OFFSET
1,1
COMMENTS
Numbers n such that m^2 + m/5 < n <= m^2 + 2m/5 for some positive integer m. - Robert Israel, Oct 31 2016
LINKS
Robert Israel, Table of n, a(n) for n = 1..10000
FORMULA
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
MAPLE
seq(seq(i, i=ceil((n+1/10)^2)..floor((n+2/10)^2)), n=1..40); # Robert Israel, Oct 31 2016
MATHEMATICA
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 *)
Select[Range[600], NumberDigit[Sqrt[#], -1]==1&] (* Harvey P. Dale, Jun 17 2025 *)
PROG
(PARI) isok(n) = my(d = digits(10*frac(sqrt(n))\1)); #d && d[1] == 1; \\ Michel Marcus, Oct 30 2016
(Python)
from math import isqrt
def aupto(lim): return [k for k in range(lim+1) if isqrt(100*k)%10 == 1]
print(aupto(536)) # Michael S. Branicky, Oct 17 2021, edited Jan 11 2026
CROSSREFS
KEYWORD
nonn,easy,base
AUTHOR
Patrick De Geest, Sep 15 1998
EXTENSIONS
Offset corrected by Michel Marcus, Oct 30 2016
STATUS
approved