A283207 a(n) = a(floor(n/a(n-1))) + a(floor(n/a(n-2))) with a(1) = a(2) = 2.

2, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 8, 6, 4, 6, 6, 4, 8, 6, 6, 8, 6, 6, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8

Offset: 1

Author

Altug Alkan, Mar 03 2017

Keywords

• nonn

Comments

For the first 10^6 terms, the maximum value of a(n) is 64 and the values of b(n) = least k such that a(k) = 2*n are 1, 3, 13, 12, 105, 97, 126, 96, 1681, 1552, 1746, 1537, 1734, 1926, 4050, 1536, 53793, 49665, 53890, 49185, 53862, 57024, 55616, 49153, 55488, 55302, 81249, 61446, 83619, 115214, 162000, 49152; note that b(2^n) = 3*2^((n+2)*(n-1)/2) for n = 1 to 5.

This sequence is a_{1,2}(n) where a_{r,s}(n) = a_{r,s}(floor(n/a_{r,s}(n-r))) + a_{r,s}(floor(n/a_{r,s}(n-s))) with a_{r,s}(n) = 2 for n <= s (r < s). - Altug Alkan, Jun 28 2020

Examples

			a(5) = 4 because a(5) = a(floor(5/a(4))) + a(floor(5/a(3))) = a(floor(5/4)) + a(floor(5/4)) = a(1) + a(1) = 4.
		

Links

• Altug Alkan, Table of n, a(n) for n = 1..10000
• Altug Alkan, Alternative Graph of A283207
• Altug Alkan, Line plot of (a(floor(n/a(n-1))), a(floor(n/a(n-2)))) for n <= 2^15
• Rémy Sigrist, Scatterplot of the first 2^34 terms
• Rémy Sigrist, Line plot of (a(floor(n/a(n-1))), a(floor(n/a(n-2)))) for n <= 12855108032

Crossrefs

Cf. A005185, A130535.

Programs

Maple

A:= Vector(100):
A[1]:= 2: A[2]:= 2:
for n from 3 to 100 do A[n]:= A[floor(n/A[n-1])] + A[floor(n/A[n-2])] od:
convert(A,list); # Robert Israel, Jun 23 2020

Mathematica

a[1] = a[2] = 2; a[n_] := a[n] = a[Floor[n/a[n - 1]]] + a[Floor[n/a[n - 2]]]; Array[a, 120] (* Michael De Vlieger, Mar 06 2017 *)

PARI

a=vector(100); a[1]=a[2]=2; for(n=3, #a, a[n]=a[n\a[n-1]]+a[n\a[n-2]]); a

Python

from functools import lru_cache
@lru_cache(maxsize=None)
def A283207(n):
    return 2 if n <= 2 else A283207(n//A283207(n-1)) + A283207(n//A283207(n-2)) # Chai Wah Wu, Jun 23 2020