A070864 - OEIS

A070864

a(1) = a(2) = 1; a(n) = 2 + a(n - a(n-1)).

1, 1, 3, 3, 3, 5, 3, 5, 5, 5, 7, 5, 7, 5, 7, 7, 7, 9, 7, 9, 7, 9, 7, 9, 9, 9, 11, 9, 11, 9, 11, 9, 11, 9, 11, 11, 11, 13, 11, 13, 11, 13, 11, 13, 11, 13, 11, 13, 13, 13, 15, 13, 15, 13, 15, 13, 15, 13, 15, 13, 15, 13, 15, 15, 15, 17, 15, 17, 15, 17, 15, 17, 15, 17, 15, 17, 15, 17, 15

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OFFSET

1,3

REFERENCES

S. Wolfram, A New Kind of Science, Wolfram Media, 2002; p. 129.

LINKS

T. D. Noe, Table of n, a(n) for n = 1..1000

Nick Hobson, Python program for this sequence

Eric Weisstein's World of Mathematics, Wolfram Sequences

Index entries for Hofstadter-type sequences

FORMULA

Conjecture. Let a(1)=a(2)=1 and for n > 2 let k = floor(sqrt(n+1))-1 and d=n-k(k+2). Then, if (d is 0, 1, or 2) OR (d=0 mod 2), a(n)=2k+1; otherwise a(n)=2k+3. This has been verified for n <= 15000. Thus the asymptotic behavior appears to be a(n) ~ floor(sqrt(n+1)). - John W. Layman, May 21 2002

By induction, a(1)=a(2)=1, a(3)=a(4)=a(5)=3 and for k >= 3 we obtain the following formulas for the 2k-1 consecutive values from a(k^2-2k+2) up to a(k^2+1): a(k^2+1) = a(k^2) = 2k-1, if 1 <= i <= 2k-3 then a(k^2-i) = 2k-2-(-1)^i, hence asymptotically a(n) ~ 2*sqrt(n). - Benoit Cloitre, Jul 28 2002

a(n) = 2*floor(n^(1/2)) + r where r is in {-1,1}. More precisely, let g(n) = round(sqrt(n)) - floor(sqrt(n+1)-1/sqrt(n+1)); then for n >= 1 we get: a(2*n) = 2*floor(sqrt(2*n)) - 2*g(ceiling(n/2)) + 1 and something similar for a(2*n+1). - Benoit Cloitre, Mar 06 2009

a(n) = 2*floor(n^(1/2)) - (-1)^(n + ceiling(n^(1/2))) for n > 0. - Branko Curgus, Feb 10 2011

EXAMPLE

If k = 4, a(4^2+1) = a(17) = a(16) = 2*4 - 1 = 7, a(15) = 2*4 - 2 - (-1)^1 = 7, a(14) = 2*4 - 2 - (-1)^2 = 5, a(13)=7, a(12)=5, a(11)=7.

MATHEMATICA

a[1]=a[2]=1; a[n_]:= a[n]= 2 + a[n -a[n-1]]; Table[a[n], {n, 80}]

PROG

(Sage)

@CachedFunction

def a(n): # A070864

if (n<3): return 1

else: return a(n - a(n-1)) + 2

[a(n) for n in (1..80)] # G. C. Greubel, Mar 28 2022