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.

Previous Showing 21-30 of 34 results. Next

A120477 Apply partial sum operator 5 times to partition numbers.

Original entry on oeis.org

1, 6, 22, 63, 155, 343, 702, 1352, 2480, 4370, 7445, 12323, 19894, 31421, 48675, 74111, 111099, 164221, 239656, 345670, 493243, 696861, 975518, 1353971, 1864315, 2547941, 3457972, 4662273, 6247169, 8322010, 11024775, 14528914, 19051697
Offset: 0

Views

Author

Jonathan Vos Post, Jul 21 2006

Keywords

Comments

In general, if g.f. = 1/(1-x)^m * Product_{k>=1} 1/(1-x^k), then a(n) ~ 2^(m/2 - 2) * 3^((m-1)/2) * n^(m/2 - 1) * exp(Pi*sqrt(2*n/3)) / Pi^m. - Vaclav Kotesovec, Oct 30 2015

Crossrefs

Column k=6 of A292508.

Programs

  • Maple
    with(combinat): g:=1/(1-x)^5/product(1-x^k,k=1..50): gser:=series(g,x=0,40): seq(coeff(gser,x,n),n=0..37); # Emeric Deutsch, Jul 24 2006
  • Mathematica
    nmax = 50; CoefficientList[Series[1/((1-x)^5 * Product[1-x^k, {k, 1, nmax}]), {x, 0, nmax}], x] (* Vaclav Kotesovec, Oct 30 2015 *)

Formula

G.f.: 1/((1-x)^5*Product_{k>=1} (1-x^k)). - Emeric Deutsch, Jul 24 2006
a(n) ~ 9*sqrt(2)*n^(3/2) * exp(Pi*sqrt(2*n/3)) / Pi^5. - Vaclav Kotesovec, Oct 30 2015

Extensions

More terms from Emeric Deutsch, Jul 24 2006

A299779 Triangle read by rows: T(n,k) is the total number of cliques of size k in all partitions of all positive integers <= n.

Original entry on oeis.org

1, 2, 1, 5, 1, 1, 9, 3, 1, 1, 17, 5, 2, 1, 1, 28, 9, 4, 2, 1, 1, 47, 14, 7, 3, 2, 1, 1, 73, 24, 10, 6, 3, 2, 1, 1, 114, 35, 17, 9, 5, 3, 2, 1, 1, 170, 55, 25, 14, 8, 5, 3, 2, 1, 1, 253, 80, 38, 20, 13, 7, 5, 3, 2, 1, 1, 365, 118, 55, 31, 18, 12, 7, 5, 3, 2, 1, 1, 525, 167, 80, 44, 27, 17, 11, 7, 5, 3, 2, 1, 1
Offset: 1

Views

Author

Omar E. Pol, Apr 04 2018

Keywords

Comments

Column k gives the partial sums of the k-th column of triangle A197126.

Examples

			Triangle begins:
    1;
    2,   1;
    5,   1,  1;
    9,   3,  1,  1;
   17,   5,  2,  1,  1;
   28,   9,  4,  2,  1,  1;
   47,  14,  7,  3,  2,  1,  1;
   73,  24, 10,  6,  3,  2,  1,  1;
  114,  35, 17,  9,  5,  3,  2,  1,  1;
  170,  55, 25, 14,  8,  5,  3,  2,  1,  1;
  253,  80, 38, 20, 13,  7,  5,  3,  2,  1,  1;
  365, 118, 55, 31, 18, 12,  7,  5,  3,  2,  1,  1;
  525, 167, 80, 44, 27, 17, 11,  7,  5,  3,  2,  1,  1;
...
		

Crossrefs

Column 1 gives A000097.
Row sums give A014153.

Programs

  • Maple
    b:= proc(n, p, k) option remember; `if`(n=0, [1, 0], `if`(p<1, [0, 0],
          add((l->`if`(m=k, l+[0, l[1]], l))(b(n-p*m, p-1, k)), m=0..n/p)))
        end:
    T:= proc(n, k) option remember;
          b(n$2, k)[2]+`if`(nAlois P. Heinz, Apr 27 2018
  • Mathematica
    b[n_, p_, k_] := b[n, p, k] = If[n == 0, {1, 0}, If[p < 1, {0, 0}, Sum[ Function[l, If[m==k, l+{0, l[[1]]}, l]][b[n-p*m, p-1, k]], {m, 0, n/p}]]];
    T[n_, k_] := b[n, n, k][[2]] + If[n < k, 0, T[n-1, k]];
    Table[Table[T[n, k], {k, 1, n}], {n, 1, 20}] // Flatten (* Jean-François Alcover, Dec 07 2019, after Alois P. Heinz *)

Formula

T(n,k) = Sum_{j=k..n} A197126(j,k).
T(2n+1,n+1) = A000041(n). - Alois P. Heinz, Apr 27 2018
Sum_{k=1..n} k * T(n,k) = A284870(n). - Alois P. Heinz, May 14 2018

A103650 G.f.: x^2/((1-x^2)^2*Product_{i>0}(1-x^i)).

Original entry on oeis.org

0, 1, 1, 4, 5, 12, 16, 31, 42, 72, 98, 155, 210, 315, 423, 610, 812, 1136, 1498, 2047, 2674, 3585, 4642, 6125, 7865, 10240, 13046, 16791, 21237, 27060, 33993, 42933, 53591, 67155, 83332, 103687, 127956, 158196, 194217, 238720, 291663, 356582
Offset: 1

Views

Author

Vladeta Jovovic, Mar 26 2005

Keywords

Comments

Let pi be a partition of n and b(pi,k) = Sum p, where p runs over all distinct parts p of pi whose multiplicities are >=k. Let T(n,k) = Sum b(pi,k), when pi runs over all partitions pi of n. G.f. for T(n,k) is x^k/((1-x^k)^2*Product_{i>0}(1-x^i)). a(n) = T(n,2).

Examples

			Partitions of 4 are [1, 1, 1, 1], [1, 1, 2], [2, 2], [1, 3], [4] and a(4) = 1 + 1 + 2 + 0 + 0 = 4.
		

Crossrefs

Cf. A014153.

Programs

  • Mathematica
    Drop[ CoefficientList[ Series[ x^2/((1 - x^2)^2*Product[(1 - x^i), {i, 50}]), {x, 0, 42}], x], 1] (* Robert G. Wilson v, Mar 29 2005 *)
    Table[Sum[PartitionsP[k]*(n-k)*(1 + (-1)^(n-k))/4, {k, 0, n}], {n, 1, 50}] (* Vaclav Kotesovec, Jul 30 2016 *)

Formula

a(n) = Sum_{k>0} k * A264404(n,k). - Alois P. Heinz, Nov 29 2015
For n>2, a(n) is the Euler transform of [1,3,1,1,1,1,...]. - Benedict W. J. Irwin, Jul 29 2016
a(n) ~ sqrt(3) * exp(Pi*sqrt(2*n/3)) / (8*Pi^2). - Vaclav Kotesovec, Jul 30 2016

Extensions

More terms from Robert G. Wilson v, Mar 29 2005

A116930 Sum of parts, counted without multiplicities, in all partitions of n into odd parts.

Original entry on oeis.org

1, 1, 4, 5, 10, 14, 22, 31, 44, 61, 82, 111, 145, 191, 245, 316, 399, 506, 631, 788, 973, 1200, 1468, 1792, 2174, 2630, 3167, 3802, 4547, 5422, 6445, 7638, 9029, 10642, 12515, 14679, 17181, 20061, 23379, 27185, 31554, 36551, 42268, 48787, 56224, 64681, 74300
Offset: 1

Views

Author

Emeric Deutsch, Feb 27 2006

Keywords

Examples

			a(5) = 10 because the partitions of 5 into odd parts are [5], [3,1,1] and [1,1,1,1,1], with sum of the parts, counted without multiplicities 5 + (3+1) + 1 = 10.
a(5) = 10: There are three partitions of 5 into distinct parts, namely [5], [4,1], and [3,2]. We have phi(5) + phi(4) + phi(1) + phi(3) + phi(2) = 4 + 2 + 1 + 2 + 1 = 10. - _Peter Bala_, Dec 26 2013
		

Crossrefs

Programs

  • Maple
    f:=x*(1+x^2)/(1-x^2)^2/product(1-x^(2*j-1),j=1..40): fser:=series(f,x=0,55): seq(coeff(fser,x^n),n=1..49);
    # second Maple program:
    b:= proc(n, i) option remember; `if`(n=0, [1, 0],
          `if`(n>i*(i+1)/2, 0, b(n, i-1)+(p-> p+[0, p[1]
           *numtheory[phi](i)])(b(n-i, min(n-i, i-1)))))
        end:
    a:= n-> b(n$2)[2]:
    seq(a(n), n=1..50);  # Alois P. Heinz, Aug 15 2021
  • Mathematica
    b[n_, i_] := b[n, i] = If[n == 0, {1, 0},
         If[n > i (i + 1)/2, {0, 0}, b[n, i - 1] +
         With[{p = b[n - i, Min[n-i, i-1]]}, p + {0, p[[1]]*EulerPhi[i]}]]];
    a[n_] := b[n, n][[2]];
    Table[a[n], {n, 1, 50}] (* Jean-François Alcover, Mar 13 2022, after Alois P. Heinz *)

Formula

a(n) = Sum_{k=1..n} k * A116929(n,k).
G.f.: x(1+x^2)/[(1-x^2)^2*product(1-x^(2*j-1),j=1..infinity)].
a(n) = Sum_{parts k in all partitions of n into distinct parts} phi(k), where phi(k) is the Euler totient function (see A000010). An example is given below. - Peter Bala, Dec 26 2013
a(n) ~ 3^(3/4) * exp(Pi*sqrt(n/3)) * n^(1/4) / (2*Pi^2). - Vaclav Kotesovec, Jun 11 2025

A213180 Sum over all partitions lambda of n of Sum_{p:lambda} p^m(p,lambda), where m(p,lambda) is the multiplicity of part p in lambda.

Original entry on oeis.org

0, 1, 3, 7, 16, 28, 59, 91, 170, 269, 450, 655, 1162, 1602, 2527, 3793, 5805, 8034, 12660, 17131, 26484, 37384, 53738, 73504, 114683, 153613, 221225, 313339, 453769, 609179, 927968, 1223909, 1804710, 2522264, 3539835, 4855420, 7439870, 9765555, 14009545
Offset: 0

Views

Author

Alois P. Heinz, Feb 27 2013

Keywords

Examples

			a(6) = 59: (1^6) + (2+1^4) + (2^2+1^2) + (2^3) + (3+1^3) + (3+2+1) + (3^2) + (4+1^2) + (4+2) + (5+1) + (6) = 1+3+5+8+4+6+9+5+6+6+6 = 59.
		

Crossrefs

Cf. A000070 (Sum 1), A006128 (Sum m), A014153 (Sum p), A024786 (Sum floor(1/m)), A066183 (Sum p^2*m), A066186 (Sum p*m), A073336 (Sum floor(m/p)), A116646 (Sum delta(m,2)), A117524 (Sum delta(m,3)), A103628 (Sum delta(m,1)*p), A117525 (Sum delta(m,2)*p), A197126, A213191.

Programs

  • Maple
    b:= proc(n, p) option remember; `if`(n=0, [1, 0], `if`(p<1, [0, 0],
          add((l->`if`(m=0, l, l+[0, l[1]*p^m]))(b(n-p*m, p-1)), m=0..n/p)))
        end:
    a:= n-> b(n, n)[2]:
    seq(a(n), n=0..40);
  • Mathematica
    b[n_, p_] := b[n, p] = If[n==0, {1, 0}, If[p<1, {0, 0}, Sum[Function[l, If[m==0, l, l+{0, l[[1]]*p^m}]][b[n-p*m, p-1]], {m, 0, n/p}]]]; a[n_] := b[n, n][[2]]; Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Feb 15 2017, translated from Maple *)

Formula

From Vaclav Kotesovec, May 24 2018: (Start)
a(n) ~ c * 3^(n/3), where
c = 5.0144820680945600131204662934686439430547... if mod(n,3)=0
c = 4.6144523178014379613985400559486878971522... if mod(n,3)=1
c = 4.5237761454818383598444208605033385016299... if mod(n,3)=2
(End)

A270143 a(n) = Sum_{k=0..n} (-1)^(k+1) * k * A000041(n-k).

Original entry on oeis.org

0, 1, -1, 3, -2, 6, -3, 11, -4, 19, -4, 31, -2, 50, 3, 79, 15, 122, 38, 187, 78, 284, 146, 426, 257, 635, 431, 939, 701, 1377, 1110, 2007, 1718, 2906, 2613, 4178, 3914, 5971, 5781, 8482, 8440, 11976, 12191, 16816, 17438, 23483, 24730, 32615, 34794, 45070
Offset: 0

Views

Author

Vaclav Kotesovec, Mar 12 2016

Keywords

Comments

Convolution of A000041 and A181983.

Crossrefs

Programs

  • Mathematica
    Table[Sum[(-1)^(n-k+1)*PartitionsP[k]*(n-k), {k, 0, n}], {n, 0, 100}]
    nmax = 100; CoefficientList[Series[x/(1 + x)^2 * Product[1/(1 - x^k), {k, 1, nmax}], {x, 0, nmax}], x]

Formula

a(n) = Sum_{k=0..n} (-1)^(n-k+1) * (n-k) * A000041(k).
a(n) ~ A000041(n)/4.
a(n) ~ exp(Pi*sqrt(2*n/3)) / (16*n*sqrt(3)).
G.f.: x/(1+x)^2 * Product_{k>=1} 1/(1-x^k).

A141157 Triangle read by rows, A000012 * A140207.

Original entry on oeis.org

1, 2, 1, 3, 2, 2, 4, 3, 4, 3, 5, 4, 6, 6, 5, 6, 5, 8, 9, 10, 7, 7, 6, 10, 12, 15, 14, 11, 8, 7, 12, 15, 20, 21, 22, 15, 9, 8, 14, 18, 25, 28, 33, 30, 22, 10, 9, 16, 21, 30, 35, 44, 45, 44, 30, 11, 10, 18, 24, 35, 42, 55, 60, 66, 60, 42, 12, 11, 20, 27, 40, 49, 66, 75, 88, 90, 84, 56
Offset: 0

Views

Author

Keywords

Comments

Right border = partition numbers, A000041.
Row sums = A014153: (1, 3, 7, 14, 26, 45, 75,...).

Examples

			First few rows of the triangle are:
1;
2, 1;
3, 2, 2;
4, 3, 4, 3;
5, 4, 6, 6, 5;
6, 5, 8, 9, 10, 7;
7, 6, 10, 12, 15, 14, 11;
8, 7, 12, 15, 20, 21, 22, 15;
9, 8, 14, 18, 25, 28, 33, 30, 22;
10, 9, 16, 21, 30, 35, 44, 45, 44, 30;
...
		

Crossrefs

Formula

Triangle read by rows, A000012 * A140207; equivalent to taking partial sums of triangle A140207 terms, by columns.

Extensions

a(60) split and more terms from Georg Fischer, May 29 2023

A213679 Total sum of parts <= n of multiplicity 0 in all partitions of n.

Original entry on oeis.org

0, 0, 3, 11, 36, 79, 186, 345, 672, 1163, 2026, 3273, 5388, 8301, 12912, 19349, 28961, 42071, 61253, 86921, 123404, 171972, 239020, 327386, 447743, 604255, 813645, 1084657, 1441643, 1899450, 2496510, 3255653, 4234822, 5472953, 7053217, 9038784, 11554020
Offset: 0

Views

Author

Alois P. Heinz, Mar 04 2013

Keywords

Examples

			The partitions of n=4 are [1,1,1,1], [2,1,1], [2,2], [3,1], [4].  Parts <= 4 with multiplicity m=0 sum up to (2+3+4)+(3+4)+(1+3+4)+(2+4)+(1+2+3) = 36, thus a(4) = 36.
		

Crossrefs

Column k=0 of A222730.

Programs

  • Maple
    b:= proc(n, p) option remember; `if`(n=0 and p=0, [1, 0], `if`(p<1, [0$2],
          add((l->`if`(m=0, l+[0, l[1]*p], l))(b(n-p*m, p-1)), m=0..n/p)))
        end:
    a:= n-> b(n, n)[2]:
    seq(a(n), n=0..55);
  • Mathematica
    b[n_, p_] := b[n, p] = If[n == 0 && p == 0, {1, 0}, If[p == 0, Array[0&, n+2], Sum[Function[l, ReplacePart[l, m+2 -> p*l[[1]] + l[[m+2]]]][Join[b[n-p*m, p-1], Array[0&, p*m]]], {m, 0, n/p}]]]; a[n_] := b[n, n][[2]]; Table[a[n], {n, 0, 55}] (* Jean-François Alcover, Jan 24 2014, after Alois P. Heinz *)

Formula

a(n) = A000217(n)*A000041(n)-A014153(n-1).

A359279 Irregular triangle T(n,k) (n>=1, k>=1) read by rows in which the length of row n equals the partition number A000041(n-1) and every column k gives the positive triangular numbers A000217.

Original entry on oeis.org

1, 3, 6, 1, 10, 3, 1, 15, 6, 3, 1, 1, 21, 10, 6, 3, 3, 1, 1, 28, 15, 10, 6, 6, 3, 3, 1, 1, 1, 1, 36, 21, 15, 10, 10, 6, 6, 3, 3, 3, 3, 1, 1, 1, 1, 45, 28, 21, 15, 15, 10, 10, 6, 6, 6, 6, 3, 3, 3, 3, 1, 1, 1, 1, 1, 1, 1, 55, 36, 28, 21, 21, 15, 15, 10, 10, 10, 10, 6, 6, 6, 6, 3, 3, 3, 3, 3, 3, 3, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 1

Views

Author

Omar E. Pol, Dec 23 2022

Keywords

Comments

All divisors of the largest partition into consecutive parts of all terms in row n are also all parts of all partitions of n.

Examples

			Triangle begins:
   1;
   3;
   6,  1;
  10,  3,  1;
  15,  6,  3,  1,  1;
  21, 10,  6,  3,  3,  1,  1;
  28, 15, 10,  6,  6,  3,  3, 1, 1, 1, 1;
  36, 21, 15, 10, 10,  6,  6, 3, 3, 3, 3, 1, 1, 1, 1;
  45, 28, 21, 15, 15, 10, 10, 6, 6, 6, 6, 3, 3, 3, 3, 1, 1, 1, 1, 1, 1, 1;
...
From _Omar E. Pol_, Feb 28 2023: (Start)
For n = 4 the fourth row is [10, 3, 1]. The largest partition into consecutive parts of every term are respectively [4, 3, 2, 1], [2, 1], [1]. The divisors of these parts are [(1, 2, 4), (1, 3), (1, 2), (1)], [(1, 2), (1)], [1]. These 12 divisors are also all parts of all partitions of 4. They are  [(4), (2, 2), (3, 1), (2, 1, 1), (1, 1, 1, 1)]. (End)
		

Crossrefs

Row sums give A014153 (convolution of A000041 and A000027).
This sequence has the same row sums as A176206, A299779 and A359350.

Programs

  • Mathematica
    A359279[rowmax_]:=Table[Flatten[Table[ConstantArray[(n-m)(n-m+1)/2,PartitionsP[m]-PartitionsP[m-1]],{m,0,n-1}]],{n,rowmax}];
    A359279[10] (* Generates 10 rows *) (* Paolo Xausa, Mar 06 2023 *)
  • PARI
    A359279(rowmax)=vector(rowmax,n,concat(vector(n,m,vector(numbpart(m-1)-numbpart(m-2),i,(n-m+1)*(n-m+2)/2))));
    A359279(10) \\ Generates 10 rows - Paolo Xausa, Mar 06 2023

Formula

T(n,k) = A000217(A336811(n,k)).

A359350 Irregular triangle T(n,k) (n >= 1, k >= 1) read by rows: row n is constructed by replacing A336811(n,k) with the largest partition into consecutive parts of A000217(A336811(n,k)).

Original entry on oeis.org

1, 2, 1, 3, 2, 1, 1, 4, 3, 2, 1, 2, 1, 1, 5, 4, 3, 2, 1, 3, 2, 1, 2, 1, 1, 1, 6, 5, 4, 3, 2, 1, 4, 3, 2, 1, 3, 2, 1, 2, 1, 2, 1, 1, 1, 7, 6, 5, 4, 3, 2, 1, 5, 4, 3, 2, 1, 4, 3, 2, 1, 3, 2, 1, 3, 2, 1, 2, 1, 2, 1, 1, 1, 1, 1, 8, 7, 6, 5, 4, 3, 2, 1, 6, 5, 4, 3, 2, 1, 5, 4, 3, 2, 1, 4, 3, 2, 1, 4, 3, 2, 1, 3, 2, 1
Offset: 1

Views

Author

Omar E. Pol, Dec 27 2022

Keywords

Comments

All divisors of all terms in row n are also all parts of all partitions of n.
The terms of row n listed in nonincreasing order give the n-th row of A176206.
The number of k's in row n is equal to A000041(n-k), 1 <= k <= n.
The number of terms >= k in row n is equal to A000070(n-k), 1 <= k <= n.
The number of k's in the first n rows (or in the first A014153(n-1) terms of the sequence) is equal to A000070(n-k), 1 <= k <= n.
The number of terms >= k in the first n rows (or in the first A014153(n-1) terms of the sequence) is equal to A014153(n-k), 1 <= k <= n.
Row n is constructed replacing A336811(n,k) with the largest partition into consecutive parts of A359279(n,k).

Examples

			Triangle begins:
  1;
  2, 1;
  3, 2, 1, 1;
  4, 3, 2, 1, 2, 1, 1;
  5, 4, 3, 2, 1, 3, 2, 1, 2, 1, 1, 1;
  6, 5, 4, 3, 2, 1, 4, 3, 2, 1, 3, 2, 1, 2, 1, 2, 1, 1, 1;
  ...
Or also the triangle begins:
  [1];
  [2, 1];
  [3, 2, 1],          [1];
  [4, 3, 2, 1],       [2, 1],       [1];
  [5, 4, 3, 2, 1],    [3, 2, 1],    [2, 1],    [1],    [1];
  [6, 5, 4, 3, 2, 1], [4, 3, 2, 1], [3, 2, 1], [2, 1], [2, 1], [1], [1];
  ...
For n = 3 the third row is [3, 2, 1, 1]. The divisors of these terms are [1, 3], [1, 2], [1], [1]. These six divisors are also all parts of all partitions of 3. They are [3], [2, 1], [1, 1, 1].
		

Crossrefs

Row sums give A014153 (convolution of A000041 and A000027).
Row lengths give A000070.
Row n has A000041(n-1) blocks.
This triangle has the same row sums as A176206, A299779 and A359279.

Programs

  • Mathematica
    A359350row[n_]:=Flatten[Table[ConstantArray[Range[n-m,1,-1],PartitionsP[m]-PartitionsP[m-1]],{m,0,n-1}]];Array[A359350row,10] (* Paolo Xausa, Sep 01 2023 *)
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