A274055 Relative of Hofstadter Q-sequence: a(n) = n for 1 <= n <= 42; a(n) = a(n-a(n-1)) + a(n-a(n-2)) for n > 42.

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 3, 43, 44, 5, 45, 6, 7, 46, 48, 10, 8, 48, 52, 12, 49, 14, 54, 11, 53, 57, 16, 13, 17, 15, 56, 20, 20

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.

This sequence eventually settles into a pattern resembling A272610.

Links

• Nathan Fox, Table of n, a(n) for n = 1..40000
• N. Fox, Hofstadter-like Sequences over Nonstandard Integers", Talk given at the Rutgers Experimental Mathematics Seminar, November 10 2016.

Crossrefs

Cf. A005185, A272610, A272611, A272612, A272613, A278056, A278057, A278058, A278059, A278060, A278061, A278062, A278063, A278064, A278065.

Formula

If the index is between 77 and 89 (inclusive), then a(5n) = 3, a(5n+1) = 5, a(5n+2) = 88n-1188, a(5n+3) = 5, a(5n+4) = 88.

If the index is between 95 and 397 (inclusive), then a(5n) = 396n-6820, a(5n+1) = 3, a(5n+2) = 396, a(5n+3) = 3, a(5n+4) = 5.

If the index is between 403 and 24860 (inclusive), then a(5n) = 24860, a(5n+1) = 3, a(5n+2) = 5, a(5n+3) = 24860n-1939476, a(5n+4) = 5.

If the index is at least 24863, then a(5n) = 24860*A272613(n-4972), a(5n+1) = 4, a(5n+2) = 5*A272611(n-4972), a(5n+3) = 5*A272611(n-4971), a(5n+4) = 5*A272612(n-4971). This pattern lasts as long as A272611 exists (which is conjectured to be forever).