A043276 -id:A043276 - cp's OEIS frontend

A227185 The largest part in the unordered partition encoded in the runlengths of the binary expansion of n.

0, 1, 1, 2, 2, 1, 2, 3, 3, 2, 1, 2, 3, 2, 3, 4, 4, 3, 2, 3, 2, 1, 2, 3, 4, 3, 2, 3, 4, 3, 4, 5, 5, 4, 3, 4, 3, 2, 3, 4, 3, 2, 1, 2, 3, 2, 3, 4, 5, 4, 3, 4, 3, 2, 3, 4, 5, 4, 3, 4, 5, 4, 5, 6, 6, 5, 4, 5, 4, 3, 4, 5, 4, 3, 2, 3, 4, 3, 4, 5, 4, 3, 2, 3, 2, 1, 2

Offset: 0

Antti Karttunen, Jul 05 2013

The bijective encoding of nonordered partitions via compositions (ordered partitions) present in the binary expansion of n is explained in A227184.

It appears that a(4n+2) = a(2n+1). - Ralf Stephan, Jul 20 2013

			12 has binary expansion "1100", for which the lengths of runs (consecutive blocks of 0- or 1-bits) are [2,2]. Converting this to a partition in the manner explained in A227184 gives the partition {2+3}. Its largest part is 3, thus a(12)=3, which is actually the first time when this sequence differs from A043276.

Antti Karttunen, Table of n, a(n) for n = 0..8192

For all n, A005811(n) = a(A129594(n)). Cf. also A136480 (for n>= 1, gives the smallest part) and A227183, A227184, A226062, A092339, A227147.

a(n) gives the rightmost nonzero term on the n-th row of A227189.

Table[Function[b, Max@ Accumulate@ Prepend[If[Length@ b > 1, Rest[b] - 1, {}], First@ b] - Boole[n == 0]]@ Map[Length, Split@ Reverse@ IntegerDigits[ n, 2]], {n, 0, 120}] // Flatten (* Michael De Vlieger, May 09 2017 *)

(define (A227185 n) (if (zero? n) n (+ 1 (- (A029837 (+ 1 n)) (A005811 n)))))
(define (A227185v2 n) (if (zero? n) n (car (reverse (binexp_to_ascpart n))))) ;; Alternative definition, using the auxiliary functions given in A227184.

Defining formula:
a(0)=0; and for n>=1, a(n) = A029837(n+1) - (A005811(n)-1). [Because the largest part in the unordered partition in this encoding scheme is computed as (c_1 + (c_2-1) + (c_3-1) + ... + (c_k-1)) where c_1 .. c_k are the parts of the k-part composition that sum together as c_1 + c_2 + ... + c_k = A029837(n+1) (the binary width of n), so we subtract from the total binary width of n the number of runs (A005811) minus 1.]
Equivalently: a(n) = A092339(n)+1 for n>0.
a(n) = A005811(A129594(n)). [This just states the fact that when conjugating a partition, the largest part of an old partition will be the number of the parts in the new, conjugated partition.]

A043279 Maximal run length in base 5 representation of n.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 2, 3, 2, 2, 2, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 3, 2, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 2

Offset: 1

Clark Kimberling

Winston de Greef, Table of n, a(n) for n = 1..10000

Cf. A007091.

Cf. A043276-A043290 for base-2 to base-16 analogs.

A043279[n_]:=Max[Map[Length,Split[IntegerDigits[n,5]]]];Array[A043279,100] (* Paolo Xausa, Sep 27 2023 *)

A043279(n, b=5)={my(m,c=1); while(n>0, n%b==(n\=b)%b && c++ && next; m=max(m, c); c=1); m} \\ M. F. Hasler, Jul 23 2013

A175597 Minimal run length in binary representation of n.

Original entry on oeis.org

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

Offset: 1

Juri-Stepan Gerasimov, Jul 21 2010

Cf. A043276.

A175597 := proc(n) local dgs,odig,runl,thisr,i ; dgs := convert(n,base,2) ; odig := -1 ; runl := nops(dgs)+1 ; thisr := 0 ; for i from 1 to nops(dgs) do d := op(i,dgs) ; if i = 1 then thisr := 1; else if d <> odig then if thisr < runl then runl := thisr ; end if; thisr := 1 ; else thisr := thisr+1 ; end if; end if; odig := d ; end do: if thisr < runl then runl := thisr ; end if; return runl ; end proc: seq(A175597(n),n=1..130) ; # R. J. Mathar, Jul 26 2010

A043281 Maximal run length in base-7 representation of n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 2, 3, 2, 2, 2, 2, 2, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1

Offset: 1

Clark Kimberling

Winston de Greef, Table of n, a(n) for n = 1..10000

Cf. A007093 (base 7).

Cf. A043276-A043290 for base-2 to base-16 analogs.

Max[Length/@Split[IntegerDigits[#,7]]]&/@Range[100] (* Harvey P. Dale, Mar 30 2016 *)

A043281(n, b=7)={my(m,c=1); while(n>0, n%b==(n\=b)%b&&c++&&next; m=max(m, c); c=1); m} \\ M. F. Hasler, Jul 23 2013

from itertools import groupby
from sympy.ntheory.factor_ import digits
def A043281(n): return max(len(list(g)) for k, g in groupby(digits(n,7)[1:])) # Chai Wah Wu, Mar 09 2023

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

A166535 Positive integers whose binary representation does not contain a run of more than three consecutive 0's or 1's.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90

Offset: 1

John W. Layman, Oct 16 2009

nonn

A179970 is a subsequence. - Reinhard Zumkeller, Aug 04 2010

There are A000073(n+2) terms with binary length n. - Rémy Sigrist, Sep 30 2022

R. Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A000073, A063037, A000975.

Select[Range[100],Max[Length/@Split[IntegerDigits[#,2]]]<4&] (* Harvey P. Dale, Aug 01 2020 *)

tribonacci(n) = ([0,1,0; 0,0,1; 1,1,1]^n)[2,1]
a(n) = { if (n<=4, return (n), my (s=1); for (i=1, oo, my (f=tribonacci(i+2)); i
f (nRémy Sigrist, Sep 30 2022

It appears (but has not yet been proved) that the terms of a(n) can be computed recursively as follows. Let {c(i)} be defined for i >= 5 by c(i)=2c(i-1)+1 if i is congruent to 2 mod 4, else c(i)=2c(i-1)-1. I.e., {c(i)}={1,3,5,9,17,35,...} for i=5,6,7,... . Let a(n)=n for n=1,2,...,7. For n>7, choose k so that s(k) < n < s(k+1), where s(k) = Sum_{j=3..k} t(j) with t(j) being the j-th term of the tribonacci sequence A000073 (with initial terms t(0)=0, t(1)=0, t(2)=1). Then a(n) = c(k) + 2a(s(k)) - a(2s(k)+1). This has been verified for n up to 2400.

{i: A043276(i) <= 3}. - R. J. Mathar, Jun 04 2021

A249144 a(0) = 0, after which a(n) gives the total number of runs of the same length as the rightmost run in the binary representation of a(n-1) [i.e., A136480(a(n-1))] among the binary expansions of all previous terms, including the runs in a(n-1) itself.

Original entry on oeis.org

0, 1, 2, 4, 1, 6, 7, 1, 8, 2, 11, 3, 4, 5, 17, 19, 7, 4, 8, 5, 25, 26, 29, 31, 1, 32, 2, 35, 12, 14, 37, 41, 45, 49, 50, 52, 22, 57, 58, 61, 63, 1, 64, 2, 67, 25, 69, 73, 76, 32, 3, 33, 80, 4, 34, 87, 14, 92, 35, 36, 38, 99, 42, 105, 108, 47, 5, 114, 116, 49, 119, 23, 24, 25, 123, 54, 126, 127, 1, 128, 2

Offset: 0

Antti Karttunen, Oct 22 2014

Inspired by A248034.

			a(0) = 0 (by definition), and 0 is also '0' in binary.
For n = 1, we see that in a(0) there is one run of length 1, which is total number of runs of length 1 so far in terms a(0) .. a(n-1), thus a(1) = 1.
For n = 2, we see that the rightmost run of a(1) = 1 ('1' also in binary) has occurred two times in total (once in a(0) and a(1)), thus a(2) = 2.
For n = 3, we see that the rightmost run of a(2) = 2 ('10' in binary) is one bit long, and so far there has occurred four such runs in total (namely once in a(0) and a(1), twice in a(2)), thus a(3) = 4.
For n = 4, we see that the rightmost run of a(3) = 4 ('100' in binary) is two bits long, and it is so far the first and only two-bit run in the sequence, thus a(4) = 1.
For n = 5, we see that the rightmost run of a(4) = 1 ('1' in binary) is one bit long, and so far there has occurred 6 such one-bit runs in terms a(0) .. a(4), thus a(5) = 6.
For n = 6, we see that the rightmost run of a(5) = 6 ('110' in binary) is one bit long, and so far there has occurred 7 such one bit runs in terms a(0) .. a(5), thus a(6) = 7.

Antti Karttunen, Table of n, a(n) for n = 0..10000

Cf. A249143, A249146, A249148, A248034.

A043280 Maximal run length in base 6 representation of n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 2, 3, 2, 2, 2, 2, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 2, 3, 2, 2, 2, 1

Offset: 1

Clark Kimberling

Winston de Greef, Table of n, a(n) for n = 1..10000

Cf. A007092.

Cf. A043276-A043290 for base-2 to base-16 analogs.

A043280[n_]:=Max[Map[Length,Split[IntegerDigits[n,6]]]];Array[A043280,100] (* Paolo Xausa, Sep 27 2023 *)

A043280(n, b=6)={my(m,c=1); while(n>0, n%b==(n\=b)%b&&c++&&next; m=max(m, c); c=1); m} \\ M. F. Hasler, Jul 23 2013

A043282 Maximal run length in base 8 representation of n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 1, 2, 3, 2, 2, 2, 2, 2, 2, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 1

Clark Kimberling

Winston de Greef, Table of n, a(n) for n = 1..10000

Cf. A007094 (base 8).

Cf. A043276-A043290 for base-2 to base-16 analogs.

A043282[n_]:=Max[Map[Length,Split[IntegerDigits[n,8]]]];Array[A043282,100] (* Paolo Xausa, Sep 27 2023 *)

A043282(n, b=8)={my(m, c=1); while(n>0, n%b==(n\=b)%b && c++ && next; m=max(m, c); c=1); m} \\ M. F. Hasler, Jul 23 2013

from itertools import groupby
def A043282(n): return max(len(list(g)) for k, g in groupby(oct(n)[2:])) # Chai Wah Wu, Mar 09 2023

A043285 Maximal run length in base 11 representation of n.

Original entry on oeis.org

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

Offset: 1

Clark Kimberling

Antti Karttunen, Table of n, a(n) for n = 1..16384

Cf. A043276-A043290 for base-2 to base-16 analogs.

(Max[Length/@Split[IntegerDigits[#,11]]])&/@Range[90] (* Harvey P. Dale, Jun 04 2013 *)

A043285(n, b=11){=my(m, c=1); while(n>0, n%b==(n\=b)%b&&c++&&next; m=max(m, c); c=1); m} \\ - M. F. Hasler, Jul 23 2013

More terms from Antti Karttunen, Nov 20 2017

cp's OEIS Frontend

A227185 The largest part in the unordered partition encoded in the runlengths of the binary expansion of n.

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A043279 Maximal run length in base 5 representation of n.

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A175597 Minimal run length in binary representation of n.

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A043281 Maximal run length in base-7 representation of n.

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

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A166535 Positive integers whose binary representation does not contain a run of more than three consecutive 0's or 1's.

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A249144 a(0) = 0, after which a(n) gives the total number of runs of the same length as the rightmost run in the binary representation of a(n-1) [i.e., A136480(a(n-1))] among the binary expansions of all previous terms, including the runs in a(n-1) itself.

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A043280 Maximal run length in base 6 representation of n.

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A043282 Maximal run length in base 8 representation of n.