A101211 -id:A101211 - cp's OEIS frontend

A044918 Positive integers whose base-2 run lengths form a palindrome.

1, 2, 3, 5, 7, 9, 10, 12, 15, 17, 21, 27, 31, 33, 38, 42, 45, 51, 52, 56, 63, 65, 73, 85, 93, 99, 107, 119, 127, 129, 142, 150, 153, 165, 170, 178, 189, 195, 204, 212, 219, 231, 232, 240, 255, 257, 273, 297, 313, 325, 341, 365, 381

Offset: 1

Clark Kimberling

This sequence exactly contains those positive integers in A006995 (positive binary palindromes) together with the terms of A035928 (those positive integers n where reversing the order of the binary digits produces the binary complement of n). - Leroy Quet, Sep 14 2009

Also the indices of the compositions that are palindromic. For the definition of the index of a composition see A298644. For example, 93 is in the sequence since its binary form is 1011101 and the composition [1,1,3,1,1] is palindromic. On the other hand, 132 is not in the sequence since its binary form is 10000100 and the composition [1,4,1,2] is not palindromic. The command c(n) from the Maple program yields the composition having index n. - Emeric Deutsch, Jan 28 2018

Rémy Sigrist, Table of n, a(n) for n = 1..10000
Rémy Sigrist, Logarithmic scatterplot of (n, a(n+1)-a(n)) for n = 1..9999

Cf. A006995, A035928. - Leroy Quet, Sep 14 2009

Cf. A298644, A101211. - Emeric Deutsch, Jan 28 2018

Runs:=proc(L) local j,r,i,k:j:=1: r[j]:=L[1]: for i from 2 to nops(L) do if L[i]=L[i-1] then r[j]:=r[j], L[i] else j:=j+1: r[j]:=L[i] end if end do: [seq([r[k]],k=1..j)] end proc: RunLengths:=proc(L) map(nops,Runs(L)) end  proc: c:=proc(n) ListTools:-Reverse(convert(n,base,2)): RunLengths(`%`) end proc: A:={}: for n from 1 to 500 do crev(n):=[seq(c(n)[1+ nops(c(n))-j],j=1..nops(c(n)))] od:  for n from 1 to 500 do if c(n)=crev(n) then A:=A union {n} else fi od: A; # most of the Maple program is due to W. Edwin Clark. # Emeric Deutsch, Jan 28 2018

Position[Array[Length /@ Split@ IntegerDigits[#, 2] &, 400], ? PalindromeQ, 1] // Flatten (* _Michael De Vlieger, Jan 28 2018 *)

ispal(v) = {for(i=1, #v\2, if (v[i] != v[#v-i+1], return(0));); return(1);}
isok(n) = {b = binary(n); lastb = b[1]; vrun = vector(1); vrun[1] = 1; for (i=2, #b, if (b[i] != lastb, vrun = concat(vrun, 1); lastb = b[i];, vrun[#vrun]++;)); return (ispal(vrun));} \\ Michel Marcus, Jul 10 2013

A296656 Triangle whose n-th row is the concatenated sequence of all Lyndon compositions of n in reverse-lexicographic order.

Original entry on oeis.org

1, 2, 3, 1, 2, 4, 1, 3, 1, 1, 2, 5, 2, 3, 1, 4, 1, 2, 2, 1, 1, 3, 1, 1, 1, 2, 6, 2, 4, 1, 5, 1, 3, 2, 1, 2, 3, 1, 1, 4, 1, 1, 2, 2, 1, 1, 1, 3, 1, 1, 1, 1, 2, 7, 3, 4, 2, 5, 2, 2, 3, 1, 6, 1, 4, 2, 1, 3, 3, 1, 2, 4, 1, 2, 2, 2, 1, 2, 1, 3, 1, 1, 5, 1, 1, 3, 2

Offset: 1

Gus Wiseman, Dec 18 2017

			Triangle of Lyndon compositions begins:
(1),
(2),
(3),(12),
(4),(13),(112),
(5),(23),(14),(122),(113),(1112),
(6),(24),(15),(132),(123),(114),(1122),(1113),(11112),
(7),(34),(25),(223),(16),(142),(133),(124),(1222),(1213),(115),(1132),(1123),(11212),(1114),(11122),(11113),(111112).

Cf. A000740, A001037, A001045, A008965, A059966, A060223, A066099, A101211, A102659, A124734, A185700, A228369, A281013, A294859, A296302, A296373.

LyndonQ[q_]:=Array[OrderedQ[{q,RotateRight[q,#]}]&,Length[q]-1,1,And]&&Array[RotateRight[q,#]&,Length[q],1,UnsameQ];
Table[Sort[Select[Join@@Permutations/@IntegerPartitions[n],LyndonQ],OrderedQ[PadRight[{#2,#1}]]&],{n,7}]

Row n is a concatenation of A059966(n) Lyndon words with total length A000740(n).

A318927 Take the binary expansion of n, starting with the most significant bit, and concatenate the lengths of the runs.

Original entry on oeis.org

1, 11, 2, 12, 111, 21, 3, 13, 121, 1111, 112, 22, 211, 31, 4, 14, 131, 1211, 122, 1112, 11111, 1121, 113, 23, 221, 2111, 212, 32, 311, 41, 5, 15, 141, 1311, 132, 1212, 12111, 1221, 123, 1113

Offset: 1

N. J. A. Sloane, Sep 09 2018

Obviously this compressed notation is useful only for n < 1023. A101211 is a version which works for all n.

			n, binary, run lengths, -> a(n)
1, [1], [1] -> 1
2, [1, 0], [1, 1] -> 11
3, [1, 1], [2] -> 2
4, [1, 0, 0], [1, 2] -> 12
5, [1, 0, 1], [1, 1, 1] -> 111
6, [1, 1, 0], [2, 1] -> 21
7, [1, 1, 1], [3] -> 3
...

Rémy Sigrist, Table of n, a(n) for n = 1..1022

Cf. A101211 (without concatenation, as rows), A227736 (rows reversed), A318926 (reverse concatenation).

Cf. A382255 (Heinz numbers instead of concatenation of the run lengths).

Array[FromDigits@ Flatten[IntegerDigits@ Length[#] & /@ Split@ IntegerDigits[#, 2]] &, 40] (* Michael De Vlieger, Feb 17 2022 *)

a(n) = { my(d=[], r); while(n, n>>=r=valuation(n+n%2, 2); d=concat(digits(r), d)); fromdigits(d) } \\ Rémy Sigrist, Feb 17 2022, edited by M. F. Hasler, Mar 11 2025

from itertools import groupby
def A318927(n): return int(''.join(str(len(list(g))) for k, g in groupby(bin(n)[2:]))) # Chai Wah Wu, Mar 11 2022

A043284 Maximal run length in base-10 representation of n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2

Offset: 1

Clark Kimberling

The first term larger than 2 is a(111) = 3. - M. F. Hasler, Jul 21 2013

Paolo Xausa, Table of n, a(n) for n = 1..10000

Cf. A043276-A043290 for base-2 to base-16 analogs.
Cf. A030556-A030561, A030575-A030580 (related to base-6 run lengths).
Cf. A227186, A227188, A101211, A005811 (related to base-2 run lengths).

A043284[n_]:=Max[Map[Length,Split[IntegerDigits[n]]]];Array[A043284,100] (* Paolo Xausa, Sep 27 2023 *)

A043284(n)={my(m,c=1);while(n>0,n%10==(n\=10)%10 && c++ && next;m=max(m,c);c=1);m} \\ M. F. Hasler, Jul 23 2013

For n < 111, a(n) = 1 except for a(n) = 2 when n==0 (mod 11) or n = 100. - M. F. Hasler, Jul 21 2013

Data completed up to a(100), first difference with A083230, by M. F. Hasler, Oct 18 2019

A175930 Concatenation of run lengths in binary expansion of n, written in base 2, then converted to base 10.

Original entry on oeis.org

1, 3, 2, 6, 7, 5, 3, 7, 13, 15, 14, 10, 11, 7, 4, 12, 15, 27, 26, 30, 31, 29, 15, 11, 21, 23, 22, 14, 15, 9, 5, 13, 25, 31, 30, 54, 55, 53, 27, 31, 61, 63, 62, 58, 59, 31, 28, 20, 23, 43, 42, 46, 47, 45, 23, 15, 29, 31, 30, 18, 19, 11, 6, 14, 27, 51, 50, 62, 63, 61, 31, 55, 109, 111

Offset: 1

Dylan Hamilton, Oct 23 2010

			6 = 110, two runs, lengths 2 and 1, so we write down 101 and convert it to base 10, getting 5. So a(6) = 5.

Rémy Sigrist, Table of n, a(n) for n = 1..8192

Cf. A004755, A101211, A321226.

f[n_] := FromDigits[Flatten[IntegerDigits[Length /@ Split[IntegerDigits[n, \ 2]], 2]], 2]
Array[f, NUMBER OF TERMS]

a(n) = my (b=[]); while (n, my (x=valuation(n+(n%2), 2)); b = concat(binary(x), b); n \= 2^x); fromdigits(b, 2) \\ Rémy Sigrist, Jul 02 2019

from itertools import groupby
def a(n):
    c = "".join(bin(len(list(g)))[2:] for k, g in groupby(bin(n)[2:]))
    return int(c, 2)
print([a(n) for n in range(1, 75)]) # Michael S. Branicky, Oct 02 2021

From Rémy Sigrist, Jul 02 2019: (Start)
a(2^k-1) = k for any k > 0.
a(2^k) = A004755(k) for any k > 0. (End)

Edited by N. J. A. Sloane, Oct 23 2010

A280998 Numbers with a prime number of 1's in their binary reflected Gray code representation.

Original entry on oeis.org

2, 4, 5, 6, 8, 9, 11, 12, 13, 14, 16, 17, 19, 21, 23, 24, 25, 27, 28, 29, 30, 32, 33, 35, 37, 39, 41, 43, 45, 47, 48, 49, 51, 53, 55, 56, 57, 59, 60, 61, 62, 64, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 96, 97, 99, 101, 103

Offset: 1

Indranil Ghosh, Jan 12 2017

From Emeric Deutsch, Jan 28 2018: (Start)
Also the indices of the compositions that have a prime number of parts. For the definition of the index of a composition see A298644.
For example, 27 is in the sequence since its binary form is 11011 and the composition [2,1,2] has 3 parts.
On the other hand, 58 is not in the sequence since its binary form is 111010 and the composition [3,1,1,1] has 4 parts.
The command c(n) from the Maple program yields the composition having index n. (End)

			27 is in the sequence because the binary reflected Gray code representation of 27 is 10110 which has 3 1's, and 3 is prime.

Indranil Ghosh, Table of n, a(n) for n = 1..10001
Wikipedia, Gray code.

Cf. A014550, A052294, A005811, A298644, A101211.

Runs := proc (L) local j, r, i, k: j := 1: r[j] := L[1]:
for i from 2 to nops(L) do if L[i] = L[i-1] then r[j] := r[j], L[i] else j := j+1:
r[j] := L[i] end if end do: [seq([r[k]], k = 1 .. j)] end proc:
RunLengths := proc (L) map(nops, Runs(L)) end proc:
c := proc (n) ListTools:-Reverse(convert(n, base, 2)): RunLengths(%) end proc:
A := {}: for n to 175 do if isprime(nops(c(n))) = true then A := `union`(A, {n}) else end if end do: A;
# most of the program is due to W. Edwin Clark. # Emeric Deutsch, Jan 28 2018

Select[Range[100], PrimeQ[DigitCount[BitXor[#, Floor[#/2]], 2, 1]] &] (* Amiram Eldar, May 01 2021 *)

is(n)=isprime(hammingweight(bitxor(n, n>>1))) \\ Charles R Greathouse IV, Jan 12 2017

A361644 Irregular triangle T(n, k), n >= 0, k = 1..max(1, 2^(A005811(n)-1)), read by rows; the n-th row lists the integers with the same binary length as n and whose partial sums of run lengths are included in those of n.

Original entry on oeis.org

0, 1, 2, 3, 3, 4, 7, 4, 5, 6, 7, 6, 7, 7, 8, 15, 8, 9, 14, 15, 8, 9, 10, 11, 12, 13, 14, 15, 8, 11, 12, 15, 12, 15, 12, 13, 14, 15, 14, 15, 15, 16, 31, 16, 17, 30, 31, 16, 17, 18, 19, 28, 29, 30, 31, 16, 19, 28, 31, 16, 19, 20, 23, 24, 27, 28, 31

Offset: 0

Rémy Sigrist, Mar 19 2023

In other words, the n-th row contains the numbers k with the same binary length as n and for any i >= 0, if the i-th bit and the (i+1)-th bit in k are different then they are also different in n (i = 0 corresponding to the least significant bit).

The value m appears 2^A092339(m) times in the triangle (see A361674).

			Triangle begins (in decimal and in binary):
  n   n-th row      bin(n)  n-th row in binary
  --  ------------  ------  ------------------
   0  0                  0  0
   1  1                  1  1
   2  2, 3              10  10, 11
   3  3                 11  11
   4  4, 7             100  100, 111
   5  4, 5, 6, 7       101  100, 101, 110, 111
   6  6, 7             110  110, 111
   7  7                111  111
   8  8, 15           1000  1000, 1111
   9  8, 9, 14, 15    1001  1000, 1001, 1110, 1111
.
For n = 9:
- the binary expansion of 9 is "1001",
- the corresponding run lengths are 1, 2, 1,
- so the 9th row contains the values with the following run lengths:
      1, 2, 1  ->   9 ("1001" in binary)
      1,  2+1  ->   8 ("1000" in binary)
      1+2,  1  ->  14 ("1110" in binary)
       1+2+1   ->  15 ("1111" in binary)

Rémy Sigrist, Table of n, a(n) for n = 0..9841 (rows for n = 0..511 flattened)
Index entries for sequences related to binary expansion of n

Cf. A003817, A005811, A092339, A101211, A225081, A342126, A361645, A361646, A361674, A361676.

row(n) = { my (r = []); while (n, my (v = valuation(n+n%2, 2)); n \= 2^v; r = concat(v, r)); my (s = [if (#r, 2^r[1]-1, 0)]); for (k = 2, #r, s = concat(s * 2^r[k], [(h+1)*2^r[k]-1|h<-s]);); vecsort(s); }

T(n, 1) = A342126(n).

T(n, max(1, 2^(A005811(n)-1))) = A003817(n).

A371256 The run lengths transform of the ternary expansion of n corresponds to the run lengths transform of the binary expansion of a(n).

Original entry on oeis.org

0, 1, 1, 2, 3, 2, 2, 2, 3, 4, 5, 5, 6, 7, 6, 5, 5, 4, 4, 5, 5, 5, 4, 5, 6, 6, 7, 8, 9, 9, 10, 11, 10, 10, 10, 11, 12, 13, 13, 14, 15, 14, 13, 13, 12, 11, 10, 10, 10, 11, 10, 9, 9, 8, 8, 9, 9, 10, 11, 10, 10, 10, 11, 11, 10, 10, 9, 8, 9, 10, 10, 11, 12, 13, 13

Offset: 0

Rémy Sigrist, Mar 16 2024

For any v >= 0, the value v appears 2^A005811(v) times in the sequence.

			The first terms, alongside the ternary expansion of n and the binary expansion of a(n), are:
  n   a(n)  ter(n)  bin(a(n))
  --  ----  ------  ---------
   0     0       0          0
   1     1       1          1
   2     1       2          1
   3     2      10         10
   4     3      11         11
   5     2      12         10
   6     2      20         10
   7     2      21         10
   8     3      22         11
   9     4     100        100
  10     5     101        101
  11     5     102        101
  12     6     110        110
  13     7     111        111
  14     6     112        110
  15     5     120        101

Rémy Sigrist, Table of n, a(n) for n = 0..6561

See A371263 for a similar sequence.

Cf. A004488, A005811, A005823, A005836, A101211, A166242, A371257.

a(n) = { my (r = [], d, l, v = 0); while (n, d = n%3; l = 0; while ((n%3)==d, n\=3; l++;); r = concat(l, r);); for (k = 1, #r, v = (v+k%2)*2^r[k]-k%2); v }

a(A005823(n)) = n - 1.
a(A005836(n)) = n - 1.
a(A004488(n)) = a(n).
abs(a(n+1) - a(n)) <= 1.

A318926 Take the binary expansion of n, starting with the least significant bit, and concatenate the lengths of the runs.

Original entry on oeis.org

1, 11, 2, 21, 111, 12, 3, 31, 121, 1111, 211, 22, 112, 13, 4, 41, 131, 1121, 221, 2111, 11111, 1211, 311, 32, 122, 1112, 212, 23, 113, 14, 5, 51, 141, 1131, 231, 2121, 11121, 1221, 321, 3111, 12111, 111111, 21111, 2211, 11211, 1311, 411, 42, 132, 1122, 222, 2112, 11112, 1212, 312, 33, 123

Offset: 1

N. J. A. Sloane, Sep 09 2018

Obviously this compressed notation is useful only for n < 2047. A227736 is a version which works for all n. [Corrected by M. F. Hasler, Mar 12 2025]

			n, binary, run lengths, -> a(n)
1, [1], [1] -> 1
2, [0, 1], [1, 1] -> 11
3, [1, 1], [2] ->  2
4, [0, 0, 1], [2, 1] -> 21
5, [1, 0, 1], [1, 1, 1] -> 111
6, [0, 1, 1], [1, 2] -> 12
7, [1, 1, 1], [3] -> 3
8, [0, 0, 0, 1], [3, 1] ->  31,
...
From _M. F. Hasler_, Mar 12 2025: (Start)
For n = 1023 = 2^10-1, n = '1'*10 in binary, so there is only one run of length 10, whence a(n) = 10. This value cannot occur at any other index n.
For n = 1024 = 2^10, n = '1'+'0'*10 in binary, so the run lengths, from right to left, are [10, 1], whence a(n) = 101. The only other index n for which this value occurs is n = 2^101-1.
For n = 1025 = 2^10+1, n = '1'+'0'*9+'1' in binary, so a(n) = 191. This values occurs for the second time as a(n = 2^19), for the third time for a(n = 2^92-2), and for the 4th and last time as a(n = 2^191-1).
Similarly, a(1026) = 1181 appears for the second time at n = 2^19 + 1 = 524289;
  a(1027) = 281 occurs a 2nd, 3rd and 4th time at n = 2^28, (2^81-1)*2 and 2^281-1.
The first duplicate value occurs as a(2047 = 2^11-1) = 11 = a(2). (End)

Paolo Xausa, Table of n, a(n) for n = 1..10000
Claude Lenormand, Deux transformations sur les mots, Preprint, 5 pages, Nov 17 2003. Apparently unpublished. This is a scanned copy of the version that the author sent to me in 2003. - _N. J. A. Sloane_, Sep 09 2018. See Procedure 1.

Cf. A227736 (run lengths in rows instead of concatenation), A101211 (rows in reverse order), A318927 (concatenation in reverse order).

A318926[n_] := FromDigits[Flatten[IntegerDigits[Map[Length, Split[Reverse[IntegerDigits[n, 2]]]]]]];
Array[A318926, 100] (* Paolo Xausa, Mar 16 2025 *)

A318926(n)=eval(strjoin(Vecrev(A101211_row(n)))); \\ M. F. Hasler, Mar 11 2025

from itertools import groupby
def A318926(n): return int(''.join(str(len(list(g))) for k, g in groupby(bin(n)[:1:-1]))) # Chai Wah Wu, Mar 11 2022

More terms from M. F. Hasler, Mar 12 2025

A335835 Sort the run lengths in binary expansion of n in descending order (with multiplicities).

Original entry on oeis.org

0, 1, 2, 3, 6, 5, 6, 7, 14, 13, 10, 13, 12, 13, 14, 15, 30, 29, 26, 25, 26, 21, 26, 29, 28, 25, 26, 25, 28, 29, 30, 31, 62, 61, 58, 57, 50, 53, 50, 57, 58, 53, 42, 53, 50, 53, 58, 61, 60, 57, 50, 51, 50, 53, 50, 57, 56, 57, 58, 57, 60, 61, 62, 63, 126, 125

Offset: 0

Rémy Sigrist, Jun 26 2020

This sequence preserves the number of runs (A005811) and the length (A070939) of the binary representation of a number.

			For n = 72:
- the binary representation of 72 is "1001000",
- the corresponding run lengths are: 1, 2, 1, 3,
- in descending order: 3, 2, 1, 1,
- so the binary representation of a(72) is "1110010",
- and a(72) = 114.

Rémy Sigrist, Table of n, a(n) for n = 0..8191
Index entries for sequences related to binary expansion of n

Cf. A005811, A037016 (fixed points), A070939, A101211, A335834.

torl(n) = { my (rr=[]); while (n, my (r=valuation(n+(n%2), 2)); rr = concat(r, rr); n\=2^r); rr }
fromrl(rr) = { my (v=0); for (k=1, #rr, v = (v+(k%2))*2^rr[k]-(k%2)); v }
a(n) = { fromrl(vecsort(torl(n),,4)) }

a(a(n)) = a(n).

cp's OEIS Frontend

A044918 Positive integers whose base-2 run lengths form a palindrome.

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A296656 Triangle whose n-th row is the concatenated sequence of all Lyndon compositions of n in reverse-lexicographic order.

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A318927 Take the binary expansion of n, starting with the most significant bit, and concatenate the lengths of the runs.

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A043284 Maximal run length in base-10 representation of n.

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A175930 Concatenation of run lengths in binary expansion of n, written in base 2, then converted to base 10.

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A280998 Numbers with a prime number of 1's in their binary reflected Gray code representation.

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A361644 Irregular triangle T(n, k), n >= 0, k = 1..max(1, 2^(A005811(n)-1)), read by rows; the n-th row lists the integers with the same binary length as n and whose partial sums of run lengths are included in those of n.

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