A069010 -id:A069010 - cp's OEIS frontend

A087207 A binary representation of the primes that divide a number, shown in decimal.

0, 1, 2, 1, 4, 3, 8, 1, 2, 5, 16, 3, 32, 9, 6, 1, 64, 3, 128, 5, 10, 17, 256, 3, 4, 33, 2, 9, 512, 7, 1024, 1, 18, 65, 12, 3, 2048, 129, 34, 5, 4096, 11, 8192, 17, 6, 257, 16384, 3, 8, 5, 66, 33, 32768, 3, 20, 9, 130, 513, 65536, 7, 131072, 1025, 10, 1, 36, 19, 262144, 65, 258

Offset: 1

Mitch Cervinka (puritan(AT)planetkc.com), Oct 26 2003

The binary representation of a(n) shows which prime numbers divide n, but not the multiplicities. a(2)=1, a(3)=10, a(4)=1, a(5)=100, a(6)=11, a(10)=101, a(30)=111, etc.
For n > 1, a(n) gives the (one-based) index of the column where n is located in array A285321. A008479 gives the other index. - Antti Karttunen, Apr 17 2017
From Antti Karttunen, Jun 18 & 20 2017: (Start)
A268335 gives all n such that a(n) = A248663(n); the squarefree numbers (A005117) are all the n such that a(n) = A285330(n) = A048675(n).
For all n > 1 for which the value of A285331(n) is well-defined, we have A285331(a(n)) <= floor(A285331(n)/2), because then n is included in the binary tree A285332 and a(n) is one of its ancestors (in that tree), and thus must be at least one step nearer to its root than n itself.
Conjecture: Starting at any n and iterating the map n -> a(n), we will always reach 0 (see A288569). This conjecture is equivalent to the conjecture that at any n that is neither a prime nor a power of two, we will eventually hit a prime number (which then becomes a power of two in the next iteration). If this conjecture is false then sequence A285332 cannot be a permutation of natural numbers. On the other hand, if the conjecture is true, then A285332 must be a permutation of natural numbers, because all primes and powers of 2 occur in definite positions in that tree. This conjecture also implies the conjectures made in A019565 and A285320 that essentially claim that there are neither finite nor infinite cycles in A019565.
If there are any 2-cycles in this sequence, then both terms of the cycle should be present in A286611 and the larger one should be present in A286612.
(End)
Binary rank of the distinct prime indices of n, where the binary rank of an integer partition y is given by Sum_i 2^(y_i-1). For all prime indices (with multiplicity) we have A048675. - Gus Wiseman, May 25 2024

			a(38) = 129 because 38 = 2*19 = prime(1)*prime(8) and 129 = 2^0 + 2^7 (in binary 10000001).
a(140) = 13, binary 1101 because 140 is divisible by the first, third and fourth primes and 2^(1-1) + 2^(3-1) + 2^(4-1) = 13.

N. J. A. Sloane, Table of n, a(n) for n = 1..10000 [First 1000 terms from _T. D. Noe_]
Index entries for sequences related to binary expansion of n
Index entries for sequences computed from indices in prime factorization

For partial sums see A288566.
Cf. A000040, A000120, A001221, A005117, A008479, A019565, A055396, A285320, A285321, A285329, A285330, A285332.
Sequences with related definitions: A007947, A008472, A027748, A048675, A248663, A276379 (same sequence shown in base 2), A288569, A289271, A297404.
Cf. A286608 (numbers n for which a(n) < n), A286609 (n for which a(n) > n), and also A286611, A286612.
A003986, A003961, A059896 are used to express relationship between terms of this sequence.
Related to A267116 via A225546.
Positions of particular values are: A000079\{1} (1), A000244\{1} (2), A033845 (3), A000351\{1} (4), A033846 (5), A033849 (6), A143207 (7), A000420\{1} (8), A033847 (9), A033850 (10), A033851 (12), A147576 (14), A147571 (15), A001020\{1} (16), A033848 (17).
A048675 gives binary rank of prime indices.
A061395 gives greatest prime index, least A055396.
A112798 lists prime indices, length A001222, reverse A296150, sum A056239.
Binary indices (listed A048793):
- length A000120, complement A023416
- min A001511, opposite A000012
- sum A029931, product A096111
- max A029837 or A070939, opposite A070940
- complement A368494, sum A359400
- opposite complement A371571, sum A359359
- opposite A371572, sum A230877
Cf. A000720, A005940, A018819, A023506, A071814, A225620, A277319, A277905, A304818, A372689, A372890.

a087207 = sum . map ((2 ^) . (subtract 1) . a049084) . a027748_row
-- Reinhard Zumkeller, Jul 16 2013

a[n_] := Total[ 2^(PrimePi /@ FactorInteger[n][[All, 1]] - 1)]; a[1] = 0; Table[a[n], {n, 1, 69}] (* Jean-François Alcover, Dec 12 2011 *)

a(n) = {if (n==1, 0, my(f=factor(n), v = []); forprime(p=2, vecmax(f[,1]), v = concat(v, vecsearch(f[,1], p)!=0);); fromdigits(Vecrev(v), 2));} \\ Michel Marcus, Jun 05 2017

A087207(n)=vecsum(apply(p->1<M. F. Hasler, Jun 23 2017

from sympy import factorint, primepi
def a(n):
    return sum(2**primepi(i - 1) for i in factorint(n))
print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Jun 06 2017

(definec (A087207 n) (if (= 1 n) 0 (+ (A000079 (+ -1 (A055396 n))) (A087207 (A028234 n))))) ;; This uses memoization-macro definec
(define (A087207 n) (A048675 (A007947 n))) ;; Needs code from A007947 and A048675. - Antti Karttunen, Jun 19 2017

Additive with a(p^e) = 2^(i-1) where p is the i-th prime. - Vladeta Jovovic, Oct 29 2003
a(n) gives the m such that A019565(m) = A007947(n). - Naohiro Nomoto, Oct 30 2003
A000120(a(n)) = A001221(n); a(n) = Sum(2^(A049084(p)-1): p prime-factor of n). - Reinhard Zumkeller, Nov 30 2003
G.f.: Sum_{k>=1} 2^(k-1)*x^prime(k)/(1-x^prime(k)). - Franklin T. Adams-Watters, Sep 01 2009
From Antti Karttunen, Apr 17 2017, Jun 19 2017 & Dec 06 2018: (Start)
a(n) = A048675(A007947(n)).
a(1) = 0; for n > 1, a(n) = 2^(A055396(n)-1) + a(A028234(n)).
A000035(a(n)) = 1 - A000035(n). [a(n) and n are of opposite parity.]
A248663(n) <= a(n) <= A048675(n). [XOR-, OR- and +-variants.]
a(A293214(n)) = A218403(n).
a(A293442(n)) = A267116(n).
A069010(a(n)) = A287170(n).
A007088(a(n)) = A276379(n).
A038374(a(n)) = A300820(n) for n >= 1.
(End)
From Peter Munn, Jan 08 2020: (Start)
a(A059896(n,k)) = a(n) OR a(k) = A003986(a(n), a(k)).
a(A003961(n)) = 2*a(n).
a(n^2) = a(n).
a(n) = A267116(A225546(n)).
a(A225546(n)) = A267116(n).
(End)

More terms from Don Reble, Ray Chandler and Naohiro Nomoto, Oct 28 2003

Name clarified by Antti Karttunen, Jun 18 2017

A060130 Number of nonzero digits in factorial base representation (A007623) of n; minimum number of transpositions needed to compose each permutation in the lists A060117 & A060118.

Original entry on oeis.org

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

Offset: 0

Antti Karttunen, Mar 02 2001

nonn

			19 = 3*(3!) + 0*(2!) + 1*(1!), thus it is written as "301" in factorial base (A007623). The count of nonzero digits in that representation is 2, so a(19) = 2.

Antti Karttunen, Table of n, a(n) for n = 0..40320
Index entries for sequences related to factorial base representation

Cf. A007623, A034968, A055091, A060117, A060118, A060128, A060129, A060131, A060502, A257687, A275734, A275735, A276076.
Cf. A227130 (positions of even terms), A227132 (of odd terms).
Cf. also A225901, A232094, A257694, A257695.
The topmost row and the leftmost column in array A230415, the left edge of triangle A230417.
Differs from similar A267263 for the first time at n=30.

A060130(n) = count_nonfixed(convert(PermUnrank3R(n), 'disjcyc'))-nops(convert(PermUnrank3R(n), 'disjcyc')) or nops(fac_base(n))-nops(positions(0, fac_base(n)))
fac_base := n -> fac_base_aux(n, 2); fac_base_aux := proc(n, i) if(0 = n) then RETURN([]); else RETURN([op(fac_base_aux(floor(n/i), i+1)), (n mod i)]); fi; end;
count_nonfixed := l -> convert(map(nops, l), `+`);
positions := proc(e, ll) local a, k, l, m; l := ll; m := 1; a := []; while(member(e, l[m..nops(l)], 'k')) do a := [op(a), (k+m-1)]; m := k+m; od; RETURN(a); end;
# For procedure PermUnrank3R see A060117

Block[{nn = 105, r}, r = MixedRadix[Reverse@ Range[2, -1 + SelectFirst[Range@ 12, #! > nn &]]]; Array[Count[IntegerDigits[#, r], k_ /; k > 0] &, nn, 0]] (* Michael De Vlieger, Dec 30 2017 *)

(define (A060130 n) (let loop ((n n) (i 2) (s 0)) (cond ((zero? n) s) (else (loop (quotient n i) (+ 1 i) (+ s (if (zero? (remainder n i)) 0 1)))))))
;; Two other implementations, that use memoization-macro definec:
(definec (A060130 n) (if (zero? n) n (+ 1 (A060130 (A257687 n)))))
(definec (A060130 n) (if (zero? n) n (+ (A257511 n) (A060130 (A257684 n)))))
;; Antti Karttunen, Dec 30 2017

a(0) = 0; for n > 0, a(n) = 1 + a(A257687(n)).
a(0) = 0; for n > 0, a(n) = A257511(n) + a(A257684(n)).
a(n) = A060129(n) - A060128(n).
a(n) = A084558(n) - A257510(n).
a(n) = A275946(n) + A275962(n).
a(n) = A275948(n) + A275964(n).
a(n) = A055091(A060119(n)).
a(n) = A069010(A277012(n)) = A000120(A275727(n)).
a(n) = A001221(A275733(n)) = A001222(A275733(n)).
a(n) = A001222(A275734(n)) = A001222(A275735(n)) = A001221(A276076(n)).
a(n) = A046660(A275725(n)).
a(A225901(n)) = a(n).
A257511(n) <= a(n) <= A034968(n).
A275806(n) <= a(n).
a(A275804(n)) = A060502(A275804(n)). [A275804 gives all the positions where this coincides with A060502.]
a(A276091(n)) = A260736(A276091(n)). [A276091 gives all the positions where this coincides with A260736.]

Example-section added, name edited, the old Maple-code moved away from the formula-section, and replaced with all the new formulas by Antti Karttunen, Dec 30 2017

A328592 Numbers whose binary expansion has all different lengths of runs of 1's.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 7, 8, 11, 12, 13, 14, 15, 16, 19, 22, 23, 24, 25, 26, 28, 29, 30, 31, 32, 35, 38, 39, 44, 46, 47, 48, 49, 50, 52, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 67, 70, 71, 76, 78, 79, 88, 92, 94, 95, 96, 97, 98, 100, 103, 104, 110, 111, 112, 113, 114

Offset: 1

Gus Wiseman, Oct 20 2019

nonn

Also numbers whose binary indices have different lengths of runs of successive parts. A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.

The complement is {5, 9, 10, 17, 18, 20, 21, 27, ...}.

			The sequence of terms together with their binary expansions and binary indices begins:
   0:     0 ~ {}
   1:     1 ~ {1}
   2:    10 ~ {2}
   3:    11 ~ {1,2}
   4:   100 ~ {3}
   6:   110 ~ {2,3}
   7:   111 ~ {1,2,3}
   8:  1000 ~ {4}
  11:  1011 ~ {1,2,4}
  12:  1100 ~ {3,4}
  13:  1101 ~ {1,3,4}
  14:  1110 ~ {2,3,4}
  15:  1111 ~ {1,2,3,4}
  16: 10000 ~ {5}
  19: 10011 ~ {1,2,5}
  22: 10110 ~ {2,3,5}
  23: 10111 ~ {1,2,3,5}
  24: 11000 ~ {4,5}
  25: 11001 ~ {1,4,5}
  26: 11010 ~ {2,4,5}

The version for prime indices is A130091.
The binary expansion of n has A069010(n) runs of 1's.
The lengths of runs of 1's in the binary expansion of n are row n of A245563.
Numbers whose binary expansion has equal lengths of runs of 1's are A164707.
Cf. A000120, A003714, A014081, A098859, A121016, A328593, A328595, A328596.

Select[Range[0,100],UnsameQ@@Length/@Split[Join@@Position[Reverse[IntegerDigits[#,2]],1],#2==#1+1&]&]

A052331 Inverse of A052330; A binary encoding of Fermi-Dirac factorization of n, shown in decimal.

Original entry on oeis.org

0, 1, 2, 4, 8, 3, 16, 5, 32, 9, 64, 6, 128, 17, 10, 256, 512, 33, 1024, 12, 18, 65, 2048, 7, 4096, 129, 34, 20, 8192, 11, 16384, 257, 66, 513, 24, 36, 32768, 1025, 130, 13, 65536, 19, 131072, 68, 40, 2049, 262144, 258, 524288, 4097, 514, 132, 1048576, 35

Offset: 1

Christian G. Bower, Dec 15 1999

nonn

Every number can be represented uniquely as a product of numbers of the form p^(2^k), sequence A050376. This sequence is a binary representation of this factorization, with a(p^(2^k)) = 2^(i-1), where i is the index (A302778) of p^(2^k) in A050376. Additive with a(p^e) = sum a(p^(2^e_k)) where e = sum(2^e_k) is the binary representation of e and a(p^(2^k)) is as described above. - Franklin T. Adams-Watters, Oct 25 2005 - Index offset corrected by Antti Karttunen, Apr 17 2018

			n = 84 has Fermi-Dirac factorization A050376(5) * A050376(3) * A050376(2) = 7*4*3. Thus a(84) = 2^(5-1) + 2^(3-1) + 2^(2-1) = 16 + 4 + 2 = 22 ("10110" in binary = A182979(84)). - _Antti Karttunen_, Apr 17 2018

Antti Karttunen, Table of n, a(n) for n = 1..4096
Index entries for sequences that are permutations of the natural numbers

Cf. A050376, A052330, A064547, A302024, A302029, A302776, A302778, A302785, A302786, A302787, A302790, A302784.
Cf. A182979 (same sequence shown in binary).
One less than A064358.
Cf. also A156552.

A052331=a(n)={for(i=1,#n=factor(n)~,n[2,i]>1||next; m=binary(n[2,i]); n=concat(n,Mat(vector(#m-1,j,[n[1,i]^2^(#m-j),m[j]]~)));n[2,i]%=2); n||return(0); m=vecsort(n[1,]); forprime(p=1,m[#m],my(j=0);while(p^2^j>1} \\ M. F. Hasler, Apr 08 2015

up_to_e = 8192;
v050376 = vector(up_to_e);
ispow2(n) = (n && !bitand(n,n-1));
i = 0; for(n=1,oo,if(ispow2(isprimepower(n)), i++; v050376[i] = n); if(i == up_to_e,break));
A052331(n) = { my(s=0,e); while(n > 1, fordiv(n, d, if(((n/d)>1)&&ispow2(isprimepower(n/d)), e = vecsearch(v050376, n/d); if(!e, print("v050376 too short!"); return(1/0)); s += 2^(e-1); n = d; break))); (s); }; \\ Antti Karttunen, Apr 12 2018

a(1)=0; a(n*A050376(k)) = a(n) + 2^k for a(n) < 2^k, k=0, 1, ... - Thomas Ordowski, Mar 23 2005
From Antti Karttunen, Apr 13 2018: (Start)
a(1) = 0; for n > 1, a(n) = A000079(A302785(n)-1) + a(A302776(n)).
For n > 1, a(n) = A000079(A302786(n)-1) * A302787(n).
a(n) = A064358(n)-1.
A000120(a(n)) = A064547(n).
A069010(a(n)) = A302790(n).
(End)

A014081 a(n) is the number of occurrences of '11' in the binary expansion of n.

Original entry on oeis.org

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

Offset: 0

Simon Plouffe

a(n) takes the value k for the first time at n = 2^(k+1)-1. Cf. A000225. - Robert G. Wilson v, Apr 02 2009

a(n) = A213629(n,3) for n > 2. - Reinhard Zumkeller, Jun 17 2012

			The binary expansion of 15 is 1111, which contains three occurrences of 11, so a(15)=3.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
J.-P. Allouche, On an Inequality in a 1970 Paper of R. L. Graham, INTEGERS 21A (2021), #A2.
Jean-Paul Allouche and Jeffrey Shallit, Sums of digits and the Hurwitz zeta function, in: K. Nagasaka and E. Fouvry (eds.), Analytic Number Theory, Lecture Notes in Mathematics, Vol. 1434, Springer, Berlin, Heidelberg, 1990, pp. 19-30.
John Brillhart and L. Carlitz, Note on the Shapiro Polynomials, Proceedings of the American Mathematical Society, volume 25, number 1, May 1970, pages 114-118 (see A001782 for a scanned copy), with a(n) = exponent in theorem 4.
Helmut Prodinger, Generalizing the sum of digits function, SIAM J. Algebraic Discrete Methods, Vol. 3, No. 1 (1982), pp. 35-42. MR0644955 (83f:10009). [See B_2(11,n) on p. 35. - _N. J. A. Sloane_, Apr 06 2014]
Michel Rigo and Manon Stipulanti, Revisiting regular sequences in light of rational base numeration systems, arXiv:2103.16966 [cs.FL], 2021. Mentions this sequence.
Bartosz Sobolewski and Lukas Spiegelhofer, Block occurrences in the binary expansion, arXiv:2309.00142 [math.NT], 2023.
Ralf Stephan, Some divide-and-conquer sequences with (relatively) simple ordinary generating functions, 2004.
Ralf Stephan, Table of generating functions.
Eric Weisstein's World of Mathematics, Digit Block.
Eric Weisstein's World of Mathematics, Rudin-Shapiro Sequence.
Index entries for sequences related to binary expansion of n

Cf. A014082, A033264, A037800, A056973, A000225, A213629, A000120, A069010, A100046.
First differences give A245194.
A245195 gives 2^a(n).

import Data.Bits ((.&.))
a014081 n = a000120 (n .&. div n 2)  -- Reinhard Zumkeller, Jan 23 2012

# To count occurrences of 11..1 (k times) in binary expansion of v:
cn := proc(v, k) local n, s, nn, i, j, som, kk;
som := 0;
kk := convert(cat(seq(1, j = 1 .. k)),string);
n := convert(v, binary);
s := convert(n, string);
nn := length(s);
for i to nn - k + 1 do
if substring(s, i .. i + k - 1) = kk then som := som + 1 fi od;
som; end; # This program no longer worked. Corrected by N. J. A. Sloane, Apr 06 2014.
[seq(cn(n,2),n=0..300)];
# Alternative:
A014081 := proc(n) option remember;
  if n mod 4 <= 1 then procname(floor(n/4))
elif n mod 4 = 2 then procname(n/2)
else 1 + procname((n-1)/2)
fi
end proc:
A014081(0):= 0:
map(A014081, [$0..1000]); # Robert Israel, Sep 04 2015

f[n_] := Count[ Partition[ IntegerDigits[n, 2], 2, 1], {1, 1}]; Table[ f@n, {n, 0, 104}] (* Robert G. Wilson v, Apr 02 2009 *)
Table[SequenceCount[IntegerDigits[n,2],{1,1},Overlaps->True],{n,0,120}] (* Harvey P. Dale, Jun 06 2022 *)

A014081(n)=sum(i=0,#binary(n)-2,bitand(n>>i,3)==3)  \\ M. F. Hasler, Jun 06 2012

a(n) = hammingweight(bitand(n, n>>1)) ;
vector(105, i, a(i-1))  \\ Gheorghe Coserea, Aug 30 2015

def a(n): return sum([((n>>i)&3==3) for i in range(len(bin(n)[2:]) - 1)]) # Indranil Ghosh, Jun 03 2017

from re import split
def A014081(n): return sum(len(d)-1 for d in split('0+', bin(n)[2:]) if d != '') # Chai Wah Wu, Feb 04 2022

a(4n) = a(4n+1) = a(n), a(4n+2) = a(2n+1), a(4n+3) = a(2n+1) + 1. - Ralf Stephan, Aug 21 2003
G.f.: (1/(1-x)) * Sum_{k>=0} t^3/((1+t)*(1+t^2)), where t = x^(2^k). - Ralf Stephan, Sep 10 2003
a(n) = A000120(n) - A069010(n). - Ralf Stephan, Sep 10 2003
Sum_{n>=1} A014081(n)/(n*(n+1)) = A100046 (Allouche and Shallit, 1990). - Amiram Eldar, Jun 01 2021

A164707 A positive integer n is included if all runs of 1's in binary n are of the same length.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 17, 18, 20, 21, 24, 27, 28, 30, 31, 32, 33, 34, 36, 37, 40, 41, 42, 48, 51, 54, 56, 60, 62, 63, 64, 65, 66, 68, 69, 72, 73, 74, 80, 81, 82, 84, 85, 96, 99, 102, 108, 112, 119, 120, 124, 126, 127, 128, 129, 130, 132, 133, 136

Offset: 1

Leroy Quet, Aug 23 2009

Clarification: A binary number consists of "runs" completely of 1's alternating with runs completely of 0's. No two or more runs all of the same digit are adjacent.

This sequence contains in part positive integers that each contain one run of 1's. For those members of this sequence each with at least two runs of 1's, see A164709.

			From _Gus Wiseman_, Oct 31 2019: (Start)
The sequence of terms together with their binary expansions and binary indices begins:
   1:      1 ~ {1}
   2:     10 ~ {2}
   3:     11 ~ {1,2}
   4:    100 ~ {3}
   5:    101 ~ {1,3}
   6:    110 ~ {2,3}
   7:    111 ~ {1,2,3}
   8:   1000 ~ {4}
   9:   1001 ~ {1,4}
  10:   1010 ~ {2,4}
  12:   1100 ~ {3,4}
  14:   1110 ~ {2,3,4}
  15:   1111 ~ {1,2,3,4}
  16:  10000 ~ {5}
  17:  10001 ~ {1,5}
  18:  10010 ~ {2,5}
  20:  10100 ~ {3,5}
  21:  10101 ~ {1,3,5}
  24:  11000 ~ {4,5}
  27:  11011 ~ {1,2,4,5}
(End)

Ivan Neretin, Table of n, a(n) for n = 1..10000

Cf. A164708, A164709, A164710.
The version for prime indices is A072774.
The binary expansion of n has A069010(n) runs of 1's.
Numbers whose runs are all of different lengths are A328592.
Partitions with equal multiplicities are A047966.
Numbers whose binary expansion is aperiodic are A328594.
Numbers whose reversed binary expansion is a necklace are A328595.
Numbers whose reversed binary expansion is a Lyndon word are A328596.
Cf. A000120, A003714, A014081, A065609, A070939, A121016, A245563, A275692.

isA164707 := proc(n) local bdg,arl,lset ; bdg := convert(n,base,2) ; lset := {} ; arl := -1 ; for p from 1 to nops(bdg) do if op(p,bdg) = 1 then if p = 1 then arl := 1 ; else arl := arl+1 ; end if; else if arl > 0 then lset := lset union {arl} ; end if; arl := 0 ; end if; end do ; if arl > 0 then lset := lset union {arl} ; end if; return (nops(lset) <= 1 ); end proc: for n from 1 to 300 do if isA164707(n) then printf("%d,",n) ; end if; end do; # R. J. Mathar, Feb 27 2010

Select[Range@ 140, SameQ @@ Map[Length, Select[Split@ IntegerDigits[#, 2], First@ # == 1 &]] &] (* Michael De Vlieger, Aug 20 2017 *)

foreach(1..140){
    %runs=();
    $runs{$}++ foreach split /0+/, sprintf("%b",$);
    print "$_, " if 1==keys(%runs);
}
# Ivan Neretin, Nov 09 2015

Extended beyond 42 by R. J. Mathar, Feb 27 2010

A287170 a(n) = number of runs of consecutive prime numbers among the prime divisors of n.

Original entry on oeis.org

0, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 2, 1, 1, 1, 2, 2, 2, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 2, 2, 1, 1, 2, 2, 2, 2, 1, 1, 1, 2, 2, 1, 2, 2, 1, 2, 2, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 2, 2

Offset: 1

Rémy Sigrist, Jun 04 2017

nonn

a(n) = 0 iff n = 1.
a(n) = 1 iff n belongs to A073491.
a(p) = 1 for any prime p.
a(A002110(n)) = 1 for any n > 0.
a(n!) = 1 for any n > 1.
a(A066205(n)) = n for any n > 0.
a(n) = a(A007947(n)) for any n > 0.
a(n) = a(A003961(n)) for any n > 0.
a(n*m) <= a(n) + a(m) for any n > 0 and m > 0.
Each number n can be uniquely represented as a product of a(n) distinct terms from A073491; this representation is minimal relative to the number of terms.

			See illustration of the first terms in the Links section.
The prime indices of 18564 are {1,1,2,4,6,7}, which separate into maximal gapless submultisets {1,1,2}, {4}, {6,7}, so a(18564) = 3; this corresponds to the ordered factorization 18564 = 12 * 7 * 221. - _Gus Wiseman_, Sep 03 2022

Rémy Sigrist, Table of n, a(n) for n = 1..10000
Rémy Sigrist, Illustration of the first terms
Rémy Sigrist, Scatterplot of the ordinal transform of the first 1000000 terms

Cf. A002110, A003961, A007947, A069010, A087207.
Positions of first appearances are A066205.
These are the row-lengths of A356226 and A356234. Other statistics are:
- length: A287170 (this sequence)
- minimum: A356227
- maximum: A356228
- bisected length: A356229
- standard composition: A356230
- Heinz number: A356231
- positions of first appearances: A356603 or A356232 (sorted)
A001222 counts prime factors, distinct A001221.
A003963 multiplies together the prime indices.
A056239 adds up the prime indices, row sums of A112798.
A073491 lists numbers with gapless prime indices, complement A073492.
Cf. A000005, A053251, A055932, A061395, A073493, A107428, A132747, A137921, A193829, A286470.

Table[Length[Select[First/@If[n==1,{},FactorInteger[n]],!Divisible[n,NextPrime[#]]&]],{n,30}] (* Gus Wiseman, Sep 03 2022 *)

a(n) = my (f=factor(n)); if (#f~==0, return (0), return (#f~ - sum(i=1, #f~-1, if (primepi(f[i,1])+1 == primepi(f[i+1,1]), 1, 0))))

from sympy import factorint, primepi
def a087207(n):
    f=factorint(n)
    return sum([2**primepi(i - 1) for i in f])
def a069010(n): return sum(1 for d in bin(n)[2:].split('0') if len(d)) # this function from Chai Wah Wu
def a(n): return a069010(a087207(n)) # Indranil Ghosh, Jun 06 2017

a(n) = A069010(A087207(n))

A245563 Table read by rows: row n gives list of lengths of runs of 1's in binary expansion of n, starting with low-order bits.

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane, Aug 10 2014

A formula for A071053(n) depends on this table.

			Here are the run lengths for the numbers 0 through 21:
0, []
1, [1]
2, [1]
3, [2]
4, [1]
5, [1, 1]
6, [2]
7, [3]
8, [1]
9, [1, 1]
10, [1, 1]
11, [2, 1]
12, [2]
13, [1, 2]
14, [3]
15, [4]
16, [1]
17, [1, 1]
18, [1, 1]
19, [2, 1]
20, [1, 1]
21, [1, 1, 1]

Reinhard Zumkeller, Rows n = 0..1000 of triangle, flattened
Index entries for sequences related to binary expansion of n

Row sums = A000120 (the binary weight).
Cf. A245562, A071053.
Cf. A030308, A227736.
Row lengths are A069010.
The version for prime indices (instead of binary indices) is A124010.
Numbers with distinct run-lengths are A328592.
Numbers with equal run-lengths are A164707.
Cf. A003714, A014081, A328166, A328594, A328595, A328596.

import Data.List (group)
a245563 n k = a245563_tabf !! n !! k
a245563_row n = a245563_tabf !! n
a245563_tabf = [0] : map
   (map length . (filter ((== 1) . head)) . group) (tail a030308_tabf)
-- Reinhard Zumkeller, Aug 10 2014

for n from 0 to 128 do
lis:=[]; t1:=convert(n,base,2); L1:=nops(t1); out1:=1; c:=0;
for i from 1 to L1 do
if out1 = 1 and t1[i] = 1 then out1:=0; c:=c+1;
elif out1 = 0 and t1[i] = 1 then c:=c+1;
elif out1 = 1 and t1[i] = 0 then c:=c;
elif out1 = 0 and t1[i] = 0 then lis:=[op(lis),c]; out1:=1; c:=0;
fi;
if i = L1 and c>0 then lis:=[op(lis),c]; fi;
od:
lprint(n,lis);
od:

Join@@Table[Length/@Split[Join@@Position[Reverse[IntegerDigits[n,2]],1],#2==#1+1&],{n,0,100}] (* Gus Wiseman, Nov 03 2019 *)

from re import split
A245563_list = [0]
for n in range(1,100):
    A245563_list.extend(len(d) for d in split('0+',bin(n)[:1:-1]) if d != '')
# Chai Wah Wu, Sep 07 2014

A121016 Numbers whose binary expansion is properly periodic.

Original entry on oeis.org

3, 7, 10, 15, 31, 36, 42, 45, 54, 63, 127, 136, 153, 170, 187, 204, 221, 238, 255, 292, 365, 438, 511, 528, 561, 594, 627, 660, 682, 693, 726, 759, 792, 825, 858, 891, 924, 957, 990, 1023, 2047, 2080, 2145, 2184, 2210, 2275, 2340, 2405, 2457, 2470, 2535

Offset: 1

Jacob A. Siehler, Sep 08 2006

A finite sequence is aperiodic if its cyclic rotations are all different. - Gus Wiseman, Oct 31 2019

			For example, 204=(1100 1100)_2 and 292=(100 100 100)_2 belong to the sequence, but 30=(11110)_2 cannot be split into repeating periods.
From _Gus Wiseman_, Oct 31 2019: (Start)
The sequence of terms together with their binary expansions and binary indices begins:
   3:         11 ~ {1,2}
   7:        111 ~ {1,2,3}
   10:      1010 ~ {2,4}
   15:      1111 ~ {1,2,3,4}
   31:     11111 ~ {1,2,3,4,5}
   36:    100100 ~ {3,6}
   42:    101010 ~ {2,4,6}
   45:    101101 ~ {1,3,4,6}
   54:    110110 ~ {2,3,5,6}
   63:    111111 ~ {1,2,3,4,5,6}
  127:   1111111 ~ {1,2,3,4,5,6,7}
  136:  10001000 ~ {4,8}
  153:  10011001 ~ {1,4,5,8}
  170:  10101010 ~ {2,4,6,8}
  187:  10111011 ~ {1,2,4,5,6,8}
  204:  11001100 ~ {3,4,7,8}
  221:  11011101 ~ {1,3,4,5,7,8}
  238:  11101110 ~ {2,3,4,6,7,8}
  255:  11111111 ~ {1,2,3,4,5,6,7,8}
  292: 100100100 ~ {3,6,9}
(End)

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

A020330 is a subsequence.
Numbers whose binary expansion is aperiodic are A328594.
Numbers whose reversed binary expansion is Lyndon are A328596.
Numbers whose binary indices have equal run-lengths are A164707.
Cf. A000120, A003714, A014081, A065609, A069010, A275692, A328595.

PeriodicQ[n_, base_] := Block[{l = IntegerDigits[n, base]}, MemberQ[ RotateLeft[l, # ] & /@ Most@ Divisors@ Length@l, l]]; Select[ Range@2599, PeriodicQ[ #, 2] &]

is(n)=n=binary(n);fordiv(#n,d,for(i=1,#n/d-1, for(j=1,d, if(n[j]!=n[j+i*d], next(3)))); return(d<#n)) \\ Charles R Greathouse IV, Dec 10 2013

A384877 Irregular triangle read by rows where row k lists the lengths of maximal anti-runs (increasing by more than 1) in the binary indices of n.

Original entry on oeis.org

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

Offset: 0

Gus Wiseman, Jun 17 2025

A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.

			The binary indices of 182 are {2,3,5,6,8}, with maximal anti-runs ((2),(3,5),(6,8)) so row 182 is (1,2,2).
Triangle begins:
   0: ()
   1: (1)
   2: (1)
   3: (1,1)
   4: (1)
   5: (2)
   6: (1,1)
   7: (1,1,1)
   8: (1)
   9: (2)
  10: (2)
  11: (1,2)
  12: (1,1)
  13: (2,1)
  14: (1,1,1)
  15: (1,1,1,1)

Row-sums are A000120.
Positions of rows of the form (1,1,...) are A023758.
Positions of first appearances of each distinct row appear to be A052499.
For runs instead of anti-runs we have A245563, reverse A245562.
Row-lengths are A384890.
A355394 counts partitions without a neighborless part, singleton case A355393.
A356606 counts strict partitions without a neighborless part, complement A356607.
A384175 counts subsets with all distinct lengths of maximal runs, complement A384176.
Cf. A044813, A048793, A069010, A164707, A243815, A246029, A328592, A384177, A384877, A384879, A384893.

bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
Table[Length/@Split[bpe[n],#2!=#1+1&],{n,0,100}]

cp's OEIS Frontend

A087207 A binary representation of the primes that divide a number, shown in decimal.

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A060130 Number of nonzero digits in factorial base representation (A007623) of n; minimum number of transpositions needed to compose each permutation in the lists A060117 & A060118.

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A328592 Numbers whose binary expansion has all different lengths of runs of 1's.

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A052331 Inverse of A052330; A binary encoding of Fermi-Dirac factorization of n, shown in decimal.

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A014081 a(n) is the number of occurrences of '11' in the binary expansion of n.

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A164707 A positive integer n is included if all runs of 1's in binary n are of the same length.

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