A007932 -id:A007932 - cp's OEIS frontend

A045670 Digital sum (in base 10) of numbers in base 3 of the alternate number system.

1, 2, 3, 2, 3, 4, 3, 4, 5, 4, 5, 6, 3, 4, 5, 4, 5, 6, 5, 6, 7, 4, 5, 6, 5, 6, 7, 6, 7, 8, 5, 6, 7, 6, 7, 8, 7, 8, 9, 4, 5, 6, 5, 6, 7, 6, 7, 8, 5, 6, 7, 6, 7, 8, 7, 8, 9, 6, 7, 8, 7, 8, 9, 8, 9, 10, 5, 6, 7, 6, 7, 8, 7, 8, 9, 6, 7, 8, 7, 8, 9, 8, 9, 10, 7, 8, 9, 8, 9, 10

Offset: 1

Bob Forslund (forslund(AT)tbaytel.net)

The alternate number system has no need for the digit zero.

Robert R. Forslund, A Logical Alternative to the Existing Positional Number System, Southwest Journal of Pure and Applied Mathematics. Vol. 1 1995 pp. 27-29.

Andrew Howroyd, Table of n, a(n) for n = 1..10000
R. R. Forslund, A logical alternative to the existing positional number system, Southwest Journal of Pure and Applied Mathematics, 1995 Vol. 1.
R. R. Forslund, Positive Integer Pages
EMIS, Mirror site for Southwest Journal of Pure and Applied Mathematics

Cf. A007931, A007932, A007953.

a(n)={my(k=3); while(n>k, n-=k; k*=3); my(d=digits(n+k-1,3)); vecsum(d) + #d - 2} \\ Andrew Howroyd, Jan 27 2020

a(n) = A007953(A007932(n)). - Andrew Howroyd, Jan 27 2020

Terms a(25) and beyond from Andrew Howroyd, Jan 27 2020

A282234 a(n) = Fibonacci(n) represented in bijective base-3 numeration.

Original entry on oeis.org

1, 1, 2, 3, 12, 22, 111, 133, 321, 1231, 2322, 11323, 22122, 111222, 211121, 323113, 1311311, 3112131, 12131212, 22321113, 112222332, 212321222, 332321331, 1323113323, 3133213131, 12311111231, 23221332132, 113233221133, 221232331112, 1112313322322

Offset: 1

Alois P. Heinz, Feb 09 2017

Alois P. Heinz, Table of n, a(n) for n = 1..1000
Wikipedia, Bijective numeration

Column k=3 of A214679.

Cf. A000045, A007932.

a:= proc(n) local b, d, l, m; l:= NULL;
      b, m:= 3, combinat[fibonacci](n);
      while m>0 do  d:= irem(m, b, 'm');
        if d=0 then d:=b; m:=m-1 fi;
        l:= d, l
      od; parse(cat(l))
    end:
seq(a(n), n=0..35);

a(n) = A007932(A000045(n)).

A334055 Iteration count (or -1) corresponding to A334054(n).

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 1, 1, 3, 1, 4, 10, 3, 1, 1, 1, 1, 3, 1, 3, 7, 2, 9, 3, 1, 7, 3, 3, 4, -1, 3, 3, -1, 4, 1, -1, 5, 9, -1, -1, 7, 3, 1, 2, 3, 1, 5, 9, 5, -1, 3, 1, 4, 3, 3, 4, -1, 3, 4, -1, 3, 1, -1, 5, 7, -1, 3, 1, 1, -1, 2, 9, 11, 3, 8, 4, 10, 3, 4, -1, -1, 3, -1, -1, 3, 3, -1, 8, 10

Offset: 1

Scott R. Shannon, Sep 07 2020

sign

This is the number of iterations for the starting number, containing only digits 1,2 and 3 (see A007932) to reappear in the iterative cycle of its own 'Look and Say' description.
See A334054 for further details, a list of the number that do reappear in their iterative cycle, and a proof that the number 233 can never reappear in its cycle.
For all 1092 numbers up to six digits long containing digits 1,2,3 there are 397 numbers which reappear in their own iterative cycle, while 695 do not and can be proven never will.
In the same range, numbers 121322 and 213223 take 33 cycles before reappearing. The final string in the later case has 49470 digits. See the link file for details of the other numbers.

			The numbers containing only digits 1,2 and 3 are given in A007932.
a(1) = 1 as the number 1 take one iteration to reappear: 1->11 which contains '1' as a substring.
a(4) = 2 as the number 11 takes two iterations to reappear: 11->21->1211 which contains '11' as a substring.
a(9) = 3 as the number 23 takes three iterations to reappear: 23->1213->11121113->31123113 which contains '23' as a substring.
a(11) = 4 as the number 32 takes four iterations to reappear: 32->1312->11131112->31133112->1321232112 which contains '32' as a substring.
a(12) = 10 as the number 33 takes ten iterations to reappear: 33->23->1213->11121113->31123113->132112132113->11131221121113122113->311311222112311311222113->1321132132211213211321322113->11131221131211132221121113122113121113222113->3113112221131112311332211231131122211311123113322113 which contains '33' as a substring.

Scott R. Shannon, Details of all 1092 numbers up to six digits long.

Cf. A334054, A005150, A007932, A119566, A007932, A045918.

A334054 Lexicographically earliest numbers containing only digits 1,2,3 that appear again in the iterative cycle of their own 'Look and Say' description (Cf. A005150).

Original entry on oeis.org

1, 2, 3, 11, 12, 13, 21, 22, 23, 31, 32, 33, 111, 112, 113, 121, 122, 123, 131, 132, 133, 211, 212, 213, 221, 222, 223, 231, 232, 311, 312, 321, 322, 331, 332, 1112, 1113, 1121, 1122, 1123, 1131, 1132, 1133, 1211, 1213, 1221, 1222, 1223, 1231, 1232, 1311, 1312, 1321, 1322, 1331, 1332, 2111

Offset: 1

Scott R. Shannon, Sep 07 2020

nonn

Take a number that contains only digits 1,2,3 and then describe the given number with its 'Look and Say' string, see A005150. Repeat this process until the starting number appears in the resulting string. The sequence lists the numbers that eventually reappear.
A334055 gives the number of iterations for the starting number to reappear and details of the final string for all recurrent numbers.
The number 233 is the first number that does not appear in its iterative Look and Say cycle. See examples below.

			1 is a term as the Look and Say description of 1 is '11', which contains '1' as a substring.
111 is a term as the Look and Say description of 111 is '31'. Repeating this process leads to '1311' and then '111321', which contains '111' as a substring.
1112 is a term as the Look and Say description of 1112 is '3112'. Repeating this process leads to '132112', '1113122112', '311311222112', '13211321322112', '1113122113121113222112', and then '31131122211311123113322112', which contains '1112' as a substring.
233 is not a term. If it were, it would have to have a parent string that could produce 233 from its Look and Say description, and one can show that that is not possible. The '33' part cannot be interpreted as 'three 3's' as '333' cannot be inside the parent string. To prove that, consider that the string '333' can be interpreted in two ways. 1. "x number of 3's followed by three 3's" which is not possible as that would be written as "x+three 3's". 2. "three 3's followed by three x's", where x cannot be a 3, else it would be written as "six 3's". So it has to be either '333111' or '333222'. But the first case has to be interpreted as either "x 3's followed by three 3's followed by one 1 ..." which is not possible as that would be written as "x+three 3's", else it has to be interpreted as "three 3's followed by three 1's followed by one 1", which is not possible as that would be written as "... four 1's". Likewise for '333222', hence '333' has no valid parent string. Therefore '233' must be interpreted as "two 3's followed by 3 x's", where the x's cannot be 3's. Thus the two possible strings are '33111' or '33222'. But the '33' part of these strings cannot be "three 3's" as shown, so the first string has to be interpreted as "x 3's followed by three 1's followed by one 1", but that is not possible as the 1's would be combined. Similarly for the second string. Hence '233' has no parent string, thus it cannot reappear in the iterative process of this sequence.
A similar argument can prove that numbers like 313, 323, 1111, 2232, 2313, and with more effort, 22213, 112321, 213321, do not have any parent string and thus do not appear in this sequence.

Cf. A334055, A005150, A119566, A007932, A045918.

cp's OEIS Frontend

A045670 Digital sum (in base 10) of numbers in base 3 of the alternate number system.

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A282234 a(n) = Fibonacci(n) represented in bijective base-3 numeration.

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A334055 Iteration count (or -1) corresponding to A334054(n).

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A334054 Lexicographically earliest numbers containing only digits 1,2,3 that appear again in the iterative cycle of their own 'Look and Say' description (Cf. A005150).

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