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

[lfdahl]

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 $a_{2013}$?

 

[Albert1]

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 $a_{2013}$?

Spoiler

1+2+3+4+------+62=1953
$\therefore a_{1953}=62$
2013-1953=60<63
and we have:
$a_{2013}=63$

 

[kaliprasad]

Spoiler

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

 

[lfdahl]

Thankyou kaliprasad and Albert! Your anwers are correct. Good job!:)