a_8 = a_9 = a_{10} = 4$, and so on. That is,
$a_n ∶ 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, . . . . $ What is $a_{2013}$?
1+2+3+4+------+62=1953
$\therefore a_{1953}=62$
2013-1953=60<63
and we have:
$a_{2013}=63$
 
Last edited:
the 1st term is 1, there are 2's then 3 3's so on

there are n n's after n-1 so 1st n is at n(n-1)/2 and last n at n(n+1)/2

2013 > 62 * 63/2 but < 63 * 64/2 so ans is 63

as 62 finishes at 63 * 62/2 or 1953 and 63 frm 1954 to 2016 positions
 
Thankyou kaliprasad and Albert! Your anwers are correct. Good job!:)
 
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