|
|
A336112
|
|
a(n) is the least number k such that the Sum_{i=0..k} sqrt(k) equals or exceeds n.
|
|
0
|
|
|
0, 1, 2, 3, 3, 4, 4, 5, 5, 6, 6, 7, 7, 7, 8, 8, 8, 9, 9, 9, 10, 10, 10, 11, 11, 11, 12, 12, 12, 12, 13, 13, 13, 14, 14, 14, 14, 15, 15, 15, 15, 16, 16, 16, 16, 17, 17, 17, 17, 18, 18, 18, 18, 19, 19, 19, 19, 19, 20, 20, 20, 20, 21, 21, 21, 21, 21, 22, 22, 22, 22, 23, 23, 23, 23
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,3
|
|
COMMENTS
|
Let c = (9/4)^(1/3) = (3/2)^(2/3) ~ 1.310370697..., then a(n) ~ c*n^(2/3).
a(10^k) for k>= 0: 1, 6, 28, 131, 608, 2823, 13104, 60822, 282311, 1310371, 6082202, 28231081, 131037070, 608220200, ..., .
|
|
LINKS
|
|
|
FORMULA
|
a(k*n) ~ k^(2/3)*a(n).
|
|
EXAMPLE
|
a(0) = 0 since the sqrt(0) = 0;
a(1) = 1 since the sqrt(0) + sqrt(1) = 1;
a(2) = 2 since the sqrt(0) + sqrt(1) + sqrt(2) ~ 2.41421... which exceeds 2;
a(3) = 3 since the sqrt(0) + sqrt(1) + sqrt(2) + sqrt(3) ~ 4.146264... which easily exceeds 3;
a(4) = 3 because the sqrt(0) + sqrt(1) + sqrt(2) + sqrt(3) ~ 4.146264... which barely exceeds 4; etc.
|
|
MATHEMATICA
|
f[n_] := Block[{k = s = 0}, While[s < n, k++; s = s + Sqrt@k]; k]; Array[f, 75, 0]
|
|
PROG
|
(PARI) a(n) = my(s=0, k=0); while ((s+=sqrt(k)) < n, k++); k; \\ Michel Marcus, Jul 09 2020
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|