A278066 Relative of Hofstadter Q-sequence: a(1) = 5, a(2) = 2; thereafter a(n) = a(n-a(n-1)) + a(n-a(n-2)).

5, 2, 5, 2, 5, 7, 2, 12, 2, 12, 2, 12, 7, 4, 14, 14, 10, 14, 7, 14, 6, 26, 10, 4, 20, 33, 2, 33, 2, 33, 2, 33, 2, 38, 2, 38, 2, 38, 7, 4, 40, 40, 10, 40, 7, 14, 6, 78, 10, 4, 46, 85, 2, 85, 2, 85, 2, 85, 2, 85, 2, 85, 2, 85, 2, 85, 2, 85, 2, 85, 2, 85, 2, 85, 2, 85, 2

Offset: 1

Author

Nathan Fox, Nov 13 2016

Keywords

• nonn

Comments

In calculating terms of this sequence, use the convention that a(n)=0 for n<=0.

Most terms in this sequence alternate between 2 and a term of A275163. These runs are separated by 18 other terms, and each run is approximately twice as long as the previous.

Links

• Nathan Fox, Table of n, a(n) for n = 1..10000
• Nathan Fox, Hofstadter-like Sequences over Nonstandard Integers", Talk given at the Rutgers Experimental Mathematics Seminar, November 10 2016.

Crossrefs

Cf. A005185, A275163, A278067, A278068.

Programs

Mathematica

a[1] = 5; a[2] = 2; a[n_] := a[n] = If[n < 1, 0, a[n-a[n-1]] + a[n-a[n-2]]];
Array[a, 100] (* Paolo Xausa, May 30 2024 *)

Formula

a(1) = 5, a(2) = 2, a(3) = 5, a(4) = 2, a(5) = 5, a(6) = 7, a(7) = 2; thereafter, for k>=0,

a(A275163(k)+1)=A275163(k)+5

a(A275163(k)+2)=2

a(A275163(k)+3)=A275163(k)+5

a(A275163(k)+4)=2

a(A275163(k)+5)=A275163(k)+5

a(A275163(k)+6)=7

a(A275163(k)+7)=4

a(A275163(k)+8)=A275163(k)+7

a(A275163(k)+9)=A275163(k)+7

a(A275163(k)+10)=10

a(A275163(k)+11)=A275163(k)+7

a(A275163(k)+12)=7

a(A275163(k)+13)=14

a(A275163(k)+14)=6

a(A275163(k)+15)=2*A275163(k)+12

a(A275163(k)+16)=10

a(A275163(k)+17)=4

a(A275163(k)+18)=A275163(k)+13

a(A275163(k)+i)=A275163(k+1), i odd, 19<=i<A275163(k+1)-A275163(k)

a(A275163(k)+i)=2, i odd, 20<=i<=A275163(k+1)-A275163(k).