A322667 a(n) is the smallest positive integer k such that floor((k + 1)^2/10^n) - floor(k^2/10^n) = 2.

7, 59, 531, 5099, 50316, 500999, 5003162, 50009999, 500031622, 5000099999, 50000316227, 500000999999, 5000003162277, 50000009999999, 500000031622776, 5000000099999999, 50000000316227766, 500000000999999999, 5000000003162277660, 50000000009999999999

Offset

1,1

Comments

For n >= 2, note that when k < 5*10^(n-1) we have (k + 1)^2 - k^2 = 2*k + 1 < 10^n, so a(n) >= 5*10^(n-1). For 0 <= t < sqrt(10^n), floor((5*10^(n-1) + t)^2/10^n) = 25*10^(n-2) + t; for t = ceiling(sqrt(10^n)), floor((5*10^(n-1) + t)^2/10^n) = 25*10^(n-2) + t + 1. Take n = 3 as an example. When k < 500, (k + 1)^2 - k^2 < 1000, so a(3) >= 500. Since 31^2 = 961 < 1000, 32^2 = 1024 > 1000, (500 + t)^2 is successively 251001, 252004, ..., 281961, 283024, so a(3) = 500 + 31 = 531.

Links

Table of n, a(n) for n=1..20.

Formula

a(n) = 5*10^(n-1) + ceiling(10^(n/2)) - 1 for n >= 2.

Example

floor(7^2/10) = 4, floor(8^2/10) = 6, and 7 is the smallest k such that floor((k + 1)^2/10) - floor(k^2/10) = 2, so a(1) = 7.
floor(59^2/10) = 34, floor(60^2/10) = 36, and 59 is the smallest k such that floor((k + 1)^2/100) - floor(k^2/100) = 2, so a(2) = 59.

Prog

(PARI) a(n) = if(n==1, 7, 5*10^(n-1) + ceil(10^(n/2)) - 1)

Crossrefs

Cf. A017936, A317774, A322666.

Sequence in context: A359691 A218201 A015568 * A363105 A101487 A210397

Adjacent sequences: A322664 A322665 A322666 * A322668 A322669 A322670

Keyword

nonn

Author

Jianing Song, Dec 22 2018

Status

approved