A094761 a(n) = n + (square excess of n).

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

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+1-1) = 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), 528-530.

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*n-sqrtint(n)^2 \\ Charles R Greathouse IV, Jul 01 2012

Crossrefs

Cf. A053186, A094763, A094764, A094765.

Sequence in context: A063259 A333717 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