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a(n) = n + (square excess of n).
Conjecture (verified up to 727): the numbers not in this sequence are those of A008865. [From _- _R. J. Mathar_, Jan 23 2009]
(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)
except when (n+14) so, cause of (2) and (3) is a , the sequence has blocks of even and odd numbers starting with an even or odd square, where a(n) collapses back to (nm^2 and of length 2m+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:
(6) but, cause 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
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.
f[n_] := 2 n - (Floor@ Sqrt@ n)^2; Table[f@ n, {n, 0, 71}] [From _(* _Robert G. Wilson v_, Jan 23 2009] *)
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f[n_] := 2 n - (Floor@ Sqrt@ n)^2; Table[f@ n, {n, 0, 71}] [From _Robert G. Wilson, v (rgwv(AT)rgwv.com), _, Jan 23 2009]
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S. H. Weintraub, An interesting recursion, Amer. Math. Monthly, 111 (No. 6, 2004), 528-530.
S. H. Weintraub, <a href="http://www.lehighjstor.edu/~shw2org/preprintsstable/recursion.pdf4145074">An interesting recursion</a>, Amer. Math. Monthly, 111 (No. 6, 2004), 528-530.
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