A334732 -id:A334732 - cp's OEIS frontend

A330889 a(n) is the total number of parts in all partitions of n into consecutive parts that differ by 3.

1, 1, 1, 1, 3, 1, 3, 1, 3, 1, 3, 4, 3, 1, 6, 1, 3, 4, 3, 1, 6, 5, 3, 4, 3, 5, 6, 1, 3, 8, 3, 1, 6, 5, 8, 4, 3, 5, 6, 6, 3, 8, 3, 1, 11, 5, 3, 4, 3, 10, 12, 1, 3, 8, 8, 1, 12, 5, 3, 9, 3, 5, 12, 1, 8, 8, 3, 1, 12, 17, 3, 4, 3, 5, 17, 1, 10, 8, 3, 6, 12, 5, 3, 11, 8, 5, 12, 1, 3, 13

Offset: 1

Omar E. Pol, Apr 30 2020

The one-part partition n = n is included in the count.

For the relation to pentagonal numbers see also A330888.

			For n = 21 there are three partitions of 21 into consecutive parts that differ by 3, including 21 as a partition. They are [21], [12, 9] and [10, 7, 4]. The number of parts of these partitions are 1, 2 and 3 respectively. The total number of parts is 1 + 2 + 3 = 6, so a(27) = 6.

Seiichi Manyama, Table of n, a(n) for n = 1..10000

Row sums of A330888.
Column k=3 of A334466.
Sequences of the same family whose consecutive parts differs by k are: A000203 (k=0), A204217 (k=1), A066839 (k=2), this sequence (k=3), A334464 (k=4), A334732 (k=5), A334949 (k=6).
Cf. A338730.

A330889 := proc(n)
    local a;
    a := 0 ;
    for k from 1 do
        if n>= A000325(k) then
            a := a+A330888(n,k);
        else
            return a;
        end if;
    end do:
end proc: # R. J. Mathar, Oct 02 2020

nmax = 100;
CoefficientList[Sum[n x^(n(3n-1)/2-1)/(1-x^n), {n, 1, nmax}]+O[x]^nmax, x] (* Jean-François Alcover, Nov 30 2020 *)
Table[Sum[If[n > 3*k*(k-1)/2 && IntegerQ[n/k - 3*(k-1)/2], k, 0], {k, Divisors[2*n]}], {n, 1, 100}] (* Vaclav Kotesovec, Oct 23 2024 *)

my(N=66, x='x+O('x^N)); Vec(sum(k=1, N, k*x^(k*(3*k-1)/2)/(1-x^k))) \\ Seiichi Manyama, Dec 04 2020

Conjecture: G.f.: Sum_{n>=1} n*x^(n*(3*n-1)/2)/(1-x^n).
Proof from Matthew C. Russell, Nov 21 2020:
The summation variable n runs over the number of parts in the partition.
For fixed n, the smallest such partition is:
1 + 4 + 7 + ... + (3n-2).
The above sum is equal to n * (3*n-1) / 2. That's where the x^(n*(3*n-1)/2) factor comes from.
Then we want to (add 1 to every part), (add 2 to every part), etc. to get 2 + 5 + 8 + ..., 3 + 6 + 9 + ..., which corresponds to adding n, 2*n, 3*n, etc. to the base partition. So we divide by (1 - x^n).
Multiply by n (to count the total number of parts) and we are done. QED
Sum_{k=1..n} a(k) ~ (2/3)^(3/2) * n^(3/2). - Vaclav Kotesovec, Oct 23 2024

More terms from R. J. Mathar, Oct 02 2020

A334949 a(n) is the total number of parts in all partitions of n into consecutive parts that differ by 6.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 4, 3, 1, 6, 1, 3, 4, 3, 1, 6, 1, 3, 4, 3, 1, 6, 1, 3, 4, 7, 1, 6, 1, 7, 4, 3, 1, 10, 1, 3, 4, 7, 1, 6, 1, 7, 4, 3, 1, 10, 1, 3, 4, 7, 6, 6, 1, 7, 4, 8, 1, 10, 1, 3, 9, 7, 1, 6, 1, 12, 4, 3, 1, 10, 6, 3, 4, 7, 1, 11, 1, 7, 4

Offset: 1

Omar E. Pol, May 27 2020

nonn

The one-part partition n = n is included in the count.

For the relation to the octagonal numbers see also A334947.

			For n = 24 there are three partitions of 24 into consecutive parts that differ by 6, including 24 as a valid partition. They are [24], [15, 9] and [14, 8, 2]. There are 1, 2 and 3 parts respectively, hence the total number of parts is 1 + 2 + 3 = 6, so a(24) = 6.

Seiichi Manyama, Table of n, a(n) for n = 1..10000

Row sums of A334947.
Column k=6 of A334466.
Sequences of the same family whose consecutive parts differs by k are: A000203 (k=0), A204217 (k=1), A066839 (k=2), A330889 (k=3), A334464 (k=4), A334732 (k=5), this sequence (k=6).
Cf. A000567, A334946, A334947, A334948, A334953.

nmax = 100;
CoefficientList[Sum[n x^(n(3n-2)-1)/(1-x^n), {n, 1, nmax}]+O[x]^nmax, x] (* Jean-François Alcover, Nov 30 2020 *)
Table[Sum[If[n > 3*k*(k-1), k, 0], {k, Divisors[n]}], {n, 1, 100}] (* Vaclav Kotesovec, Oct 22 2024 *)

my(N=66, x='x+O('x^N)); Vec(sum(k=1, N, k*x^(k*(3*k-2))/(1-x^k))) \\ Seiichi Manyama, Dec 04 2020

G.f.: Sum_{n>=1} n*x^(n*(3*n-2))/(1-x^n). (For proof, see A330889. - Omar E. Pol, Nov 22 2020)

Sum_{k=1..n} a(k) ~ 2 * n^(3/2) / 3^(3/2). - Vaclav Kotesovec, Oct 23 2024

A334466 Square array read by antidiagonals upwards: T(n,k) is the total number of parts in all partitions of n into consecutive parts that differ by k, with n >= 1, k >= 0.

Original entry on oeis.org

1, 3, 1, 4, 1, 1, 7, 3, 1, 1, 6, 1, 1, 1, 1, 12, 3, 3, 1, 1, 1, 8, 4, 1, 1, 1, 1, 1, 15, 3, 3, 3, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 18, 6, 3, 3, 3, 1, 1, 1, 1, 1, 12, 5, 4, 1, 1, 1, 1, 1, 1, 1, 1, 28, 3, 3, 3, 3, 3, 1, 1, 1, 1, 1, 1, 14, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 24, 3, 6, 3, 3, 3, 3, 1

Offset: 1

Omar E. Pol, May 01 2020

The one-part partition n = n is included in the count.
The column k is related to (k+2)-gonal numbers, assuming that 2-gonals are the nonnegative numbers, 3-gonals are the triangular numbers, 4-gonals are the squares, 5-gonals are the pentagonal numbers, and so on.
Note that the number of parts for T(n,0) = A000203(n), equaling the sum of the divisors of n.
For fixed k>0, Sum_{j=1..n} T(j,k) ~ 2^(3/2) * n^(3/2) / (3*sqrt(k)). - Vaclav Kotesovec, Oct 23 2024

			Square array starts:
   n\k|   0  1  2  3  4  5  6  7  8  9 10 11 12
   ---+---------------------------------------------
   1  |   1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
   2  |   3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
   3  |   4, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
   4  |   7, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
   5  |   6, 3, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
   6  |  12, 4, 3, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, ...
   7  |   8, 3, 1, 3, 1, 3, 1, 1, 1, 1, 1, 1, 1, ...
   8  |  15, 1, 3, 1, 3, 1, 3, 1, 1, 1, 1, 1, 1, ...
   9  |  13, 6, 4, 3, 1, 3, 1, 3, 1, 1, 1, 1, 1, ...
  10  |  18, 5. 3. 1. 3. 1, 3, 1, 3, 1, 1, 1, 1, ...
  11  |  12, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 1, 1, ...
  12  |  28, 4, 6, 4, 3, 1, 3, 1, 3, 1, 3, 1, 1, ...
  ...
For n = 9 we have that:
For k = 0 the partitions of 9 into consecutive parts that differ by 0 (or simply: the partitions of 9 into equal parts) are [9], [3,3,3], [1,1,1,1,1,1,1,1,1]. In total there are 13 parts, so T(9,0) = 13.
For k = 1 the partitions of 9 into consecutive parts that differ by 1 (or simply: the partitions of 9 into consecutive parts) are [9], [5,4], [4,3,2]. In total there are six parts, so T(9,1) = 6.
For k = 2 the partitions of 9 into consecutive parts that differ by 2 are [9], [5, 3, 1]. In total there are four parts, so T(9,2) = 4.

Columns k: A000203 (k=0), A204217 (k=1), A066839 (k=2), A330889 (k=3), A334464 (k=4), A334732 (k=5), A334949 (k=6), A377300 (k=7), A377301 (k=8).
Triangles whose row sums give the column k: A127093 (k=0), A285914 (k=1), A330466 (k=2) (conjectured), A330888 (k=3), A334462 (k=4), A334540 (k=5), A339947 (k=6).
Sequences of number of partitions related to column k: A000005 (k=0), A001227 (k=1), A038548 (k=2), A117277 (k=3), A334461 (k=4), A334541 (k=5), A334948 (k=6).
Tables of partitions related to column k: A010766 (k=0), A286001 (k=1), A332266 (k=2), A334945 (k=3), A334618 (k=4).
Polygonal numbers related to column k: A001477 (k=0), A000217 (k=1), A000290 (k=2), A000326 (k=3), A000384 (k=4), A000566 (k=5), A000567 (k=6).
Cf. A051735, A237048, A303300, A330887, A334460, A334465, A334946.
Cf. A038040, A245579, A060872, A334467, A327262, A334733, A334953.
Cf. A057145, A323345.

nmax = 14;
col[k_] := col[k] = CoefficientList[Sum[n x^(n(k n - k + 2)/2)/(1 - x^n), {n, 1, nmax}] + O[x]^(nmax+1), x];
T[n_, k_] := col[k][[n+1]];
Table[T[n-k, k], {n, 1, nmax}, {k, 0, n-1}] // Flatten (* Jean-François Alcover, Nov 30 2020 *)

The g.f. for column k is Sum_{n>=1} n*x^(n*(k*n-k+2)/2)/(1-x^n). (For proof, see A330889. - N. J. A. Sloane, Nov 21 2020)

A334464 a(n) is the total number of parts in all partitions of n into consecutive parts that differ by 4.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 4, 3, 1, 6, 1, 3, 4, 3, 1, 6, 1, 3, 4, 7, 1, 6, 1, 7, 4, 3, 1, 10, 1, 3, 4, 7, 1, 6, 1, 7, 9, 3, 1, 10, 1, 8, 4, 7, 1, 6, 6, 7, 4, 3, 1, 15, 1, 3, 4, 7, 6, 12, 1, 7, 4, 8, 1, 16, 1, 3, 9, 7, 1, 12, 1, 12, 4, 3, 1, 16, 6, 3, 4, 7, 1, 17, 8, 7, 4

Offset: 1

Omar E. Pol, May 05 2020

nonn

The one-part partition n = n is included in the count.

For the relation to hexagonal numbers see also A334462.

			For n = 28 there are three partitions of 28 into consecutive parts that differ by 4, including 28 as a valid partition. They are [28], [16, 12] and [13, 9, 5, 1]. The number of parts of these partitions are 1, 2, 4 respectively. The total number of parts is 1 + 2 + 4 = 7, so a(28) = 7.

Seiichi Manyama, Table of n, a(n) for n = 1..10000

Row sums of A334462.
Column k=4 of A334466.
Cf. A000384.
Sequences of the same family whose consecutive parts differs by k are: A000203 (k=0), A204217 (k=1), A066839 (k=2), A330889 (k=3), this sequence (k=4), A334732 (k=5), A334949 (k=6).
Cf. A334461.

nmax = 100;
CoefficientList[Sum[n x^(n(2n-1)-1)/(1-x^n), {n, 1, nmax}]+O[x]^nmax, x] (* Jean-François Alcover, Nov 30 2020 *)
Table[Sum[If[n > 2*k*(k-1), k, 0], {k, Divisors[n]}], {n, 1, 100}] (* Vaclav Kotesovec, Oct 22 2024 *)

my(N=66, x='x+O('x^N)); Vec(sum(k=1, N, k*x^(k*(2*k-1))/(1-x^k))) \\ Seiichi Manyama, Dec 04 2020

G.f.: Sum_{n>=1} n*x^(n*(2*n-1))/(1-x^n). (For proof, see A330889. - N. J. A. Sloane, Nov 21 2020)

Sum_{k=1..n} a(k) ~ sqrt(2) * n^(3/2) / 3. - Vaclav Kotesovec, Oct 23 2024

A334467 Square array read by antidiagonals upwards: T(n,k) is the sum of all parts of all partitions of n into consecutive parts that differ by k, with n >= 1, k >= 0.

Original entry on oeis.org

1, 4, 1, 6, 2, 1, 12, 6, 2, 1, 10, 4, 3, 2, 1, 24, 10, 8, 3, 2, 1, 14, 12, 5, 4, 3, 2, 1, 32, 14, 12, 10, 4, 3, 2, 1, 27, 8, 7, 6, 5, 4, 3, 2, 1, 40, 27, 16, 14, 12, 5, 4, 3, 2, 1, 22, 20, 18, 8, 7, 6, 5, 4, 3, 2, 1, 72, 22, 20, 18, 16, 14, 6, 5, 4, 3, 2, 1, 26, 24, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1

Offset: 1

Omar E. Pol, May 05 2020

			Array begins:
     k  0   1   2   3   4   5   6   7   8   9  10
   n +------------------------------------------------
   1 |  1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1, ...
   2 |  4,  2,  2,  2,  2,  2,  2,  2,  2,  2,  2, ...
   3 |  6,  6,  3,  3,  3,  3,  3,  3,  3,  3,  3, ...
   4 | 12,  4,  8,  4,  4,  4,  4,  4,  4,  4,  4, ...
   5 | 10, 10,  5, 10,  5,  5,  5,  5,  5,  5,  5, ...
   6 | 24, 12, 12,  6, 12,  6,  6,  6,  6,  6,  6, ...
   7 | 14, 14,  7, 14,  7, 14,  7,  7,  7,  7,  7, ...
   8 | 32,  8, 16,  8, 16,  8, 16,  8,  8,  8,  8, ...
   9 | 27, 27, 18, 18,  9, 18,  9, 18,  9,  9,  9, ...
  10 | 40, 20, 20, 10, 20, 20, 20, 10, 20, 10, 10, ...
...

Columns k: A038040 (k=0), A245579 (k=1), A060872 (k=2), A334463 (k=3), A327262 (k=4), A334733 (k=5), A334953 (k=6).
Every diagonal starting with 1 gives A000027.
Sequences of number of parts related to column k: A000203 (k=0), A204217 (k=1), A066839 (k=2) (conjectured), A330889 (k=3), A334464 (k=4), A334732 (k=5), A334949 (k=6).
Sequences of number of partitions related to column k: A000005 (k=0), A001227 (k=1), A038548 (k=2), A117277 (k=3), A334461 (k=4), A334541 (k=5), A334948 (k=6).
Polygonal  numbers related to column k: A001477 (k=0), A000217 (k=1), A000290 (k=2), A000326 (k=3), A000384 (k=4), A000566 (k=5), A000567 (k=6).
Cf. A245579, A323345, A334466.

nmax = 13;
col[k_] := col[k] = CoefficientList[Sum[x^(n(k n - k + 2)/2 - 1)/(1 - x^n), {n, 1, nmax}] + O[x]^nmax, x];
T[n_, k_] := n col[k][[n]];
Table[T[n-k, k], {n, 1, nmax}, {k, 0, n-1}] // Flatten (* Jean-François Alcover, Nov 30 2020 *)

T(n,k) = n*A323345(n,k).

cp's OEIS Frontend

A330889 a(n) is the total number of parts in all partitions of n into consecutive parts that differ by 3.

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A334949 a(n) is the total number of parts in all partitions of n into consecutive parts that differ by 6.

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A334466 Square array read by antidiagonals upwards: T(n,k) is the total number of parts in all partitions of n into consecutive parts that differ by k, with n >= 1, k >= 0.

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A334464 a(n) is the total number of parts in all partitions of n into consecutive parts that differ by 4.

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A334467 Square array read by antidiagonals upwards: T(n,k) is the sum of all parts of all partitions of n into consecutive parts that differ by k, with n >= 1, k >= 0.

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