

A283885


Relative of Hofstadter Qsequence: a(n) = max(0, n+3442) for n <= 0; a(n) = a(na(n1)) + a(na(n2)) + a(na(n3)) for n > 0.


5



6, 3443, 3444, 3445, 9, 3446, 3447, 3448, 12, 3449, 3450, 3451, 15, 3452, 3453, 17, 3455, 18, 3455, 3457, 3458, 22, 21, 6895, 6889, 9, 18, 6904, 6907, 3451, 22, 3472, 3477, 3455, 27, 36, 3479, 6894, 3446, 39, 3480, 3486, 3450, 42
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OFFSET

1,1


COMMENTS

Sequences like this are more naturally considered with the first nonzero term in position 1. But this sequence would then match A000027 for its first 3442 terms.
Most terms in this sequence appear in long period5 quasilinear runs. These runs are separated by 11943 other terms, and each run is approximately six times as long as the previous.
The first such run that falls into a predictable pattern begins at index 90682, though there are other similar patterns earlier.


LINKS

Nathan Fox, Table of n, a(n) for n = 1..100000


FORMULA

If the index is between 67 and 3443 (inclusive), then a(7n) = 7n+2, a(7n+1) = 7n+3444, a(7n+2) = 7n+3446, a(7n+3) = 7, a(7n+4) = 2n+6929, a(7n+5) = n+6877, a(7n+6) = 3440.
For nonnegative integers i, if 1 <= 5n + r <= (487329/5)*6^(i+1)  88639/5, then
a((487329/5)*6^i  28924/5 + 5n) = 5
a((487329/5)*6^i  28924/5 + 5n + 1) = (1461987/5)*6^i  52797/5 + 3n
a((487329/5)*6^i  28924/5 + 5n + 2) = 3
a((487329/5)*6^i  28924/5 + 5n + 3) = (487329/5)*6^i  28909/5 + 5n
a((487329/5)*6^i  28924/5 + 5n + 4) = (1461987/5)*6^i  52792/5 + 3n.


MAPLE

A283885:=proc(n) option remember: if n <= 0 then max(0, n+3442): else A283885(nA283885(n1)) + A283885(nA283885(n2)) + A283885(nA283885(n3)): fi: end:


CROSSREFS

Cf. A005185, A267501, A274058, A278055, A278066, A283884, A283886, A283887, A283888.
Sequence in context: A030245 A247965 A172955 * A198706 A198625 A198653
Adjacent sequences: A283882 A283883 A283884 * A283886 A283887 A283888


KEYWORD

nonn


AUTHOR

Nathan Fox, Mar 19 2017


STATUS

approved



