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 A331492 Numbers k such that the digits of k^(1/5) begin with k. 6
 0, 1, 17, 315, 316, 5623, 99999, 100000, 1778279, 31622776, 562341324, 562341325, 9999999999, 10000000000, 177827941003, 3162277660168, 56234132519034, 999999999999999, 1000000000000000, 17782794100389227, 17782794100389228, 316227766016837932, 316227766016837933 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,3 COMMENTS The following algorithm will generate all numbers k such that the digits of k^(1/b) begins with k: For each integer m >= 0, compute r = floor(10^(bm/(b-1)). Let s <= r be the largest integer >= 0 such that (r-s)*10^(bm) < (r-s+1)^b. Then r, r-1, ... r-s are such numbers k and there are no other such numbers. LINKS Chai Wah Wu, Table of n, a(n) for n = 1..1140 EXAMPLE 5623^(1/5) = 5.6233305990931... which starts with the digits 5623, so 5623 is in the sequence. CROSSREFS Cf. A307371, A307588, A307600. Sequence in context: A197526 A282965 A142428 * A244764 A089571 A196455 Adjacent sequences:  A331489 A331490 A331491 * A331493 A331494 A331495 KEYWORD nonn,base AUTHOR Chai Wah Wu, Jan 18 2020 STATUS approved

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Last modified November 28 03:29 EST 2021. Contains 349400 sequences. (Running on oeis4.)