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 11-18 of 18 results.

A284003 a(n) = A007913(A283477(n)) = A019565(A006068(n)).

Original entry on oeis.org

1, 2, 6, 3, 30, 15, 5, 10, 210, 105, 35, 70, 7, 14, 42, 21, 2310, 1155, 385, 770, 77, 154, 462, 231, 11, 22, 66, 33, 330, 165, 55, 110, 30030, 15015, 5005, 10010, 1001, 2002, 6006, 3003, 143, 286, 858, 429, 4290, 2145, 715, 1430, 13, 26, 78, 39, 390, 195, 65, 130, 2730, 1365, 455, 910, 91, 182, 546, 273, 510510, 255255, 85085, 170170, 17017
Offset: 0

Views

Author

Antti Karttunen, Mar 18 2017

Keywords

Comments

A squarefree analog of A302783. Each term is either a divisor or a multiple of the next one. In contrast to A302033 at each step the previous term can be multiplied (or divided), not just by a single prime, but possibly by a product of several distinct ones, A019565(A000975(k)). E.g., a(3) = 3, a(4) = 2*5*a(3) = 30. - Antti Karttunen, Apr 17 2018

Crossrefs

Programs

Formula

a(n) = A007913(A283477(n)).
Other identities. For all n >= 0:
A048675(a(n)) = A006068(n).
A046523(a(n)) = A284004(n).
It seems that A001222(a(n)) = A209281(n).
a(n) = A019565(A006068(n)) = A302033(A064707(n)). - Antti Karttunen, Apr 16 2018

Extensions

Name amended with a second formula by Antti Karttunen, Apr 16 2018

A342768 a(n) = A342767(n, n).

Original entry on oeis.org

1, 2, 3, 8, 5, 12, 7, 32, 27, 20, 11, 48, 13, 28, 45, 128, 17, 108, 19, 80, 63, 44, 23, 192, 125, 52, 243, 112, 29, 180, 31, 512, 99, 68, 175, 432, 37, 76, 117, 320, 41, 252, 43, 176, 405, 92, 47, 768, 343, 500, 153, 208, 53, 972, 275, 448, 171, 116, 59, 720
Offset: 1

Views

Author

Rémy Sigrist, Apr 02 2021

Keywords

Comments

This sequence has similarities with A087019.
These are the positions of first appearances of each positive integer in A346701, and also in A346703. - Gus Wiseman, Aug 09 2021

Examples

			For n = 42:
- 42 = 2 * 3 * 7, so:
          2 3 7
        x 2 3 7
        -------
          2 3 7
        2 3 3
    + 2 2 2
    -----------
      2 2 3 3 7
- hence a(42) = 2 * 2 * 3 * 3 * 7 = 252.
		

Crossrefs

The sum of prime indices of a(n) is 2*A056239(n) - A061395(n).
The version for even indices is A129597(n) = 2*a(n) for n > 1.
The sorted version is A346635.
These are the positions of first appearances in A346701 and in A346703.
A001221 counts distinct prime factors.
A001222 counts prime factors with multiplicity.
A027193 counts partitions of odd length, ranked by A026424.
A209281 adds up the odd bisection of standard compositions (even: A346633).
A346697 adds up the odd bisection of prime indices (reverse: A346699).

Programs

  • Mathematica
    Table[n^2/FactorInteger[n][[-1,1]],{n,100}] (* Gus Wiseman, Aug 09 2021 *)
  • PARI
    See Links section.

Formula

a(n) = n iff n = 1 or n is a prime number.
a(p^k) = p^(2*k-1) for any k > 0 and any prime number p.
A007947(a(n)) = A007947(n).
A001222(a(n)) = 2*A001222(n) - 1 for any n > 1.
From Gus Wiseman, Aug 09 2021: (Start)
A001221(a(n)) = A001221(n).
If g = A006530(n) is the greatest prime factor of n, then a(n) = n^2/g.
a(n) = A129597(n)/2.
(End)

A346635 Numbers whose division (or multiplication) by their greatest prime factor yields a perfect square. Numbers k such that k*A006530(k) is a perfect square.

Original entry on oeis.org

1, 2, 3, 5, 7, 8, 11, 12, 13, 17, 19, 20, 23, 27, 28, 29, 31, 32, 37, 41, 43, 44, 45, 47, 48, 52, 53, 59, 61, 63, 67, 68, 71, 73, 76, 79, 80, 83, 89, 92, 97, 99, 101, 103, 107, 108, 109, 112, 113, 116, 117, 124, 125, 127, 128, 131, 137, 139, 148, 149, 151, 153
Offset: 1

Views

Author

Gus Wiseman, Aug 10 2021

Keywords

Comments

This is the sorted version of A342768(n) = position of first appearance of n in A346701 (but A346703 works also).

Examples

			The terms together with their prime indices begin:
     1: {}          31: {11}            71: {20}
     2: {1}         32: {1,1,1,1,1}     73: {21}
     3: {2}         37: {12}            76: {1,1,8}
     5: {3}         41: {13}            79: {22}
     7: {4}         43: {14}            80: {1,1,1,1,3}
     8: {1,1,1}     44: {1,1,5}         83: {23}
    11: {5}         45: {2,2,3}         89: {24}
    12: {1,1,2}     47: {15}            92: {1,1,9}
    13: {6}         48: {1,1,1,1,2}     97: {25}
    17: {7}         52: {1,1,6}         99: {2,2,5}
    19: {8}         53: {16}           101: {26}
    20: {1,1,3}     59: {17}           103: {27}
    23: {9}         61: {18}           107: {28}
    27: {2,2,2}     63: {2,2,4}        108: {1,1,2,2,2}
    28: {1,1,4}     67: {19}           109: {29}
    29: {10}        68: {1,1,7}        112: {1,1,1,1,4}
		

Crossrefs

Removing 1 gives a subset of A026424.
The unsorted even version is A129597.
The unsorted version is A342768(n) = A342767(n,n).
Except the first term, the even version is 2*a(n).
A000290 lists squares.
A001221 counts distinct prime factors.
A001222 counts all prime factors.
A006530 gives the greatest prime factor.
A061395 gives the greatest prime index.
A027193 counts partitions of odd length.
A056239 adds up prime indices, row sums of A112798.
A209281 = odd bisection sum of standard compositions (even: A346633).
A316524 = alternating sum of prime indices (sign: A344617, rev.: A344616).
A325534 counts separable partitions, ranked by A335433.
A325535 counts inseparable partitions, ranked by A335448.
A344606 counts alternating permutations of prime indices.
A346697 = odd bisection sum of prime indices (weights of A346703).
A346699 = odd bisection sum of reversed prime indices (weights of A346701).

Programs

  • Maple
    filter:= proc(n) issqr(n/max(numtheory:-factorset(n))) end proc:
    filter(1):= true:
    select(filter, [$1..200]); # Robert Israel, Nov 26 2022
  • Mathematica
    sqrQ[n_]:=IntegerQ[Sqrt[n]];
    Select[Range[100],sqrQ[#*FactorInteger[#][[-1,1]]]&]
  • PARI
    isok(m) = (m==1) || issquare(m/vecmax(factor(m)[,1])); \\ Michel Marcus, Aug 12 2021

Formula

a(n) = A129597(n)/2 for n > 1.

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

Views

Author

Gus Wiseman, Aug 12 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 odd bisection of the n-th row of A066099.

Examples

			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.
		

Crossrefs

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.

Programs

  • Mathematica
    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}]

Formula

A029837(a(n)) = A209281(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

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).

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)

A284004 a(n) = A046523(A284003(n)).

Original entry on oeis.org

1, 2, 6, 2, 30, 6, 2, 6, 210, 30, 6, 30, 2, 6, 30, 6, 2310, 210, 30, 210, 6, 30, 210, 30, 2, 6, 30, 6, 210, 30, 6, 30, 30030, 2310, 210, 2310, 30, 210, 2310, 210, 6, 30, 210, 30, 2310, 210, 30, 210, 2, 6, 30, 6, 210, 30, 6, 30, 2310, 210, 30, 210, 6, 30, 210, 30, 510510, 30030, 2310, 30030, 210, 2310, 30030, 2310, 30, 210, 2310, 210, 30030, 2310, 210, 2310
Offset: 0

Views

Author

Antti Karttunen, Mar 18 2017

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Times @@ MapIndexed[Prime[First@ #2]^#1 &, Reverse@ Sort@ FactorInteger[#][[All, -1]]] - Boole[# == 1] &@ Apply[Times, FactorInteger[#] /. {p_, e_} /; e > 0 :> Times @@ (p^Mod[e, 2])] &[Times @@ Map[#1^#2 & @@ # &, FactorInteger[#] /. {p_, e_} /; e == 1 :> {Times @@ Prime@ Range@ PrimePi@ p, e}] &[Times @@ Prime@ Flatten@ Position[#, 1] &@ Reverse@ IntegerDigits[n, 2]]], {n, 0, 52}] (* Michael De Vlieger, Mar 18 2017 *)
  • PARI
    \\ Code for A284003 given under that entry.
    A046523(n) = my(f=vecsort(factor(n)[, 2], , 4), p); prod(i=1, #f, (p=nextprime(p+1))^f[i]) \\ From Charles R Greathouse IV, Aug 17 2011
    A284004(n) = A046523(A284003(n));
    
  • Scheme
    (define (A284004 n) (A046523 (A284003 n)))

Formula

a(n) = A046523(A284003(n)).
a(n) = A002110(A001222(A284003(n))) = A002110(A209281(n+1)). [Latter so far only conjectured.]

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.

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).

A089215 Thue-Morse sequence on the integers.

Original entry on oeis.org

1, 2, 3, 2, 4, 3, 2, 3, 5, 4, 3, 4, 2, 3, 4, 3, 6, 5, 4, 5, 3, 4, 5, 4, 2, 3, 4, 3, 5, 4, 3, 4, 7, 6, 5, 6, 4, 5, 6, 5, 3, 4, 5, 4, 6, 5, 4, 5, 2, 3, 4, 3, 5, 4, 3, 4, 6, 5, 4, 5, 3, 4, 5, 4, 8, 7, 6, 7, 5, 6, 7, 6, 4, 5, 6, 5, 7, 6, 5, 6, 3, 4, 5, 4, 6, 5, 4, 5, 7, 6, 5, 6, 4, 5, 6, 5, 2, 3, 4, 3, 5, 4, 3, 4, 6
Offset: 1

Views

Author

Benoit Cloitre, Dec 10 2003

Keywords

Comments

Sequence is S(infinity) where S(1)={1,2} and S(n+1)=S(n)S'(n) where S'(n) is obtained from S(n) by substituting an element x of S(n) with M(n)-x where M(n)=2+Max{S(n)}.
For comparison, the Thue-Morse sequence on alphabet {1,2} is constructed as follows: S(1)={1,2} and S(n+1)=S(n)S'(n) where S'(n) is obtained from S(n) by substituting an element x of S(n) with 3-x.

Examples

			S(1)={1,2} then M(1)=4 and S'(1)={4-1,4-2}={3,2}. So S(2)={1,2,3,2}. M(2)=5 so S(3)={1,2,3,2}{5-1,5-2,5-3,5-2} and sequence begins 1,2,3,2,4,3,2,3,...
		

Crossrefs

Programs

  • PARI
    a(n) = n--; my(s=1,h); while((h = n>>s), n=bitxor(n,h); s<<=1); hammingweight(n) + 1; \\ Kevin Ryde, Jun 25 2022

Formula

Sum_{k=1..n} a(k) is asymptotic to C*n*log(n) with C=0.8....
a(n) = A209281(n) + 1. - Kevin Ryde, Jun 25 2022
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