A136480 -id:A136480 - cp's OEIS frontend

A227738 Irregular table read by rows: each row n (n>=1) lists the positions where the runs of bits change between 0's and 1's in the binary expansion of n, when scanning it from the least significant to the most significant end.

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

Offset: 1

Antti Karttunen, Jul 25 2013

Row n has A005811(n) terms.
As a sequence, seems to have a particular fractal structure, probably allowing additional formulas.
Row n lists the positions of 1-bits in the binary expansion of the Gray code for n, A003188(n), when 1 is the rightmost position. A003188(17) = 25 = 11001_2 gives row 17: 1,4,5. - Alois P. Heinz, Feb 01 2023

			Table begins as:
  Row  n in    Terms on
   n   binary  that row
   1      1    1;
   2     10    1,2;
   3     11    2;
   4    100    2,3;
   5    101    1,2,3;
   6    110    1,3;
   7    111    3;
   8   1000    3,4;
   9   1001    1,3,4;
  10   1010    1,2,3,4;
  11   1011    2,3,4;
  12   1100    2,4;
  13   1101    1,2,4;
  14   1110    1,4;
  15   1111    4;
  16  10000    4,5;
etc.
The terms also give the partial sums of runlengths, when the binary expansion of n is scanned from the least significant to the most significant end.

Antti Karttunen, The rows 1..1023 of the table, flattened
Wikipedia, Gray code

Each row n (n>=1) contains the initial A005811(n) nonzero terms from the beginning of row n of A227188. A227192(n) gives the sum of terms on row n. A136480 gives the first column.
Cf. also A227188, A227736, A227739.
A318926 is a compressed version. If the order is reversed we get A101211 and A318927.
Cf. A003188, A360287.

T:= n-> (l-> seq(`if`(l[i]=1, i, [][]), i=1..nops(l)))(
                 Bits[Split](Bits[Xor](n, iquo(n, 2)))):
seq(T(n), n=1..50);  # Alois P. Heinz, Feb 01 2023

Table[Rest@FoldList[Plus,0,Length/@Split[Reverse[IntegerDigits[n,2]]]],{n,34}]//Flatten (* Wouter Meeussen, Aug 31 2013 *)

a(n) = A227188(A227737(n),A227740(n)).

Alternatively, if A227740(n) is 0, then a(n) = A227736(n), otherwise a(n) = a(n-1) + A227736(n). [Each row gives cumulative sums of the runlengths of binary representation of n]

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

Original entry on oeis.org

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

Offset: 0

Antti Karttunen, Jul 05 2013

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

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

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

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

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

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

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

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

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

A227186 Array A(n,k) read by antidiagonals: the length of the (k+1)-th run (k>=0) of binary digits of n, first run starting from the least significant bit of n.

Original entry on oeis.org

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

Offset: 0

Antti Karttunen, Jul 06 2013

A(n,k) is set to zero if there are less than k+1 runs.

The irregular table A101211 gives the nonzero terms of each row in reverse order. The terms on row n sum to A029837(n+1). The product of nonzero terms on row n>0 is A167489(n). Number of nonzero terms on each row: A005811.

			The top-left corner of the array:
0, 0, 0, 0, 0, ... (0, in binary 0, has no runs (by convention), thus at first we have all-0 sequence)
1, 0, 0, 0, 0, ... (1, in binary 1, has one run of length 1)
1, 1, 0, 0, 0, ... (2, in binary 10, has two runs of length 1 both)
2, 0, 0, 0, 0, ... (3, in binary 11, has one run of length 2)
2, 1, 0, 0, 0, ... (4, in binary 100, the rightmost run of length 2 given first, then the second run of length 1)
1, 1, 1, 0, 0, ... (5, in binary 101, has three runs of one bit each)
1, 2, 0, 0, 0, ...
3, 0, 0, 0, 0, ...
3, 1, 0, 0, 0, ...
1, 2, 1, 0, 0, ...
1, 1, 1, 1, 0, ...
2, 1, 1, 0, 0, ...
2, 2, 0, 0, 0, ...
1, 1, 2, 0, 0, ...
1, 3, 0, 0, 0, ...
4, 0, 0, 0, 0, ...

Antti Karttunen, The first 141 antidiagonals of the table, flattened

Used to compute A227183. Cf. also A163575, A227188, A227189.

A227186 := proc(n,k)
    local bdgs,ru,i,b,a;
    bdgs := convert(n,base,2) ;
    if nops(bdgs) = 0 then
        return 0 ;
    end if;
    ru := 0 ;
    i := 1 ;
    b := op(i,bdgs) ;
    a := 1 ;
    for i from 2 to nops(bdgs) do
        if op(i,bdgs) <> op(i-1,bdgs) then
            if ru = k then
                return a;
            end if;
            a := 1 ;
            ru := ru+1 ;
        else
            a := a+1 ;
        end if;
    end do:
    if ru =k then
        a ;
    else
        0 ;
    end if;
end proc: # R. J. Mathar, Jul 23 2013

A227186(n,k)=while(k>=0,for(c=1,n,bittest(n,0)==bittest(n\=2,0)&next;k&break;return(c));n||return;k--) \\ To let A(0,0)=1 add "!n||!" in front of while(...). TO DO: add default value k=-1 and implement "flattened" sequence, such that A227186(n) yields a(n). M. Hasler, Jul 21 2013

(define (A227186 n) (A227186bi (A002262 n) (A025581 n)))
(define (A227186bi n k) (cond ((< (A005811 n) (+ 1 k)) 0) ((zero? k) (A136480 n)) (else (A227186bi (A163575 n) (- k 1)))))

A(n,0) = A136480(n), n>0.

A227188 Square array A(n,k) read by antidiagonals: the one-based bit-index where the (k+1)-st run in the binary expansion of n ends, as read from the least significant end.

Original entry on oeis.org

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

Offset: 0

Antti Karttunen, Jul 06 2013

A(n,k) is set to zero if there are fewer runs in n than k+1.
Equally, when A005811(n) > 1, A(n,k) gives the zero-based bit-index where the (k+2)-th run in the binary expansion of n starts, counted from the least significant end.
Each row gives the partial sums of the terms on the corresponding row in A227186, up to the first zero.

			The top-left corner of the array:
row #  row starts as
   0    0, 0, 0, 0, 0, ...
   1    1, 0, 0, 0, 0, ...
   2    1, 2, 0, 0, 0, ...
   3    2, 0, 0, 0, 0, ...
   4    2, 3, 0, 0, 0, ...
   5    1, 2, 3, 0, 0, ...
   6    1, 3, 0, 0, 0, ...
   7    3, 0, 0, 0, 0, ...
   8    3, 4, 0, 0, 0, ...
   9    1, 3, 4, 0, 0, ...
  10    1, 2, 3, 4, 0, ...
  11    2, 3, 4, 0, 0, ...
  12    2, 4, 0, 0, 0, ...
  13    1, 2, 4, 0, 0, ...
  14    1, 4, 0, 0, 0, ...
  15    4, 0, 0, 0, 0, ...
  16    4, 5, 0, 0, 0, ...
etc.
For example, for n = 8, whose binary expansion is "1000", we get the run lengths 3 and 1 (scanning from the right), partial sums of which are 3 and 4, thus row 8 begins as A(8,0)=3, A(8,1)=4, A(8,2)=0, ...

Antti Karttunen, The first 141 antidiagonals of the table, flattened

Cf. A227192 (row sums). Number of nonzero terms on each row: A005811.

Cf. also A227186, A227189, A163575.

A227188 := proc(n,k)
    local bdgs,ru,i,b,a;
    bdgs := convert(n,base,2) ;
    if nops(bdgs) = 0 then
        return 0 ;
    end if;
    ru := 0 ;
    i := 1 ;
    b := op(i,bdgs) ;
    for i from 2 to nops(bdgs) do
        if op(i,bdgs) <> op(i-1,bdgs) then
            if ru = k then
                return i-1;
            end if;
            ru := ru+1 ;
        end if;
    end do:
    if ru =k then
        nops(bdgs) ;
    else
        0 ;
    end if;
end proc: # R. J. Mathar, Jul 23 2013

Table[PadRight[Rest@FoldList[Plus,0,Length/@Split[Reverse[IntegerDigits[j,2]]]],i+1-j][[i+1-j]],{i,0,12},{j,0,i}] (* Wouter Meeussen, Aug 31 2013 *)

(define (A227188 n) (A227188bi (A002262 n) (A025581 n)))
(define (A227188bi n k) (cond ((< (A005811 n) (+ 1 k)) 0) ((zero? k) (A136480 n)) (else (+ (A136480 n) (A227188bi (A163575 n) (- k 1))))))

A(n,0) = A136480(n), n>0.

A341694 Square array T(n, k) read by antidiagonals upwards, n, k > 0: T(n, k) = A227736(n, k) for k = 1..A005811(n), and T(n, k) = T(n, k - A005811(n)) + ... + T(n, k-1) for k > A005811(n).

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 2, 2, 2, 1, 1, 1, 2, 3, 1, 1, 1, 3, 2, 5, 1, 3, 2, 1, 4, 2, 8, 1, 3, 3, 3, 3, 7, 2, 13, 1, 1, 1, 3, 5, 5, 11, 2, 21, 1, 1, 2, 4, 3, 8, 9, 18, 2, 34, 1, 2, 1, 1, 5, 3, 13, 17, 29, 2, 55, 1, 2, 1, 1, 4, 9, 3, 21, 31, 47, 2, 89, 1

Offset: 1

Rémy Sigrist, Feb 17 2021

This table contains all Fibonacci sequences of order m > 0 with positive terms:
- order 1 corresponds to constant sequences (n in A126646),
- order 2 corresponds to Fibonacci-like sequences (n in A043569),
- order 3 corresponds to tribonacci-like sequences (n in A043570),
- order 4 corresponds to tetranacci-like sequences (n in A043571).
For any n > 0, the row A341746(n) corresponds to the n-th row from which the first term has been removed.

			Array T(n, k) begins:
  n\k|  1  2  3  4  5   6   7   8   9   10   11   12   13    14
  ---+---------------------------------------------------------
    1|  1  1  1  1  1   1   1   1   1    1    1    1    1     1 --> A000012
    2|  1  1  2  3  5   8  13  21  34   55   89  144  233   377 --> A000045
    3|  2  2  2  2  2   2   2   2   2    2    2    2    2     2 --> A007395
    4|  2  1  3  4  7  11  18  29  47   76  123  199  322   521 --> A000032
    5|  1  1  1  3  5   9  17  31  57  105  193  355  653  1201 --> A000213
    6|  1  2  3  5  8  13  21  34  55   89  144  233  377   610 --> A000045
    7|  3  3  3  3  3   3   3   3   3    3    3    3    3     3 --> A010701
    8|  3  1  4  5  9  14  23  37  60   97  157  254  411   665 --> A104449
    9|  1  2  1  4  7  12  23  42  77  142  261  480  883  1624 --> A275778
   10|  1  1  1  1  4   7  13  25  49   94  181  349  673  1297 --> A000288

Rémy Sigrist, Table of n, a(n) for n = 1..11325
Rémy Sigrist, Colored representation of the table for n, k = 1..1024
Rémy Sigrist, PARI program for A341694
Wikipedia, Fibonacci numbers of higher order
Index entries for sequences related to Fibonacci numbers

Cf. A005811, A043569, A043570, A043571, A126646, A136480, A227736, A341699, A341746.

Cf. A000012, A000032, A000045, A000213, A000288, A007395, A010701, A104449, A275778.

PARI
```
See Links section.
```

T(A341746(n), k) = T(n, k+1).

T(n, 1) = A136480(n).

A092054 Base-2 logarithm of the sum of numerator and denominator of the convergents of the continued fraction expansion [1; 1/2, 1/3, 1/4, ..., 1/n, ...].

Original entry on oeis.org

1, 2, 4, 6, 7, 8, 11, 14, 15, 16, 18, 20, 21, 22, 26, 30, 31, 32, 34, 36, 37, 38, 41, 44, 45, 46, 48, 50, 51, 52, 57, 62, 63, 64, 66, 68, 69, 70, 73, 76, 77, 78, 80, 82, 83, 84, 88, 92, 93, 94, 96, 98, 99, 100, 103, 106, 107, 108, 110, 112, 113, 114, 120, 126, 127, 128, 130

Offset: 1

Paul D. Hanna, Feb 19 2004

nonn

Consider the convergents of the continued fraction expansion [1; 1/2, 1/3, 1/4, ..., 1/n, ...]. The numerators of the convergents are A001902 (successive denominators of Wallis's product approximation to Pi/2) and the denominators of the convergents are A092053. The sum of the numerators and the denominators equals a power of 2: A001902(n) + A092053(n) = 2^a(n).
Also, a(n-1) is the number of the comparisons that Floyd's heap-construction algorithm will use, in the worst case, to create an n-element heap. See Wikipedia link, section "Building a heap". - Marek A. Suchenek, Mar 16 2014:
First differences appear to be essentially A136480. - Chris Boyd, Jan 14 2016

			a(4)=6 since [1; 1/2, 1/3, 1/4] = 1 + 1/(1/2 + 1/(1/3 + 1/(1/4))) = 45/19; and the sum of the numerator and denominator of 45/19 equals 45 + 19 = 2^6.

Marek A. Suchenek, Elementary Yet Precise Worst-Case Analysis of Floyd's Heap-Construction Program, Fundamenta Informaticae 120 (2012), pp 75--92.
Wikipedia, Binary heap

Cf. A001902, A092053, A136480.

{a(n)=local(A);CF=contfracpnqn(vector(n,k,1/k)); A=length(binary(numerator(1+CF[1,1]/CF[2,1])))-1}

2^a(n) = A001902(n) + A092053(n).
It appears that a(n) = Sum_{k=1..n} A001511(floor((k+1)/2)). Equivalently, a(n) = 2n + 1 - A000120(n) - A000120(n+1) = A011371(n) + A011371(n+1). - Franklin T. Adams-Watters, Feb 02 2006
a(n-1) = 2*n - 2*A000120(n) - A007814(n); see Suchenek link for a proof. - Marek A. Suchenek, Mar 16 2014

A336962 Right-rotate run-lengths of consecutive equal digits in binary representation of n.

Original entry on oeis.org

0, 1, 2, 3, 6, 5, 4, 7, 14, 11, 10, 13, 12, 9, 8, 15, 30, 23, 22, 27, 26, 21, 20, 29, 28, 19, 18, 25, 24, 17, 16, 31, 62, 47, 46, 55, 54, 45, 44, 59, 58, 43, 42, 53, 52, 41, 40, 61, 60, 39, 38, 51, 50, 37, 36, 57, 56, 35, 34, 49, 48, 33, 32, 63, 126, 95, 94

Offset: 0

Rémy Sigrist, Aug 09 2020

This sequence is a permutation of the nonnegative integers, with inverse A336963.

			The first terms, in decimal and in binary, are:
  n   a(n)  bin(n)  bin(a(n))
  --  ----  ------  ---------
   0     0       0          0
   1     1       1          1
   2     2      10         10
   3     3      11         11
   4     6     100        110
   5     5     101        101
   6     4     110        100
   7     7     111        111
   8    14    1000       1110
   9    11    1001       1011
  10    10    1010       1010
  11    13    1011       1101
  12    12    1100       1100
  13     9    1101       1001
  14     8    1110       1000
  15    15    1111       1111

Rémy Sigrist, Table of n, a(n) for n = 0..8191
Rémy Sigrist, Colored scatterplot of the first 2^16 terms (where the color is function of A136480(n))
Index entries for sequences related to binary expansion of n
Index entries for sequences that are permutations of the natural numbers

Cf. A038572, A101211, A136480, A140690, A336963 (inverse).

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

a(n) = n iff n = 0 or n belongs to A140690.

A361670 Squarefree part of the n-th triangular number.

Original entry on oeis.org

1, 3, 6, 10, 15, 21, 7, 1, 5, 55, 66, 78, 91, 105, 30, 34, 17, 19, 190, 210, 231, 253, 69, 3, 13, 39, 42, 406, 435, 465, 31, 33, 561, 595, 70, 74, 703, 741, 195, 205, 861, 903, 946, 110, 115, 1081, 282, 6, 1, 51, 1326, 1378, 159, 165, 385, 399, 1653, 1711, 1770, 1830, 1891, 217, 14, 130, 2145, 2211, 2278

Offset: 1

R. J. Mathar, Mar 20 2023

a(n) / A083481(n) is either 2 or 1/2 depending on A136480(n) being even or odd, which is indicated by A039963(n).

a(n) = 1 for n>0 in A001108. - Michel Marcus, Mar 22 2023

Cf. A000217, A007913, A083481 (of oblong), A361671 (of tetrahedral).

Cf. A001108, A039963.

a:= n-> mul(i[1]^irem(i[2], 2), i=ifactors(n*(n+1)/2)[2]):
seq(a(n), n=1..70);  # Alois P. Heinz, Mar 20 2023

a(n) = core(n*(n+1)/2); \\ Michel Marcus, Mar 22 2023

from sympy.ntheory.factor_ import core
def A361670(n): return core(n*(n+1)>>1) # Chai Wah Wu, Mar 20 2023

a(n) = A007913(A000217(n)).

A344346 Numbers k which have an odd number of trailing zeros in their binary reflected Gray code A014550(k).

Original entry on oeis.org

3, 4, 11, 12, 15, 16, 19, 20, 27, 28, 35, 36, 43, 44, 47, 48, 51, 52, 59, 60, 63, 64, 67, 68, 75, 76, 79, 80, 83, 84, 91, 92, 99, 100, 107, 108, 111, 112, 115, 116, 123, 124, 131, 132, 139, 140, 143, 144, 147, 148, 155, 156, 163, 164, 171, 172, 175, 176, 179, 180

Offset: 1

Amiram Eldar, May 15 2021

Numbers k such that A050605(k-1) is odd.
Numbers k such that A136480(k) is even.
The asymptotic density of this sequence is 1/3.

			3 is a term since its Gray code, 10, has 1 trailing zero, and 1 is odd.
15 is a term since its Gray code, 1000, has 3 trailing zeros, and 3 is odd.

Amiram Eldar, Table of n, a(n) for n = 1..10000
Eric Weisstein's World of Mathematics, Gray Code.
Wikipedia, Gray code.
Index entries for 2-automatic sequences.

Cf. A014550, A050605, A081706, A136480.

Similar sequences: A001950 (Zeckendorf), A036554 (binary), A145204 (ternary), A217319 (quaternary), A232745 (factorial), A342050 (primorial).

Select[Range[180], OddQ @ IntegerExponent[# * (# + 1)/2, 2] &]

def A344346(n):
    def f(x):
        c, s = (n+1>>1)+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
    m, k = n, f(n)
    while m != k: m, k = k, f(k)
    return (m<<2)-(n&1) # Chai Wah Wu, Jan 29 2025

a(n) = A081706(n) + 1. - Hugo Pfoertner, May 16 2021

A378992 a(n) = A011371(n) - A048881(n); The exponent of the highest power of 2 dividing the n-th factorial minus the exponent of the highest power of 2 dividing n-th Catalan number.

Original entry on oeis.org

0, 0, 0, 1, 2, 2, 2, 4, 6, 6, 6, 7, 8, 8, 8, 11, 14, 14, 14, 15, 16, 16, 16, 18, 20, 20, 20, 21, 22, 22, 22, 26, 30, 30, 30, 31, 32, 32, 32, 34, 36, 36, 36, 37, 38, 38, 38, 41, 44, 44, 44, 45, 46, 46, 46, 48, 50, 50, 50, 51, 52, 52, 52, 57, 62, 62, 62, 63, 64, 64, 64, 66, 68, 68, 68, 69, 70, 70, 70, 73, 76, 76, 76

Offset: 0

Antti Karttunen, Dec 16 2024

Apparently, after the initial three 0's, only terms of A092054 occur, every other as a single copy, and every other in a batch of 3 duplicated terms.

Antti Karttunen, Table of n, a(n) for n = 0..20000
Index entries for sequences related to binary expansion of n

Partial sums of A050605.
Cf. A000108, A000120, A000142, A007814, A011371, A048881.
Cf. also A092054, A204988, A050603, A136480.

A378992[n_] := n - DigitCount[n, 2, 1] - DigitCount[n + 1, 2, 1] + 1;
Array[A378992, 100, 0] (* or *)
MapIndexed[#2[[1]] - # &, Total[Partition[DigitCount[Range[0, 100], 2, 1], 2, 1], {2}]] (* Paolo Xausa, Dec 28 2024 *)

A378992(n) = (1+(n-hammingweight(n)-hammingweight(1+n)));

a(n) = A007814(A000142(n)) - A007814(A000108(n)) = A011371(n) - A048881(n).

a(0) = 0; for n > 0, a(n) = A050605(n-1) + a(n-1), where A050605(n) = A007814(n+1)+A007814(n+2)-1.

cp's OEIS Frontend

A227738 Irregular table read by rows: each row n (n>=1) lists the positions where the runs of bits change between 0's and 1's in the binary expansion of n, when scanning it from the least significant to the most significant end.

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

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A227186 Array A(n,k) read by antidiagonals: the length of the (k+1)-th run (k>=0) of binary digits of n, first run starting from the least significant bit of n.

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A227188 Square array A(n,k) read by antidiagonals: the one-based bit-index where the (k+1)-st run in the binary expansion of n ends, as read from the least significant end.

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A341694 Square array T(n, k) read by antidiagonals upwards, n, k > 0: T(n, k) = A227736(n, k) for k = 1..A005811(n), and T(n, k) = T(n, k - A005811(n)) + ... + T(n, k-1) for k > A005811(n).

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A092054 Base-2 logarithm of the sum of numerator and denominator of the convergents of the continued fraction expansion [1; 1/2, 1/3, 1/4, ..., 1/n, ...].

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A336962 Right-rotate run-lengths of consecutive equal digits in binary representation of n.

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