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.

Showing 1-10 of 11 results. Next

A296122 Number of twice-partitions of n with no repeated partitions.

Original entry on oeis.org

1, 1, 2, 5, 10, 20, 40, 77, 157, 285, 552, 1018, 1921, 3484, 6436, 11622, 21082, 37550, 67681, 119318, 211792, 372003, 653496, 1137185, 1986234, 3429650, 5935970, 10205907, 17537684, 29958671, 51189932, 86967755, 147759421, 249850696, 422123392, 710495901
Offset: 0

Views

Author

Gus Wiseman, Dec 05 2017

Keywords

Comments

a(n) is the number of sequences of distinct integer partitions whose sums are weakly decreasing and add up to n.

Examples

			The a(4) = 10 twice-partitions: (4), (31), (22), (211), (1111), (3)(1), (21)(1), (111)(1), (2)(11), (11)(2).
		

Crossrefs

Programs

  • Maple
    b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0, add(j!*
          binomial(combinat[numbpart](i), j)*b(n-i*j, i-1), j=0..n/i)))
        end:
    a:= n-> b(n$2):
    seq(a(n), n=0..40);  # Alois P. Heinz, Dec 06 2017
  • Mathematica
    Table[Length[Join@@Table[Select[Tuples[IntegerPartitions/@p],UnsameQ@@#&],{p,IntegerPartitions[n]}]],{n,15}]
    (* Second program: *)
    b[n_, i_] := b[n, i] = If[n == 0, 1, If[i < 1, 0, Sum[j!*
         Binomial[PartitionsP[i], j]*b[n - i*j, i - 1], {j, 0, n/i}]]];
    a[n_] := b[n, n];
    a /@ Range[0, 40] (* Jean-François Alcover, May 19 2021, after Alois P. Heinz *)

Extensions

a(15)-a(34) from Robert G. Wilson v, Dec 06 2017

A277130 Number of planar branching factorizations of n.

Original entry on oeis.org

0, 1, 1, 2, 1, 3, 1, 6, 2, 3, 1, 14, 1, 3, 3, 24, 1, 14, 1, 14, 3, 3, 1, 78, 2, 3, 6, 14, 1, 25, 1, 112, 3, 3, 3, 110, 1, 3, 3, 78, 1, 25, 1, 14, 14, 3, 1, 464, 2, 14, 3, 14, 1, 78, 3, 78, 3, 3, 1, 206, 1, 3, 14, 568, 3, 25, 1, 14, 3, 25, 1, 850, 1, 3, 14, 14
Offset: 1

Views

Author

Michel Marcus, Oct 01 2016

Keywords

Comments

A planar branching factorization of n is either the number n itself or a sequence of at least two planar branching factorizations, one of each factor in an ordered factorization of n. - Gus Wiseman, Sep 11 2018

Examples

			From _Gus Wiseman_, Sep 11 2018: (Start)
The a(12) = 14 planar branching factorizations:
  12,
  (2*6), (3*4), (4*3), (6*2), (2*2*3), (2*3*2), (3*2*2),
  (2*(2*3)), (2*(3*2)), (3*(2*2)), ((2*2)*3), ((2*3)*2), ((3*2)*2).
(End)
		

Crossrefs

Programs

  • C
    #include 
    #include 
    #include 
    #define MAX 10000
    /* Number of planar branching factorizations of n. */
    #define lu unsigned long
    lu nbr[MAX]; /* number of branching */
    lu a, b, d, e; /* temporary variables */
    lu n; lu m, p; // factors of n
    lu x; // square root of n
    void main(unsigned argc, char *argv[])
    {
      memset(nbr, 0, MAX*sizeof(lu));
      for (b=0, n=1; nDaniel Mondot, May 19 2017 */
  • Mathematica
    ordfacs[n_]:=If[n<=1,{{}},Join@@Table[(Prepend[#1,d]&)/@ordfacs[n/d],{d,Rest[Divisors[n]]}]]
    otfs[n_]:=Prepend[Join@@Table[Tuples[otfs/@f],{f,Select[ordfacs[n],Length[#]>1&]}],n];
    Table[Length[otfs[n]],{n,20}] (* Gus Wiseman, Sep 11 2018 *)

Formula

a(prime^n) = A118376(n). a(product of n distinct primes) = A319122(n). - Gus Wiseman, Sep 11 2018

Extensions

Terms a(65) and beyond from Daniel Mondot, May 19 2017

A295281 Number of complete strict tree-factorizations of n > 1.

Original entry on oeis.org

1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 1, 1, 4, 1, 0, 1, 1, 1, 3, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 0, 1, 4, 1, 1, 1, 4, 1, 6, 1, 1, 1, 1, 1, 4, 1, 1, 0, 1, 1, 9, 1, 1, 1, 1, 1, 9, 1
Offset: 2

Views

Author

Gus Wiseman, Nov 19 2017

Keywords

Comments

A strict tree-factorization (see A295279 for definition) is complete if its leaves are all prime numbers.
From Andrew Howroyd, Nov 18 2018: (Start)
a(n) depends only on the prime signature of n.
This sequence is very similar but not identical to the number of complete orderless identity tree-factorizations of n. The first difference is at n=900 (square of three primes). Here a(n) = 191 whereas the other sequence would have 197. (End)

Examples

			The a(72) = 6 complete strict tree-factorizations are: 2*3*(2*(2*3)), 2*(2*3*(2*3)), 2*(2*(3*(2*3))), 2*(3*(2*(2*3))), 3*(2*(2*(2*3))), (2*3)*(2*(2*3)).
		

Crossrefs

Programs

  • Mathematica
    postfacs[n_]:=If[n<=1,{{}},Join@@Table[Map[Prepend[#,d]&,Select[postfacs[n/d],Min@@#>=d&]],{d,Rest[Divisors[n]]}]];
    sftc[n_]:=Prepend[Join@@Function[fac,Tuples[sftc/@fac]]/@Select[postfacs[n],And[Length[#]>1,UnsameQ@@#]&],n];
    Table[Length[Select[sftc[n],FreeQ[#,_Integer?(!PrimeQ[#]&)]&]],{n,2,100}]
  • PARI
    seq(n)={my(v=vector(n), w=vector(n)); v[1]=1; for(k=2, n, w[k]=v[k]+isprime(k); forstep(j=n\k*k, k, -k, v[j]+=w[k]*v[j/k])); w[2..n]} \\ Andrew Howroyd, Nov 18 2018

Formula

a(product of n distinct primes) = A000311(n).
Positions of zeros are proper prime powers A025475. Positions of nonzero entries are A085971.

A295632 Write 1/Product_{n > 1}(1 - 1/n^s) in the form Product_{n > 1}(1 + a(n)/n^s).

Original entry on oeis.org

1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1
Offset: 2

Views

Author

Gus Wiseman, Nov 24 2017

Keywords

Comments

First negative entry is a(1024) = -4.

Crossrefs

Programs

  • Mathematica
    nn=100;
    facs[n_]:=If[n<=1,{{}},Join@@Table[Map[Prepend[#,d]&,Select[facs[n/d],Min@@#>=d&]],{d,Rest[Divisors[n]]}]];
    Solve[Table[Length[facs[n]]==Sum[Times@@a/@f,{f,Select[facs[n],UnsameQ@@#&]}],{n,2,nn}],Table[a[n],{n,2,nn}]][[1,All,2]]

A319118 Number of multimin tree-factorizations of n.

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 1, 6, 2, 2, 1, 8, 1, 2, 2, 24, 1, 6, 1, 8, 2, 2, 1, 42, 2, 2, 6, 8, 1, 8, 1, 112, 2, 2, 2, 38, 1, 2, 2, 42, 1, 8, 1, 8, 8, 2, 1, 244, 2, 6, 2, 8, 1, 24, 2, 42, 2, 2, 1, 58, 1, 2, 8, 568, 2, 8, 1, 8, 2, 8, 1, 268, 1, 2, 6, 8, 2, 8, 1, 244, 24
Offset: 1

Views

Author

Gus Wiseman, Sep 10 2018

Keywords

Comments

A multimin factorization of n is an ordered factorization of n into factors greater than 1 such that the sequence of minimal primes dividing each factor is weakly increasing. A multimin tree-factorization of n is either the number n itself or a sequence of multimin tree-factorizations, one of each factor in a multimin factorization of n with at least two factors.

Examples

			The a(12) = 8 multimin tree-factorizations:
  12,
  (2*6), (4*3), (6*2), (2*2*3),
  (2*(2*3)), ((2*2)*3), ((2*3)*2).
Or as series-reduced plane trees of multisets:
  112,
  (1,12), (11,2), (12,1), (1,1,2),
  (1,(1,2)), ((1,1),2), ((1,2),1).
		

Crossrefs

Programs

  • Mathematica
    facs[n_]:=If[n<=1,{{}},Join@@Table[(Prepend[#1,d]&)/@Select[facs[n/d],Min@@#1>=d&],{d,Rest[Divisors[n]]}]];
    mmftrees[n_]:=Prepend[Join@@(Tuples[mmftrees/@#]&/@Select[Join@@Permutations/@Select[facs[n],Length[#]>1&],OrderedQ[FactorInteger[#][[1,1]]&/@#]&]),n];
    Table[Length[mmftrees[n]],{n,100}]

Formula

a(prime^n) = A118376(n).
a(product of n distinct primes) = A005804(n).

A319137 Number of strict planar branching factorizations of n.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 3, 1, 3, 1, 9, 1, 3, 3, 7, 1, 9, 1, 9, 3, 3, 1, 37, 1, 3, 3, 9, 1, 25, 1, 21, 3, 3, 3, 57, 1, 3, 3, 37, 1, 25, 1, 9, 9, 3, 1, 161, 1, 9, 3, 9, 1, 37, 3, 37, 3, 3, 1, 153, 1, 3, 9, 75, 3, 25, 1, 9, 3, 25, 1, 345, 1, 3, 9, 9, 3, 25, 1, 161
Offset: 1

Views

Author

Gus Wiseman, Sep 11 2018

Keywords

Comments

A strict planar branching factorization of n is either the number n itself or a sequence of at least two strict planar branching factorizations, one of each factor in a strict ordered factorization of n.

Examples

			The a(12) = 9 trees:
  12,
  (2*6), (3*4), (4*3),(6*2),
  (2*(2*3)), (2*(3*2)), ((2*3)*2), ((3*2)*2).
		

Crossrefs

Programs

  • Mathematica
    ordfacs[n_]:=If[n<=1,{{}},Join@@Table[(Prepend[#1,d]&)/@ordfacs[n/d],{d,Rest[Divisors[n]]}]]
    sotfs[n_]:=Prepend[Join@@Table[Tuples[sotfs/@f],{f,Select[ordfacs[n],And[Length[#]>1,UnsameQ@@#]&]}],n];
    Table[Length[sotfs[n]],{n,100}]

Formula

a(prime^n) = A319123(n + 1).
a(product of n distinct primes) = A319122(n).

A319138 Number of complete strict planar branching factorizations of n.

Original entry on oeis.org

0, 1, 1, 0, 1, 2, 1, 0, 0, 2, 1, 4, 1, 2, 2, 0, 1, 4, 1, 4, 2, 2, 1, 8, 0, 2, 0, 4, 1, 18, 1, 0, 2, 2, 2, 28, 1, 2, 2, 8, 1, 18, 1, 4, 4, 2, 1, 16, 0, 4, 2, 4, 1, 8, 2, 8, 2, 2, 1, 84, 1, 2, 4, 0, 2, 18, 1, 4, 2, 18, 1, 112, 1, 2, 4, 4, 2, 18, 1, 16, 0, 2, 1
Offset: 1

Views

Author

Gus Wiseman, Sep 11 2018

Keywords

Comments

A strict planar branching factorization of n is either the number n itself or a sequence of at least two strict planar branching factorizations, one of each factor in a strict ordered factorization of n. A strict planar branching factorization is complete if the leaves are all prime numbers.

Examples

			The a(12) = 4 trees: (2*(2*3)), (2*(3*2)), ((2*3)*2), ((3*2)*2).
		

Crossrefs

Programs

  • Mathematica
    ordfacs[n_]:=If[n<=1,{{}},Join@@Table[(Prepend[#1,d]&)/@ordfacs[n/d],{d,Rest[Divisors[n]]}]]
    sotfs[n_]:=Prepend[Join@@Table[Tuples[sotfs/@f],{f,Select[ordfacs[n],And[Length[#]>1,UnsameQ@@#]&]}],n];
    Table[Length[Select[sotfs[n],FreeQ[#,_Integer?(!PrimeQ[#]&)]&]],{n,100}]

Formula

a(prime^n) = A000007(n - 1).
a(product of n distinct primes) = A032037(n).

A295635 Write 2 - Zeta(s) in the form 1/Product_{n > 1}(1 + a(n)/n^s).

Original entry on oeis.org

1, 1, 2, 1, 2, 1, 2, 2, 2, 1, 4, 1, 2, 2, 6, 1, 4, 1, 4, 2, 2, 1, 8, 2, 2, 2, 4, 1, 6, 1, 6, 2, 2, 2, 12, 1, 2, 2, 8, 1, 6, 1, 4, 4, 2, 1, 16, 2, 4, 2, 4, 1, 8, 2, 8, 2, 2, 1, 16, 1, 2, 4, 10, 2, 6, 1, 4, 2, 6, 1, 24, 1, 2, 4, 4, 2, 6, 1, 16, 6, 2, 1, 16, 2, 2
Offset: 2

Views

Author

Gus Wiseman, Nov 24 2017

Keywords

Crossrefs

Programs

  • Mathematica
    nn=100;
    facs[n_]:=If[n<=1,{{}},Join@@Table[Map[Prepend[#,d]&,Select[facs[n/d],Min@@#>=d&]],{d,Rest[Divisors[n]]}]];
    -Solve[Table[-1==Sum[Times@@a/@f,{f,facs[n]}],{n,2,nn}],Table[a[n],{n,2,nn}]][[1,All,2]]

A295636 Write 2 - Zeta(s) in the form Product_{n > 1}(1 - a(n)/n^s).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 4, 1, 2, 2, 3, 1, 4, 1, 4, 2, 2, 1, 8, 1, 2, 2, 4, 1, 6, 1, 6, 2, 2, 2, 8, 1, 2, 2, 8, 1, 6, 1, 4, 4, 2, 1, 16, 1, 4, 2, 4, 1, 8, 2, 8, 2, 2, 1, 16, 1, 2, 4, 8, 2, 6, 1, 4, 2, 6, 1, 24, 1, 2, 4, 4, 2, 6, 1, 16, 3, 2, 1, 16, 2, 2, 2
Offset: 2

Views

Author

Gus Wiseman, Nov 24 2017

Keywords

Crossrefs

Programs

  • Mathematica
    nn=100;
    facs[n_]:=If[n<=1,{{}},Join@@Table[Map[Prepend[#,d]&,Select[facs[n/d],Min@@#>=d&]],{d,Rest[Divisors[n]]}]];
    -Solve[Table[-1==Sum[Times@@a/@f,{f,Select[facs[n],UnsameQ@@#&]}],{n,2,nn}],Table[a[n],{n,2,nn}]][[1,All,2]]

Formula

a(n) = Sum_t (-1)^(v(t)-1) where the sum is over all strict tree-factorizations of n (see A295279 for definition) and v(t) is the number of nodes (branchings and leaves) in t.

A316784 Number of orderless identity tree-factorizations of n.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 4, 1, 2, 2, 3, 1, 4, 1, 4, 2, 2, 1, 10, 1, 2, 2, 4, 1, 8, 1, 6, 2, 2, 2, 13, 1, 2, 2, 10, 1, 8, 1, 4, 4, 2, 1, 26, 1, 4, 2, 4, 1, 10, 2, 10, 2, 2, 1, 28, 1, 2, 4, 13, 2, 8, 1, 4, 2, 8, 1, 46, 1, 2, 4, 4, 2, 8, 1, 26, 3, 2, 1
Offset: 1

Views

Author

Gus Wiseman, Jul 13 2018

Keywords

Comments

A factorization of n is a finite nonempty multiset of positive integers greater than 1 with product n. An orderless identity tree-factorization of n is either (case 1) the number n itself or (case 2) a finite set of two or more distinct orderless identity tree-factorizations, one of each factor in a factorization of n.
a(n) depends only on the prime signature of n. - Andrew Howroyd, Nov 18 2018

Examples

			The a(24)=10 orderless identity tree-factorizations:
  24
  (4*6)
  (3*8)
  (2*12)
  (2*3*4)
  (4*(2*3))
  (3*(2*4))
  (2*(2*6))
  (2*(3*4))
  (2*(2*(2*3)))
		

Crossrefs

Programs

  • Mathematica
    postfacs[n_]:=If[n<=1,{{}},Join@@Table[Map[Prepend[#,d]&,Select[postfacs[n/d],Min@@#>=d&]],{d,Rest[Divisors[n]]}]];
    oltsfacs[n_]:=If[n<=1,{{}},Prepend[Select[Union@@Function[q,Sort/@Tuples[oltsfacs/@q]]/@DeleteCases[postfacs[n],{n}],UnsameQ@@#&],n]];
    Table[Length[oltsfacs[n]],{n,100}]
  • PARI
    seq(n)={my(v=vector(n), w=vector(n)); w[1]=v[1]=1; for(k=2, n, w[k]=v[k]+1; forstep(j=n\k*k, k, -k, my(i=j, e=0); while(i%k==0, i/=k; e++; v[j] += binomial(w[k], e)*v[i]))); w} \\ Andrew Howroyd, Nov 18 2018

Formula

a(p^n) = A300660(n) for prime p. - Andrew Howroyd, Nov 18 2018
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