cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Previous Showing 21-30 of 54 results. Next

A345914 Numbers k such that the k-th composition in standard order (row k of A066099) has reverse-alternating sum >= 0.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 7, 8, 10, 11, 12, 13, 14, 15, 16, 19, 20, 21, 22, 24, 26, 27, 28, 30, 31, 32, 35, 36, 37, 38, 40, 41, 42, 43, 44, 46, 47, 48, 50, 51, 52, 53, 54, 55, 56, 58, 59, 60, 61, 62, 63, 64, 67, 69, 70, 72, 73, 74, 76, 79, 80, 82, 83, 84, 86, 87, 88
Offset: 1

Views

Author

Gus Wiseman, Jul 04 2021

Keywords

Comments

The reverse-alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(k-i) y_i.
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.

Examples

			The sequence of terms together with the corresponding compositions begins:
     0: ()           19: (3,1,1)        40: (2,4)
     1: (1)          20: (2,3)          41: (2,3,1)
     2: (2)          21: (2,2,1)        42: (2,2,2)
     3: (1,1)        22: (2,1,2)        43: (2,2,1,1)
     4: (3)          24: (1,4)          44: (2,1,3)
     6: (1,2)        26: (1,2,2)        46: (2,1,1,2)
     7: (1,1,1)      27: (1,2,1,1)      47: (2,1,1,1,1)
     8: (4)          28: (1,1,3)        48: (1,5)
    10: (2,2)        30: (1,1,1,2)      50: (1,3,2)
    11: (2,1,1)      31: (1,1,1,1,1)    51: (1,3,1,1)
    12: (1,3)        32: (6)            52: (1,2,3)
    13: (1,2,1)      35: (4,1,1)        53: (1,2,2,1)
    14: (1,1,2)      36: (3,3)          54: (1,2,1,2)
    15: (1,1,1,1)    37: (3,2,1)        55: (1,2,1,1,1)
    16: (5)          38: (3,1,2)        56: (1,1,4)
		

Crossrefs

The version for prime indices is A000027, counted by A000041.
These compositions are counted by A116406.
The case of non-Heinz numbers of partitions is A119899, counted by A344608.
The version for Heinz numbers of partitions is A344609, counted by A344607.
These are the positions of terms >= 0 in A344618.
The version for unreversed alternating sum is A345913.
The opposite (k <= 0) version is A345916.
The strict (k > 0) case is A345918.
The complement is A345920, counted by A294175.
A011782 counts compositions.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A236913 counts partitions of 2n with reverse-alternating sum <= 0.
A316524 gives the alternating sum of prime indices (reverse: A344616).
A344610 counts partitions by sum and positive reverse-alternating sum.
A344611 counts partitions of 2n with reverse-alternating sum >= 0.
A345197 counts compositions by sum, length, and alternating sum.
Standard compositions: A000120, A066099, A070939, A228351, A124754, A344618.
Compositions of n, 2n, or 2n+1 with alternating/reverse-alternating sum k:
- k = 0: counted by A088218, ranked by A344619/A344619.
- k = 1: counted by A000984, ranked by A345909/A345911.
- k = -1: counted by A001791, ranked by A345910/A345912.
- k = 2: counted by A088218, ranked by A345925/A345922.
- k = -2: counted by A002054, ranked by A345924/A345923.
- k >= 0: counted by A116406, ranked by A345913/A345914.
- k <= 0: counted by A058622(n-1), ranked by A345915/A345916.
- k > 0: counted by A027306, ranked by A345917/A345918.
- k < 0: counted by A294175, ranked by A345919/A345920.
- k != 0: counted by A058622, ranked by A345921/A345921.
- k even: counted by A081294, ranked by A053754/A053754.
- k odd: counted by A000302, ranked by A053738/A053738.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join@@Position[ Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    sats[y_]:=Sum[(-1)^(i-Length[y])*y[[i]],{i,Length[y]}];
    Select[Range[0,100],sats[stc[#]]>=0&]

A345915 Numbers k such that the k-th composition in standard order (row k of A066099) has alternating sum <= 0.

Original entry on oeis.org

0, 3, 6, 10, 12, 13, 15, 20, 24, 25, 27, 30, 36, 40, 41, 43, 46, 48, 49, 50, 51, 53, 54, 55, 58, 60, 61, 63, 72, 80, 81, 83, 86, 92, 96, 97, 98, 99, 101, 102, 103, 106, 108, 109, 111, 116, 120, 121, 123, 126, 136, 144, 145, 147, 150, 156, 160, 161, 162, 163
Offset: 1

Views

Author

Gus Wiseman, Jul 08 2021

Keywords

Comments

The alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i.
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.

Examples

			The sequence of terms together with the corresponding compositions begins:
     0: ()
     3: (1,1)
     6: (1,2)
    10: (2,2)
    12: (1,3)
    13: (1,2,1)
    15: (1,1,1,1)
    20: (2,3)
    24: (1,4)
    25: (1,3,1)
    27: (1,2,1,1)
    30: (1,1,1,2)
    36: (3,3)
    40: (2,4)
    41: (2,3,1)
		

Crossrefs

The version for Heinz numbers of partitions is A028260 (counted by A027187).
These compositions are counted by A058622.
These are the positions of terms <= 0 in A124754.
The reverse-alternating version is A345916.
The opposite (k >= 0) version is A345917.
The strictly negative (k < 0) version is A345919.
A000041 counts partitions of 2n with alternating sum 0, ranked by A000290.
A011782 counts compositions.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A236913 counts partitions of 2n with reverse-alternating sum <= 0.
A316524 gives the alternating sum of prime indices (reverse: A344616).
A345197 counts compositions by sum, length, and alternating sum.
Standard compositions: A000120, A066099, A070939, A228351, A124754, A344618.
Compositions of n, 2n, or 2n+1 with alternating/reverse-alternating sum k:
- k = 0: counted by A088218, ranked by A344619/A344619.
- k = 1: counted by A000984, ranked by A345909/A345911.
- k = -1: counted by A001791, ranked by A345910/A345912.
- k = 2: counted by A088218, ranked by A345925/A345922.
- k = -2: counted by A002054, ranked by A345924/A345923.
- k >= 0: counted by A116406, ranked by A345913/A345914.
- k <= 0: counted by A058622(n-1), ranked by A345915/A345916.
- k > 0: counted by A027306, ranked by A345917/A345918.
- k < 0: counted by A294175, ranked by A345919/A345920.
- k != 0: counted by A058622, ranked by A345921/A345921.
- k even: counted by A081294, ranked by A053754/A053754.
- k odd: counted by A000302, ranked by A053738/A053738.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
    Select[Range[0,100],ats[stc[#]]<=0&]

A345916 Numbers k such that the k-th composition in standard order (row k of A066099) has reverse-alternating sum <= 0.

Original entry on oeis.org

0, 3, 5, 9, 10, 13, 15, 17, 18, 23, 25, 29, 33, 34, 36, 39, 41, 43, 45, 46, 49, 50, 53, 55, 57, 58, 61, 63, 65, 66, 68, 71, 75, 77, 78, 81, 85, 89, 90, 95, 97, 98, 103, 105, 109, 113, 114, 119, 121, 125, 129, 130, 132, 135, 136, 139, 141, 142, 145, 147, 149
Offset: 1

Views

Author

Gus Wiseman, Jul 08 2021

Keywords

Comments

The reverse-alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(k-i) y_i.
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.

Examples

			The sequence of terms together with the corresponding compositions begins:
     0: ()
     3: (1,1)
     5: (2,1)
     9: (3,1)
    10: (2,2)
    13: (1,2,1)
    15: (1,1,1,1)
    17: (4,1)
    18: (3,2)
    23: (2,1,1,1)
    25: (1,3,1)
    29: (1,1,2,1)
    33: (5,1)
    34: (4,2)
    36: (3,3)
		

Crossrefs

The version for Heinz numbers of partitions is A000290.
These compositions are counted by A058622.
These are the positions of terms <= 0 in A344618.
The opposite (k >= 0) version is A345914.
The version for unreversed alternating sum is A345915.
The strictly negative (k < 0) version is A345920.
A011782 counts compositions.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A236913 counts partitions of 2n with reverse-alternating sum <= 0.
A316524 gives the alternating sum of prime indices (reverse: A344616).
A344611 counts partitions of 2n with reverse-alternating sum >= 0.
A345197 counts compositions by sum, length, and alternating sum.
Standard compositions: A000120, A066099, A070939, A228351, A124754, A344618.
Compositions of n, 2n, or 2n+1 with alternating/reverse-alternating sum k:
- k = 0: counted by A088218, ranked by A344619/A344619.
- k = 1: counted by A000984, ranked by A345909/A345911.
- k = -1: counted by A001791, ranked by A345910/A345912.
- k = 2: counted by A088218, ranked by A345925/A345922.
- k = -2: counted by A002054, ranked by A345924/A345923.
- k >= 0: counted by A116406, ranked by A345913/A345914.
- k <= 0: counted by A058622(n-1), ranked by A345915/A345916.
- k > 0: counted by A027306, ranked by A345917/A345918.
- k < 0: counted by A294175, ranked by A345919/A345920.
- k != 0: counted by A058622, ranked by A345921/A345921.
- k even: counted by A081294, ranked by A053754/A053754.
- k odd: counted by A000302, ranked by A053738/A053738.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    sats[y_]:=Sum[(-1)^(i-Length[y])*y[[i]],{i,Length[y]}];
    Select[Range[0,100],sats[stc[#]]<=0&]

A039004 Numbers whose base-4 representation has the same number of 1's and 2's.

Original entry on oeis.org

0, 3, 6, 9, 12, 15, 18, 24, 27, 30, 33, 36, 39, 45, 48, 51, 54, 57, 60, 63, 66, 72, 75, 78, 90, 96, 99, 102, 105, 108, 111, 114, 120, 123, 126, 129, 132, 135, 141, 144, 147, 150, 153, 156, 159, 165, 177, 180, 183, 189, 192, 195, 198, 201, 204, 207, 210, 216, 219
Offset: 1

Views

Author

Keywords

Comments

Numbers such that sum (-1)^k*b(k) = 0 where b(k)=k-th binary digit of n (see A065359). - Benoit Cloitre, Nov 18 2003
Conjecture: a(C(2n,n)-1) = 4^n - 1. (A000984 is C(2n,n)). - Gerald McGarvey, Nov 18 2007
From Russell Jay Hendel, Jun 23 2015: (Start)
We prove the McGarvey conjecture (A) a(e(n,n)-1) = 4^n-1, with e(n,m) = A034870(n,m) = binomial(2n,m), the even rows of Pascal's triangle. By the comment from Hendel in A034870, we have the function s(n,k) = #{n-digit, base-4 numbers with n-k more 1-digits than 2-digits}. As shown in A034870, (B) #s(n,k)= e(n,k) with # indicating cardinality, that is, e(n,k) = binomial(2n,k) gives the number of n-digit, base-4 numbers with n-k more 1-digits than 2-digits.
We now show that (B) implies (A). By definition, s(n,n) contains the e(n,n) = binomial(2n,n) numbers with an equal number of 1-digits and 2-digits. The biggest n-digit, base-4 number is 333...3 (n copies of 3). Since 333...33 has zero 1-digits and zero 2-digits it follows that 333...333 is a member of s(n,n) and hence it is the biggest member of s(n,n). But 333...333 (n copies of 3) in base 4 has value 4^n-1. Since A039004 starts with index 0 (that is, 0 is the 0th member of A039004), it immediately follows that 4^n-1 is the (e(n,n)-1)st member of A039004, proving the McGarvey conjecture. (End)
Also numbers whose alternating sum of binary expansion is 0, i.e., positions of zeros in A345927. These are numbers whose binary expansion has the same number of 1's at even positions as at odd positions. - Gus Wiseman, Jul 28 2021

Crossrefs

A subset of A001969 (evil numbers).
A base-2 version is A031443 (digitally balanced numbers).
Positions of 0's in A065359 and A345927.
Positions of first appearances are A086893.
The version for standard compositions is A344619.
A000120 and A080791 count binary digits, with difference A145037.
A003714 lists numbers with no successive binary indices.
A011782 counts compositions.
A030190 gives the binary expansion of each nonnegative integer.
A070939 gives the length of an integer's binary expansion.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A101211 lists run-lengths in binary expansion:
- row-lengths: A069010
- reverse: A227736
- ones only: A245563
A138364 counts compositions with alternating sum 0:
- bisection: A001700/A088218
- complement: A058622
A328594 lists numbers whose binary expansion is aperiodic.
A345197 counts compositions by length and alternating sum.

Programs

  • Fortran
    c See link in A139351.
  • Maple
    N:= 1000: # to get all terms up to N, which should be divisible by 4
    B:= Array(0..N-1):
    d:= ceil(log[4](N));
    S:= Array(0..N-1,[seq(op([0,1,-1,0]),i=1..N/4)]):
    for i from 1 to d do
      B:= B + S;
      S:= Array(0..N-1,i-> S[floor(i/4)]);
    od:
    select(t -> B[t]=0, [$0..N-1]); # Robert Israel, Jun 24 2015
  • Mathematica
    ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
    Select[Range[0,100],ats[IntegerDigits[#,2]]==0&] (* Gus Wiseman, Jul 28 2021 *)
  • PARI
    for(n=0,219,if(sum(i=1,length(binary(n)),(-1)^i*component(binary(n),i))==0,print1(n,",")))
    

Formula

Conjecture: there is a constant c around 5 such that a(n) is asymptotic to c*n. - Benoit Cloitre, Nov 24 2002
That conjecture is false. The number of members of the sequence from 0 to 4^d-1 is binomial(2d,d) which by Stirling's formula is asymptotic to 4^d/sqrt(Pi*d). If Cloitre's conjecture were true we would have 4^d-1 asymptotic to c*4^d/sqrt(Pi*d), a contradiction. - Robert Israel, Jun 24 2015

A346697 Sum of the odd-indexed parts (odd bisection) of the multiset of prime indices of n.

Original entry on oeis.org

0, 1, 2, 1, 3, 1, 4, 2, 2, 1, 5, 3, 6, 1, 2, 2, 7, 3, 8, 4, 2, 1, 9, 2, 3, 1, 4, 5, 10, 4, 11, 3, 2, 1, 3, 3, 12, 1, 2, 2, 13, 5, 14, 6, 5, 1, 15, 4, 4, 4, 2, 7, 16, 3, 3, 2, 2, 1, 17, 3, 18, 1, 6, 3, 3, 6, 19, 8, 2, 5, 20, 4, 21, 1, 5, 9, 4, 7, 22, 5, 4, 1
Offset: 1

Views

Author

Gus Wiseman, Aug 01 2021

Keywords

Comments

A prime index of n is a number m such that prime(m) divides n. The multiset of prime indices of n is row n of A112798.

Examples

			The prime indices of 1100 are {1,1,3,3,5}, so a(1100) = 1 + 3 + 5 = 9.
The prime indices of 2100 are {1,1,2,3,3,4}, so a(2100) = 1 + 2 + 3 = 6.
		

Crossrefs

The version for standard compositions is A209281(n+1) (even: A346633).
Subtracting the even version gives A316524 (reverse: A344616).
The even version is A346698.
The reverse version is A346699.
The even reverse version is A346700.
A000120 and A080791 count binary digits 1 and 0, with difference A145037.
A000302 counts compositions with odd alternating sum, ranked by A053738.
A001414 adds up prime factors, row sums of A027746.
A029837 adds up parts of standard compositions (alternating: A124754).
A056239 adds up prime indices, row sums of A112798.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A325534 counts separable partitions, ranked by A335433.
A325535 counts inseparable partitions, ranked by A335448.
A344606 counts alternating permutations of prime indices.

Programs

  • Mathematica
    primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
    Table[Total[First/@Partition[Append[primeMS[n],0],2]],{n,100}]

Formula

a(n) = A056239(n) - A346698(n).
a(n) = A316524(n) + A346698(n).
a(n odd omega) = A346699(n).
a(n even omega) = A346700(n).
A344616(n) = A346699(n) - A346700(n).

A060142 Ordered set S defined by these rules: 0 is in S and if x is in S then 2x+1 and 4x are in S.

Original entry on oeis.org

0, 1, 3, 4, 7, 9, 12, 15, 16, 19, 25, 28, 31, 33, 36, 39, 48, 51, 57, 60, 63, 64, 67, 73, 76, 79, 97, 100, 103, 112, 115, 121, 124, 127, 129, 132, 135, 144, 147, 153, 156, 159, 192, 195, 201, 204, 207, 225, 228, 231, 240, 243, 249, 252, 255, 256, 259, 265, 268, 271
Offset: 0

Views

Author

Clark Kimberling, Mar 05 2001

Keywords

Comments

After expelling 0 and 1, the numbers 4x occupy same positions in S that 1 occupies in the infinite Fibonacci word (A003849).
a(A026351(n)) = A219608(n); a(A004957(n)) = 4 * a(n). - Reinhard Zumkeller, Nov 26 2012
Apart from the initial term, this lists the indices of the 1's in A086747. - N. J. A. Sloane, Dec 05 2019
From Gus Wiseman, Jun 10 2020: (Start)
Numbers k such that the k-th composition in standard order has all odd parts, or numbers k such that A124758(k) is odd. 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. For example, the sequence of all compositions into odd parts begins:
0: () 57: (1,1,3,1) 135: (5,1,1,1)
1: (1) 60: (1,1,1,3) 144: (3,5)
3: (1,1) 63: (1,1,1,1,1,1) 147: (3,3,1,1)
4: (3) 64: (7) 153: (3,1,3,1)
7: (1,1,1) 67: (5,1,1) 156: (3,1,1,3)
9: (3,1) 73: (3,3,1) 159: (3,1,1,1,1,1)
12: (1,3) 76: (3,1,3) 192: (1,7)
15: (1,1,1,1) 79: (3,1,1,1,1) 195: (1,5,1,1)
16: (5) 97: (1,5,1) 201: (1,3,3,1)
19: (3,1,1) 100: (1,3,3) 204: (1,3,1,3)
25: (1,3,1) 103: (1,3,1,1,1) 207: (1,3,1,1,1,1)
28: (1,1,3) 112: (1,1,5) 225: (1,1,5,1)
31: (1,1,1,1,1) 115: (1,1,3,1,1) 228: (1,1,3,3)
33: (5,1) 121: (1,1,1,3,1) 231: (1,1,3,1,1,1)
36: (3,3) 124: (1,1,1,1,3) 240: (1,1,1,5)
39: (3,1,1,1) 127: (1,1,1,1,1,1,1) 243: (1,1,1,3,1,1)
48: (1,5) 129: (7,1) 249: (1,1,1,1,3,1)
51: (1,3,1,1) 132: (5,3) 252: (1,1,1,1,1,3)
(End)
Numbers whose binary representation has the property that every run of consecutive 0's has even length. - Harry Richman, Jan 31 2024

Examples

			From _Harry Richman_, Jan 31 2024: (Start)
In the following, dots are used for zeros in the binary representation:
   n  binary(a(n))  a(n)
   0:    .......     0
   1:    ......1     1
   2:    .....11     3
   3:    ....1..     4
   4:    ....111     7
   5:    ...1..1     9
   6:    ...11..    12
   7:    ...1111    15
   8:    ..1....    16
   9:    ..1..11    19
  10:    ..11..1    25
  11:    ..111..    28
  12:    ..11111    31
  13:    .1....1    33
  14:    .1..1..    36
  15:    .1..111    39
  16:    .11....    48
  17:    .11..11    51
  18:    .111..1    57
  19:    .1111..    60
  20:    .111111    63
  21:    1......    64
  22:    1....11    67
(End)
		

Crossrefs

Cf. A003714 (no consecutive 1's in binary expansion).
Odd partitions are counted by A000009.
Numbers with an odd number of 1's in binary expansion are A000069.
Numbers whose binary expansion has odd length are A053738.
All of the following pertain to compositions in standard order (A066099):
- Length is A000120.
- Compositions without odd parts are A062880.
- Sum is A070939.
- Product is A124758.
- Strict compositions are A233564.
- Heinz number is A333219.
- Number of distinct parts is A334028.

Programs

  • Haskell
    import Data.Set (singleton, deleteFindMin, insert)
    a060142 n = a060142_list !! n
    a060142_list = 0 : f (singleton 1) where
       f s = x : f (insert (4 * x) $ insert (2 * x + 1) s') where
           (x, s') = deleteFindMin s
    -- Reinhard Zumkeller, Nov 26 2012
    
  • Mathematica
    Take[Nest[Union[Flatten[# /. {{i_Integer -> i}, {i_Integer -> 2 i + 1}, {i_Integer -> 4 i}}]] &, {1}, 5], 32]  (* Or *)
    Select[Range[124], FreeQ[Length /@ Select[Split[IntegerDigits[#, 2]], First[#] == 0 &], ?OddQ] &] (* _Birkas Gyorgy, May 29 2012 *)
  • PARI
    is(n)=if(n<3, n<2, if(n%2,is(n\2),n%4==0 && is(n/4))) \\ Charles R Greathouse IV, Oct 21 2013

Extensions

Corrected by T. D. Noe, Nov 01 2006
Definition simplified by Charles R Greathouse IV, Oct 21 2013

A030300 Runs have lengths 2^n, n >= 0.

Original entry on oeis.org

1, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 1

Views

Author

Keywords

Comments

An example of a sequence with property that the fraction of 1's in the first n terms does not converge to a limit. - N. J. A. Sloane, Sep 24 2007
Image, under the coding sending a,d,e -> 1 and b,c -> 0, of the fixed point, starting with a, of the morphism a -> ab, b -> cd, c -> ee, d -> eb, e -> cc. - Jeffrey Shallit, May 14 2016
This sequence taken as digits of a base-b fraction is g(1/b) = Sum_{n>=1} a(n)/b^n = b/(b-1) * Sum_{k>=0} (-1)^k/b^(2^k) per the generating function below. With initial 0, it is binary expansion .01001111 = A275975. With initial 0 and digits 2*a(n), it is ternary expansion .02002222 = A160386. These and in general g(1/b) for any integer b>=2 are among forms which Kempner showed are transcendental. - Kevin Ryde, Sep 07 2019

Crossrefs

Cf. A030301. Partial sums give A079947.
Characteristic function of A053738.

Programs

  • Maple
    f0 := n->[seq(0,i=1..2^n)]; f1 := n->[seq(1,i=1..2^n)]; s := []; for i from 0 to 4 do s := [op(s), op(f1(2*i)), op(f0(2*i+1))]; od: A030300 := s;
  • Mathematica
    nMax = 6; Table[1 - Mod[n, 2], {n, 0, nMax}, {2^n}] // Flatten (* Jean-François Alcover, Oct 20 2016 *)
  • PARI
    a(n) = if(n, !(logint(n,2)%2)); /* Kevin Ryde, Aug 02 2019 */
    
  • Python
    def A030300(n): return n.bit_length()&1 # Chai Wah Wu, Jan 30 2023

Formula

a(n) = A065359(n) + A083905(n).
a(n) = (1/2)*(1+(-1)^floor(log_2(n))). - Benoit Cloitre, Feb 22 2003
G.f.: 1/(1-x) * Sum_{k>=0} (-1)^k*x^2^k. - Ralf Stephan, Jul 12 2003
a(n) = 1 - a(floor(n/2)). - Vladeta Jovovic, Aug 04 2003
a(n) = A115253(2n, n) mod 2. - Paul Barry, Jan 18 2006
a(n) = 1 - A030301(n). - Antti Karttunen, Oct 10 2017

A350951 Number of odd parts minus number of odd conjugate parts in the integer partition with Heinz number n.

Original entry on oeis.org

0, 0, -2, 2, -2, 0, -4, 2, 0, 0, -4, 0, -6, -2, 0, 4, -6, 0, -8, 0, -2, -2, -8, 2, 2, -4, -2, -2, -10, 0, -10, 4, -2, -4, 0, 2, -12, -6, -4, 2, -12, -2, -14, -2, -2, -6, -14, 2, 0, 2, -4, -4, -16, 0, 0, 0, -6, -8, -16, 2, -18, -8, -4, 6, -2, -2, -18, -4, -6, 0
Offset: 1

Views

Author

Gus Wiseman, Mar 14 2022

Keywords

Comments

The Heinz number of a partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k). This gives a bijective correspondence between positive integers and integer partitions.
All terms are even.

Examples

			The prime indices of 78 are (6,2,1), with conjugate (3,2,1,1,1,1), so a(78) = 1 - 5 = -4.
		

Crossrefs

The version comparing even with odd parts is A195017.
The version comparing even with odd conjugate parts is A350849.
The version comparing even conjugate with odd conjugate parts is A350941.
The version comparing odd with even conjugate parts is A350942.
Positions of 0's are A350944, even rank case A345196, counted by A277103.
The version comparing even with even conjugate parts is A350950.
There are four individual statistics:
- A257991 counts odd parts, conjugate A344616.
- A257992 counts even parts, conjugate A350847.
There are five other possible pairings of statistics:
- A325698: # of even parts = # of odd parts, counted by A045931.
- A349157: # of even parts = # of odd conjugate parts, counted by A277579.
- A350848: # of even conj parts = # of odd conj parts, counted by A045931.
- A350943: # of even conjugate parts = # of odd parts, counted by A277579.
- A350945: # of even parts = # of even conjugate parts, counted by A350948.
There are three possible double-pairings of statistics:
- A350946, counted by A351977.
- A350949, counted by A351976.
- A351980, counted by A351981.
The case of all four statistics equal is A350947, counted by A351978.
A056239 adds up prime indices, counted by A001222, row sums of A112798.
A103919 counts partitions by number of odd parts.
A116482 counts partitions by number of even parts.
A122111 represents partition conjugation using Heinz numbers.
A316524 gives the alternating sum of prime indices.

Programs

  • Mathematica
    primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
    conj[y_]:=If[Length[y]==0,y,Table[Length[Select[y,#>=k&]],{k,1,Max[y]}]];
    Table[Count[primeMS[n],?OddQ]-Count[conj[primeMS[n]],?OddQ],{n,100}]

Formula

a(n) = A257991 - A344616(n).
a(A122111(n)) = -a(n), where A122111 represents partition conjugation.

A346703 Product of primes at odd positions in the weakly increasing list (with multiplicity) of prime factors of n.

Original entry on oeis.org

1, 2, 3, 2, 5, 2, 7, 4, 3, 2, 11, 6, 13, 2, 3, 4, 17, 6, 19, 10, 3, 2, 23, 4, 5, 2, 9, 14, 29, 10, 31, 8, 3, 2, 5, 6, 37, 2, 3, 4, 41, 14, 43, 22, 15, 2, 47, 12, 7, 10, 3, 26, 53, 6, 5, 4, 3, 2, 59, 6, 61, 2, 21, 8, 5, 22, 67, 34, 3, 14, 71, 12, 73, 2, 15, 38
Offset: 1

Views

Author

Gus Wiseman, Aug 08 2021

Keywords

Examples

			The prime factors of 108 are (2,2,3,3,3), with odd bisection (2,3,3), with product 18, so a(108) = 18.
The prime factors of 720 are (2,2,2,2,3,3,5), with odd bisection (2,2,3,5), with product 60, so a(720) = 60.
		

Crossrefs

Positions of 2's are A001747.
Positions of primes are A037143 (complement: A033942).
The even reverse version appears to be A329888.
Positions of first appearances are A342768.
The sum of prime indices of a(n) is A346697(n), reverse: A346699.
The reverse version is A346701.
The even version is A346704.
A001221 counts distinct prime factors.
A001222 counts all prime factors.
A056239 adds up prime indices, row sums of A112798.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A209281 (shifted) adds up the odd bisection of standard compositions.
A316524 gives the alternating sum of prime indices (reverse: A344616).
A335433/A335448 rank separable/inseparable partitions.
A344606 counts alternating permutations of prime indices.
A344617 gives the sign of the alternating sum of prime indices.
A346633 adds up the even bisection of standard compositions.
A346698 gives the sum of the even bisection of prime indices.
A346700 gives the sum of the even bisection of reversed prime indices.

Programs

  • Mathematica
    Table[Times@@First/@Partition[Append[Flatten[Apply[ConstantArray,FactorInteger[n],{1}]],0],2],{n,100}]

Formula

a(n) * A346704(n) = n.
A056239(a(n)) = A346697(n).

A345908 Traces of the matrices (A345197) counting integer compositions by length and alternating sum.

Original entry on oeis.org

1, 1, 0, 1, 3, 3, 6, 15, 24, 43, 92, 171, 315, 629, 1218, 2313, 4523, 8835, 17076, 33299, 65169
Offset: 0

Views

Author

Gus Wiseman, Jul 26 2021

Keywords

Comments

The matrices (A345197) count the integer compositions of n of length k with alternating sum i, where 1 <= k <= n, and i ranges from -n + 2 to n in steps of 2. Here, the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. So a(n) is the number of compositions of n of length (n + s)/2, where s is the alternating sum of the composition.

Examples

			The a(0) = 1 through a(7) = 15 compositions of n = 0..7 of length (n + s)/2 where s = alternating sum (empty column indicated by dot):
  ()  (1)  .  (2,1)  (2,2)    (2,3)    (2,4)      (2,5)
                     (1,1,2)  (1,2,2)  (1,3,2)    (1,4,2)
                     (2,1,1)  (2,2,1)  (2,3,1)    (2,4,1)
                                       (1,1,3,1)  (1,1,3,2)
                                       (2,1,2,1)  (1,2,3,1)
                                       (3,1,1,1)  (2,1,2,2)
                                                  (2,2,2,1)
                                                  (3,1,1,2)
                                                  (3,2,1,1)
                                                  (1,1,1,1,3)
                                                  (1,1,2,1,2)
                                                  (1,1,3,1,1)
                                                  (2,1,1,1,2)
                                                  (2,1,2,1,1)
                                                  (3,1,1,1,1)
		

Crossrefs

Traces of the matrices given by A345197.
Diagonals and antidiagonals of the same matrices are A346632 and A345907.
Row sums of A346632.
A011782 counts compositions.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A316524 gives the alternating sum of prime indices (reverse: A344616).
Other diagonals are A008277 of A318393 and A055884 of A320808.
Compositions of n, 2n, or 2n+1 with alternating/reverse-alternating sum k:
- k = 0: counted by A088218, ranked by A344619/A344619.
- k = 1: counted by A000984, ranked by A345909/A345911.
- k = -1: counted by A001791, ranked by A345910/A345912.
- k = 2: counted by A088218, ranked by A345925/A345922.
- k = -2: counted by A002054, ranked by A345924/A345923.
- k >= 0: counted by A116406, ranked by A345913/A345914.
- k <= 0: counted by A058622(n-1), ranked by A345915/A345916.
- k > 0: counted by A027306, ranked by A345917/A345918.
- k < 0: counted by A294175, ranked by A345919/A345920.
- k != 0: counted by A058622, ranked by A345921/A345921.
- k even: counted by A081294, ranked by A053754/A053754.
- k odd: counted by A000302, ranked by A053738/A053738.

Programs

  • Mathematica
    ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
    Table[Length[Select[Join@@Permutations/@IntegerPartitions[n],Length[#]==(n+ats[#])/2&]],{n,0,15}]
Previous Showing 21-30 of 54 results. Next