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A034097
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Fractional part of square root of a(n) starts with digit 1.
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2
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10, 17, 27, 38, 51, 66, 67, 83, 84, 103, 104, 124, 125, 147, 148, 172, 173, 174, 199, 200, 201, 229, 230, 231, 260, 261, 262, 293, 294, 295, 328, 329, 330, 331, 365, 366, 367, 368, 405, 406, 407, 408, 446, 447, 448, 449, 489, 490, 491, 492, 534, 535, 536
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OFFSET
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1,1
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COMMENTS
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Numbers n such that m^2 + m/5 < n <= m^2 + 2m/5 for some positive integer m. - Robert Israel, Oct 31 2016
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LINKS
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FORMULA
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For any a(n) in this sequence, letting k = integer part of sqrt( a(n) ), a(n) belongs to a run of round( 1/25 + 1/5 k ) consecutive terms that differ by one (conjectured). E.g. round( 1/25 + 1/5 floor( sqrt( 330 ) ) ) =4 and 330 is in the run 328, 329, 330, 331, consecutive terms of the series differing by one. - Dimitri Papadopoulos, Oct 28 2016
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MAPLE
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seq(seq(i, i=ceil((n+1/10)^2)..floor((n+2/10)^2)), n=1..40); # Robert Israel, Oct 31 2016
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MATHEMATICA
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t={}; For[i=1, i<5001, i++, If[N[Sqrt[i]]-IntegerPart[Sqrt[i]]>=.1&&N[Sqrt[i]]-IntegerPart[Sqrt[i]]<.2, t=Append[t, i]]]; t (* generates the sequence*)
For[i = 1, i < 5001, i++, If[N[Sqrt[i]] - IntegerPart[Sqrt[i]] >= .1 &&N[Sqrt[i]] - IntegerPart[Sqrt[i]] < .2, tt = {i}; For[j = 1, j < Round[N[1/25 + 1/5 IntegerPart[Sqrt[i]]]], j++, tt = Append[tt, i + j]]; Print[tt]; i = i + Round[N[1/25 + 1/5 IntegerPart[Sqrt[i]]]]]] (* generates the sequence with consecutive terms that differ by one grouped together*) (* Dimitri Papadopoulos, Oct 28 2016 *)
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PROG
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(PARI) isok(n) = my(d = digits(10*frac(sqrt(n))\1)); #d && d[1] == 1; \\ Michel Marcus, Oct 30 2016
(Python)
from math import sqrt
def aupto(lim): return [k for k in range(lim+1) if int(10*sqrt(k))%10 == 1]
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CROSSREFS
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KEYWORD
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nonn,easy,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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