

A228530


Summarize the previous two terms!


1



1, 11, 31, 3113, 3133, 3153, 215315, 31123335, 41225335, 3132631435, 313263243516, 413283242536, 31527334253618, 3152733435261728, 4152832445263728, 3172634445263738, 2162636435363738, 2142934425663728, 216273442566272819, 319233542556372829
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OFFSET

1,2


COMMENTS

a(k) is found by counting the frequency of the digits in terms a(k1) and a(k2). Digits with zero frequency are not counted.
At n=54 the sequence enters a cycle of 46 terms so that for n>=100 we have a(k) = a(k46)]. [Lars Blomberg, Jan 04 2014]


LINKS

Lars Blomberg, Table of n, a(n) for n = 1..145 containing the beginning and two full cycles.
Original reddit post where this sequence was found.


EXAMPLE

For n=5, a(5) is found by counting the frequency of the digits in the last two terms; there are three 1s and three 3s, so you get "three one three three", or 3133.


CROSSREFS

Like A005151, but uses the previous two terms instead of just the previous term.
Sequence in context: A143765 A023306 A068839 * A177360 A060857 A045982
Adjacent sequences: A228527 A228528 A228529 * A228531 A228532 A228533


KEYWORD

nonn,base


AUTHOR

Edison Y. He, Sep 14 2013


EXTENSIONS

Corrected a(8)a(15), added a(16)a(20) by Lars Blomberg, Jan 04 2014


STATUS

approved



