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 A322667 a(n) is the smallest positive integer k such that floor((k + 1)^2/10^n) - floor(k^2/10^n) = 2. 2
 7, 59, 531, 5099, 50316, 500999, 5003162, 50009999, 500031622, 5000099999, 50000316227, 500000999999, 5000003162277, 50000009999999, 500000031622776, 5000000099999999, 50000000316227766, 500000000999999999, 5000000003162277660, 50000000009999999999 (list; graph; refs; listen; history; text; internal format)
 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 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: A059705 A218201 A015568 * A101487 A210397 A099659 Adjacent sequences:  A322664 A322665 A322666 * A322668 A322669 A322670 KEYWORD nonn AUTHOR Jianing Song, Dec 22 2018 STATUS approved

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Last modified April 15 07:59 EDT 2021. Contains 342975 sequences. (Running on oeis4.)