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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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.
Cf. A000984, A001700, A028260, A066207, A120452, A195017, A257991, A257992, A344607, A344609, A345958, A349155.

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)).

A129597 Central diagonal of array A129595.

Original entry on oeis.org

1, 4, 6, 16, 10, 24, 14, 64, 54, 40, 22, 96, 26, 56, 90, 256, 34, 216, 38, 160, 126, 88, 46, 384, 250, 104, 486, 224, 58, 360, 62, 1024, 198, 136, 350, 864, 74, 152, 234, 640, 82, 504, 86, 352, 810, 184, 94, 1536, 686, 1000, 306, 416, 106, 1944, 550, 896, 342
Offset: 1

Views

Author

Antti Karttunen, May 01 2007, based on Marc LeBrun's Jan 11 2006 message on SeqFan mailing list

Keywords

Comments

These are the positions of first appearances of each positive integer in A346704. - Gus Wiseman, Oct 16 2021

Links

Crossrefs

a(n) = A129595(n,n).
The sum of prime indices of a(n) is 2*A056239(n) - A061395(n) + 1 for n > 1.
The version for odd indices is A342768(n) = a(n)/2 for n > 1.
Except the first term, the sorted version is 2*A346635.
These are the positions of first appearances in A346704.
A001221 counts distinct prime factors.
A001222 counts prime factors with multiplicity.
A027187 counts partitions of even length, ranked by A028260.
A346633 adds up the even bisection of standard compositions (odd: A209281).
A346698 adds up the even bisection of prime indices (reverse: A346699).
Cf. A000290, A006530, A037143, A329888, A344606, A345957, A346697, A346700, A346701.

Programs

  • Mathematica
    Table[If[n==1,1,2*n^2/FactorInteger[n][[-1,1]]],{n,100}] (* Gus Wiseman, Aug 10 2021 *)
  • PARI
    A129597(n) = if(1==n, n, my(f=factor(n)); (2*n*n)/f[#f~, 1]); \\ Antti Karttunen, Oct 16 2021

Formula

From Gus Wiseman, Aug 10 2021: (Start)
For n > 1, A001221(a(n)) = A099812(n).
If g = A006530(n) is the greatest prime factor of n > 1, then a(n) = 2n^2/g.
a(n) = A100484(A000720(n)) = 2n iff n is prime.
a(n > 1) = 2*A342768(n).
(End)

A346705 The a(n)-th composition in standard order is the 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, 4, 2, 1, 3, 0, 1, 2, 1, 4, 2, 1, 3, 8, 4, 2, 5, 1, 3, 6, 3, 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, 0, 1, 2, 1, 4, 2, 1, 3, 8, 4, 2, 5, 1, 3, 6, 3, 16, 8, 4, 9, 2, 5
Offset: 0

Views

Author

Gus Wiseman, Aug 19 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.
a(n) is the row number in A066099 of the even bisection (even-indexed terms) of the n-th row of A066099.

Examples

			Composition number 741 in standard order is (2,1,1,3,2,1), with even bisection (1,3,1), which is composition number 25 in standard order, so a(741) = 25.

Crossrefs

Length of the a(n)-th standard composition is A000120(n)/2 rounded down.
Positions of first appearances appear to be A088698, sorted: A277335.
The version for reversed prime indices appears to be A329888, sums A346700.
Sum of the a(n)-th standard composition is A346633.
An unordered reverse version for odd bisection is A346701, sums A346699.
The version for odd bisection is A346702, sums A209281(n+1).
An unordered version for odd bisection is A346703, sums A346697.
An unordered version is A346704, sums A346698.
A011782 counts compositions.
A029837 gives length of binary expansion, or sometimes A070939.
A066099 lists compositions in standard order.
A097805 counts compositions by alternating sum.
Cf. A000302, A025047, A088218, A124754, A124767, A228351, A290258, A290259, A344618, A345197.

Programs

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

Formula

A029837(a(n)) = A346633(n).
Previous Showing 11-13 of 13 results.

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