A246029 -id:A246029 - cp's OEIS frontend

A181819 Prime shadow of n: a(1) = 1; for n>1, if n = Product prime(i)^e(i), then a(n) = Product prime(e(i)).

1, 2, 2, 3, 2, 4, 2, 5, 3, 4, 2, 6, 2, 4, 4, 7, 2, 6, 2, 6, 4, 4, 2, 10, 3, 4, 5, 6, 2, 8, 2, 11, 4, 4, 4, 9, 2, 4, 4, 10, 2, 8, 2, 6, 6, 4, 2, 14, 3, 6, 4, 6, 2, 10, 4, 10, 4, 4, 2, 12, 2, 4, 6, 13, 4, 8, 2, 6, 4, 8, 2, 15, 2, 4, 6, 6, 4, 8, 2, 14, 7, 4, 2, 12, 4, 4, 4, 10, 2, 12, 4, 6, 4, 4, 4, 22, 2, 6, 6, 9, 2, 8, 2, 10, 8

Offset: 1

Matthew Vandermast, Dec 07 2010

a(n) depends only on prime signature of n (cf. A025487). a(m) = a(n) iff m and n have the same prime signature, i.e., iff A046523(m) = A046523(n).

Because A046523 (the smallest representative of prime signature of n) and this sequence are functions of each other as A046523(n) = A181821(a(n)) and a(n) = a(A046523(n)), it implies that for all i, j: a(i) = a(j) <=> A046523(i) = A046523(j) <=> A101296(i) = A101296(j), i.e., that equivalence-class-wise this is equal to A101296, and furthermore, applying any function f on this sequence gives us a sequence b(n) = f(a(n)) whose equivalence class partitioning is equal to or coarser than that of A101296, i.e., b is then a sequence that depends only on the prime signature of n (the multiset of exponents of its prime factors), although not necessarily in a very intuitive way. - Antti Karttunen, Apr 28 2022

			20 = 2^2*5 has the exponents (2,1) in its prime factorization. Accordingly, a(20) = prime(2)*prime(1) = A000040(2)*A000040(1) = 3*2 = 6.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Index entries for sequences computed from exponents in factorization of n

Column 1 of A353510 (see also A325239 and A325277).
Cf. A000040, A000265, A001511, A001222, A003963, A005361, A007814, A008578, A028234, A046523, A056239, A064553, A064989, A067029, A101296 (restricted growth sequence transform), A108951, A122111, A124010, A124859, A156552, A181820, A181821, A182850, A182855, A182857 (also A323014), A115621, A101296, A238690, A238745, A238747, A238748, A246029, A304465, A304647, A305732, A305733, A320118, A323022, A325501, A325502, A325507, A325508, A325755 (A353566), A325756, A328830 [= a(a(n))], A328835, A351564 (characteristic function of A130091), A351944, A351946, A353379.
Left inverse of A304660.

a181819 = product . map a000040 . a124010_row
-- Reinhard Zumkeller, Mar 26 2012

A181819 := proc(n)
    local a;
    a := 1;
    for pf in ifactors(n)[2] do
        a := a*ithprime(pf[2]) ;
    end do:
    a ;
end proc:
seq(A181819(n),n=1..80) ; # R. J. Mathar, Jan 09 2019
# second Maple program:
a:= n-> mul(ithprime(i[2]), i=ifactors(n)[2]):
seq(a(n), n=1..105);  # Alois P. Heinz, Nov 26 2024

{1}~Join~Table[Times @@ Prime@ Map[Last, FactorInteger@ n], {n, 2, 120}] (* Michael De Vlieger, Feb 07 2016 *)

a(n) = {my(f=factor(n)); prod(k=1, #f~, prime(f[k,2]));} \\ Michel Marcus, Nov 16 2015

;; With memoization-macro definec.
(definec (A181819 n) (cond ((= 1 n) 1) (else (* (A000040 (A067029 n)) (A181819 (A028234 n))))))
;; Antti Karttunen, Feb 05 2016

;; With memoization-macro definec.
(definec (A181819 n) (cond ((= 1 n) 1) ((even? n) (* (A000040 (A007814 n)) (A181819 (A000265 n)))) (else (A181819 (A064989 n)))))
;; Antti Karttunen, Feb 07 2016

From Antti Karttunen, Feb 07 2016: (Start)
a(1) = 1; for n > 1, a(n) = A000040(A067029(n)) * a(A028234(n)).
a(1) = 1; for n > 1, a(n) = A008578(A001511(n)) * a(A064989(n)).
Other identities. For all n >= 1:
a(A124859(n)) = A122111(a(n)) = A238745(n). - from Matthew Vandermast's formulas for the latter sequence.
(End)
a(n) = A246029(A156552(n)). - Antti Karttunen, Oct 15 2016
From Antti Karttunen, Apr 28 & Apr 30 2022: (Start)
A181821(a(n)) = A046523(n) and a(A046523(n)) = a(n). [See comments]
a(n) = A329900(A124859(n)) = A319626(A124859(n)).
a(n) = A246029(A156552(n)).
a(a(n)) = A328830(n).
a(A304660(n)) = n.
a(A108951(n)) = A122111(n).
a(A185633(n)) = A322312(n).
a(A025487(n)) = A181820(n).
a(A276076(n)) = A275735(n) and a(A276086(n)) = A328835(n).
As the sequence converts prime exponents to prime indices, it effects the following mappings:
A001221(a(n)) = A071625(n). [Number of distinct indices --> Number of distinct exponents]
A001222(a(n)) = A001221(n). [Number of indices (i.e., the number of prime factors with multiplicity) --> Number of exponents (i.e., the number of distinct prime factors)]
A056239(a(n)) = A001222(n). [Sum of indices --> Sum of exponents]
A066328(a(n)) = A136565(n). [Sum of distinct indices --> Sum of distinct exponents]
A003963(a(n)) = A005361(n). [Product of indices --> Product of exponents]
A290103(a(n)) = A072411(n). [LCM of indices --> LCM of exponents]
A156061(a(n)) = A290107(n). [Product of distinct indices --> Product of distinct exponents]
A257993(a(n)) = A134193(n). [Index of the least prime not dividing n --> The least number not among the exponents]
A055396(a(n)) = A051904(n). [Index of the least prime dividing n --> Minimal exponent]
A061395(a(n)) = A051903(n). [Index of the greatest prime dividing n --> Maximal exponent]
A008966(a(n)) = A351564(n). [All indices are distinct (i.e., n is squarefree) --> All exponents are distinct]
A007814(a(n)) = A056169(n). [Number of occurrences of index 1 (i.e., the 2-adic valuation of n) --> Number of occurrences of exponent 1]
A056169(a(n)) = A136567(n). [Number of unitary prime divisors --> Number of exponents occurring only once]
A064989(a(n)) = a(A003557(n)) = A295879(n). [Indices decremented after <--> Exponents decremented before]
Other mappings:
A007947(a(n)) = a(A328400(n)) = A329601(n).
A181821(A007947(a(n))) = A328400(n).
A064553(a(n)) = A000005(n) and A000005(a(n)) = A182860(n).
A051903(a(n)) = A351946(n).
A003557(a(n)) = A351944(n).
A258851(a(n)) = A353379(n).
A008480(a(n)) = A309004(n).
a(A325501(n)) = A325507(n) and a(A325502(n)) = A038754(n+1).
a(n!) = A325508(n).
(End)

Name "Prime shadow" (coined by Gus Wiseman in A325755) prefixed to the definition by Antti Karttunen, Apr 27 2022

A156552 Unary-encoded compressed factorization of natural numbers.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 8, 7, 6, 9, 16, 11, 32, 17, 10, 15, 64, 13, 128, 19, 18, 33, 256, 23, 12, 65, 14, 35, 512, 21, 1024, 31, 34, 129, 20, 27, 2048, 257, 66, 39, 4096, 37, 8192, 67, 22, 513, 16384, 47, 24, 25, 130, 131, 32768, 29, 36, 71, 258, 1025, 65536, 43, 131072, 2049, 38, 63, 68, 69, 262144

Offset: 1

Leonid Broukhis, Feb 09 2009

The primes become the powers of 2 (2 -> 1, 3 -> 2, 5 -> 4, 7 -> 8); the composite numbers are formed by taking the values for the factors in the increasing order, multiplying them by the consecutive powers of 2, and summing. See the Example section.
From Antti Karttunen, Jun 27 2014: (Start)
The odd bisection (containing even terms) halved gives A244153.
The even bisection (containing odd terms), when one is subtracted from each and halved, gives this sequence back.
(End)
Question: Are there any other solutions that would satisfy the recurrence r(1) = 0; and for n > 1, r(n) = Sum_{d|n, d>1} 2^A033265(r(d)), apart from simple variants 2^k * A156552(n)? See also A297112, A297113. - Antti Karttunen, Dec 30 2017

			For 84 = 2*2*3*7 -> 1*1 + 1*2 + 2*4 + 8*8 =  75.
For 105 = 3*5*7 -> 2*1 + 4*2 + 8*4 = 42.
For 137 = p_33 -> 2^32 = 4294967296.
For 420 = 2*2*3*5*7 -> 1*1 + 1*2 + 2*4 + 4*8 + 8*16 = 171.
For 147 = 3*7*7 = p_2 * p_4 * p_4 -> 2*1 + 8*2 + 8*4 = 50.

David A. Corneth, Table of n, a(n) for n = 1..10000 (first 1024 terms from Antti Karttunen)
Hans Havermann, Factorization of the first 10000 terms, in format [[primes], [exponents]]
A puzzle by Sergey Orlov (in Russian)
Index entries for sequences related to binary expansion of n
Index entries for sequences that are permutations of the natural numbers
Index entries for sequences computed from indices in prime factorization

One less than A005941.
Inverse permutation: A005940 with starting offset 0 instead of 1.
Cf. A000079, A000120, A001222, A052126, A054429, A061395, A064216, A064989, A003188, A243071, A243065-A243066, A244153, A243354, A112798, A125106, A056239, A161511.
Cf. also A297106, A297112 (Möbius transform), A297113, A153013, A290308, A300827, A323243, A323244, A323247, A324201, A324812 (n for which a(n) is a square), A324813, A324822, A324823, A324398, A324713, A324815, A324819, A324865, A324866, A324867.
Other related permutations: A253551, A253792, A253564, A253791, A277195, A297163, A297164, A297165, A297166, A302023, A305418, A322863, A322864.

Table[Floor@ Total@ Flatten@ MapIndexed[#1 2^(#2 - 1) &, Flatten[ Table[2^(PrimePi@ #1 - 1), {#2}] & @@@ FactorInteger@ n]], {n, 67}] (* Michael De Vlieger, Sep 08 2016 *)

a(n) = {my(f = factor(n), p2 = 1, res = 0); for(i = 1, #f~, p = 1 << (primepi(f[i, 1]) - 1); res += (p * p2 * (2^(f[i, 2]) - 1)); p2 <<= f[i, 2]); res}; \\ David A. Corneth, Mar 08 2019

A064989(n) = {my(f); f = factor(n); if((n>1 && f[1,1]==2), f[1,2] = 0); for (i=1, #f~, f[i,1] = precprime(f[i,1]-1)); factorback(f)};
A156552(n) = if(1==n, 0, if(!(n%2), 1+(2*A156552(n/2)), 2*A156552(A064989(n)))); \\ (based on the given recurrence) - Antti Karttunen, Mar 08 2019

# Program corrected per instructions from Leonid Broukhis. - Antti Karttunen, Jun 26 2014
# However, it gives correct answers only up to n=136, before corruption by a wrap-around effect.
# Note that the correct answer for n=137 is A156552(137) = 4294967296.
$max = $ARGV[0];
$pow = 0;
foreach $i (2..$max) {
@a = split(/ /, `factor $i`);
shift @a;
$shift = 0;
$cur = 0;
while ($n = int shift @a) {
$prime{$n} = 1 << $pow++ if !defined($prime{$n});
$cur |= $prime{$n} << $shift++;
}
print "$cur, ";
}
print "\n";
(Scheme, with memoization-macro definec from Antti Karttunen's IntSeq-library, two different implementations)
(definec (A156552 n) (cond ((= n 1) 0) (else (+ (A000079 (+ -2 (A001222 n) (A061395 n))) (A156552 (A052126 n))))))
(definec (A156552 n) (cond ((= 1 n) (- n 1)) ((even? n) (+ 1 (* 2 (A156552 (/ n 2))))) (else (* 2 (A156552 (A064989 n))))))
;; Antti Karttunen, Jun 26 2014

from sympy import primepi, factorint
def A156552(n): return sum((1<Chai Wah Wu, Mar 10 2023

From Antti Karttunen, Jun 26 2014: (Start)
a(1) = 0, a(n) = A000079(A001222(n)+A061395(n)-2) + a(A052126(n)).
a(1) = 0, a(2n) = 1+2*a(n), a(2n+1) = 2*a(A064989(2n+1)). [Compare to the entanglement recurrence A243071].
For n >= 0, a(2n+1) = 2*A244153(n+1). [Follows from the latter clause of the above formula.]
a(n) = A005941(n) - 1.
As a composition of related permutations:
a(n) = A003188(A243354(n)).
a(n) = A054429(A243071(n)).
For all n >= 1, A005940(1+a(n)) = n and for all n >= 0, a(A005940(n+1)) = n. [The offset-0 version of A005940 works as an inverse for this permutation.]
This permutations also maps between the partition-lists A112798 and A125106:
A056239(n) = A161511(a(n)). [The sums of parts of each partition (the total sizes).]
A003963(n) = A243499(a(n)). [And also the products of those parts.]
(End)
From Antti Karttunen, Oct 09 2016: (Start)
A161511(a(n)) = A056239(n).
A029837(1+a(n)) = A252464(n). [Binary width of terms.]
A080791(a(n)) = A252735(n). [Number of nonleading 0-bits.]
A000120(a(n)) = A001222(n). [Binary weight.]
For all n >= 2, A001511(a(n)) = A055396(n).
For all n >= 2, A000120(a(n))-1 = A252736(n). [Binary weight minus one.]
A252750(a(n)) = A252748(n).
a(A250246(n)) = A252754(n).
a(A005117(n)) = A277010(n). [Maps squarefree numbers to a permutation of A003714, fibbinary numbers.]
A085357(a(n)) = A008966(n). [Ditto for their characteristic functions.]
For all n >= 0:
a(A276076(n)) = A277012(n).
a(A276086(n)) = A277022(n).
a(A260443(n)) = A277020(n).
(End)
From Antti Karttunen, Dec 30 2017: (Start)
For n > 1, a(n) = Sum_{d|n, d>1} 2^A033265(a(d)). [See comments.]
More linking formulas:
A106737(a(n)) = A000005(n).
A290077(a(n)) = A000010(n).
A069010(a(n)) = A001221(n).
A136277(a(n)) = A181591(n).
A132971(a(n)) = A008683(n).
A106400(a(n)) = A008836(n).
A268411(a(n)) = A092248(n).
A037011(a(n)) = A010052(n) [conjectured, depends on the exact definition of A037011].
A278161(a(n)) = A046951(n).
A001316(a(n)) = A061142(n).
A277561(a(n)) = A034444(n).
A286575(a(n)) = A037445(n).
A246029(a(n)) = A181819(n).
A278159(a(n)) = A124859(n).
A246660(a(n)) = A112624(n).
A246596(a(n)) = A069739(n).
A295896(a(n)) = A053866(n).
A295875(a(n)) = A295297(n).
A284569(a(n)) = A072411(n).
A286574(a(n)) = A064547(n).
A048735(a(n)) = A292380(n).
A292272(a(n)) = A292382(n).
A244154(a(n)) = A048673(n), a(A064216(n)) = A244153(n).
A279344(a(n)) = A279339(n), a(A279338(n)) = A279343(n).
a(A277324(n)) = A277189(n).
A037800(a(n)) = A297155(n).
For n > 1, A033265(a(n)) = 1+A297113(n).
(End)
From Antti Karttunen, Mar 08 2019: (Start)
a(n) = A048675(n) + A323905(n).
a(A324201(n)) = A000396(n), provided there are no odd perfect numbers.
The following sequences are derived from or related to the base-2 expansion of a(n):
A000265(a(n)) = A322993(n).
A002487(a(n)) = A323902(n).
A005187(a(n)) = A323247(n).
A324288(a(n)) = A324116(n).
A323505(a(n)) = A323508(n).
A079559(a(n)) = A323512(n).
A085405(a(n)) = A323239(n).
The following sequences are obtained by applying to a(n) a function that depends on the prime factorization of its argument, which goes "against the grain" because a(n) is the binary code of the factorization of n, which in these cases is then factored again:
A000203(a(n)) = A323243(n).
A033879(a(n)) = A323244(n) = 2*a(n) - A323243(n),
A294898(a(n)) = A323248(n).
A000005(a(n)) = A324105(n).
A000010(a(n)) = A324104(n).
A083254(a(n)) = A324103(n).
A001227(a(n)) = A324117(n).
A000593(a(n)) = A324118(n).
A001221(a(n)) = A324119(n).
A009194(a(n)) = A324396(n).
A318458(a(n)) = A324398(n).
A192895(a(n)) = A324100(n).
A106315(a(n)) = A324051(n).
A010052(a(n)) = A324822(n).
A053866(a(n)) = A324823(n).
A001065(a(n)) = A324865(n) = A323243(n) - a(n),
A318456(a(n)) = A324866(n) = A324865(n) OR a(n),
A318457(a(n)) = A324867(n) = A324865(n) XOR a(n),
A318458(a(n)) = A324398(n) = A324865(n) AND a(n),
A318466(a(n)) = A324819(n) = A323243(n) OR 2*a(n),
A318467(a(n)) = A324713(n) = A323243(n) XOR 2*a(n),
A318468(a(n)) = A324815(n) = A323243(n) AND 2*a(n).
(End)

More terms from Antti Karttunen, Jun 28 2014

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}]

A384879 Numbers whose binary indices have all distinct lengths of maximal anti-runs (increasing by more than 1).

Original entry on oeis.org

1, 2, 4, 5, 8, 9, 10, 11, 13, 16, 17, 18, 19, 20, 21, 22, 25, 26, 32, 33, 34, 35, 36, 37, 38, 40, 41, 42, 43, 44, 49, 50, 52, 53, 64, 65, 66, 67, 68, 69, 70, 72, 73, 74, 75, 76, 80, 81, 82, 83, 84, 85, 86, 88, 97, 98, 100, 101, 104, 105, 106, 128, 129, 130

Offset: 1

Gus Wiseman, Jun 17 2025

nonn

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 813 are {1,3,4,6,9,10}, with maximal anti-runs ((1,3),(4,6,9),(10)), with lengths (2,3,1), so 813 is in the sequence.
The terms together with their binary expansions and binary indices begin:
    1:       1 ~ {1}
    2:      10 ~ {2}
    4:     100 ~ {3}
    5:     101 ~ {1,3}
    8:    1000 ~ {4}
    9:    1001 ~ {1,4}
   10:    1010 ~ {2,4}
   11:    1011 ~ {1,2,4}
   13:    1101 ~ {1,3,4}
   16:   10000 ~ {5}
   17:   10001 ~ {1,5}
   18:   10010 ~ {2,5}
   19:   10011 ~ {1,2,5}
   20:   10100 ~ {3,5}
   21:   10101 ~ {1,3,5}
   22:   10110 ~ {2,3,5}
   25:   11001 ~ {1,4,5}
   26:   11010 ~ {2,4,5}

Subsets of this type are counted by A384177, for runs A384175 (complement A384176).
These are the indices of strict rows in A384877, see A384878, A245563, A245562, A246029.
A000120 counts binary indices.
A098859 counts Wilf partitions (distinct multiplicities), complement A336866.
A356606 counts strict partitions without a neighborless part, complement A356607.
A384890 counts maximal anti-runs in binary indices, runs A069010.
Cf. A023758, A044813, A048793, A164707, A242882, A243815, A325325, A328592, A384879, A384884, A384886, A384893.

bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
Select[Range[100],UnsameQ@@Length/@Split[bpe[#],#2!=#1+1&]&]

A385816 The number k such that the k-th composition in standard order lists the maximal anti-run lengths of the binary indices of n. Standard composition number of row n of A384877.

Original entry on oeis.org

0, 1, 1, 3, 1, 2, 3, 7, 1, 2, 2, 6, 3, 5, 7, 15, 1, 2, 2, 6, 2, 4, 6, 14, 3, 5, 5, 13, 7, 11, 15, 31, 1, 2, 2, 6, 2, 4, 6, 14, 2, 4, 4, 12, 6, 10, 14, 30, 3, 5, 5, 13, 5, 9, 13, 29, 7, 11, 11, 27, 15, 23, 31, 63, 1, 2, 2, 6, 2, 4, 6, 14, 2, 4, 4, 12, 6, 10, 14

Offset: 0

Gus Wiseman, Jul 15 2025

nonn

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 k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.
If the k-th composition in standard order is y, then the standard composition number of y is defined to be k.

			The binary indices of 181 are {1,3,5,6,8}, with maximal anti-runs ((1,3,5),(6,8)), with lengths (3,2), which is the 18th composition in standard order, so a(181) = 18.

Gus Wiseman, Statistics, classes, and transformations of standard compositions

The reverse version is A209859.
Sorted positions of first appearances are A247648.
These are standard composition numbers of rows of A384877 (duplicates removed A385886).
For runs instead of anti-runs the reverse is A385887 (duplicates removed A232559).
For runs instead of anti-runs we have A385889 (duplicates removed A385818).
A245563 lists run lengths of binary indices (ranks A246029), reverse A245562.
Cf. A000120, A029931, A044813, A048793, A052499, A069010, A348366, A384175, A384878, A384879, A384893.

bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
stcinv[q_]:=Total[2^(Accumulate[Reverse[q]])]/2;
stcinv/@Table[Length/@Split[bpe[n],#2!=#1+1&],{n,0,100}]

A385886 Irregular triangle read by rows listing the lengths of maximal anti-runs (sequences of distinct consecutive elements increasing by more than 1) of binary indices, duplicate rows removed.

Original entry on oeis.org

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

Offset: 0

Gus Wiseman, Jul 14 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.

This is the triangle A384877, except all duplicates after the first instance of each composition are removed. It lists all compositions in order of their first appearance as a row of A384877.

			The binary indices of 27 are {1,2,4,5}, with maximal anti-runs ((1),(2,4),(5)), with lengths (1,2,1). After removing duplicates, this is our row 10.
The binary indices of 53 are {1,3,5,6}, with maximal anti-runs ((1,3,5),(6)), with lengths (3,1). After removing duplicates, this is our row 16.
Triangle begins:
   0: .
   1: 1
   2: 1 1
   3: 2
   4: 1 1 1
   5: 1 2
   6: 2 1
   7: 1 1 1 1
   8: 3
   9: 1 1 2
  10: 1 2 1
  11: 2 1 1
  12: 1 1 1 1 1
  13: 1 3
  14: 2 2
  15: 1 1 1 2
  16: 3 1
  17: 1 1 2 1
  18: 1 2 1 1
  19: 2 1 1 1
  20: 1 1 1 1 1 1

In the following references, "before" is short for "before removing duplicate rows".
Positions of singleton rows appear to be A001906 = A055588 - 1.
Positions of rows of the form (1,1,...) appear to be A001911-2, before A023758.
Row sums appear to be A200648, before A000120.
Row lengths appear to be A200649, before A384890.
Standard composition numbers of each row appear to be A348366.
Before we had A384877, ranks A385816, firsts A052499.
For runs instead of anti-runs we have A385817, see A245563, A245562, A246029.
Cf. A044813, A048793, A069010, A164707, A242882, A328592, A384175, A384177, A384878, A384879, A384893.

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

A209859 Rewrite the binary expansion of n from the most significant end, 1 -> 1, 0+1 (one or more zeros followed by one) -> 0, drop the trailing zeros of the original n.

Original entry on oeis.org

0, 1, 1, 3, 1, 2, 3, 7, 1, 2, 2, 5, 3, 6, 7, 15, 1, 2, 2, 5, 2, 4, 5, 11, 3, 6, 6, 13, 7, 14, 15, 31, 1, 2, 2, 5, 2, 4, 5, 11, 2, 4, 4, 9, 5, 10, 11, 23, 3, 6, 6, 13, 6, 12, 13, 27, 7, 14, 14, 29, 15, 30, 31, 63, 1, 2, 2, 5, 2, 4, 5, 11, 2, 4, 4, 9, 5, 10, 11, 23, 2, 4, 4, 9, 4, 8, 9, 19, 5, 10, 10, 21, 11, 22, 23, 47, 3, 6, 6, 13, 6, 12, 13, 27, 6, 12, 12, 25, 13

Offset: 0

Antti Karttunen, Mar 24 2012

This is the number k such that the k-th composition in standard order is the reversed sequence of lengths of the maximal anti-runs of the binary indices of n. Here, the binary indices of n are row n of A048793, and the k-th composition in standard order is row k of A066099. For example, the binary indices of 100 are {3,6,7}, with maximal anti-runs ((3,6),(7)), with reversed lengths (1,2), which is the 6th composition in standard order, so a(100) = 6. - Gus Wiseman, Jul 27 2025

			102 in binary is 1100110, we rewrite it from the left so that first two 1's stay same ("11"), then "001" is rewritten to "0", the last 1 to "1", and we ignore the last 0, thus getting 1101, which is binary expansion of 13, thus a(102) = 13.

Gus Wiseman, Statistics, classes, and transformations of standard compositions

This is an "inverse" of A071162, i.e. a(A071162(n)) = n for all n. Bisection: A209639. Used to construct permutation A209862.
Removing duplicates appears to give A358654.
Sorted positions of firsts appearances appear to be A247648+1.
A245563 lists run-lengths of binary indices (ranks A246029), reverse A245562.
A384175 counts subsets with all distinct lengths of maximal runs, complement A384176.
Cf. A000120, A023758, A029931, A044813, A048793, A052499, A069010, A348366, A384878, A384879, A384893.
Cf. A384877, A385816, A385887, A385889.

bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
stcinv[q_]:=Total[2^(Accumulate[Reverse[q]])]/2;
Table[stcinv[Reverse[Length/@Split[bpe[n],#2!=#1+1&]]],{n,0,100}] (* Gus Wiseman, Jul 25 2025 *)

import re
def a(n): return int(re.sub("0+1", "0", bin(n)[2:].rstrip("0")), 2) if n else 0
print([a(n) for n in range(109)])  # Michael S. Branicky, Jul 25 2025

(define (A209859 n) (let loop ((n n) (s 0) (i (A053644 n))) (cond ((zero? n) s) ((> i n) (if (> (/ i 2) n) (loop n s (/ i 2)) (loop (- n (/ i 2)) (* 2 s) (/ i 4)))) (else (loop (- n i) (+ (* 2 s) 1) (/ i 2))))))

a(n) = a(A000265(n)).

A385889 The number k such that the k-th composition in standard order is the sequence of lengths of maximal runs of binary indices of n.

Original entry on oeis.org

0, 1, 1, 2, 1, 3, 2, 4, 1, 3, 3, 5, 2, 6, 4, 8, 1, 3, 3, 5, 3, 7, 5, 9, 2, 6, 6, 10, 4, 12, 8, 16, 1, 3, 3, 5, 3, 7, 5, 9, 3, 7, 7, 11, 5, 13, 9, 17, 2, 6, 6, 10, 6, 14, 10, 18, 4, 12, 12, 20, 8, 24, 16, 32, 1, 3, 3, 5, 3, 7, 5, 9, 3, 7, 7, 11, 5, 13, 9, 17, 3

Offset: 0

Gus Wiseman, Jul 16 2025

nonn

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 k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

			The binary indices of 27 are {1,2,4,5}, with maximal runs ((1,2),(4,5)), with lengths (2,2), which is the 10th composition in standard order, so a(27) = 10.
The binary indices of 100 are {3,6,7}, with maximal runs ((3),(6,7)), with lengths (1,2), which is the 6th composition in standard order, so a(100) = 6.

Gus Wiseman, Statistics, classes, and transformations of standard compositions

Sorted positions of firsts appearances appear to be A247648+1.
After removing duplicates we get A385818.
The reverse version is A385887.
A245563 lists run lengths of binary indices (ranks A246029), reverse A245562.
A384877 lists anti-run lengths of binary indices (ranks A385816), reverse A209859.
Cf. A000120, A023758, A029931, A044813, A048793, A069010, A348366, A384175, A384878, A385817.

bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
stcinv[q_]:=Total[2^(Accumulate[Reverse[q]])]/2;
Table[stcinv[Length/@Split[bpe[n],#2==#1+1&]],{n,0,100}]

A385817 Irregular triangle read by rows listing the lengths of maximal runs (sequences of consecutive elements increasing by 1) of binary indices, duplicate rows removed.

Original entry on oeis.org

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

Offset: 0

Gus Wiseman, Jul 14 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.

This is the triangle A245563, except all duplicates after the first instance of each composition are removed. It lists all compositions in order of their first appearance as a row of A245563.

			The binary indices of 53 are {1,3,5,6}, with maximal runs ((1),(3),(5,6)), with lengths (1,1,2). After removing duplicates, this is our row 16.
Triangle begins:
   0: .
   1: 1
   2: 2
   3: 1 1
   4: 3
   5: 2 1
   6: 1 2
   7: 4
   8: 1 1 1
   9: 3 1
  10: 2 2
  11: 1 3
  12: 5
  13: 2 1 1
  14: 1 2 1
  15: 4 1
  16: 1 1 2
  17: 3 2
  18: 2 3
  19: 1 4
  20: 6
  21: 1 1 1 1

In the following references, "before" is short for "before removing duplicate rows".
Positions of singleton rows appear to be A000071 = A000045-1, before A023758.
Positions of firsts appearances appear to be A001629.
Positions of rows of the form (1,1,...) appear to be A055588 = A001906+1.
First term of each row appears to be A083368.
Row sums appear to be A200648, before A000120.
Row lengths after the first row appear to be A200650+1, before A069010 = A037800+1.
Before the removals we had A245563 (except first term), see A245562, A246029, A328592.
For anti-run ranks we have A385816, before A348366, firsts A052499.
Standard composition numbers of rows are A385818, before A385889.
For anti-runs we have A385886, before A384877, firsts A384878.
Cf. A044813, A048793, A164707, A200649, A242882, A243815, A384175, A384890.

DeleteDuplicates[Table[Length/@Split[Join@@Position[Reverse[IntegerDigits[n,2]],1],#2==#1+1&],{n,0,100}]]

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A253081 Partial sums of A246029.

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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.

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A384879 Numbers whose binary indices have all distinct lengths of maximal anti-runs (increasing by more than 1).

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A385816 The number k such that the k-th composition in standard order lists the maximal anti-run lengths of the binary indices of n. Standard composition number of row n of A384877.

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A385886 Irregular triangle read by rows listing the lengths of maximal anti-runs (sequences of distinct consecutive elements increasing by more than 1) of binary indices, duplicate rows removed.

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A209859 Rewrite the binary expansion of n from the most significant end, 1 -> 1, 0+1 (one or more zeros followed by one) -> 0, drop the trailing zeros of the original n.

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