Find-the-2023rd-term-in-the-sequence-2-3-5-6-7-8-10-11-12-13-14-15-17-18-obtained-by-subtracting-integer-squares-from-natural-numbers-

Question Number 202306 by hardmath last updated on 24/Dec/23

Find the 2023rd term in the sequence 2,3,5,6,7,8,10,11,12,13,14,15,17,18,… obtained by subtracting integer squares from natural numbers.

Answered by AST last updated on 24/Dec/23

1, 2, 3, \dots, 2023 c o n t a i n s 44 s q u a r e s

\Rightarrow 2023 i s t h e (2023 - 44 = 1979) t h t e r m

2023 - 1979 t h t e r m

2023 + 44 - 2022 n d t e r m; s i n c e 45^{2} = 2025

\Rightarrow 2023 r d t e r m = 2023 + 45 = 2068

Commented by hardmath last updated on 24/Dec/23

thank you dear professor

Answered by mr W last updated on 24/Dec/23

1, 2, 3, 4, 5, \dots, n

1^{2}, 2^{2}, 3^{2}, \dots, k^{2} w i t h k^{2} < n, i . e . k = ⌊ \sqrt{n} ⌋

m = n - k = n - ⌊ \sqrt{n} ⌋

t h a t m e a n s t h e m^{t h} t e r m i s n .

⌊ \sqrt{n} ⌋ ⩽ \sqrt{n}

m = n - ⌊ \sqrt{n} ⌋ ⩾ n - \sqrt{n}

n - \sqrt{n} - m ⩽ 0

\Rightarrow \sqrt{n} ⩽ \frac{1 + \sqrt{1 + 4 m}}{2}

\Rightarrow n ⩽ \frac{{(\sqrt{4 m + 1} + 1)}^{2}}{4}

i . e . t h e m^{t h} t e r m i s

a_{m} = ⌊ \frac{{(\sqrt{4 m + 1} + 1)}^{2}}{4} ⌋

e x a m p l e s :

m = 2023 :

a_{2023} = ⌊ \frac{{(\sqrt{4 \times 2023 + 1} + 1)}^{2}}{4} ⌋ = 2068

m = 2025 :

a_{2025} = ⌊ \frac{{(\sqrt{4 \times 2025 + 1} + 1)}^{2}}{4} ⌋ = 2070

m = 1000 :

a_{1000} = ⌊ \frac{{(\sqrt{4 \times 1000 + 1} + 1)}^{2}}{4} ⌋ = 1032

m = 3000 :

a_{3000} = ⌊ \frac{{(\sqrt{4 \times 3000 + 1} + 1)}^{2}}{4} ⌋ = 3055

Commented by hardmath last updated on 24/Dec/23

perfect solution tahnkyou dear professor