A318921 -id:A318921 - cp's OEIS frontend

A175046 Write n in binary, then increase each run of 0's by one 0, and increase each run of 1's by one 1. a(n) is the decimal equivalent of the result.

3, 12, 7, 24, 51, 28, 15, 48, 99, 204, 103, 56, 115, 60, 31, 96, 195, 396, 199, 408, 819, 412, 207, 112, 227, 460, 231, 120, 243, 124, 63, 192, 387, 780, 391, 792, 1587, 796, 399, 816, 1635, 3276, 1639, 824, 1651, 828, 415, 224, 451, 908, 455, 920, 1843, 924

Offset: 1

Leroy Quet, Dec 02 2009

A318921 expands the runs in a similar way, and A318921(a(n)) = A001477(n). - Andrew Weimholt, Sep 08 2018
From Chai Wah Wu, Nov 18 2018: (Start)
Let f(k) = Sum_{i=2^k..2^(k+1)-1} a(i), i.e., the sum ranges over all numbers with a (k+1)-bit binary expansion. Thus f(0) = a(1) = 3 and f(1) = a(2) + a(3) = 19.
Then f(k) = 20*6^(k-1) - 2^(k-1) for k > 0.
Proof: by summing over the recurrence relations for a(n) (see formula section), we get f(k+2) = Sum_{i=2^k..2^(k+1)-1} (f(4i) + f(4i+1) + f(4i+2) + f(4i+3)) = Sum_{i=2^k..2^(k+1)-1} (6*a(2i) + 6*a(2i+1) + 4) = 6*f(k+1) + 2^(k+2). Solving this first-order recurrence relation with the initial condition f(1) = 19 shows that f(k) = 20*6^(k-1)-2^(k-1) for k > 0.
(End)

			6 in binary is 110. Increase each run by one digit to get 11100, which is 28 in decimal. So a(6) = 28.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
N. J. A. Sloane, Coordination Sequences, Planing Numbers, and Other Recent Sequences (II), Experimental Mathematics Seminar, Rutgers University, Jan 31 2019, Part I, Part 2, Slides. (Mentions this sequence)
Chai Wah Wu, Record values in appending and prepending bitstrings to runs of binary digits, arXiv:1810.02293 [math.NT], 2018.

Cf. A175047, A175048, A324127 (partial sums).
Cf. also A030308, A007088, A182512, A318921.
For records see A319422, A319423, A319424.

import Data.List (group)
a175046 = foldr (\b v -> 2 * v + b) 0 .
          concatMap (\bs@(b:_) -> b : bs) . group . a030308_row
-- Reinhard Zumkeller, Jul 05 2013

a[n_] := (Append[#, #[[1]]]& /@ Split[IntegerDigits[n, 2]]) // Flatten // FromDigits[#, 2]&;
Array[a, 60] (* Jean-François Alcover, Nov 12 2018 *)

A175046(n)={for(i=2,#n=binary (n*2+bittest (n,0)),n[i]!=n[i-1]&&n[i-1]*=[1,1]);fromdigits(concat(n),2)} \\ M. F. Hasler, Sep 08 2018

from re import split
def A175046(n):
    return int(''.join(d+'1' if '1' in d else d+'0' for d in split('(0+)|(1+)',bin(n)[2:]) if d != '' and d != None),2) # Chai Wah Wu, Sep 24 2018

def a(n):
    b = bin(n)[2:]
    return int(b.replace("01", "001").replace("10", "110") + b[-1], 2)
print([a(n) for n in range(1, 55)]) # Michael S. Branicky, Dec 07 2021

2n+1 <= a(n) < 2*(n+1/n)^2; a(n) mod 4 = 3*(n mod 2). - M. F. Hasler, Sep 08 2018
a(n) <= (9*n^2 + 12*n)/5, with equality iff n = (2/3)*(4^k-1) = A182512(k) for some k, i.e., n = 10101...10 in binary. - Conjectured by N. J. A. Sloane, Sep 09 2018, proved by M. F. Hasler, Sep 12 2018
From M. F. Hasler, Sep 12 2018: (Start)
Proof of N. J. A. Sloane's formula: For given (binary) length L(n) = floor(log_2(n)+1), the length of a(n) is maximal, L(a(n)) = 2*L(n), if and only if n's bits are alternating, i.e., n in A020988 (if even) or in A002450 (if odd).
For n = A020988(k) (= k times '10' in base 2) = (4^k - 1)*2/3, one has a(n) = A108020(k) (= k times '1100' in base 2) = (16^k - 1)*4/5. This yields a(n)/n = (4^k + 1)*6/5 = (n*9 + 12)/5, i.e., the given upper bound.
For n = A002450(k) = (4^k - 1)/3, one gets a(n) = A182512(k) = (16^k - 1)/5, whence a(n)/n = (4^k + 1)*3/5 = (n*9 + 6)/5, smaller than the bound.
If L(a(n)) < 2 L(n) - 1, then log_2(a(n)) < floor(log_2(a(n))+1) = L(a(n)) <= 2*L(n) - 2 = 2*floor(log_2(n)+1)-2 = 2*floor(log_2(n)) <= 2*log_2(n), whence a(n) < n^2.
It remains to consider the case L(a(n)) = 2 L(n) - 1. There are two possibilities:
If n = 10..._2, then n >= 2^(L(n)-1) and a(n) = 1100..._2 < 1101_2 * 2^(L(a(n))-4) = 13*2^(2*L(n)-5), so a(n)/n^2 < 13*2^(-5+2) = 13/8 = 1.625 < 9/5 = 1.8.
If n = 11..._2, then n >= 3*2^(L(n)-2) and a(n) = 111..._2 < 2^L(a(n)) = 2^(2*L(n)-1), so a(n)/n^2 < 2^(-1+4)/9 = 8/9 < 1 < 9/5.
This shows that a(n)/n^2 <= 9/5 + 12/(5*n) always holds, with equality iff n is in A020988; and a(n)/n^2 < 13/8 if n is not in A020988 or A002450. (End)
From M. F. Hasler, Sep 10 2018: (Start)
Right inverse of A318921: A318921 o A175046 = id (= A001477).
a(A020988(k)) = A108020(k); a(A002450(k)) = A182512(k); a(A000225(k)) = A000225(k+1) (achieves the lower bound a(n) >= 2n + 1) for all k >= 0. (End)
From David A. Corneth, Sep 20 2018: (Start)
a(4*k)   = 2*a(2*k).
a(4*k+1) = 4*a(2*k) + 3.
a(4*k+2) = 4*a(2*k+1).
a(4*k+3) = 2*a(2*k+1) + 1. (End)

Extended by Ray Chandler, Dec 18 2009

A329747 Runs-resistance of the sequence of prime indices of n.

Original entry on oeis.org

0, 0, 0, 1, 0, 2, 0, 1, 1, 2, 0, 3, 0, 2, 2, 1, 0, 3, 0, 3, 2, 2, 0, 3, 1, 2, 1, 3, 0, 2, 0, 1, 2, 2, 2, 2, 0, 2, 2, 3, 0, 2, 0, 3, 3, 2, 0, 3, 1, 3, 2, 3, 0, 3, 2, 3, 2, 2, 0, 4, 0, 2, 3, 1, 2, 2, 0, 3, 2, 2, 0, 3, 0, 2, 3, 3, 2, 2, 0, 3, 1, 2, 0, 4, 2, 2, 2, 3, 0, 3, 2, 3, 2, 2, 2, 3, 0, 3, 3, 2, 0, 2, 0, 3, 2, 2, 0, 3, 0, 2, 2, 3, 0, 2, 2, 3, 3, 2, 2, 4

Offset: 1

Gus Wiseman, Nov 21 2019

nonn

First differs from A304455 at a(90) = 3, A304455(90) = 4.
For the operation of taking the sequence of run-lengths of a finite sequence, runs-resistance is defined as the number of applications required to reach a singleton.
A prime index of n is a number m such that prime(m) divides n. The sequence of prime indices of n is row n of A112798.

			We have (1,2,2,3) -> (1,2,1) -> (1,1,1) -> (3), so a(90) = 3.

Antti Karttunen, Table of n, a(n) for n = 1..65537
Claude Lenormand, Deux transformations sur les mots, Preprint, 5 pages, Nov 17 2003.
Index entries for sequences related to prime indices in the factorization of n.

The version for partitions is A329746.
The version for compositions is A329744.
The version for binary words is A329767.
The version for binary expansion is A318928.
Cf. A008578 (positions of 0's), A056239, A112798, A329745, A329750.

primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
runsres[q_]:=Length[NestWhileList[Length/@Split[#]&,q,Length[#]>1&]]-1;
Table[runsres[primeMS[n]],{n,50}]

pis_to_runs(n) = { my(runs=List([]), f=factor(n)); for(i=1,#f~,while(f[i,2], listput(runs,primepi(f[i,1])); f[i,2]--)); (runs); };
runlengths(lista) = if(!#lista, lista, if(1==#lista, List([1]), my(runs=List([]), rl=1); for(i=1, #lista, if((i< #lista) && (lista[i]==lista[i+1]), rl++, listput(runs,rl); rl=1)); (runs)));
A329747(n) = { my(runs=pis_to_runs(n)); for(i=0,oo,if(#runs<=1, return(i), runs = runlengths(runs))); }; \\ Antti Karttunen, Jan 20 2025

More terms from Antti Karttunen, Jan 20 2025

A026465 Length of n-th run of identical symbols in the Thue-Morse sequence A010060 (or A001285).

Original entry on oeis.org

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

Offset: 1

Clark Kimberling

It appears that the sequence can be calculated by any of the following methods:
(1) Start with 1 and repeatedly replace 1 with 1, 2, 1 and 2 with 1, 2, 2, 2, 1;
(2) a(1) = 1, all terms are either 1 or 2 and, for n > 0, a(n) = 1 if the length of the n-th run of 2's is 1; a(n) = 2 if the length of the n-th run of consecutive 2's is 3, with each run of 2's separated by a run of two 1's;
(3) replace each 3 in A080426 with 2. - John W. Layman, Feb 18 2003
Number of representations of n as a sum of Jacobsthal numbers (1 is allowed twice as a part). Partial sums are A003159. With interpolated zeros, g.f. is (Product_{k>=1} (1 + x^A078008(k)))/2. - Paul Barry, Dec 09 2004
In other words, the consecutive 0's or 1's in A010060 or A010059. - Robin D. Saunders (saunders_robin_d(AT)hotmail.com), Sep 06 2006
From Carlo Carminati, Feb 25 2011: (Start)
The sequence (starting with the second term) can also be calculated by the following method:
Apply repeatedly to the string S_0 = [2] the following algorithm: take a string S, double it, if the last figure is 1, just add the last figure to the previous one, if the last figure is greater than one, decrease it by one unit and concatenate a figure 1 at the end. (This algorithm is connected with the interpretation of the sequence as a continued fraction expansion.) (End)
This sequence, starting with the second term, happens to be the continued fraction expansion of the biggest cluster point of the set {x in [0,1]: F^k(x) >= x, for all k in N}, where F denotes the Farey map (see A187061). - Carlo Carminati, Feb 28 2011
Starting with the second term, the fixed point of the substitution 2 -> 211, 1 -> 2. - Carlo Carminati, Mar 03 2011
It appears that this sequence contains infinitely many distinct palindromic subsequences. - Alexander R. Povolotsky, Oct 30 2016
From Michel Dekking, Feb 13 2019: (Start)
Let tau defined by tau(0) = 01, tau(1) = 10 be the Thue-Morse morphism, with fixed point A010060. Consecutive runs in A010060 are 0, 11, 0, 1, 00, 1, 1, ..., which are coded by their lengths 1, 2, 1, 1, 2, ... Under tau^2 consecutive runs are mapped to consecutive runs:
      tau^2(0) = 0110, tau^2(1) = 1001,
      tau^2(00) = 01100110, tau^2(11) = 10011001.
The reason is that (by definition of a run!) runs of 0's and runs of 1's alternate in the sequence of runs, and this is inherited by the image of these runs under tau^2.
Under tau^2 the runs of length 1 are mapped to the sequence 1,2,1 of run lengths, and the runs of length 2 are mapped to the sequence 1,2,2,2,1 of run lengths. This proves John Layman's conjecture number (1): it follows that (a(n)) is fixed point of the morphism alpha
      alpha: 1 -> 121, 2 -> 12221.
Since alpha(1) and alpha(2) are both palindromes, this also proves Alexander Povolotsky's conjecture.
(End)

N. J. A. Sloane, Table of n, a(n) for n = 1..10000
G. Allouche, Jean-Paul Allouche and Jeffrey Shallit, Kolam indiens, dessins sur le sable aux îles Vanuatu, courbe de Sierpinski et morphismes de monoïde, Ann. Inst. Fourier (Grenoble), Vol. 56, No. 7 (2006), pp. 2115-2130.
Jean-Paul Allouche, On the morphism 1 -> 121, 2 -> 12221, CNRS France, 2024. See pp. 1-3, 7.
Jean-Paul Allouche, On the morphism 1 -> 121, 2 -> 12221, Preprint, 2024 [Local copy, with permission]
Jean-Paul Allouche, Andre Arnold, Jean Berstel, Srecko Brlek, William Jockusch, Simon Plouffe and Bruce E. Sagan, A sequence related to that of Thue-Morse, Discrete Math., Vol. 139, No. 1-3 (1995), pp. 455-461.
Claudio Bonanno, Carlo Carminati, Stefano Isola and Giulio Tiozzo, Dynamics of continued fractions and kneading sequences of unimodal maps, arXiv:1012.2131 [math.DS], 2010-2012.
Srećko Brlek, Enumeration of factors in the Thue-Morse word, Discrete Applied Math., Vol. 24, No. 1-3 (1989), pp. 83-96.
Julien Cassaigne, Limit values of the recurrence quotient of Sturmian sequences, Theoret. Comput. Sci., Vol. 218, No. 1 (1999), pp. 3-12.
Artūras Dubickas, On the distance from a rational power to the nearest integer, Journal of Number Theory, Vol. 117, No. 1 (March 2006), pp. 222-239.
Artūras Dubickas, On a sequence related to that of Thue-Morse and its applications, Discrete Math., Vol. 307, No. 9-10 (2007), pp. 1082-1093. MR2292537 (2008b:11086).
Cor Kraaikamp, Thomas A. Schmidt and Wolfgang Steiner, Natural extensions and entropy of alpha-continued fractions, arXiv:1011.4283 [math.DS], 2010-2012.
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 page 2.
Kevin Ryde, PARI/GP Code
Jeffrey Shallit, Automaticity IV: Sequences, sets, and diversity, J. Théor. Nombres Bordeaux, Vol. 8, No. 2 (1996), pp. 347-367. See page 354.

Cf. A010060, A001285, A101615, A026490 (run lengths).
A080426 is an essentially identical sequence with another set of constructions.
Cf. A104248 (bisection odious), A143331 (bisection evil), A003159 (partial sums).
Cf. A187061, A363361 (as continued fraction).

import Data.List (group)
a026465 n = a026465_list !! (n-1)
a026465_list = map length $ group a010060_list
-- Reinhard Zumkeller, Jul 15 2014

# From Carlo Carminati, Feb 25 2011:
## period-doubling routine:
double:=proc(SS)
NEW:=[op(S), op(S)]:
if op(nops(NEW),NEW)=1
then NEW:=[seq(op(j,NEW), j=1..nops(NEW)-2),op(nops(NEW)-1,NEW)+1]:
else NEW:=[seq(op(j,NEW), j=1..nops(NEW)-1),op(nops(NEW)-1,NEW)-1,1]:
fi:
end proc:
# 10 loops of the above routine generate the first 1365 terms of the sequence
# (except for the initial term):
S:=[2]:
for j from 1 to 10  do S:=double(S); od:
S;
# From N. J. A. Sloane, Dec 31 2013:
S:=[b]; M:=14;
for n from 1 to M do T:=subs({b=[b,a,a], a=[b]}, S);
    S := map(x->op(x),T); od:
T:=subs({a=1,b=2},S): T:=[1,op(T)]: [seq(T[n],n=1..40)];

Length /@ Split@ Nest[ Flatten@ Join[#, # /. {1 -> 2, 2 -> 1}] &, {1}, 7]
NestList[ Flatten[# /. {1 -> {2}, 2 -> {1, 1, 2}}] &, {1}, 7] // Flatten (* Robert G. Wilson v, May 20 2014 *)

PARI
```
\\ See links.
```

def A026465(n):
    if n==1: return 1
    def iterfun(f,n=0):
        m, k = n, f(n)
        while m != k: m, k = k, f(k)
        return m
    def f(x):
        c, s = x, bin(x)[2:]
        l = len(s)
        for i in range(l&1^1,l,2):
            c -= int(s[i])+int('0'+s[:i],2)
        return c
    return iterfun(lambda x:f(x)+n,n)-iterfun(lambda x:f(x)+n-1,n-1) # Chai Wah Wu, Jan 29 2025

a(1) = 1; for n > 1, a(n) = A003159(n) - A003159(n-1). - Benoit Cloitre, May 31 2003
G.f.: Product_{k>=1} (1 + x^A001045(k)). - Paul Barry, Dec 09 2004
Asymptotic mean: lim_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 3/2. - Amiram Eldar, Jan 16 2022

Corrected and extended by John W. Layman, Feb 18 2003

Definition revised by N. J. A. Sloane, Dec 30 2013

A227736 Irregular table read by rows: the first entry of n-th row is length of run of rightmost identical bits (either 0 or 1, equal to n mod 2), followed by length of the next run of bits, etc., in the binary representation of n, when scanned from the least significant to the most significant end.

Original entry on oeis.org

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

Offset: 1

Antti Karttunen, Jul 25 2013

Row n has A005811(n) terms. In rows 2^(k-1)..2^k-1 we have all the compositions (ordered partitions) of k. Other orderings of compositions: A101211 (same with rows reversed), A066099, A108244 and A124734.

Each row n >= 1 contains the initial A005811(n) nonzero terms from the beginning of row n of A227186. A070939(n) gives the sum of terms on row n, while A167489(n) gives the product of its terms. A136480 gives the first column. A101211 lists the terms of each row in reverse order.

			Table begins as:
  Row  n in    Terms on
   n   binary  that row
   1      1    1;
   2     10    1,1;
   3     11    2;
   4    100    2,1;
   5    101    1,1,1;
   6    110    1,2;
   7    111    3;
   8   1000    3,1;
   9   1001    1,2,1;
  10   1010    1,1,1,1;
  11   1011    2,1,1;
  12   1100    2,2;
  13   1101    1,1,2;
  14   1110    1,3;
  15   1111    4;
  16  10000    4,1;
etc. with the terms of row n appearing in reverse order compared how the runs of the same length appear in the binary expansion of n (Cf. A101211).
From _Omar E. Pol_, Sep 08 2013: (Start)
Illustration of initial terms:
  ---------------------------------------
  k   m     Diagram        Composition
  ---------------------------------------
  .          _
  1   1     |_|_           1;
  2   1     |_| |          1, 1,
  2   2     |_ _|_         2;
  3   1     |_  | |        2, 1,
  3   2     |_|_| |        1, 1, 1,
  3   3     |_|   |        1, 2,
  3   4     |_ _ _|_       3;
  4   1     |_    | |      3, 1,
  4   2     |_|_  | |      1, 2, 1,
  4   3     |_| | | |      1, 1, 1, 1,
  4   4     |_ _|_| |      2, 1, 1,
  4   5     |_  |   |      2, 2,
  4   6     |_|_|   |      1, 1, 2,
  4   7     |_|     |      1, 3,
  4   8     |_ _ _ _|_     4;
  5   1     |_      | |    4, 1,
  5   2     |_|_    | |    1, 3, 1,
  5   3     |_| |   | |    1, 1, 2, 1,
  5   4     |_ _|_  | |    2, 2, 1,
  5   5     |_  | | | |    2, 1, 1, 1,
  5   6     |_|_| | | |    1, 1, 1, 1, 1,
  5   7     |_|   | | |    1, 2, 1, 1,
  5   8     |_ _ _|_| |    3, 1, 1,
  5   9     |_    |   |    3, 2,
  5  10     |_|_  |   |    1, 2, 2,
  5  11     |_| | |   |    1, 1, 1, 2,
  5  12     |_ _|_|   |    2, 1, 2,
  5  13     |_  |     |    2, 3,
  5  14     |_|_|     |    1, 1, 3,
  5  15     |_|       |    1, 4,
  5  16     |_ _ _ _ _|    5;
.
Also irregular triangle read by rows in which row k lists the compositions of k, k >= 1.
Triangle begins:
 [1];
 [1,1], [2];
 [2,1], [1,1,1], [1,2],[3];
 [3,1], [1,2,1], [1,1,1,1], [2,1,1], [2,2], [1,1,2], [1,3], [4];
 [4,1], [1,3,1], [1,1,2,1], [2,2,1], [2,1,1,1], [1,1,1,1,1], [1,2,1,1], [3,1,1], [3,2], [1,2,2], [1,1,1,2], [2,1,2], [2,3], [1,1,3], [1,4], [5];
Row k has length A001792(k-1).
Row sums give A001787(k), k >= 1.
(End)

Antti Karttunen, The rows 1..1023 of the table, flattened
Mikhail Kurkov, Comments on A227736.
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. A227738 and also A227739 for similar table for unordered partitions.
Cf. A030308, A245562, A245563.
Cf. A101211 (rows in reversed order).

import Data.List (group)
a227736 n k = a227736_tabf !! (n-1) !! (k-1)
a227736_row n = a227736_tabf !! (n-1)
a227736_tabf = map (map length . group) $ tail a030308_tabf
-- Reinhard Zumkeller, Aug 11 2014

Array[Length /@ Reverse@ Split@ IntegerDigits[#, 2] &, 34] // Flatten (* Michael De Vlieger, Dec 11 2020 *)

apply( {A227736_row(n, r=[1], b=n%2)=while(n\=2, n%2==b && r[#r]++ || [b=1-b, r=concat(r,1)]); r}, [1..22]) \\ M. F. Hasler, Mar 11 2025

def A227736_row(n): return[len(list(g))for _,g in groupby(bin(n)[:1:-1])]
from itertools import groupby # M. F. Hasler, Mar 11 2025

(define (A227736 n) (A227186bi (A227737 n) (A227740 n))) ;; The Scheme-function for A227186bi has been given in A227186.

a(n) = A227186(A227737(n), A227740(n)).

a(n) = A101211(A227741(n)).

A319411 Triangle read by rows: T(n,k) = number of binary vectors of length n with runs-resistance k (1 <= k <= n).

Original entry on oeis.org

2, 2, 2, 2, 2, 4, 2, 4, 6, 4, 2, 2, 12, 12, 4, 2, 6, 30, 18, 8, 0, 2, 2, 44, 44, 32, 4, 0, 2, 6, 82, 76, 74, 16, 0, 0, 2, 4, 144, 138, 172, 52, 0, 0, 0, 2, 6, 258, 248, 350, 156, 4, 0, 0, 0, 2, 2, 426, 452, 734, 404, 28, 0, 0, 0, 0, 2, 10, 790, 752, 1500, 938, 104, 0, 0, 0, 0, 0

Offset: 1

N. J. A. Sloane, Sep 20 2018

"Runs-resistance" is defined in A318928.
Row sums are 2,4,8,16,... (the binary vectors may begin with 0 or 1).
This is similar to A329767 but without the k = 0 column and with a different row n = 1. - Gus Wiseman, Nov 25 2019

			Triangle begins:
2,
2, 2,
2, 2, 4,
2, 4, 6, 4,
2, 2, 12, 12, 4,
2, 6, 30, 18, 8, 0,
2, 2, 44, 44, 32, 4, 0,
2, 6, 82, 76, 74, 16, 0, 0,
2, 4, 144, 138, 172, 52, 0, 0, 0,
2, 6, 258, 248, 350, 156, 4, 0, 0, 0,
2, 2, 426, 452, 734, 404, 28, 0, 0, 0, 0,
2, 10, 790, 752, 1500, 938, 104, 0, 0, 0, 0, 0,
...
Lenormand gives the first 20 rows.
The calculation of row 4 is as follows.
We may assume the first bit is a 0, and then double the answers.
vector / runs / steps to reach a single number:
0000 / 4 / 1
0001 / 31 -> 11 -> 2 / 3
0010 / 211 -> 12 -> 11 -> 2 / 4
0011 / 22 -> 2 / 2
0100 / 112 -> 21 -> 11 -> 2 / 4
0101 / 1111 -> 4 / 2
0110 / 121 -> 111 -> 3 / 3
0111 / 13 -> 11 -> 2 / 3
and we get 1 (once), 2 (twice), 3 (three times) and 4 (twice).
So row 4 is: 2,4,6,4.

Hiroaki Yamanouchi, Table of n, a(n) for n = 1..1830
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.

Row sums are A000079.
Column k = 2 is 2 * A032741 = A319410.
Column k = 3 is 2 * A329745 (because runs-resistance 2 for compositions corresponds to runs-resistance 3 for binary words).
The version for compositions is A329744.
The version for partitions is A329746.
The number of nonzero entries in row n > 0 is A319412(n).
The runs-resistance of the binary expansion of n is A318928.
Cf. A096365, A225485, A245563, A319413, A319414, A319415, A325280, A329747, A329750, A329767, A329870.

runsresist[q_]:=If[Length[q]==1,1,Length[NestWhileList[Length/@Split[#]&,q,Length[#]>1&]]-1];
Table[Length[Select[Tuples[{0,1},n],runsresist[#]==k&]],{n,10},{k,n}] (* Gus Wiseman, Nov 25 2019 *)

A319416 Cuts-resistance of n: number of applications of Lernormand's "raboter" map needed to transform the binary expansion of n to the empty string.

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane, Sep 21 2018

Here we are using Lenormand's "raboter" map in a stricter sense than in A318921 and A319419. If S is a binary string with successive runs of lengths b,c,d,e,..., the "raboter" map sends S to the binary string with successive runs of lengths b-1,c-1,d-1,e-1,... Runs of length 0 are omitted (they are indicated by dots in the examples below).
To get a(n), start with S equal to the binary expansion of n beginning with the most significant bit, and keep applying the map until we reach the empty string.
After the first step, the string may start with a string of 0's: this is acceptable because we are working with strings, not binary expansions of numbers.
For example, 34 = 100010 -> .00.. = 00 -> 0. = 0 -> . (the empty string), taking 3 steps, so a(34) = 3.
Note: this is not the same as the number of applications of the map k -> A318921(k) needed to reduce the binary expansion of n to zero (because A318921 does not distinguish between 0 and the empty string).
This is also not the same as the number of applications of the map k -> A319419(k) needed to reduce the binary expansion of n to -1 (because A319419 does not distinguish between a string of 0's and a single 0).
The value k appears for the first time when n = 2^k - 1.

			n: repeatedly applying the map / number of steps = a(n)
0: 0 -> . / 1
1: 1 -> . / 1
2: 10 -> . / 1
3: 11 -> 1 -> . / 2
4: 100 -> 0 -> . / 2
5: 101 -> . / 1
6: 110 -> 1 -> . / 2
7: 111 -> 11 -> 1 -> . / 3
8: 1000 -> 00 -> 0 -> . / 3
9: 1001 -> 0 -> . / 2
10: 1010 -> . / 1
11: 1011 -> 1 -> . / 2
12: 1100 -> 10 -> . / 2
...

Rémy Sigrist, Table of n, a(n) for n = 0..16384
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, Coordination Sequences, Planing Numbers, and Other Recent Sequences (II), Experimental Mathematics Seminar, Rutgers University, Jan 31 2019, Part I, Part 2, Slides. (Mentions this sequence)

Positions of 1's are A000975.
Positions of 2's are A329862.
The version for runs-resistance is A318928.
The version for compositions is A329861.
Binary words counted by cuts-resistance are A319421 or A329860.
Cf. A318921, A319411, A319419, A319420, A329744, A329767, A329863, A329865.

degdep[q_]:=Length[NestWhileList[Join@@Rest/@Split[#]&,q,Length[#]>0&]]-1;
Table[degdep[IntegerDigits[n,2]],{n,0,50}] (* Gus Wiseman, Nov 25 2019 *)

a(n) = my (b=binary(n), w=#b); for (k=1, oo, my (ww=0); for (i=2, w, if (b[i-1]==b[i], b[ww++]=b[i])); if (ww==0, return (k), w=ww)) \\ Rémy Sigrist, Sep 23 2018

More terms from Rémy Sigrist, Sep 23 2018

A329767 Triangle read by rows where T(n,k) is the number of binary words of length n >= 0 with runs-resistance k, 0 <= k <= n.

Original entry on oeis.org

1, 2, 0, 0, 2, 2, 0, 2, 2, 4, 0, 2, 4, 6, 4, 0, 2, 2, 12, 12, 4, 0, 2, 6, 30, 18, 8, 0, 0, 2, 2, 44, 44, 32, 4, 0, 0, 2, 6, 82, 76, 74, 16, 0, 0, 0, 2, 4, 144, 138, 172, 52, 0, 0, 0, 0, 2, 6, 258, 248, 350, 156, 4, 0, 0, 0, 0, 2, 2, 426, 452, 734, 404, 28, 0, 0, 0, 0

Offset: 0

Gus Wiseman, Nov 21 2019

A composition of n is a finite sequence of positive integers with sum n.
For the operation of taking the sequence of run-lengths of a finite sequence, runs-resistance is defined as the number of applications required to reach a singleton.
Except for the k = 0 column and the n = 0 and n = 1 rows, this is the triangle appearing on page 3 of Lenormand, which is A319411. Unlike A318928, we do not here require that a(n) >= 1.
The n = 0 row is chosen to ensure that the row-sums are A000079, although the empty word arguably has indeterminate runs-resistance.

			Triangle begins:
   1
   2   0
   0   2   2
   0   2   2   4
   0   2   4   6   4
   0   2   2  12  12   4
   0   2   6  30  18   8   0
   0   2   2  44  44  32   4   0
   0   2   6  82  76  74  16   0   0
   0   2   4 144 138 172  52   0   0   0
   0   2   6 258 248 350 156   4   0   0   0
   0   2   2 426 452 734 404  28   0   0   0   0
For example, row n = 4 counts the following words:
  0000  0011  0001  0010
  1111  0101  0110  0100
        1010  0111  1011
        1100  1000  1101
              1001
              1110

Claude Lenormand, Deux transformations sur les mots, Preprint, 5 pages, Nov 17 2003.

Row sums are A000079.
Column k = 2 is A319410.
Column k = 3 is 2 * A329745.
The version for compositions is A329744.
The version for partitions is A329746.
The number of nonzero entries in row n > 0 is A319412(n).
The runs-resistance of the binary expansion of n is A318928.
Cf. A001037, A096365, A225485, A245563, A319411, A325280, A329747, A329750.

runsres[q_]:=If[Length[q]==1,0,Length[NestWhileList[Length/@Split[#]&,q,Length[#]>1&]]-1];
Table[Length[Select[Tuples[{0,1},n],runsres[#]==k&]],{n,0,10},{k,0,n}]

A329860 Triangle read by rows where T(n,k) is the number of binary words of length n with cuts-resistance k.

Original entry on oeis.org

1, 0, 2, 0, 2, 2, 0, 2, 4, 2, 0, 2, 8, 4, 2, 0, 2, 12, 12, 4, 2, 0, 2, 20, 22, 14, 4, 2, 0, 2, 28, 48, 28, 16, 4, 2, 0, 2, 44, 84, 70, 32, 18, 4, 2, 0, 2, 60, 162, 136, 90, 36, 20, 4, 2, 0, 2, 92, 276, 298, 178, 110, 40, 22, 4, 2, 0, 2, 124, 500, 564, 432, 220, 132, 44, 24, 4, 2

Offset: 0

Gus Wiseman, Nov 23 2019

For the operation of shortening all runs by 1, cuts-resistance is defined to be the number of applications required to reach an empty word.

			Triangle begins:
   1
   0   2
   0   2   2
   0   2   4   2
   0   2   8   4   2
   0   2  12  12   4   2
   0   2  20  22  14   4   2
   0   2  28  48  28  16   4   2
   0   2  44  84  70  32  18   4   2
   0   2  60 162 136  90  36  20   4   2
   0   2  92 276 298 178 110  40  22   4   2
   0   2 124 500 564 432 220 132  44  24   4   2
Row n = 4 counts the following words:
  0101  0010  0001  0000
  1010  0011  0111  1111
        0100  1000
        0110  1110
        1001
        1011
        1100
        1101

Claude Lenormand, Deux transformations sur les mots, Preprint, 5 pages, Nov 17 2003.

Column k = 2 appears to be 2 * A027383.
The version for runs-resistance is A319411 or A329767.
The cuts-resistance of the binary expansion of n is A319416(n).
The version for compositions is A329861.
Cf. A000975, A164707, A261983, A318928, A319420, A319421, A329738, A329865.

degdep[q_]:=Length[NestWhileList[Join@@Rest/@Split[#]&,q,Length[#]>0&]]-1;
Table[Length[Select[Tuples[{0,1},n],degdep[#]==k&]],{n,0,10},{k,0,n}]

For positive indices, T(n,k) = 2 * A319421(n,k).

A319421 Triangle read by rows: T(n,k) (1 <= k <= n) = one-half of the number of binary vectors of length n and cuts-resistance k.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 4, 2, 1, 1, 6, 6, 2, 1, 1, 10, 11, 7, 2, 1, 1, 14, 24, 14, 8, 2, 1, 1, 22, 42, 35, 16, 9, 2, 1, 1, 30, 81, 68, 45, 18, 10, 2, 1, 1, 46, 138, 149, 89, 55, 20, 11, 2, 1, 1, 62, 250, 282, 216, 110, 66, 22, 12, 2, 1, 1, 94, 419, 577, 422, 285, 132, 78, 24, 13, 2, 1

Offset: 1

N. J. A. Sloane, Sep 23 2018

Cuts-resistance is defined in A319416.
This triangle summarizes the data shown in A319420.
Conjecture (Sloane): Sum_{i = 1..n} i * T(n,i) = A189391(n).

			Triangle begins:
   1
   1   1
   1   2   1
   1   4   2   1
   1   6   6   2   1
   1  10  11   7   2   1
   1  14  24  14   8   2   1
   1  22  42  35  16   9   2   1
   1  30  81  68  45  18  10   2   1
   1  46 138 149  89  55  20  11   2   1
   1  62 250 282 216 110  66  22  12   2   1
   1  94 419 577 422 285 132  78  24  13   2   1
Lenormand gives first 15 rows.
For example, the "1,2,1" row here refers to the 8 vectors of length 3. There are 2 vectors of cuts-resistance 1, namely 010 and 101 (see A319416), 4 vectors of cuts-resistance 2 (100,011,001,110), and 2 of cuts-resistance 3 (000 and 111). Halving these counts we get 1,2,1

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. See page 4.
N. J. A. Sloane, Coordination Sequences, Planing Numbers, and Other Recent Sequences (II), Experimental Mathematics Seminar, Rutgers University, Jan 31 2019, Part I, Part 2, Slides. (Mentions this sequence)

Row sums are A000079.
Column k = 2 appears to be A027383.
The version for runs-resistance is A319411 or A329767.
The version for compositions is A329861.
The cuts-resistance of the binary expansion of n is A319416(n).
Cf. A000975, A164707, A189391, A261983, A318921, A318928, A319420, A329738, A329860, A329865.

degdep[q_]:=Length[NestWhileList[Join@@Rest/@Split[#]&,q,Length[#]>0&]]-1;
Table[Length[Select[Tuples[{0,1},n],First[#]==1&°dep[#]==k&]],{n,8},{k,n}] (* Gus Wiseman, Nov 25 2019 *)

T(n,k) = A329860(n,k)/2. - Gus Wiseman, Nov 25 2019

A189391 The minimum possible value for the apex of a triangle of numbers whose base consists of a permutation of the numbers 0 to n, and each number in a higher row is the sum of the two numbers directly below it.

Original entry on oeis.org

0, 1, 3, 8, 19, 44, 98, 216, 467, 1004, 2134, 4520, 9502, 19928, 41572, 86576, 179587, 372044, 768398, 1585416, 3263210, 6711176, 13775068, 28255568, 57863214, 118430584, 242061468, 494523536, 1009105372, 2058327344, 4194213448

Offset: 0

Nathaniel Johnston, Apr 20 2011

This is the Riordan transform of A000217 (triangular numbers) with the Riordan matrix (of the Bell type) A053121 (inverse of the Chebyshev S Bell matrix). See the resulting formulae below. - Wolfdieter Lang, Feb 18 2017.

Conjecture: a(n) is also half the sum of the "cuts-resistance" (see A319416, A319420, A319421) of all binary vectors of length n (see Lenormand, page 4). - N. J. A. Sloane, Sep 20 2018

			For n = 4 consider the triangle:
....19
...8  11
..5  3  8
.4  1 2  6
3  1 0 2  4
This triangle has 19 at its apex and no other such triangle with the numbers 0 - 4 on its base has a smaller apex value, so a(4) = 19.

Nathaniel Johnston, Table of n, a(n) for n = 0..1000
F. Disanto and S. Rinaldi, Symmetric convex permutominoes and involutions, PU. M. A., Vol. 22 (2011), No. 1, pp. 39-60. - From _N. J. A. Sloane_, May 04 2012
Steven R. Finch, How far might we walk at random?, arXiv:1802.04615 [math.HO], 2018.
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 20 2018

Cf. A066411, A099325, A189390, A189162, A000217, A053121, A281862.

Cf. also A319416, A319420, A319421.

m:=30; R:=PowerSeriesRing(Rationals(), m); [0] cat Coefficients(R!((2*x+Sqrt(1-4*x^2)-1)/(2*(2*x-1)^2))); // G. C. Greubel, Aug 24 2018

a:=proc(n)return add((2*n-4*k-1)*binomial(n,k),k=0..floor((n-1)/2)): end:
seq(a(n),n=0..50);

CoefficientList[Series[(2*x+Sqrt[1-4*x^2]-1) / (2*(2*x-1)^2), {x, 0, 20}], x] (* Vaclav Kotesovec, Mar 16 2014 *)

A189391(n)=sum(i=0,(n-1)\2,(2*n-4*i-1)*binomial(n,i))  \\ M. F. Hasler, Jan 24 2012

If n even, a(n) = (n+1/2)*binomial(n,n/2) - 2^(n-1); if n odd, a(n) = ((n+1)/2)*binomial(n+1,(n+1)/2) - 2^(n-1). - N. J. A. Sloane, Nov 01 2018
a(n) = Sum_{k=0..floor((n-1)/2)} (2*n-4*k-1)*binomial(n,k).
G.f.: (2*x+sqrt(1-4*x^2)-1) / (2*(2*x-1)^2). - Alois P. Heinz, Feb 09 2012
a(n) ~ 2^n * (sqrt(2n/Pi)- 1/2). - Vaclav Kotesovec, Mar 16 2014 (formula simplified by Lewis Chen, May 25 2017)
D-finite with recurrence n*a(n) + (n-5)*a(n-1) + 2*(-5*n+6)*a(n-2) + 4*(-n+8)*a(n-3) + 24*(n-3)*a(n-4) = 0. - R. J. Mathar, Jan 04 2017
From Wolfdieter Lang, Feb 18 2017:(Start)
a(n) = Sum_{m=0..n} A053121(n, m)*A000217(m), n >= 0.
G.f.: c(x^2)*Tri(x*c(x^2)), with c and Tri the g.f. of A000108 and A000217, respectively. See the explicit form of the g.f. given above by Alois P. Heinz.
(End)
2*a(n) = A152548(n)-2^n. - R. J. Mathar, Jun 17 2021

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A175046 Write n in binary, then increase each run of 0's by one 0, and increase each run of 1's by one 1. a(n) is the decimal equivalent of the result.

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A329747 Runs-resistance of the sequence of prime indices of n.

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A026465 Length of n-th run of identical symbols in the Thue-Morse sequence A010060 (or A001285).

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A227736 Irregular table read by rows: the first entry of n-th row is length of run of rightmost identical bits (either 0 or 1, equal to n mod 2), followed by length of the next run of bits, etc., in the binary representation of n, when scanned from the least significant to the most significant end.

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A319411 Triangle read by rows: T(n,k) = number of binary vectors of length n with runs-resistance k (1 <= k <= n).

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A319416 Cuts-resistance of n: number of applications of Lernormand's "raboter" map needed to transform the binary expansion of n to the empty string.

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