A145037 -id:A145037 - cp's OEIS frontend

A346633 Sum of even-indexed parts (even bisection) of the n-th composition in standard order.

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

Offset: 0

Gus Wiseman, Aug 01 2021

nonn

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.

			Composition number 741 in standard order is (2,1,1,3,2,1), so a(741) = 1 + 3 + 1 = 5.

Including odd-indexed parts gives A029837.
Subtracting from the odd version gives A124754.
Positions of zeros are A131577.
The odd-indexed version is A209281(n+1).
The version for prime indices is A346698 (reverse: A346700).
A000120 and A080791 count binary digits 1 and 0, with difference A145037.
A011782 counts compositions.
A056239 adds up prime indices, row sums of A112798.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A316524 gives the alternating sum of prime indices (reverse: A344616).
A325534 counts separable partitions, ranked by A335433.
A325535 counts inseparable partitions, ranked by A335448.
A345197 counts compositions by sum, length, and alternating sum.
Cf. A000070, A000097, A008549, A025047, A088218, A344606, A344653, A346697.

stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
Table[Total[Last/@Partition[Append[stc[n],0],2]],{n,0,100}]

a(n) = (A029837(n) - A124754(n))/2.
a(n) = A029837(n) - A209281(n + 1).
a(n) = A124754(n) + A209281(n + 1).

A372472 Number of zeros in the binary expansion of the n-th squarefree number.

Original entry on oeis.org

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

Offset: 1

Gus Wiseman, May 09 2024

			The 12th squarefree number is 17, with binary expansion (1,0,0,0,1), so a(12) = 3.

Positions of first appearances are A372473.
Restriction of A023416 to A005117.
For prime instead of squarefree we have A035103, ones A014499, bits A035100.
Counting 1's instead of 0's (so restrict A000120 to A005117) gives A372433.
For binary length we have A372475, run-lengths A077643.
A030190 gives binary expansion, reversed A030308.
A048793 lists positions of ones in reversed binary expansion, sum A029931.
A371571 lists positions of zeros in binary expansion, sum A359359.
A371572 lists positions of ones in binary expansion, sum A230877.
A372515 lists positions of zeros in reversed binary expansion, sum A359400.
Cf. A003714, A039004, A049093, A049094, A059015, A069010, A070939, A073642, A145037, A211997, A368494, A372474, A372516.

A372583 := proc(n)
    A023416(A005117(n)) ;
end proc:
seq(A372583(n),n=1..200) ; # R. J. Mathar, May 20 2024

DigitCount[Select[Range[100],SquareFreeQ],2,0]

a(n) = A023416(A005117(n)).

a(n) + A372433(n) = A070939(A005117(n)) = A372475(n).

A372541 Least k such that the k-th squarefree number has exactly n ones in its binary expansion.

Original entry on oeis.org

1, 3, 6, 11, 20, 60, 78, 157, 314, 624, 1245, 3736, 4982, 9962, 19920, 39844, 79688, 239046, 318725, 956194, 1912371, 2549834, 5099650, 15298984, 20398664, 40797327, 81594626, 163189197, 326378284, 979135127, 1305513583, 2611027094, 5222054081, 10444108051

Offset: 0

Gus Wiseman, May 09 2024

			The squarefree numbers A005117(a(n)) together with their binary expansions and binary indices begin:
       1:                   1 ~ {1}
       3:                  11 ~ {1,2}
       7:                 111 ~ {1,2,3}
      15:                1111 ~ {1,2,3,4}
      31:               11111 ~ {1,2,3,4,5}
      95:             1011111 ~ {1,2,3,4,5,7}
     127:             1111111 ~ {1,2,3,4,5,6,7}
     255:            11111111 ~ {1,2,3,4,5,6,7,8}
     511:           111111111 ~ {1,2,3,4,5,6,7,8,9}
    1023:          1111111111 ~ {1,2,3,4,5,6,7,8,9,10}
    2047:         11111111111 ~ {1,2,3,4,5,6,7,8,9,10,11}
    6143:       1011111111111 ~ {1,2,3,4,5,6,7,8,9,10,11,13}
    8191:       1111111111111 ~ {1,2,3,4,5,6,7,8,9,10,11,12,13}
   16383:      11111111111111 ~ {1,2,3,4,5,6,7,8,9,10,11,12,13,14}
   32767:     111111111111111 ~ {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15}
   65535:    1111111111111111 ~ {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16}
  131071:   11111111111111111 ~ {1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17}

Chai Wah Wu, Table of n, a(n) for n = 0..56
Wikipedia, Hamming weight.

Positions of firsts appearances in A372433.
Counting zeros instead of ones gives A372473, firsts in A372472.
For prime instead of squarefree we have A372517, firsts of A014499.
Counting bits (length) gives A372540, firsts of A372475, runs A077643.
A000120 counts ones in binary expansion (binary weight), zeros A080791.
A005117 lists squarefree numbers.
A030190 gives binary expansion, reversed A030308.
A048793 lists positions of ones in reversed binary expansion, sum A029931.
A145037, A097110 count ones minus zeros, for primes A372516, A177796.
A371571 lists positions of zeros in binary expansion, sum A359359.
A371572 lists positions of ones in binary expansion, sum A230877.
A372515 lists positions of zeros in reversed binary expansion, sum A359400.
Cf. A023416, A049093, A049094, A069010, A070939, A071403, A211997, A280296, A372474.

nn=10000;
spnm[y_]:=Max@@NestWhile[Most,y,Union[#]!=Range[0,Max@@#]&];
dcs=DigitCount[Select[Range[nn],SquareFreeQ],2,1];
Table[Position[dcs,i][[1,1]],{i,spnm[dcs-1]}]

from math import isqrt
from itertools import count
from sympy import factorint, mobius
from sympy.utilities.iterables import multiset_permutations
def A372541(n):
    if n==0: return 1
    for l in count(n):
        m = 1<Chai Wah Wu, May 10 2024

a(23)-a(33) from Chai Wah Wu, May 10 2024

A372516 Number of ones minus number of zeros in the binary expansion of the n-th prime number.

Original entry on oeis.org

0, 2, 1, 3, 2, 2, -1, 1, 3, 3, 5, 0, 0, 2, 4, 2, 4, 4, -1, 1, -1, 3, 1, 1, -1, 1, 3, 3, 3, 1, 7, -2, -2, 0, 0, 2, 2, 0, 2, 2, 2, 2, 6, -2, 0, 2, 2, 6, 2, 2, 2, 6, 2, 6, -5, -1, -1, 1, -1, -1, 1, -1, 1, 3, 1, 3, 1, -1, 3, 3, -1, 3, 5, 3, 5, 7, -1, 1, -1, 1, 1

Offset: 1

Gus Wiseman, May 13 2024

sign

Absolute value is A177718.

			The binary expansion of 83 is (1,0,1,0,0,1,1), and 83 is the 23rd prime, so a(23) = 4 - 3 = 1.

The sum instead of difference is A035100, firsts A372684 (primes A104080).
The negative version is A037861(A000040(n)).
Restriction of A145037 to the primes.
The unsigned version is A177718.
- Positions of zeros are A177796, indices of the primes A066196.
- Positions of positive terms are indices of the primes A095070.
- Positions of negative terms are indices of the primes A095071.
- Positions of negative ones are A372539, indices of the primes A095072.
- Positions of ones are A372538, indices of the primes A095073.
- Positions of nonnegative terms are indices of the primes A095074.
- Positions of nonpositive terms are indices of the primes A095075.
A000120 counts ones in binary expansion (binary weight), zeros A080791.
A030190 gives binary expansion, reversed A030308.
A035103 counts zeros in binary expansion of primes, firsts A372474.
A048793 lists binary indices, reverse A272020, sum A029931.
A070939 gives length of binary expansion.
A101211 lists run-lengths in binary expansion, row-lengths A069010.
A372471 lists the binary indices of each prime.
Cf. A000043, A003714, A005940, A059305, A061712, A066195, A071814, A211997, A372429, A372517, A372686.

Table[DigitCount[Prime[n],2,1]-DigitCount[Prime[n],2,0],{n,100}]
DigitCount[#,2,1]-DigitCount[#,2,0]&/@Prime[Range[100]] (* Harvey P. Dale, May 09 2025 *)

a(n) = A000120(A000040(n)) - A080791(A000040(n)).
a(n) = A014499(n) - A035103(n).
a(n) = A145037(A000040(n))

A360099 To get A(n,k), replace 0's in the binary expansion of n with (-1) and interpret the result in base k; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

0, 0, 1, 0, 1, -1, 0, 1, 0, 1, 0, 1, 1, 2, -1, 0, 1, 2, 3, -1, 1, 0, 1, 3, 4, 1, 1, -1, 0, 1, 4, 5, 5, 3, 1, 1, 0, 1, 5, 6, 11, 7, 5, 3, -1, 0, 1, 6, 7, 19, 13, 11, 7, -2, 1, 0, 1, 7, 8, 29, 21, 19, 13, 1, 0, -1, 0, 1, 8, 9, 41, 31, 29, 21, 14, 3, 0, 1, 0, 1, 9, 10, 55, 43, 41, 31, 43, 16, 5, 2, -1

Offset: 0

Alois P. Heinz, Jan 25 2023

The empty bit string is used as binary expansion of 0, so A(0,k) = 0.

			Square array A(n,k) begins:
   0,  0, 0,  0,  0,   0,   0,   0,   0,   0,   0, ...
   1,  1, 1,  1,  1,   1,   1,   1,   1,   1,   1, ...
  -1,  0, 1,  2,  3,   4,   5,   6,   7,   8,   9, ...
   1,  2, 3,  4,  5,   6,   7,   8,   9,  10,  11, ...
  -1, -1, 1,  5, 11,  19,  29,  41,  55,  71,  89, ...
   1,  1, 3,  7, 13,  21,  31,  43,  57,  73,  91, ...
  -1,  1, 5, 11, 19,  29,  41,  55,  71,  89, 109, ...
   1,  3, 7, 13, 21,  31,  43,  57,  73,  91, 111, ...
  -1, -2, 1, 14, 43,  94, 173, 286, 439, 638, 889, ...
   1,  0, 3, 16, 45,  96, 175, 288, 441, 640, 891, ...
  -1,  0, 5, 20, 51, 104, 185, 300, 455, 656, 909, ...

Alois P. Heinz, Antidiagonals n = 0..200, flattened

Columns k=0-6, 10 give: A062157, A145037, A006257, A147991, A147992, A153777, A147993, A359925.
Rows n=0-10 give: A000004, A000012, A023443, A000027(k+1), A165900, A002061, A165900(k+1), A002061(k+1), A083074, A152618, A062158.
Main diagonal gives A360096.
Cf. A030300, A057427.

A:= proc(n, k) option remember; local m;
     `if`(n=0, 0, k*A(iquo(n, 2, 'm'), k)+2*m-1)
    end:
seq(seq(A(n, d-n), n=0..d), d=0..12);
# second Maple program:
A:= (n, k)-> (l-> add((2*l[i]-1)*k^(i-1), i=1..nops(l)))(Bits[Split](n)):
seq(seq(A(n, d-n), n=0..d), d=0..12);

G.f. for column k satisfies g_k(x) = k*(x+1)*g_k(x^2) + x/(1+x).
A(n,k) = k*A(floor(n/2),k)+2*(n mod 2)-1 for n>0, A(0,k)=0.
A(n,k) mod 2 = A057427(n) if k is even.
A(n,k) mod 2 = A030300(n) if k is odd and n>=1.
A(2^(n+1),1) + n = 0.

A346702 The a(n)-th composition in standard order is the odd bisection of the n-th composition in standard order.

Original entry on oeis.org

0, 1, 2, 1, 4, 2, 1, 3, 8, 4, 2, 5, 1, 3, 6, 3, 16, 8, 4, 9, 2, 5, 10, 5, 1, 3, 6, 3, 12, 6, 3, 7, 32, 16, 8, 17, 4, 9, 18, 9, 2, 5, 10, 5, 20, 10, 5, 11, 1, 3, 6, 3, 12, 6, 3, 7, 24, 12, 6, 13, 3, 7, 14, 7, 64, 32, 16, 33, 8, 17, 34, 17, 4, 9, 18, 9, 36, 18

Offset: 0

Gus Wiseman, Aug 12 2021

nonn

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.

a(n) is the row number in A066099 of the odd bisection of the n-th row of A066099.

			Composition number 741 in standard order is (2,1,1,3,2,1), with odd bisection (2,1,2), which is composition number 22 in standard order, hence a(741) = 22.

Length of the a(n)-th standard composition is A000120(n)/2 rounded up.
Positions of 1's are A003945.
Positions of 2's (and zero) are A083575.
Sum of the a(n)-th standard composition is A209281(n+1).
Positions of first appearances are A290259.
The version for prime indices is A346703.
The version for even bisection is A346705, with sums A346633.
A000120 and A080791 count binary digits 1 and 0, with difference A145037.
A011782 counts compositions.
A029837 gives length of binary expansion.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A345197 counts compositions by sum, length, and alternating sum.
Cf. A000009, A000302, A000346, A025047, A088218, A124754, A294175, A346697 (reverse: A346699).

Table[Total[2^Accumulate[Reverse[First/@Partition[Append[ Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse,0],2]]]]/2,{n,0,100}]

A029837(a(n)) = A209281(n).

A300562 Condensed deep factorization of n, in binary. (Remove all trailing 0's and one trailing 1 from A300560.)

Original entry on oeis.org

0, 1, 11100, 111, 111100100, 110011100, 111110000, 1111100, 1110011, 1100111100100, 1111100100100, 1111000011100, 1110011100100, 1100111110000, 11100100111100100, 11111, 1111110000100, 11001110011, 1111110010000, 11110000111100100

Offset: 1

M. F. Hasler, Mar 08 2018

nonn

The binary representation of the deep factorization of n, A300560, is obtained by recursively replacing any factor prime(i)^e_i by the expression [i [e_i]], and finally taking '[' and ']' as binary digits 1 and 0.
This always ends in trailing 0's which can be safely removed without loss of information; then there is a final binary digit 1 that can also be dropped. The result is a(n), in binary, equal to A300563(n) when converted to decimal.
The initial a(1) = 0 results from the empty factorization of 1.
To reconstruct the deep factorization of n > 1, append a digit 1 and then as many 0's (namely: A145037(a(n))+1) as to have the same number of as 1's.

			The first term a(1) = 0 represents, by convention, the empty factorization of the number 1.
To reconstruct the full deep factorization A300560(n), append a digit 1 and then as many 0's as to balance the number of 1's:
a(2) = 1, append a 1 => 11; append two 0's => 1100 = A300560(2).
a(3) = 11100, append a 1 => 111001; append two 0's => 11100100 = A300560(3).
a(4) = 111, append a 1 => 1111; append four 0's => 11110000 = A300560(4).

Cf. A300560, A300561, A300563.

Cf. A061396, A062504, A062860.

A300562(n)=(n=eval(A300560(n)))\10^valuation(10*n+!n,10)

a(n) = A007088(A300563(n)), see there for an expression in terms of A300560. - M. F. Hasler, Mar 16 2018

A300563 Condensed deep factorization of n, A300562(n) written in decimal: floor of odd part of A300561(n) divided by 2.

Original entry on oeis.org

0, 1, 28, 7, 484, 412, 496, 124, 115, 6628, 7972, 7708, 7396, 6640, 117220, 31, 8068, 1651, 8080, 123364, 117232, 106276, 7792, 127516, 1939, 105700, 1852, 123376, 118564, 1690084, 129316, 2020, 1875748, 106372, 1985008, 30835, 127204, 106384, 1875172, 2040292, 124708

Offset: 1

M. F. Hasler, Mar 08 2018

nonn

The binary representation of the deep factorization of n, A300560, is obtained by recursively replacing any factor prime(i)^e_i by the expression [i [e_i]], and finally taking '[' and ']' as binary digits 1 and 0.
This always ends in trailing 0's which can be safely removed without loss of information; then there is a final binary digit 1 that can also be dropped. The result is A300562(n) in binary, equal to a(n) when converted to decimal.
The initial a(1) = 0 results from the empty factorization of 1.
To reconstruct the deep factorization of n > 1, take a(n)*2+1, multiply by 2^A145037(a(n)*2+1) (i.e., number of bits = 1 minus number of bits = 0), and write it in binary.

			The first term a(1) = 0 represents, by convention, the empty factorization of the number 1.
The binary-coded deep factorization is restored as follows (and a(n) calculated from this going the opposite direction):
a(2) = 1, append a bit 1 or do 1 X 2 + 1 = 3 = 11[2]. This has 2 bits 1, no bit 0 so append 2 bits 0 => A300560(2) = 1100 in binary, or 12 = A300561(2) in decimal.
a(3) = 28 = 11100[2], append a bit 1 or do 28 X 2 + 1 = 57 = 111001[2]. This has 4 bits 1 and 2 bits 0, so append two more of the latter => A300560(3) = 11100100 in binary or A300561(3) = 228 in decimal.
a(4) = 7 = 111[2], append a bit 1 or do 7 X 2 + 1 = 15 = 1111[2]. This has 4 bits 1 and no bit 0 so append 4 0's => 11110000 = A300560(4) or A300561(4) = 240 in decimal.
See A300560 for conversion of this binary coding of the deep factorization into the ordinary factorization.

Cf. A300560, A300561, A300562.

Cf. A061396, A062504, A062860; A004526, A007814.

A300563(n)=(n=A300561(n))>>(valuation(n+!n,2)+1)

a(n) = A004526(A000265(A300561(n))), where A004526 = floor(./2) and A000265(x) = x/2^A007814(x) is the odd part of x, A007814 is 2-adic valuation. - M. F. Hasler, Mar 16 2018

A301336 a(n) = total number of 1's minus total number of 0's in binary expansions of 0, ..., n.

Original entry on oeis.org

-1, 0, 0, 2, 1, 2, 3, 6, 4, 4, 4, 6, 6, 8, 10, 14, 11, 10, 9, 10, 9, 10, 11, 14, 13, 14, 15, 18, 19, 22, 25, 30, 26, 24, 22, 22, 20, 20, 20, 22, 20, 20, 20, 22, 22, 24, 26, 30, 28, 28, 28, 30, 30, 32, 34, 38, 38, 40, 42, 46, 48, 52, 56, 62, 57, 54, 51, 50, 47, 46, 45, 46, 43, 42, 41, 42

Offset: 0

Ilya Gutkovskiy, Mar 28 2018

			+---+-----+---+---+---+---+------------+
| n | bin.|1's|sum|0's|sum|    a(n)    |
+---+-----+---+---+---+---+------------+
| 0 |   0 | 0 | 0 | 1 | 1 | 0 - 1 =-1  |
| 1 |   1 | 1 | 1 | 0 | 1 | 1 - 1 = 0  |
| 2 |  10 | 1 | 2 | 1 | 2 | 2 - 2 = 0  |
| 3 |  11 | 2 | 4 | 0 | 2 | 4 - 2 = 2  |
| 4 | 100 | 1 | 5 | 2 | 4 | 5 - 4 = 1  |
| 5 | 101 | 2 | 7 | 1 | 5 | 7 - 5 = 2  |
| 6 | 110 | 2 | 9 | 1 | 6 | 9 - 6 = 3  |
+---+-----+---+---+---+---+------------+
bin. - n written in base 2;
1's - number of 1's in binary expansion of n;
0's - number of 0's in binary expansion of n;
sum - total number of 1's (or 0's) in binary expansions of 0, ..., n.

Index entries for sequences related to binary expansion of n

Cf. A000079, A000120, A000295, A000788, A001855, A023416, A037861, A059015, A083652, A145037, A268289, A301896.

a:= proc(n) option remember; `if`(n=0, -1,
      a(n-1)+add(2*i-1, i=Bits[Split](n)))
    end:
seq(a(n), n=0..75);  # Alois P. Heinz, Nov 11 2024

Accumulate[DigitCount[Range[0, 75], 2, 1] - DigitCount[Range[0, 75], 2, 0]]

def A301336(n):
    return sum(2*bin(i).count('1')-len(bin(i))+2 for i in range(n+1)) # Chai Wah Wu, Sep 03 2020

def A301336(n): return (n+1)*((n.bit_count()<<1)-(t:=(n+1).bit_length()))+(1<>j)-(r if n<<1>=m*(r:=k<<1|1) else 0)) for j in range(1,n.bit_length()+1))-2 # Chai Wah Wu, Nov 11 2024

G.f.: -1/(1 - x) + (1/(1 - x)^2)*Sum_{k>=0} x^(2^k)*(1 - x^(2^k))/(1 + x^(2^k)).
a(n) = A000788(n) - A059015(n).
a(n) = A268289(n) - 1.
a(A000079(n)) = A000295(n).

A145057 First differences of A031443.

Original entry on oeis.org

2, 7, 1, 2, 23, 2, 1, 3, 1, 2, 5, 1, 2, 4, 79, 4, 2, 1, 5, 2, 1, 3, 1, 2, 7, 2, 1, 3, 1, 2, 5, 1, 2, 4, 11, 2, 1, 3, 1, 2, 5, 1, 2, 4, 9, 1, 2, 4, 8, 287, 8, 4, 2, 1, 9, 4, 2, 1, 5, 2, 1, 3, 1, 2, 11, 4, 2, 1, 5, 2, 1, 3, 1, 2, 7, 2, 1, 3, 1, 2, 5, 1, 2, 4, 15, 4, 2, 1, 5, 2, 1, 3, 1, 2, 7, 2, 1, 3, 1, 2, 5, 1

Offset: 1

Reikku Kulon, Sep 30 2008

a(n) equals the differences between k where A145037(k) equals zero. - Reikku Kulon, Oct 02 2008

			a(3) = A031443(3) - A031443(3-1) = 10 - 9 = 1. - _David A. Corneth_, Apr 26 2022

David A. Corneth, Table of n, a(n) for n = 1..10000

Cf. A031443, A145037.

Differences[Select[Range[0, 1000], SameQ@@DigitCount[#, 2] &]] (* Paolo Xausa, Nov 16 2024 *)

cp's OEIS Frontend

A346633 Sum of even-indexed parts (even bisection) of the n-th composition in standard order.

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A372472 Number of zeros in the binary expansion of the n-th squarefree number.

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A372541 Least k such that the k-th squarefree number has exactly n ones in its binary expansion.

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A372516 Number of ones minus number of zeros in the binary expansion of the n-th prime number.

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Mathematica

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A360099 To get A(n,k), replace 0's in the binary expansion of n with (-1) and interpret the result in base k; square array A(n,k), n>=0, k>=0, read by antidiagonals.

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Maple

Formula

A346702 The a(n)-th composition in standard order is the odd bisection of the n-th composition in standard order.

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Mathematica

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A300562 Condensed deep factorization of n, in binary. (Remove all trailing 0's and one trailing 1 from A300560.)

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PARI

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A300563 Condensed deep factorization of n, A300562(n) written in decimal: floor of odd part of A300561(n) divided by 2.

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