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

1,2

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.

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).

Crossrefs

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

Sequence in context: A318892 A323404 A056064 * A167976 A024000 A181983

Adjacent sequences: A274052 A274053 A274054 * A274056 A274057 A274058

Keyword

nonn

Author

Nathan Fox, Nov 13 2016

Status

approved