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A018851
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a(n)^2 is smallest square beginning with n.
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22
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0, 1, 5, 6, 2, 23, 8, 27, 9, 3, 10, 34, 11, 37, 12, 39, 4, 42, 43, 14, 45, 46, 15, 48, 49, 5, 51, 52, 17, 54, 55, 56, 18, 58, 59, 188, 6, 61, 62, 63, 20, 203, 65, 66, 21, 213, 68, 69, 22, 7, 71, 72, 23, 73, 74, 235, 75, 24, 241, 77, 78, 247, 25, 251, 8, 81, 257, 26, 83, 263, 84, 267, 27
(list;
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refs;
listen;
history;
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internal format)
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OFFSET
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0,3
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COMMENTS
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As the graph shows, the points belong to various "curves". For each n there is a value d = d(n) such that n*10^d <= a(n)^2 < (n+1)*10^d, and so on this curve, a(n) ~ sqrt(n)*10^(d/2). The values of d(n) are given in A272677.
For a given n, d can range from 0 (if n is a square) to d_max, where it appears that d_max approx. equals 3 + floor( log_10(n/25) ). The successive points where d_max increases are given in A272678, and that entry contains more precise conjectures about the values.
For example, in the range 2600 = A272678(5) <= n < 25317 = A272678(6), d_max is 5. This is the upper curve in the graph that is seen if the "graph" button is clicked, and on this curve a(n) is about sqrt(n)*10^(5/2). (End)
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LINKS
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Zak Seidov, Blog entry [Cached copy, pdf format, with permission]
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FORMULA
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a(n) >= sqrt(n), for all n >= 0 with equality when n is a square. - Derek Orr, Jul 26 2015
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MAPLE
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f:= proc(n) local d, m;
for d from 0 do
m:= ceil(sqrt(10^d*n));
if m^2 < 10^d*(n+1) then return m fi
od
end proc:
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PROG
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(PARI) a(n)=k=1; while(k, d=digits(k^2); D=digits(n); if(#D<=#d, c=1; for(i=1, #D, if(D[i]!=d[i], c=0; break)); if(c, return(k))); k++)
vector(100, n, a(n)) \\ Derek Orr, Jul 26 2015
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CROSSREFS
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A265432 is a more complicated sequence in the same spirit.
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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