Emily Anible - cp's OEIS frontend

User: Emily Anible

Emily Anible has authored 6 sequences.

A316861 a(n) = Sum_{p in P} y(1)*y(2), where P is the set of partitions of n, and y(k) is the number of parts with multiplicity at least k in p.

0, 0, 1, 1, 4, 7, 13, 22, 38, 58, 93, 139, 208, 302, 438, 616, 869, 1200, 1650, 2239, 3026, 4038, 5374, 7081, 9292, 12103, 15704, 20236, 25992, 33191, 42237, 53490, 67524, 84860, 106341, 132736, 165212, 204928, 253518, 312629, 384585, 471734, 577276, 704584, 858078

Offset: 0

Emily Anible, Jul 15 2018

nonn

Also (1/2)*Sum_{p in P} H(1)*H(2), where P is the set of partitions of n, and H(k) is the number of k-hooks in p.

			For n=6, we sum over the partitions of 6. For each partition, we count the parts with multiplicity at least one, and those of at least two.
6............y(1)*y(2) = 1*0 = 0
5,1..........y(1)*y(2) = 2*0 = 0
4,2..........y(1)*y(2) = 2*0 = 0
4,1,1........y(1)*y(2) = 2*1 = 2
3,3..........y(1)*y(2) = 1*1 = 1
3,2,1........y(1)*y(2) = 3*0 = 0
3,1,1,1......y(1)*y(2) = 2*1 = 2
2,2,2........y(1)*y(2) = 1*1 = 1
2,2,1,1......y(1)*y(2) = 2*2 = 4
2,1,1,1,1....y(1)*y(2) = 2*1 = 2
1,1,1,1,1,1..y(1)*y(2) = 1*1 = 1
--------------------------------
Total.........................13

Alois P. Heinz, Table of n, a(n) for n = 0..1000

Cf. A000097, A116646, A000070, A024786.

b:= proc(n, i, x, y) option remember;
     `if`(n=0, x*y, `if`(i<1, 0, add(b(n-i*j, i-1,
     `if`(j>0, 1, 0)+x, `if`(j>1, 1, 0)+y), j=0..n/i)))
    end:
a:= n-> b(n$2, 0$2):
seq(a(n), n=0..55);  # Alois P. Heinz, Jul 30 2018

Array[Total[
   Count[Split@#, (_?(Length@# >= 1 &))] Count[
       Split@#, (_?(Length@# >= 2 &))] & /@
    IntegerPartitions[#]] &, 50]
(* Second program: *)
b[n_, i_, x_, y_] := b[n, i, x, y] = If[n == 0, x*y, If[i < 1, 0, Sum[b[n - i*j, i - 1, If[j > 0, 1, 0] + x, If[j > 1, 1, 0] + y], {j, 0, n/i}]]];
a[n_] := b[n, n, 0, 0];
a /@ Range[0, 55] (* Jean-François Alcover, Sep 16 2019, after Alois P. Heinz *)

seq(n)={Vec(x*(1 + x^2 + x^3)/((1 - x)^2*(1 + x)*(1 + x + x^2)*prod(i=1, n-1, 1 - x^i + O(x^n))) + O(x^n), -n)} \\ Andrew Howroyd, Jul 15 2018

G.f.: (q^3/((1-q)(1-q^2)) + q^2/(1-q^2) - q^3/(1-q^3))*Product_{j>=1} 1/(1-q^j).
a(n) = A000097(n+3) + A116646(n).
In general, Sum_{n>=0} q^n Sum_{p in P} y(s)*y(t) for s < t is given by (q^(s+t)/((1-q^s)(1-q^t)) + q^t/(1-q^t) - q^(s+t)/(1-q^(s+t))) * Product_{j>=1} 1/(1-q^j).

A304825 Sum of binomial(Y(2,p), 2) over the partitions p of n, where Y(2,p) is the number of part sizes with multiplicity 2 or greater in p.

Original entry on oeis.org

1, 1, 3, 4, 9, 12, 22, 30, 50, 68, 105, 142, 210, 281, 400, 531, 736, 967, 1311, 1707, 2274, 2935, 3851, 4930, 6389, 8116, 10402, 13121, 16658, 20872, 26275, 32719, 40880, 50613, 62807, 77343, 95389, 116874, 143331, 174789, 213251, 258903, 314367, 380079, 459462

Offset: 6

Emily Anible, May 19 2018

nonn

			For a(8), we sum over the partitions of eight. For each partition p, we take binomial(Y(2,p),2): that is, the number of parts with multiplicity at least two choose 2.
8................B(0,2) = 0
7,1..............B(0,2) = 0
6,2..............B(0,2) = 0
6,1,1............B(1,2) = 0
5,3..............B(0,2) = 0
5,2,1............B(0,2) = 0
5,1,1,1..........B(1,2) = 0
4,4..............B(1,2) = 0
4,3,1............B(0,2) = 0
4,2,2............B(1,2) = 0
4,2,1,1..........B(1,2) = 0
4,1,1,1,1........B(1,2) = 0
3,3,2............B(1,2) = 0
3,3,1,1..........B(2,2) = 1
3,2,2,1..........B(1,2) = 0
3,2,1,1,1........B(1,2) = 0
3,1,1,1,1,1......B(1,2) = 0
2,2,2,2..........B(1,2) = 0
2,2,2,1,1........B(2,2) = 1
2,2,1,1,1,1......B(2,2) = 1
2,1,1,1,1,1,1....B(1,2) = 0
1,1,1,1,1,1,1,1..B(1,2) = 0
---------------------------
Total.....................3

Cf. A024786, A302347.

b:= proc(n, i, p) option remember; `if`(n=0 or i=1,
      binomial(`if`(n>1, 1, 0)+p, 2), add(
      b(n-i*j, i-1, `if`(j>1, 1, 0)+p), j=0..n/i))
    end:
a:= n-> b(n$2, 0):
seq(a(n), n=6..60);  # Alois P. Heinz, May 19 2018

Array[Total[Binomial[Count[Split@#, _?(Length@# >= 2 &)], 2] & /@IntegerPartitions[#]] &, 50]
(* Second program: *)
b[n_, i_, p_] := b[n, i, p] = If[n == 0 || i == 1,
     Binomial[If[n > 1, 1, 0] + p, 2], Sum[
     b[n-i*j, i-1, If[j>1, 1, 0]+p], {j, 0, n/i}]];
a[n_] := b[n, n, 0];
a /@ Range[6, 60] (* Jean-François Alcover, May 30 2021, after Alois P. Heinz *)

a(n) = (A301313(n) - A024788(n))/4.

G.f.: q^6 /((1-q^2)*(1-q^4))*Product_{j>=1} 1/(1-q^j).

A302300 a(n) = Sum_{p in P} (Sum_{k_j = 1} 1)^2, where P is the set of partitions of n, and the k_j are the frequencies in p.

Original entry on oeis.org

0, 1, 1, 5, 6, 12, 21, 33, 50, 79, 116, 169, 246, 346, 487, 675, 927, 1254, 1702, 2263, 3014, 3966, 5210, 6766, 8795, 11303, 14531, 18521, 23583, 29803, 37654, 47231, 59206, 73792, 91867, 113778, 140788, 173377, 213289, 261318, 319764, 389846, 474745, 576164

Offset: 0

Emily Anible, Apr 04 2018

nonn

This sequence is part of the contribution to the b^2 term of C_{1-b,2}(q) for(1-b,2)-colored partitions - partitions in which we can label parts any of an indeterminate 1-b colors, but are restricted to using only 2 of the colors per part size. This formula is known to match the Han/Nekrasov-Okounkov hooklength formula truncated at hooks of size two up to the linear term in b.

It is of interest to enumerate and determine specific characteristics of partitions of n, considering each partition individually.

			For a(6), we sum over partitions of six. For each partition, we count 1 for each part which appears once, then square the total in each partition.
   6............1^2 = 1
   5,1..........2^2 = 4
   4,2..........2^2 = 4
   4,1,1........1^2 = 1
   3,3..........0^2 = 0
   3,2,1........3^2 = 9
   3,1,1,1......1^2 = 1
   2,2,2........0^2 = 0
   2,2,1,1......0^2 = 0
   2,1,1,1,1....1^2 = 1
   1,1,1,1,1,1..0^2 = 0
   --------------------
   Total.............21

Alois P. Heinz, Table of n, a(n) for n = 0..4000
Guo-Niu Han, The Nekrasov-Okounkov hook length formula: refinement, elementary proof, extension and applications, arXiv:0805.1398 [math.CO], 2008.
Guo-Niu Han, The Nekrasov-Okounkov hook length formula: refinement, elementary proof, extension and applications, Annales de l'institut Fourier, Tome 60 (2010) no. 1, pp. 1-29.
W. J. Keith, Restricted k-color partitions, Ramanujan Journal (2016) 40: 71.

Cf. A024786, A197126.

b:= proc(n, i, p) option remember; `if`(n=0 or i=1, (
      `if`(n=1, 1, 0)+p)^2, add(b(n-i*j, i-1,
      `if`(j=1, 1, 0)+p), j=0..n/i))
    end:
a:= n-> b(n$2, 0):
seq(a(n), n=0..60);  # Alois P. Heinz, Apr 05 2018

Array[Total@ Map[Count[Split@ #, ?(Length@ # == 1 &)]^2 &, IntegerPartitions[#]] &, 43] (* _Michael De Vlieger, Apr 05 2018 *)
b[n_, i_, p_] := b[n, i, p] = If[n == 0 || i == 1, (
     If[n == 1, 1, 0] + p)^2, Sum[b[n - i*j, i - 1,
     If[j == 1, 1, 0] + p], {j, 0, n/i}]];
a[n_] := b[n, n, 0];
a /@ Range[0, 60] (* Jean-François Alcover, Jun 06 2021, after Alois P. Heinz *)

def frequencies(partition, n):
    tot = 0
    freq_list = []
    i = 0
    for p in partition:
        freq = [0 for i in range(n+1)]
        for i in p:
            freq[i] += 1
        for f in freq:
            if f == 0:
                tot += 1
        freq_list.append(freq)
    return freq_list
def sum_square_freqs_of_one(freq_part):
    tot = 0
    for f in freq_part:
        count = 0
        for i in f:
            if i == 1:
                count += 1
        tot += count*count
    return tot
import sympy.combinatorics
def A302300(n): # rewritten by R. J. _Mathar, 2023-03-24
    a =0
    if n ==0 :
        return 0
    part = sympy.combinatorics.IntegerPartition([n])
    partlist = []
    while True:
        part = part.next_lex()
        partlist.append(part.partition)
        if len(part.partition) <=1 :
            break
    freq_part = frequencies(partlist, n)
    return sum_square_freqs_of_one(freq_part)
for n in range(20): print(A302300(n))

a(n) = Sum_{p in P} (Sum_{k_j = 1} 1)^2, where P is the set of partitions of n, and k_j are the frequencies in p.

A302347 a(n) = Sum of (Y(2,p)^2) over the partitions p of n, Y(2,p) = number of part sizes with multiplicity 2 or greater in p.

Original entry on oeis.org

0, 0, 1, 1, 3, 4, 10, 13, 25, 34, 59, 80, 127, 172, 260, 349, 505, 673, 946, 1248, 1711, 2238, 3010, 3902, 5162, 6637, 8663, 11051, 14253, 18051, 23047, 28988, 36677, 45840, 57538, 71485, 89082, 110062, 136269, 167487, 206138, 252132, 308640, 375777, 457698

Offset: 0

Emily Anible, Apr 05 2018

nonn

It is of interest to enumerate and determine specific characteristics of partitions of n, considering each partition individually.

			For a(6), we sum over partitions of six. For each partition, we count 1 for each part which appears more than once, then square the total in each partition.
6............0^2 = 0
5,1..........0^2 = 0
4,2..........0^2 = 0
4,1,1........1^2 = 1
3,3..........1^2 = 1
3,2,1........0^2 = 0
3,1,1,1......1^2 = 1
2,2,2........1^2 = 1
2,2,1,1......2^2 = 4
2,1,1,1,1....1^2 = 1
1,1,1,1,1,1..1^2 = 1
--------------------
Total.............10

Alois P. Heinz, Table of n, a(n) for n = 0..2000
Guo-Niu Han, The Nekrasov-Okounkov hook length formula: refinement, elementary proof, extension and applications, arXiv:0805.1398 [math.CO], 2008.
Guo-Niu Han, The Nekrasov-Okounkov hook length formula: refinement, elementary proof, extension and applications, Annales de l'institut Fourier, Tome 60 (2010) no. 1, pp. 1-29.
W. J. Keith, Restricted k-color partitions, Ramanujan Journal (2016) 40: 71.

Cf. A024786, A024788, A302300.

b:= proc(n, i, p) option remember; `if`(n=0 or i=1, (
      `if`(n>1, 1, 0)+p)^2, add(b(n-i*j, i-1,
      `if`(j>1, 1, 0)+p), j=0..n/i))
    end:
a:= n-> b(n$2, 0):
seq(a(n), n=0..60);  # Alois P. Heinz, Apr 05 2018

Array[Total[Count[Split@ #, ?(Length@ # > 1 &)]^2 & /@ IntegerPartitions[#]] &, 44] (* _Michael De Vlieger, Apr 07 2018 *)
b[n_, i_, p_] := b[n, i, p] = If[n == 0 || i == 1, (
     If[n > 1, 1, 0] + p)^2, Sum[b[n - i*j, i - 1,
     If[j > 1, 1, 0] + p], {j, 0, n/i}]];
a[n_] := b[n, n, 0];
a /@ Range[0, 60] (* Jean-François Alcover, Jun 06 2021, after Alois P. Heinz *)

def sum_square_freqs_greater_one(freq_list):
    tot = 0
    for f in freq_list:
        count = 0
        for i in f:
            if i > 1:
                count += 1
        tot += count*count
    return tot
def frequencies(partition, n):
    tot = 0
    freq_list = []
    i = 0
    for p in partition:
        freq = [0 for i in range(n+1)]
        for i in p:
            freq[i] += 1
        for f in freq:
            if f == 0:
                tot += 1
        freq_list.append(freq)
    return freq_list

a(n) = Sum_{p in P(n)} (H(2,p)^2 + 2*A024786 - 2*A024788), where P(n) is the set of partitions of n, and H(2,p) is the hooks of length 2 in partition p.
G.f: (q^2*(1+q^4))/((1-q^2)*(1-q^4))*Product_{j>=1} 1/(1-q^j).
a(n) ~ sqrt(3) * exp(Pi*sqrt(2*n/3)) / (8*Pi^2). - Vaclav Kotesovec, May 22 2018

A302348 a(n) = Sum_{p in P} (H(2,p)^2)/2, where P is the set of partitions of n, and H(2,p) is the number of hooks of length 2 in p.

Original entry on oeis.org

0, 0, 1, 1, 4, 5, 14, 18, 37, 50, 90, 122, 199, 270, 415, 559, 820, 1096, 1556, 2060, 2847, 3736, 5057, 6576, 8747, 11279, 14788, 18916, 24493, 31097, 39838, 50225, 63737, 79833, 100471, 125076, 156237, 193394, 239956, 295443, 364334, 446349, 547360, 667440

Offset: 0

Emily Anible, Apr 05 2018

nonn

This sequence is part of the contribution to the b^2 term of the Han/Nekrasov-Okounkov hooklength formula truncated at hooks of size two.

It is of interest to enumerate and determine specific characteristics of partitions of n, considering each partition individually.

			For a(6), we sum over partitions of six. For each partition, we count 1 for each hook of length 2, then square the total in each partition. We divide the final result in half to get a(6).
6............1^2 = 1
5,1..........1^2 = 1
4,2..........2^2 = 4
4,1,1........2^2 = 4
3,3..........2^2 = 4
3,2,1........0^2 = 0
3,1,1,1......2^2 = 4
2,2,2........2^2 = 4
2,2,1,1......2^2 = 4
2,1,1,1,1....1^2 = 1
1,1,1,1,1,1..1^2 = 1
--------------------
Total.............28/2=14

Alois P. Heinz, Table of n, a(n) for n = 0..1000
Guo-Niu Han, The Nekrasov-Okounkov hook length formula: refinement, elementary proof, extension and applications, arXiv:0805.1398 [math.CO], 2008.
Guo-Niu Han, The Nekrasov-Okounkov hook length formula: refinement, elementary proof, extension and applications, Annales de l'institut Fourier, Tome 60 (2010) no. 1, pp. 1-29.
W. J. Keith, Restricted k-color partitions, Ramanujan Journal (2016) 40: 71.

Cf. A024786, A302347.

b:= proc(n, i, p, l) option remember; `if`(n=0, p^2,
      `if`(i>n, 0, b(n, i+1, p, 1)+add(b(n-i*j, i+1, p+
      `if`(j>1, 1, 0)+l, 0), j=1..n/i)))
    end:
a:= n-> b(n, 1, 0$2)/2:
seq(a(n), n=0..50);  # Alois P. Heinz, Apr 06 2018

b[n_, i_, p_, l_] := b[n, i, p, l] = If[n == 0, p^2, If[i > n, 0, b[n, i + 1, p, 1] + Sum[b[n - i*j, i+1, p + If[j > 1, 1, 0]+l, 0], {j, 1, n/i}]]];
a[n_] := b[n, 1, 0, 0]/2;
Table[a[n], {n, 0, 50}] (* Jean-François Alcover, May 18 2018, after Alois P. Heinz *)

G.f: (q^2*(1+q^2+2*q^4))/((1-q^2)*(1-q^4)*Product_{i>0} (1-q^i)).

A301313 a(n) = Sum_{p in P} binomial(H(2,p),2), where P is the set of partitions of n, and H(2,p) = number of hooks of size 2 in p.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 6, 7, 18, 24, 49, 66, 116, 158, 255, 346, 525, 707, 1030, 1374, 1936, 2560, 3519, 4608, 6207, 8056, 10673, 13735, 17942, 22906, 29569, 37469, 47864, 60235, 76249, 95335, 119705, 148770, 185447, 229182, 283810, 348903, 429498, 525411, 643244

Offset: 0

Emily Anible, Apr 03 2018

nonn

This sequence is part of the contribution to the quadratic b^2 term of a 2-truncation of the Han/Nekrasov-Okounkov hooklength formula (2-truncation here being the limiting of hook sizes counted by the formula to only those of size 1 or 2). Exploring this sequence may lead to more general formulas regarding the hooklength formula for larger hooks, or the entire contribution to the quadratic term of the formula.

			For n=6, we sum over the partitions of 6. For each partition, we calculate binomial(number of hooks of size 2 in partition, 2):
6............binomial(1,2) = 0
5,1..........binomial(1,2) = 0
4,2..........binomial(2,2) = 1
4,1,1........binomial(2,2) = 1
3,3..........binomial(2,2) = 1
3,2,1........binomial(0,2) = 0
3,1,1,1......binomial(2,2) = 1
2,2,2........binomial(2,2) = 1
2,2,1,1......binomial(2,2) = 1
2,1,1,1,1....binomial(1,2) = 0
1,1,1,1,1,1..binomial(1,2) = 0
------------------------------
Total........................6

Vaclav Kotesovec, Table of n, a(n) for n = 0..10000 (terms 0..1900 from Alois P. Heinz)
Guo-Niu Han, The Nekrasov-Okounkov hook length formula: refinement, elementary proof, extension and applications, arXiv:0805.1398 [math.CO], 2008.
Guo-Niu Han, The Nekrasov-Okounkov hook length formula: refinement, elementary proof, extension and applications, Annales de l'institut Fourier, Tome 60 (2010) no. 1, pp. 1-29.

Cf. A024786, A138785.

b:= proc(n, i, p, l) option remember; `if`(n=0, p*(p-1)/2,
      `if`(i>n, 0, b(n, i+1, p, 1)+add(b(n-i*j, i+1, p+
      `if`(j>1, 1, 0)+l, 0), j=1..n/i)))
    end:
a:= n-> b(n, 1, 0$2):
seq(a(n), n=0..50);  # Alois P. Heinz, Apr 05 2018

b[n_, i_, p_, l_] := b[n, i, p, l] = If[n == 0, p*(p-1)/2, If[i > n, 0, b[n, i+1, p, 1] + Sum[b[n-i*j, i+1, p+If[j>1, 1, 0]+l, 0], {j, 1, n/i}]] ];
a[n_] := b[n, 1, 0, 0];
Table[a[n], {n, 0, 50}] (* Jean-François Alcover, Apr 28 2018, after Alois P. Heinz *)
Table[Sum[(2*k - 5 - (-1)^(k/2))*(1 + (-1)^k)/4 * PartitionsP[n-k], {k, 1, n}], {n, 0, 60}] (* Vaclav Kotesovec, Oct 06 2018 *)

G.f.: (q^4+3*q^6)/((1-q^2)*(1-q^4))*Product_{j>=1} 1/(1-q^j). - Emily Anible, May 18 2018

a(n) ~ sqrt(3) * exp(Pi*sqrt((2*n)/3)) / (4*Pi^2). - Vaclav Kotesovec, Oct 06 2018

a(10)-a(44) from Alois P. Heinz, Apr 03 2018

cp's OEIS Frontend

User: Emily Anible

A316861 a(n) = Sum_{p in P} y(1)*y(2), where P is the set of partitions of n, and y(k) is the number of parts with multiplicity at least k in p.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Formula

A304825 Sum of binomial(Y(2,p), 2) over the partitions p of n, where Y(2,p) is the number of part sizes with multiplicity 2 or greater in p.

Views

Author

Keywords

Examples

Crossrefs

Programs

Maple

Mathematica

Formula

A302300 a(n) = Sum_{p in P} (Sum_{k_j = 1} 1)^2, where P is the set of partitions of n, and the k_j are the frequencies in p.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Python

Formula

A302347 a(n) = Sum of (Y(2,p)^2) over the partitions p of n, Y(2,p) = number of part sizes with multiplicity 2 or greater in p.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Python

Formula

A302348 a(n) = Sum_{p in P} (H(2,p)^2)/2, where P is the set of partitions of n, and H(2,p) is the number of hooks of length 2 in p.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

A301313 a(n) = Sum_{p in P} binomial(H(2,p),2), where P is the set of partitions of n, and H(2,p) = number of hooks of size 2 in p.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Formula

Extensions