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Comma transform of A366487.
1

%I #19 Jan 05 2024 17:22:01

%S 12,35,94,15,16,28,31,34,37,41,45,55,55,55,55,61,67,74,71,89,98,97,18,

%T 19,11,11,12,13,14,15,16,17,18,19,11,11,12,13,14,15,16,17,18,22,22,24,

%U 26,28,22,22,24,26,28,22,22,24,26,28,22,22,24,26,28,22,22

%N Comma transform of A366487.

%C See A367360 for further information.

%C Let the comma sequence A121805 be known as S or C0.

%C A366487, the first differences of A121805, is the same as the comma transform of A121805; call it C1.

%C This sequence is C2 = C(C(S)), the comma transform C iterated twice.

%C C4 = C2, C5 = C2, ... once the first term (and the last term if the sequence is finite) are removed from the lower iterates of C.

%C Theorem: C^{i+2}(S) = C^i(S) for i>=2 in general and for i>=0 when all terms of S have two digits and no least significant digit is zero. See link for proof.

%C Remark. The lexicographically earliest sequence S with C(S) = S is A010850, all 11's.

%C The sequence contains 2137451 terms, with a(2137451) = 96. The next term does not exist.

%H Michael S. Branicky, <a href="/A368782/b368782.txt">Table of n, a(n) for n = 1..20000</a>

%H Michael S. Branicky, <a href="https://docs.google.com/document/d/1PXgz1eGBaUWoy1MlT0hhIkkZoLfFWAfI2TEIdsaA0c4/edit?usp=sharing">Comma Comma Proof</a>

%o (Python)

%o from itertools import islice, pairwise

%o def S(): # generator of comma sequence

%o an = 1

%o while True:

%o yield an

%o an += 10*(an%10)

%o children = [an+y for y in range(1, 10) if str(an+y)[0] == str(y)]

%o if not children: break

%o an = children[0]

%o def C(g): # generator of comma transform of sequence passed as a generator

%o yield from (10*(t%10) + int(str(u)[0]) for t, u in pairwise(g))

%o def agen(): return C(C(S()))

%o print(list(islice(agen(), 70))) # _Michael S. Branicky_, Jan 05 2024

%Y Cf. A010850, A121805, A367360, A366487.

%K nonn,base

%O 1,1

%A _Michael S. Branicky_, Jan 05 2024