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.

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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

A360071 Regular tetrangle where T(n,k,i) = number of integer partitions of n of length k with i distinct parts.

Original entry on oeis.org

1, 1, 1, 0, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 2, 0, 2, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 2, 1, 1, 1, 0, 2, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 3, 0, 3, 1, 0, 2, 1, 0, 0, 2, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0
Offset: 1

Views

Author

Gus Wiseman, Jan 28 2023

Keywords

Comments

I call this a tetrangle because it is a sequence of finite triangles. - Gus Wiseman, Jan 30 2023

Examples

			Tetrangle begins:
  1   1     1       1         1           1             1
      1 0   0 1     1 1       0 2         1 2           0 3
            1 0 0   0 1 0     0 2 0       1 1 1         0 3 1
                    1 0 0 0   0 1 0 0     0 2 0 0       0 2 1 0
                              1 0 0 0 0   0 1 0 0 0     0 2 0 0 0
                                          1 0 0 0 0 0   0 1 0 0 0 0
                                                        1 0 0 0 0 0 0
For example, finite triangle n = 5 counts the following partitions:
    (5)
     .    (41)(32)
     .   (311)(221)  .
     .     (2111)    .   .
  (11111)     .      .   .   .
		

Crossrefs

Row sums are A008284 (partitions by number of parts), reverse A058398.
First columns i = 1 are A051731.
Last columns i = k are A060016.
Column sums are A116608 (partitions by number of distinct parts).
Positive terms are counted by A360072.
A000041 counts partitions, strict A000009.
Other tetrangles: A318393, A318816, A320808, A334433, A345197.

Programs

  • Mathematica
    Table[Length[Select[IntegerPartitions[n], Length[#]==k&&Length[Union[#]]==i&]],{n,1,9},{k,1,n},{i,1,k}]

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

Original entry on oeis.org

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

Views

Author

Gus Wiseman, Aug 01 2021

Keywords

Comments

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

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

Crossrefs

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.

Programs

  • Mathematica
    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}]

Formula

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

A348614 Numbers k such that the k-th composition in standard order has sum equal to twice its alternating sum.

Original entry on oeis.org

0, 9, 11, 14, 130, 133, 135, 138, 141, 143, 148, 153, 155, 158, 168, 177, 179, 182, 188, 208, 225, 227, 230, 236, 248, 2052, 2057, 2059, 2062, 2066, 2069, 2071, 2074, 2077, 2079, 2084, 2089, 2091, 2094, 2098, 2101, 2103, 2106, 2109, 2111, 2120, 2129, 2131
Offset: 1

Views

Author

Gus Wiseman, Oct 29 2021

Keywords

Comments

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

Examples

			The terms together with their binary indices begin:
    0: ()
    9: (3,1)
   11: (2,1,1)
   14: (1,1,2)
  130: (6,2)
  133: (5,2,1)
  135: (5,1,1,1)
  138: (4,2,2)
  141: (4,1,2,1)
  143: (4,1,1,1,1)
  148: (3,2,3)
  153: (3,1,3,1)
  155: (3,1,2,1,1)
  158: (3,1,1,1,2)
		

Crossrefs

The unordered case (partitions) is counted by A000712, reverse A006330.
These compositions are counted by A262977.
Except for 0, a subset of A345917 (which is itself a subset of A345913).
A000346 = even-length compositions with alt sum != 0, complement A001700.
A011782 counts compositions.
A025047 counts wiggly compositions, ranked by A345167.
A034871 counts compositions of 2n with alternating sum 2k.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A116406 counts compositions with alternating sum >=0, ranked by A345913.
A138364 counts compositions with alternating sum 0, ranked by A344619.
A345197 counts compositions by length and alternating sum.

Programs

  • Mathematica
    ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Select[Range[0,1000],Total[stc[#]]==2*ats[stc[#]]&]

A007179 Dual pairs of integrals arising from reflection coefficients.

Original entry on oeis.org

0, 1, 1, 4, 6, 16, 28, 64, 120, 256, 496, 1024, 2016, 4096, 8128, 16384, 32640, 65536, 130816, 262144, 523776, 1048576, 2096128, 4194304, 8386560, 16777216, 33550336, 67108864, 134209536, 268435456, 536854528, 1073741824, 2147450880, 4294967296, 8589869056
Offset: 0

Views

Author

Keywords

Examples

			From _Gus Wiseman_, Feb 26 2022: (Start)
Also the number of integer compositions of n with at least one odd part. For example, the a(1) = 1 through a(5) = 16 compositions are:
  (1)  (1,1)  (3)      (1,3)      (5)
              (1,2)    (3,1)      (1,4)
              (2,1)    (1,1,2)    (2,3)
              (1,1,1)  (1,2,1)    (3,2)
                       (2,1,1)    (4,1)
                       (1,1,1,1)  (1,1,3)
                                  (1,2,2)
                                  (1,3,1)
                                  (2,1,2)
                                  (2,2,1)
                                  (3,1,1)
                                  (1,1,1,2)
                                  (1,1,2,1)
                                  (1,2,1,1)
                                  (2,1,1,1)
                                  (1,1,1,1,1)
(End)
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Column k=2 of A309748.
Odd bisection is A000302.
Even bisection is A006516 = 2^(n-1)*(2^n - 1).
The complement is counted by A077957, internal version A027383.
The internal case is A274230, even bisection A134057.
A000045(n-1) counts compositions without odd parts, non-singleton A077896.
A003242 counts Carlitz compositions.
A011782 counts compositions.
A034871, A097805, and A345197 count compositions by alternating sum.
A052952 (or A074331) counts non-singleton compositions without even parts.

Programs

  • Magma
    [Floor(2^n/2-2^(n/2)*(1+(-1)^n)/4): n in [0..40]]; // Vincenzo Librandi, Aug 20 2011
    
  • Maple
    f := n-> if n mod 2 = 0 then 2^(n-1)-2^((n-2)/2) else 2^(n-1); fi;
  • Mathematica
    LinearRecurrence[{2,2,-4},{0,1,1},30] (* Harvey P. Dale, Nov 30 2015 *)
    Table[2^(n-1)-If[EvenQ[n],2^(n/2-1),0],{n,0,15}] (* Gus Wiseman, Feb 26 2022 *)
  • PARI
    Vec(x*(1-x)/((1-2*x)*(1-2*x^2)) + O(x^50)) \\ Michel Marcus, Jan 28 2016

Formula

From Paul Barry, Apr 28 2004: (Start)
Binomial transform is (A000244(n)+A001333(n))/2.
G.f.: x*(1-x)/((1-2*x)*(1-2*x^2)).
a(n) = 2*a(n-1)+2*a(n-2)-4*a(n-3).
a(n) = 2^n/2-2^(n/2)*(1+(-1)^n)/4. (End)
G.f.: (1+x*Q(0))*x/(1-x), where Q(k)= 1 - 1/(2^k - 2*x*2^(2*k)/(2*x*2^k - 1/(1 + 1/(2*2^k - 8*x*2^(2*k)/(4*x*2^k + 1/Q(k+1)))))); (continued fraction). - Sergei N. Gladkovskii, May 22 2013
a(n) = A011782(n+2) - A077957(n) - Gus Wiseman, Feb 26 2022

A345960 Numbers whose prime indices have alternating sum 2.

Original entry on oeis.org

3, 12, 27, 30, 48, 70, 75, 108, 120, 147, 154, 192, 243, 270, 280, 286, 300, 363, 432, 442, 480, 507, 588, 616, 630, 646, 675, 750, 768, 867, 874, 972, 1080, 1083, 1120, 1144, 1200, 1323, 1334, 1386, 1452, 1470, 1587, 1728, 1750, 1768, 1798, 1875, 1920, 2028
Offset: 1

Views

Author

Gus Wiseman, Jul 12 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.
The alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. Of course, the alternating sum of prime indices is also the reverse-alternating sum of reversed prime indices.
Also numbers with odd Omega (A001222) and exactly two odd conjugate prime indices. The version for even Omega is A345962, and the union is A345961. Conjugate prime indices are listed by A321650 and ranked by A122111.

Examples

			The initial terms and their prime indices:
    3: {2}
   12: {1,1,2}
   27: {2,2,2}
   30: {1,2,3}
   48: {1,1,1,1,2}
   70: {1,3,4}
   75: {2,3,3}
  108: {1,1,2,2,2}
  120: {1,1,1,2,3}
  147: {2,4,4}
  154: {1,4,5}
  192: {1,1,1,1,1,1,2}
  243: {2,2,2,2,2}
  270: {1,2,2,2,3}
  280: {1,1,1,3,4}
  286: {1,5,6}
  300: {1,1,2,3,3}
		

Crossrefs

These partitions are counted by A000097.
The k = 0 version is A000290, counted by A000041.
The k = 1 version is A001105 (reverse: A345958).
The k > 0 version is A026424.
These multisets are counted by A120452.
These are the positions of 2's in A316524 (reverse: A344616).
The k = -1 version is A345959.
The version for reversed alternating sum is A345961.
The k = -2 version is A345962.
A000984/A345909/A345911 count/rank compositions with alternating sum 1.
A002054/A345924/A345923 count/rank compositions with alternating sum -2.
A056239 adds up prime indices, row sums of A112798.
A088218/A345925/A345922 count/rank compositions with alternating sum 2.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A325534 and A325535 count separable and inseparable partitions.
A344606 counts alternating permutations of prime indices.

Programs

  • Mathematica
    primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
    ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
    Select[Range[0,100],ats[primeMS[#]]==2&]

A349159 Numbers whose sum of prime indices is twice their alternating sum.

Original entry on oeis.org

1, 12, 63, 66, 112, 190, 255, 325, 408, 434, 468, 609, 805, 832, 931, 946, 1160, 1242, 1353, 1380, 1534, 1539, 1900, 2035, 2067, 2208, 2296, 2387, 2414, 2736, 3055, 3108, 3154, 3330, 3417, 3509, 3913, 4185, 4340, 4503, 4646, 4650, 4664, 4864, 5185, 5684, 5863
Offset: 1

Views

Author

Gus Wiseman, Nov 23 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.
The alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i.
The Heinz number of a partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k), so these are also Heinz numbers of partitions whose sum is twice their alternating sum.

Examples

			The terms and their prime indices begin:
     1: ()
    12: (2,1,1)
    63: (4,2,2)
    66: (5,2,1)
   112: (4,1,1,1,1)
   190: (8,3,1)
   255: (7,3,2)
   325: (6,3,3)
   408: (7,2,1,1,1)
   434: (11,4,1)
   468: (6,2,2,1,1)
   609: (10,4,2)
   805: (9,4,3)
   832: (6,1,1,1,1,1,1)
   931: (8,4,4)
   946: (14,5,1)
  1160: (10,3,1,1,1)
		

Crossrefs

These partitions are counted by A000712 up to 0's.
An ordered version is A348614, negative A349154.
The negative version is A348617.
The reverse version is A349160, counted by A006330 up to 0's.
A025047 counts alternating or wiggly compositions, complement A345192.
A027193 counts partitions with rev-alt sum > 0, ranked by A026424.
A034871, A097805, and A345197 count compositions by alternating sum.
A035363 = partitions with alt sum 0, ranked by A066207, complement A086543.
A056239 adds up prime indices, row sums of A112798, row lengths A001222.
A103919 counts partitions by alternating sum, reverse A344612.
A116406 counts compositions with alternating sum >= 0, ranked by A345913.
A138364 counts compositions with alternating sum 0, ranked by A344619.
A325534 counts separable partitions, ranked by A335433.
A325535 counts inseparable partitions, ranked by A335448.
A344607 counts partitions with rev-alt sum >= 0, ranked by A344609.
A346697 adds up odd-indexed prime indices.
A346698 adds up even-indexed prime indices.

Programs

  • Mathematica
    primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
    ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
    Select[Range[1000],Total[primeMS[#]]==2*ats[primeMS[#]]&]

Formula

A056239(a(n)) = 2*A316524(a(n)).
A346697(a(n)) = 3*A346698(a(n)).

A349160 Numbers whose sum of prime indices is twice their reverse-alternating sum.

Original entry on oeis.org

1, 10, 12, 39, 63, 66, 88, 112, 115, 190, 228, 255, 259, 306, 325, 408, 434, 468, 517, 544, 609, 620, 783, 793, 805, 832, 870, 931, 946, 1150, 1160, 1204, 1241, 1242, 1353, 1380, 1392, 1534, 1539, 1656, 1691, 1722, 1845, 1900, 2035, 2067, 2208, 2296, 2369
Offset: 1

Views

Author

Gus Wiseman, Nov 25 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.
The reverse-alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(k-i) y_i.
The Heinz number of a partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k), so these are also Heinz numbers of partitions whose sum is twice their reverse-alternating sum.

Examples

			The terms and their prime indices begin:
     1: ()
    10: (3,1)
    12: (2,1,1)
    39: (6,2)
    63: (4,2,2)
    66: (5,2,1)
    88: (5,1,1,1)
   112: (4,1,1,1,1)
   115: (9,3)
   190: (8,3,1)
   228: (8,2,1,1)
   255: (7,3,2)
   259: (12,4)
   306: (7,2,2,1)
   325: (6,3,3)
   408: (7,2,1,1,1)
   434: (11,4,1)
   468: (6,2,2,1,1)
		

Crossrefs

These partitions are counted by A006330 up to 0's.
The negative reverse version is A348617.
An ordered version is A349153, non-reverse A348614.
The non-reverse version is A349159.
A027193 counts partitions with rev-alt sum > 0, ranked by A026424.
A034871, A097805, A345197 count compositions by alternating sum.
A056239 adds up prime indices, row sums of A112798, row lengths A001222.
A103919 counts partitions by alternating sum, reverse A344612.
A325534 counts separable partitions, ranked by A335433.
A325535 counts inseparable partitions, ranked by A335448.
A346697 adds up odd-indexed prime indices.
A346698 adds up even-indexed prime indices.

Programs

  • Mathematica
    primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
    sats[y_]:=Sum[(-1)^(i-Length[y])*y[[i]],{i,Length[y]}];
    Select[Range[1000],Total[primeMS[#]]==2*sats[primeMS[#]]&]

Formula

A056239(a(n)) = 2*A344616(a(n)).
A346700(a(n)) = 3*A346699(a(n)).
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