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 1-4 of 4 results.

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

Views

Author

Omar E. Pol, May 01 2020

Keywords

Comments

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

Examples

			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.

Crossrefs

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.

Programs

  • Mathematica
    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 *)

Formula

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)

A334462 Irregular triangle read by rows: T(n,k) is the number of parts in the partition of n into k consecutive parts that differ by 4, n >= 1, k >= 1, and the first element of column k is in the row that is the k-th hexagonal number (A000384).

Original entry on oeis.org

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

Views

Author

Omar E. Pol, May 05 2020

Keywords

Comments

Since the trivial partition n is counted, so T(n,1) = 1.
This is also an irregular triangle read by rows: T(n,k), n >= 1, k >= 1, in which column k lists k's interleaved with k-1 zeros, and the first element of column k is in the row that is the k-th hexagonal number.
This triangle can be represented with a diagram of overlapping curves, in which every column of triangle is represented by a periodic curve.
For a general theorem about the triangles of this family see A285914.

Examples

			Triangle begins (rows 1..28):
1;
1;
1;
1;
1;
1, 2;
1, 0;
1, 2;
1, 0;
1, 2;
1, 0;
1, 2;
1, 0;
1, 2;
1, 0, 3;
1, 2, 0;
1, 0, 0;
1, 2, 3;
1, 0, 0;
1, 2, 0;
1, 0, 3;
1, 2, 0;
1, 0, 0;
1, 2, 3;
1, 0, 0;
1, 2, 0;
1, 0, 3;
1, 2, 0, 4;
...
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, so the 28th row of the triangle is [1, 2, 0, 4].

Crossrefs

Triangles of the same family where the parts differ by d are A127093 (d=0), A285914 (d=1), A330466 (d=2), A330888 (d=3), this sequence (d=4), A334540 (d=5).
Cf. A000384, A334460, A334461.

Formula

T(n,k) = k*A334460(n,k).

A334732 a(n) is the total number of parts in all partitions of n into consecutive parts that differ by 5.

Original entry on oeis.org

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

Views

Author

Omar E. Pol, May 09 2020

Keywords

Comments

The one-part partition n = n is included in the count.
For the relation to heptagonal numbers see also A334540.

Examples

			For n = 27 there are three partitions of 27 into consecutive parts that differ by 5, including 27 as a valid partition. They are [27], [16, 11] and [14, 9, 4]. The number of parts of these partitions are 1, 2, 3 respectively and the total number of parts is 1 + 2 + 3 = 6, so the a(27) = 6.

Links

Crossrefs

Row sums of A334540.
Column k=5 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), this sequence (k=5), A334949 (k=6).
Cf. A000566, A334465, A334540, A334541, A334733.

Programs

  • Maple
    A334732 := proc(n)
    local a,k;
    a := 0 ;
    for k from 1 do
        if n>= A000566(k) then
            a := a+A334540(n,k);
        else
            return a;
        end if;
    end do:
end proc:
seq(A334732(n),n=1..120) ; # R. J. Mathar, Oct 02 2020
  • Mathematica
    nmax = 100;
CoefficientList[Sum[n x^(n(5n-3)/2-1)/(1-x^n), {n, 1, nmax}]+O[x]^nmax, x] (* Jean-François Alcover, Nov 30 2020 *)
Table[Sum[If[n > 5*k*(k-1)/2 && IntegerQ[n/k - 5*(k-1)/2], k, 0], {k, Divisors[2*n]}], {n, 1, 100}] (* Vaclav Kotesovec, Oct 23 2024 *)
  • PARI
    my(N=66, x='x+O('x^N)); Vec(sum(k=1, N, k*x^(k*(5*k-3)/2)/(1-x^k))) \\ Seiichi Manyama, Dec 04 2020

Formula

G.f.: Sum_{n>=1} n*x^(n*(5*n-3)/2)/(1-x^n) (for proof see A330889).
Sum_{k=1..n} a(k) ~ 2^(3/2) * n^(3/2) / (3*sqrt(5)). - Vaclav Kotesovec, Oct 23 2024

Extensions

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

A334947 Irregular triangle read by rows: T(n,k) is the number of parts in the partition of n into k consecutive parts that differ by 6, n >= 1, k >= 1, and the first element of column k is in the row that is the k-th octagonal number (A000567).

Original entry on oeis.org

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

Views

Author

Omar E. Pol, May 27 2020

Keywords

Comments

Since the trivial partition n is counted, so T(n,1) = 1.
This is an irregular triangle read by rows: T(n,k), n >= 1, k >= 1, in which column k lists k's interleaved with k-1 zeros, and the first element of column k is in the row that is the k-th octagonal number.
This triangle can be represented with a diagram of overlapping curves, in which every column of triangle is represented by a periodic curve.
For a general theorem about the triangles of this family see A285914.

Examples

			Triangle begins (rows 1..24).
1;
1;
1;
1;
1;
1;
1;
1, 2;
1, 0;
1, 2;
1, 0;
1, 2;
1, 0;
1, 2;
1, 0;
1, 2;
1, 0;
1, 2;
1, 0;
1, 2;
1, 0, 3;
1, 2, 0;
1, 0, 0;
1, 2, 3;
...
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, so the 24th row of this triangle is [1, 2, 3].

Crossrefs

Row sums give A334949.
Triangles of the same family where the parts differ by d are A127093 (d=0), A285914 (d=1), A330466 (d=2), A330888 (d=3), A334462 (d=4), A334540 (d=5), this sequence (d=6).
Cf. A000567, A334946, A334948, A334953.

Formula

T(n,k) = k*A334946(n,k).
Showing 1-4 of 4 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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