

A063657


Numbers with property that truncated square root is unequal to rounded square root.


9



3, 7, 8, 13, 14, 15, 21, 22, 23, 24, 31, 32, 33, 34, 35, 43, 44, 45, 46, 47, 48, 57, 58, 59, 60, 61, 62, 63, 73, 74, 75, 76, 77, 78, 79, 80, 91, 92, 93, 94, 95, 96, 97, 98, 99, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 133, 134, 135, 136, 137, 138, 139, 140
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OFFSET

1,1


COMMENTS

Also: skip 1, take 0, skip 2, take 1, skip 3, take 2, ...
Integers for which the periodic part of the continued fraction for the square root of n begins with a 1.  Robert G. Wilson v, Nov 01 2001
a(n) belongs to the sequence if and only if a(n) > floor(sqrt(a(n))) * ceiling(sqrt(a(n))), i.e. a(n) in (k*(k+1),k^2), k >= 0.  Daniel Forgues, Apr 17 2011
Any integer between (k  1/2)^2 and k^2 exclusive, for k > 1, is in this sequence. If we take this sequence and remove each term that is one more than the previous term, we obtain the central polygonal numbers (A002061). If instead we remove each term that is one less than the next term, we obtain numbers that are one less than squares (A005563).  Alonso del Arte, Dec 28 2013
a(n) = A217575(n) + 1.  Reinhard Zumkeller, Jun 20 2015


LINKS

Seiichi Manyama, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Harry J. Smith)
R. B. Nelsen, Proof without Words: Consecutive sums of consecutive integers, Mathematical Association of America, p.25.


EXAMPLE

7 is in the sequence because its square root is 2.64575..., which truncates to 2 but rounds to 3.
8 is in the sequence because its square root is 2.828427..., which also truncates to 2 but rounds to 3.
9 is not in the sequence because its square root is 3 exactly, which truncates and rounds the same.
Here is the example per Lamoen's skip n, take n  1 process: starting at 0, we skip one integer (0) but take zero integers for our sequence. Then we skip two integers (1 and 2) and take one integer (3) for our sequence. Then we skip three integers (4, 5, 6) and take two integers for our sequence (7 and 8, so the sequence now stands as 3, 7, 8). Then we skip four integers (9, 10, 11, 12) and so on and so forth.
From Seiichi Manyama, Sep 19 2017: (Start)
See R. B. Nelsen's paper.
k A063656(n)  a(n)

0 0
1 1 + 2 = 3
2 4 + 5 + 6 = 7 + 8
3 9 + 10 + 11 + 12 = 13 + 14 + 15
4 16 + 17 + 18 + 19 + 20 = 21 + 22 + 23 + 24
 ...
(End)


MAPLE

A063657:=n>`if`(floor(floor(sqrt(n+1)) * (1+floor(sqrt(n+1)))/(n+1))=1, NULL, n+1); seq(A063657(n), n=1..200); # Wesley Ivan Hurt, Dec 28 2013


MATHEMATICA

Select[ Range[200], Floor[ Sqrt[ # ]] != Floor[ Sqrt[ # ] + 1/2] & ] (* or *) Select[ Range[200], First[ Last[ ContinuedFraction[ Sqrt[ # ]]]] == 1 & ]


PROG

(PARI) { n=0; for (m=0, 10^9, if (sqrt(m)%1 > .5, write("b063657.txt", n++, " ", m); if (n==1000, break)) ) } \\ Harry J. Smith, Aug 27 2009
(Haskell)
a063657 n = a063657_list !! n
a063657_list = f 0 [0..] where
f k (_:xs) = us ++ f (k + 1) (drop (k + 1) vs) where
(us, vs) = splitAt k xs
 Reinhard Zumkeller, Jun 20 2015


CROSSREFS

Cf. A063656, A004201A004202, A002620, A189151.
Cf. A217575.
Sequence in context: A028972 A153030 A031425 * A108575 A153219 A264720
Adjacent sequences: A063654 A063655 A063656 * A063658 A063659 A063660


KEYWORD

nonn


AUTHOR

Floor van Lamoen, Jul 24 2001


STATUS

approved



