

A094761


a(n) = n + (square excess of n).


2



0, 1, 3, 5, 4, 6, 8, 10, 12, 9, 11, 13, 15, 17, 19, 21, 16, 18, 20, 22, 24, 26, 28, 30, 32, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 64, 66, 68, 70, 72, 74, 76, 78
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OFFSET

0,3


COMMENTS

The trajectory of n under iteration of m > a(m) is eventually constant iff n is a perfect square.
Conjecture (verified up to 727): the numbers not in this sequence are those of A008865.  R. J. Mathar, Jan 23 2009
From Maon Wenders, Jul 01 2012: (Start)
Proof of conjecture:
(1) (n+2)^2  n^2 = n^2 + 4n + 4  n^2 = 4n + 4
(2) (n+1)^2  n^2 = n^2 + 2n + 1  n^2 = 2n + 1
(3) (n+1) + square excess of (n+1)  (n + square excess of n) = 2, except when (n+1) is a square, where a(n) collapses back to (n+1)
(4) so, cause of (2) and (3), the sequence has blocks of even and odd numbers starting with an even or odd square, m^2 and of length 2m+1:
0,
1, 3, 5,
4, 6, 8, 10, 12,
9, 11, 13, 15, 17, 19, 21,
16, 18, 20, 22, 24, 26, 28, 30, 32,
...
(5) such a block of 2m+1 numbers fills in all even or odd numbers between
n^2 and (n+2)^2
(6) but, because a block starts n^2 + 0, n^2 + 2, n^2 + 4, ..., the last number in such a block is n^2 + 2*(2n+11) = n^2 + 4n
(7) so the numbers n^2 + 4n + 2 = (n+2)^2  2 are missing.
End of proof. (End)


LINKS

Table of n, a(n) for n=0..71.
S. H. Weintraub, An interesting recursion, Amer. Math. Monthly, 111 (No. 6, 2004), 528530.


FORMULA

a(n) = n + A053186(n).


MATHEMATICA

f[n_] := 2 n  (Floor@ Sqrt@ n)^2; Table[f@ n, {n, 0, 71}] (* Robert G. Wilson v, Jan 23 2009 *)


PROG

(PARI) a(n)=2*nsqrtint(n)^2 \\ Charles R Greathouse IV, Jul 01 2012


CROSSREFS

Cf. A053186, A094763, A094764, A094765.
Sequence in context: A200324 A063259 A064425 * A114748 A182733 A107431
Adjacent sequences: A094758 A094759 A094760 * A094762 A094763 A094764


KEYWORD

nonn,easy


AUTHOR

N. J. A. Sloane, Jun 10 2004


STATUS

approved



