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 29 results. Next

A299202 Moebius function of the multiorder of integer partitions indexed by their Heinz numbers.

Original entry on oeis.org

0, 1, 1, -1, 1, -1, 1, 0, -1, -1, 1, 2, 1, -1, -1, -1, 1, 1, 1, 1, -1, -1, 1, -1, -1, -1, 0, 1, 1, 3, 1, 0, -1, -1, -1, -1, 1, -1, -1, -1, 1, 2, 1, 1, 1, -1, 1, 0, -1, 1, -1, 1, 1, -1, -1, -1, -1, -1, 1, -3, 1, -1, 2, 0, -1, 2, 1, 1, -1, 3, 1, 2, 1, -1, 1, 1, -1, 2, 1, 1, -1, -1, 1, -5, -1, -1, -1, -1, 1, -4
Offset: 1

Views

Author

Gus Wiseman, Feb 05 2018

Keywords

Comments

By convention, mu() = 0.
The Heinz number of an integer partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k).

Examples

			Heinz number of (2,1,1) is 12, so mu(2,1,1) = a(12) = 2.
		

Crossrefs

Programs

  • Mathematica
    nn=120;
    ptns=Table[If[n===1,{},Join@@Cases[FactorInteger[n]//Reverse,{p_,k_}:>Table[PrimePi[p],{k}]]],{n,nn}];
    tris=Join@@Map[Tuples[IntegerPartitions/@#]&,ptns];
    mu[y_]:=mu[y]=If[Length[y]===1,1,-Sum[Times@@mu/@t,{t,Select[tris,And[Length[#]>1,Sort[Join@@#,Greater]===y]&]}]];
    mu/@ptns

Formula

mu(y) = Sum_{g(t)=y} (-1)^d(t), where the sum is over all enriched p-trees (A289501, A299203) whose multiset of leaves is the integer partition y, and d(t) is the number of non-leaf nodes in t.

A299200 Number of twice-partitions whose domain is the integer partition with Heinz number n.

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 5, 1, 4, 3, 7, 2, 11, 5, 6, 1, 15, 4, 22, 3, 10, 7, 30, 2, 9, 11, 8, 5, 42, 6, 56, 1, 14, 15, 15, 4, 77, 22, 22, 3, 101, 10, 135, 7, 12, 30, 176, 2, 25, 9, 30, 11, 231, 8, 21, 5, 44, 42, 297, 6, 385, 56, 20, 1, 33, 14, 490, 15, 60, 15, 627, 4
Offset: 1

Views

Author

Gus Wiseman, Feb 05 2018

Keywords

Comments

The Heinz number of an integer partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k).

Examples

			The a(15) = 6 twice-partitions: (3)(2), (3)(11), (21)(2), (21)(11), (111)(2), (111)(11).
		

Crossrefs

Programs

  • Maple
    with(numtheory): with(combinat):
    a:= n-> mul(numbpart(pi(i[1]))^i[2], i=ifactors(n)[2]):
    seq(a(n), n=1..82);  # Alois P. Heinz, Jan 14 2021
  • Mathematica
    Table[Times@@Cases[FactorInteger[n],{p_,k_}:>PartitionsP[PrimePi[p]]^k],{n,100}]
  • PARI
    a(n) = {my(f = factor(n)); for (k=1, #f~, f[k, 1] = numbpart(primepi(f[k, 1]));); factorback(f);} \\ Michel Marcus, Feb 26 2018

Formula

Multiplicative with a(prime(n)) = A000041(n).

A299201 Number of twice-partitions whose composite is the integer partition with Heinz number n.

Original entry on oeis.org

1, 1, 1, 2, 1, 2, 1, 3, 2, 2, 1, 5, 1, 2, 2, 5, 1, 4, 1, 4, 2, 2, 1, 8, 2, 2, 3, 4, 1, 6, 1, 7, 2, 2, 2, 11, 1, 2, 2, 8, 1, 5, 1, 4, 4, 2, 1, 16, 2, 4, 2, 4, 1, 7, 2, 7, 2, 2, 1, 13, 1, 2, 5, 11, 2, 5, 1, 4, 2, 6, 1, 19, 1, 2, 4, 4, 2, 5, 1, 13, 5, 2, 1, 13, 2
Offset: 1

Views

Author

Gus Wiseman, Feb 05 2018

Keywords

Comments

The Heinz number of an integer partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k).

Examples

			The a(36) = 11 twice-partitions:
  (2211),
  (22)(11), (211)(2), (221)(1), (21)(21),
  (2)(2)(11), (2)(11)(2), (11)(2)(2), (22)(1)(1), (21)(2)(1),
  (2)(2)(1)(1).
		

Crossrefs

Programs

  • Mathematica
    nn=100;
    ptns=Table[If[n===1,{},Join@@Cases[FactorInteger[n]//Reverse,{p_,k_}:>Table[PrimePi[p],{k}]]],{n,nn}];
    tris=Join@@Map[Tuples[IntegerPartitions/@#]&,ptns];
    Table[Length[Select[tris,Sort[Join@@#,Greater]===y&]],{y,ptns}]

A220418 Express 1 - x - x^2 - x^3 - x^4 - ... as product (1 + g(1)*x) * (1 + g(2)*x^2) *(1 + g(3)*x^3) * ... and use a(n) = - g(n).

Original entry on oeis.org

1, 1, 2, 3, 6, 8, 18, 27, 54, 84, 186, 296, 630, 1008, 2106, 3711, 7710, 12924, 27594, 48528, 97902, 173352, 364722, 647504, 1340622, 2382660, 4918482, 9052392, 18512790, 33361776, 69273666, 127198287, 258155910, 475568220, 981288906, 1814542704, 3714566310
Offset: 1

Views

Author

Michel Marcus, Dec 14 2012

Keywords

Comments

This is the PPE (power product expansion) of A153881 (with offset 0).
When p is prime, a(p) = (2^p-2)/p (A064535).
From Petros Hadjicostas, Oct 04 2019: (Start)
This sequence appears as an example in Gingold and Knopfmacher (1995) starting at p. 1223.
In Section 3 of Gingold and Knopfmacher (1995), it is proved that, if f(z) = Product_{n >= 1} (1 + g(n))*z^n = 1/(Product_{n >= 1} (1 - h(n))*z^n), then g(2*n - 1) = h(2*n - 1) and Sum_{d|n} (1/d)*h(n/d)^d = -Sum_{d|n} (1/d)*(-g(n/d))^d. The same results were proved more than ten years later by Alkauskas (2008, 2009). [If we let a(n) = -g(n), then Alkauskas works with f(z) = Product_{n >= 1} (1 - a(n))*z^n; i.e., a(2*n - 1) = -h(2*n - 1) etc.]
The PPE of 1/(1 - x - x^2 - x^3 - x^4 - ...) is given in A290261, which is also studied in Gingold and Knopfmacher (1995, p. 1234).
(End)
The number of terms in the Zassenhaus formula exponent of order n, as computed by the algorithm by Casas, Murua & Nadinic, is equal to a(n) at least for n = 2..24. - Andrey Zabolotskiy, Apr 09 2023

Crossrefs

Programs

  • Maple
    b:= proc(n, i) option remember; `if`(n=0 or i<1, 1,
          b(n, i-1)+a(i)*b(n-i, min(n-i, i)))
        end:
    a:= proc(n) option remember; 2^n-b(n, n-1) end:
    seq(a(n), n=1..40);  # Alois P. Heinz, Jun 22 2018
  • Mathematica
    b[n_, i_] := b[n, i] = If[n == 0 || i < 1, 1, b[n, i - 1] + a[i]*b[n - i, Min[n - i, i]]];
    a[n_] := a[n] = 2^n - b[n, n - 1] ;
    Array[a, 40] (* Jean-François Alcover, Jul 09 2018, after Alois P. Heinz *)
  • PARI
    a(m) = {default(seriesprecision, m+1); gk = vector(m); pol = 1 + sum(n=1, m, -x^n); gk[1] = polcoeff( pol, 1); for (k=2, m, pol = taylor(pol/(1+gk[k-1]*x^(k-1)), x); gk[k] = polcoeff(pol, k, x);); for (k=1, m, print1(-gk[k], ", "););}

Formula

g(1) = -1 and for k > 1, g(k) satisfies Sum_{d|k} (1/d)*(-g(k/d))^d = (2^k - 1)/k, where a(k) = -g(k). - Gevorg Hmayakyan, Jun 05 2016 [Corrected by Petros Hadjicostas, Oct 04 2019. See p. 1224 in Gingold and Knopfmacher (1995).]
From Petros Hadjicostas, Oct 04 2019: (Start)
a(2*n - 1) = A290261(2*n - 1) for n >= 1 because A290261 gives the PPE of 1/(1 - x - x^2 - x^3 - ...) = (1 - x)/(1 - 2*x).
Define (A(m,n): n,m >= 1) by A(m=1,n) = -1 for n >= 1, A(m,n) = 0 for m > n >= 1 (upper triangular), and A(m,n) = A(m-1,n) - A(m-1,m-1) * A(m,n-m+1) for n >= m >= 2. Then a(n) = A(n,n). [Theorem 3 in Gingold et al. (1988).]
(End)

Extensions

Name edited by Petros Hadjicostas, Oct 04 2019

A299203 Number of enriched p-trees whose multiset of leaves is the integer partition with Heinz number n.

Original entry on oeis.org

0, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 4, 1, 1, 1, 5, 1, 3, 1, 3, 1, 1, 1, 11, 1, 1, 2, 3, 1, 5, 1, 12, 1, 1, 1, 15, 1, 1, 1, 11, 1, 4, 1, 3, 3, 1, 1, 38, 1, 3, 1, 3, 1, 9, 1, 9, 1, 1, 1, 21, 1, 1, 4, 34, 1, 4, 1, 3, 1, 5, 1, 54, 1, 1, 3, 3, 1, 4, 1, 33, 5, 1, 1, 23, 1, 1, 1, 9, 1, 20, 1, 3, 1, 1, 1, 117, 1, 3, 3, 12, 1, 4, 1, 9, 4, 1, 1, 57, 1, 4, 1, 34
Offset: 1

Views

Author

Gus Wiseman, Feb 05 2018

Keywords

Comments

By convention, a(1) = 0.
The Heinz number of an integer partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k).

Examples

			a(54) = 9: (((22)2)1), ((222)1), (((22)1)2), (((21)2)2), ((221)2), ((22)(21)), ((22)21), ((21)22), (2221).
a(40) = 11: ((31)(11)), (((31)1)1), ((3(11))1), ((311)1), (3((11)1)), (3(111)), (((11)1)3), ((111)3), ((31)11), (3(11)1), (3111).
a(36) = 15: ((22)(11)), ((2(11))2), (((11)2)2), (((21)1)2), ((211)2), (((22)1)1), (((21)2)1), ((221)1), ((21)(21)), (22(11)), (2(11)2), ((11)22), ((22)11), ((21)21), (2211).
		

Crossrefs

Programs

  • Mathematica
    nn=120;
    ptns=Table[If[n===1,{},Join@@Cases[FactorInteger[n]//Reverse,{p_,k_}:>Table[PrimePi[p],{k}]]],{n,nn}];
    tris=Join@@Map[Tuples[IntegerPartitions/@#]&,ptns];
    qci[y_]:=qci[y]=If[Length[y]===1,1,Sum[Times@@qci/@t,{t,Select[tris,And[Length[#]>1,Sort[Join@@#,Greater]===y]&]}]];
    qci/@ptns

A273866 Coefficients a(k,m) of polynomials a{k}(h) appearing in the product Product_{k >= 1} (1 - a{k}(h)*x^k) = 1 - h*x/(1-x).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 2, 2, 2, 1, 1, 3, 5, 5, 3, 1, 1, 3, 6, 7, 6, 3, 1, 1, 4, 9, 13, 13, 9, 4, 1, 1, 4, 10, 17, 20, 17, 10, 4, 1, 1, 5, 15, 30, 42, 42, 30, 15, 5, 1, 1, 5, 16, 36, 57, 66, 57, 36, 16, 5, 1
Offset: 1

Views

Author

Gevorg Hmayakyan, Jun 01 2016

Keywords

Comments

The a(k,m) form a table where each row has k-1 elements starting from 2 and the a(1,1) = 1.

Examples

			a{1}(h) = h,
a{2}(h) = h,
a{3}(h) = h^2 + h,
a{4}(h) = h^3 + h^2 + h,
a{5}(h) = h^4 + 2*h^3 + 2*h^2 + h,
a{6}(h) = h^5 + 2*h^4 + 2*h^3 + 2*h^2 + h,
a{7}(h) = h^6 + 3*h^5 + 5*h^4 + 5*h^3 + 3*h^2 + h,
a{8}(h) = h^7 + 3*h^6 + 6*h^5 + 7*h^4 + 6*h^3 + 3*h^2 + h,
a{9}(h) = h^8 + 4*h^7 + 9*h^6 + 13*h^5 + 13*h^4 + 9*h^3 + 4*h^2 + h
...
and the corresponding a(k,m) table is:
  1,
  1,
  1,  1,
  1,  1,  1,
  1,  2,  2,  1,
  1,  2,  2,  2,  1,
  1,  3,  5,  5,  3,  1,
  1,  3,  6,  7,  6,  3,  1,
  1,  4,  9, 13, 13,  9,  4,  1,
  ...
a(7,3) = 5 because there are six strict trees contributing positive one {{5,1},1}, {{4,2},1}, {{4,1},2}, {{3,2},2}, {4,{2,1}}, {{3,1},3} and there is one strict tree contributing negative one {4,2,1}. - _Gus Wiseman_, Nov 14 2016
		

Crossrefs

Programs

  • Maple
    with(ListTools), with(numtheory), with(combinat);
    L := product(1-a[k]*x^k, k = 1 .. 600);
    S := Flatten([seq(-h, i = 1 .. 100)]);
    Sabs := Flatten([seq(i, i = 1 .. 100)]);
    seq(assign(a[i] = solve(coeff(L, x^i) = `if`(is(i in Sabs), S[Search(i, Sabs)], 0), a[i])), i = 1 .. 20);
    map(coeffs, [seq(simplify(a[i]), i = 1 .. 20)]);
  • Mathematica
    strictrees[n_Integer?Positive]:=Prepend[Join@@Function[ptn,Tuples[strictrees/@ptn]]/@Select[IntegerPartitions[n],And[Length[#]>1,UnsameQ@@#]&],n];
    Table[Sum[(-1)^(Count[tree,,{0,Infinity}]-1),{tree,Select[strictrees[n],Length[Flatten[{#}]]===m&]}],{n,1,9},{m,1,n-1/.(0->1)}] (* _Gus Wiseman, Nov 14 2016 *)
    (* second program *)
    A[m_, n_] :=
      A[m, n] =
       Which[m == 1, -h, m > n >= 1, 0, True,
        A[m - 1, n] - A[m - 1, m - 1]*A[m, n - m + 1]];
    a[n_] := Expand[-A[n, n]];
    a /@ Range[1, 25] (* Petros Hadjicostas, Oct 04 2019, courtesy of Jean-François Alcover *)

Formula

a(k,m) = a(k, k-m).
For prime p: Sum_{m = 1..p-1} a(p, m) = (2^p - 2)/p.
a{k}(h) satisfies Sum_{d|k} (1/d)*(a{k/d}(h))^d = ((h+1)^k - 1)/k. [Corrected by Petros Hadjicostas, Oct 04 2019]
For prime p: a{p}(h) = ((h+1)^p - h^p - 1)/p.
See A273873 for the definition of strict tree. Then a(n,m) = Sum_t (-1)^{v(t)-1} where the sum is over all strict trees of weight n with m leaves, and v(t) is the number of nodes in t (including the leaves, which are positive integers). See example 2 and the first Mathematica program. - Gus Wiseman, Nov 14 2016

A301364 Regular triangle where T(n,k) is the number of enriched p-trees of weight n with k leaves.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 2, 4, 5, 1, 2, 6, 11, 12, 1, 3, 10, 26, 38, 34, 1, 3, 13, 39, 87, 117, 92, 1, 4, 19, 69, 181, 339, 406, 277, 1, 4, 23, 95, 303, 707, 1198, 1311, 806, 1, 5, 30, 143, 514, 1430, 2970, 4525, 4522, 2500, 1, 5, 35, 184, 762, 2446, 6124, 11627
Offset: 1

Views

Author

Gus Wiseman, Mar 19 2018

Keywords

Comments

An enriched p-tree of weight n > 0 is either a single node of weight n, or a finite sequence of two or more enriched p-trees with weakly decreasing weights summing to n.

Examples

			Triangle begins:
  1
  1   1
  1   1   2
  1   2   4   5
  1   2   6  11  12
  1   3  10  26  38  34
  1   3  13  39  87 117  92
  1   4  19  69 181 339 406 277
  ...
The T(5,4) = 11 enriched p-trees: (((21)1)1), ((2(11))1), (((11)2)1), ((211)1), ((21)(11)), (((11)1)2), ((111)2), ((21)11), (2(11)1), ((11)21), (2111).
		

Crossrefs

Programs

  • Mathematica
    eptrees[n_]:=Prepend[Join@@Table[Tuples[eptrees/@ptn],{ptn,Select[IntegerPartitions[n],Length[#]>1&]}],n];
    Table[Length[Select[eptrees[n],Count[#,_Integer,{-1}]===k&]],{n,8},{k,n}]
  • PARI
    A(n)={my(v=vector(n)); for(n=1, n, v[n] = y + polcoef(1/prod(k=1, n-1, 1 - v[k]*x^k + O(x*x^n)), n)); apply(p->Vecrev(p/y), v)}
    { my(T=A(10)); for(n=1, #T, print(T[n])) } \\ Andrew Howroyd, Aug 26 2018

A290971 Write x/(1-x) in the form Sum_{j>=1} a(j)*x^j/(1+a(j)*x^j).

Original entry on oeis.org

1, 2, 0, 6, 0, -6, 0, 54, 0, -30, 0, -114, 0, -126, 0, 4470, 0, -294, 0, -5850, 0, -2046, 0, -92418, 0, -8190, 0, -247674, 0, 2010, 0, 30229110, 0, -131070, 0, -8200914, 0, -524286, 0, -362617770, 0, 183162, 0, -354416634, 0, -8388606, 0, -53614489794, 0
Offset: 1

Views

Author

Gus Wiseman, Aug 16 2017

Keywords

Crossrefs

Programs

  • Mathematica
    nn=20;-Solve[Table[Sum[a[n/d]^d,{d,Divisors[n]}]==-1,{n,nn}],Array[a,nn]][[1,All,2]]

Formula

a(n) = -Sum_t (-1)^v(t) where the sum is over all same-trees of weight n (see A281145 for definition) and v(t) is the number of nodes (branchings and leaves) in t.

A300574 Coefficient of x^n in 1/((1-x)(1+x^3)(1-x^5)(1+x^7)(1-x^9)...).

Original entry on oeis.org

1, 1, 1, 0, 0, 1, 2, 1, 0, 0, 2, 2, 1, 0, 2, 3, 2, 0, 2, 4, 4, 0, 1, 4, 6, 2, 1, 4, 8, 4, 2, 4, 10, 6, 2, 3, 12, 10, 4, 2, 13, 14, 8, 2, 14, 18, 12, 2, 14, 22, 18, 3, 14, 26, 26, 6, 14, 29, 34, 10, 14, 32, 44, 16, 14, 34, 56, 26, 16, 34, 67, 38, 20, 34, 78, 52, 26
Offset: 0

Views

Author

Gus Wiseman, Mar 08 2018

Keywords

Comments

By Theorem 1 of Craig, the values a(n) in this list are known to be nonnegative. Combined with Theorem 2 of Seo and Yee, this shows that a(n) = |number of partitions of n into odd parts with an odd index minus the number of partitions of n into odd parts with an even index|. - William Craig, Dec 31 2021

References

  • Seunghyun Seo and Ae Ja Yee, Index of seaweed algebras and integer partitions, Electronic Journal of Combinatorics, 27:1 (2020), #P1.47. See Conjecture 1 and Theorem 2.

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[1/QPochhammer[x, -x^2], {x, 0, 100}], x]
    nmax = 100; CoefficientList[Series[Product[1/((1+x^(4*k-1))*(1-x^(4*k-3))), {k, 1, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Jun 04 2019 *)

Formula

O.g.f.: Product_{n >= 0} 1/(1 - (-1)^n x^(2n+1)).
a(n) = Sum (-1)^k where the sum is over all integer partitions of n into odd parts and k is the number of parts not congruent to 1 modulo 4.
a(n) has average order Gamma(1/4) * exp(sqrt(n/3)*Pi/2) / (2^(9/4) * 3^(1/8) * Pi^(3/4) * n^(5/8)). - Vaclav Kotesovec, Jun 04 2019

A300866 Signed recurrence over binary strict trees: a(n) = 1 - Sum_{x + y = n, 0 < x < y < n} a(x) * a(y).

Original entry on oeis.org

1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, -1, 1, 1, -2, 3, -1, -3, 8, -8, 1, 14, -26, 22, 10, -59, 90, -52, -74, 238, -291, 80, 417, -930, 915, 124, -1991, 3483, -2533, -2148, 9011, -12596, 5754, 14350, -37975, 42735, -4046, -77924, 154374, -133903, -56529, 376844, -591197, 355941, 522978, -1706239
Offset: 0

Views

Author

Gus Wiseman, Mar 13 2018

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_]:=a[n]=1-Sum[a[k]*a[n-k],{k,1,(n-1)/2}];
    Array[a,40]
  • PARI
    seq(n)={my(v=vector(n)); for(n=1, n, v[n] = 1 - sum(k=1, (n-1)\2, v[k]*v[n-k])); concat([1], v)} \\ Andrew Howroyd, Aug 27 2018
Showing 1-10 of 29 results. Next