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 1-5 of 5 results.

A175003 Triangle read by rows demonstrating Euler's pentagonal theorem for partition numbers.

Original entry on oeis.org

1, 1, 1, 2, 1, 3, 2, 5, 3, -1, 7, 5, -1, 11, 7, -2, -1, 15, 11, -3, -1, 22, 15, -5, -2, 30, 22, -7, -3, 42, 30, -11, -5, 56, 42, -15, -7, 1, 77, 56, -22, -11, 1, 101, 77, -30, -15, 2, 135, 101, -42, -22, 3, 1, 176, 135, -56, -30, 5, 1, 231, 176, -77, -42, 7, 2
Offset: 1

Views

Author

Gary W. Adamson, Apr 03 2010

Keywords

Comments

Row sums = A000041 starting with offset 1.
Sum of n-th row terms = leftmost term of next row, such that terms in each row demonstrate Euler's pentagonal theorem.
Let Q = triangle A027293 with partition numbers in each column.
Let M = a diagonalized variant of A080995 as the characteristic function of the generalized pentagonal numbers starting with offset 1: (1, 1, 0, 0, 1,...)
Sign the 1's: (++--++...) getting (1, 1, 0, 0, -1, 0, -1,...) which is the diagonal of matrix M, (as an infinite lower triangular matrix with the rest zeros).
Triangle A175003 = Q*M, with deleted zeros.
Column k starts at row A001318(k). - Omar E. Pol, Sep 21 2011
From Omar E. Pol, Apr 22 2014: (Start)
Row n has length A235963(n).
For Euler's pentagonal theorem for the sum of divisors see A238442.
Note that both of Euler's pentagonal theorems refer to generalized pentagonal numbers (A001318), not to pentagonal numbers (A000326). (End)

Examples

			Triangle begins:
    1;
    1,   1;
    2,   1;
    3,   2;
    5,   3,  -1;
    7,   5,  -1;
   11,   7,  -2,  -1;
   15,  11,  -3,  -1;
   22,  15,  -5,  -2;
   30,  22,  -7,  -3;
   42,  30, -11,  -5;
   56,  42, -15,  -7,   1;
   77,  56, -22, -11,   1;
  101,  77, -30, -15,   2;
  ...
		

Crossrefs

Formula

T(n,k) = A057077(k-1)*A000041(A195310(n,k)), n >= 1, k >= 1. - Omar E. Pol, Sep 21 2011

Extensions

Corrected and extended by Omar E. Pol, Feb 14 2013

A238442 Triangle read by rows demonstrating Euler's pentagonal theorem for the sum of divisors.

Original entry on oeis.org

1, 1, 2, 3, 1, 4, 3, 7, 4, -5, 6, 7, -1, 12, 6, -3, -7, 8, 12, -4, -1, 15, 8, -7, -3, 13, 15, -6, -4, 18, 13, -12, -7, 12, 18, -8, -6, 12, 28, 12, -15, -12, 1, 14, 28, -13, -8, 3, 24, 14, -18, -15, 4, 15, 24, 24, -12, -13, 7, 1, 31, 24, -28, -18, 6, 3
Offset: 1

Views

Author

Omar E. Pol, Feb 26 2014

Keywords

Comments

The law found by Leonhard Euler for the sum of divisors of n is that S(n) = S(n - 1) + S(n - 2) - S(n - 5) - S(n - 7) + S(n - 12) + S(n - 15) - S(n - 22) - S(n - 26) + S(n - 35) + S(n - 40) + ..., where the constants are the positive generalized pentagonal numbers, and S(0) = n, which is also a positive member of A001318.
Therefore column k lists A001318(k) together with the elements of A000203, starting at row A001318(k), but with all elements of column k multiplied by A057077(k-1).
The first element of column k is A057077(k-1)*A001318(k)which is also the last term of row A001318(k).
For Euler's pentagonal theorem for the partition numbers see A175003.
Note that both of Euler's pentagonal theorems refer to generalized pentagonal numbers (A001318), not to pentagonal numbers (A000326).

Examples

			Triangle begins:
   1;
   1,   2;
   3,   1;
   4,   3;
   7,   4,  -5;
   6,   7,  -1;
  12,   6,  -3,  -7;
   8,  12,  -4,  -1;
  15,   8,  -7,  -3;
  13,  15,  -6,  -4;
  18,  13, -12,  -7;
  12,  18,  -8,  -6,  12;
  28,  12, -15, -12,   1;
  14,  28, -13,  -8,   3;
  24,  14, -18, -15,   4,  15;
  24,  24, -12, -13,   7,   1;
  31,  24, -28, -18,   6,   3;
  18,  31, -14, -12,  12,   4;
  39,  18, -24, -28,   8,   7;
  20,  39, -24, -14,  15,   6;
  42,  20, -31, -24,  13,  12;
  32,  42, -18, -24,  18,   8, -22;
  36,  32, -39, -31,  12,  15,  -1;
  24,  36, -20, -18,  28,  13,  -3;
  60,  24, -42, -39,  14,  18,  -4;
  31,  60, -32, -20,  24,  12,  -7, -26;
  ...
For n = 21 the sum of divisors of 21 is 1 + 3 + 7 + 21 = 32. On the other hand, from Euler's Pentagonal Number Theorem we have that the sum of divisors of 21 is S_21 = S_20 + S_19 - S_16 - S_14 + S_9 + S_6, the same as the sum of the 21st row of triangle: 42 + 20 - 31 - 24 + 13 + 12 = 32, equaling the sum of divisors of 21.
For n = 22 the sum of divisors of 22 is 1 + 2 + 11 + 22 = 36. On the other hand, from Euler's Pentagonal Number Theorem we have that the sum of divisors of 22 is S_22 = S_21 + S_20 - S_17 - S_15 + S_10 + S_7 - S_0, the same as the sum of the 22nd row of triangle is 32 + 42 - 18 - 24 + 18 + 8 - 22 = 36, equaling the sum of divisors of 22. Note that S_0 = n, hence in this case S_0 = 22.
		

Crossrefs

Row sums give A000203, the sum of divisors of n.
Row n has length A235963(n).

Programs

  • Mathematica
    rows = m = 18;
    a057077[n_] := {1, 1, -1, -1}[[Mod[n, 4] + 1]];
    a001318[n_] := (1/8)((2n + 1) Mod[n, 2] + 3n^2 + 2n);
    a235963[n_] := Flatten[Table[k, {k, 0, m}, {(k+1)/(Mod[k, 2]+1)}]][[n+1]];
    T[n_, k_] := If[n == a001318[k] && k == a235963[n], a001318[k] a057077[k - 1], a057077[k - 1] DivisorSigma[1, n - a001318[k]]];
    Table[T[n, k], {n, 1, m}, {k, 1, a235963[n]}] // Flatten (* Jean-François Alcover, Nov 29 2018 *)

Formula

T(n,k) = A057077(k-1)*A001318(k), if n = A001318(k) and k = A235963(n). Otherwise T(n,k) = A057077(k-1)*A000203(n - A001318(k)), n >= 1, 1 <= k <= A235963(n).

A238132 Number of parts in all partitions of n into even number of distinct parts.

Original entry on oeis.org

0, 0, 0, 2, 2, 4, 4, 6, 6, 8, 12, 14, 18, 24, 32, 38, 50, 60, 76, 90, 110, 134, 162, 190, 228, 270, 322, 380, 446, 524, 616, 720, 838, 980, 1134, 1314, 1526, 1760, 2026, 2336, 2676, 3072, 3518, 4020, 4586, 5232, 5948, 6760, 7676, 8698, 9846, 11142, 12578
Offset: 0

Views

Author

Mircea Merca, Feb 18 2014

Keywords

Examples

			a(8)=6 because the partitions of 8 into even number of distinct parts are: 7+1, 6+2 and 5+3.
		

Crossrefs

Programs

  • Maple
    b:= proc(n, i) option remember; `if`(i*(i+1)/2n, 0, (p->
          [p[2], p[1], p[4]+p[2], p[3]+p[1]])(b(n-i, i-1)))))
        end:
    a:= n-> b(n$2)[3]:
    seq(a(n), n=0..60);  # Alois P. Heinz, Dec 27 2015
  • Mathematica
    max = 50; s = (1/2)*Product[1+x^k, {k, 1, max}]*Sum[x^k/(1+x^k), {k, 1, max}] - (1/2)*Product[1-x^k, {k, 1, max}]*Sum[x^k/(1-x^k), {k, 1, max}] + O[x]^(max+1); CoefficientList[s, x] (* Jean-François Alcover, Dec 27 2015 *)

Formula

a(n)=(1/2)*A015723(n)-(1/2)*sum{k=0..A235963(n)-1, (-1)^A110654(k)*A000005(n-A001318(k))}=A015723(n)-A238131(n).
G.f.: (1/2)*prod(k>=1, 1+x^k ) * sum(k>=1, x^k/(1+x^k) ) - (1/2)*prod(k>=1, 1-x^k) * sum(k>=1, x^k/(1-x^k) ).
a(n) ~ 3^(1/4) * log(2) * exp(Pi*sqrt(n/3)) / (4*Pi*n^(1/4)). - Vaclav Kotesovec, May 27 2018

A238131 Number of parts in all partitions of n into odd number of distinct parts.

Original entry on oeis.org

0, 1, 1, 1, 1, 1, 4, 4, 7, 10, 13, 16, 22, 25, 31, 42, 48, 59, 73, 89, 108, 132, 156, 190, 227, 271, 318, 380, 449, 526, 618, 722, 841, 980, 1138, 1321, 1526, 1760, 2028, 2333, 2683, 3070, 3517, 4017, 4584, 5228, 5948, 6757, 7673, 8696, 9845, 11132, 12577
Offset: 0

Views

Author

Mircea Merca, Feb 18 2014

Keywords

Examples

			a(8)=7 because the partitions of 8 into odd number of distinct parts are: 8, 5+2+1 and 4+3+1.
		

Crossrefs

Programs

  • Maple
    b:= proc(n, i) option remember; `if`(i*(i+1)/2n, 0, (p->
          [p[2], p[1], p[4]+p[2], p[3]+p[1]])(b(n-i, i-1)))))
        end:
    a:= n-> b(n$2)[4]:
    seq(a(n), n=0..50);  # Alois P. Heinz, Dec 27 2015
  • Mathematica
    max = 50; s = (1/2)*Product[1+x^k, {k, 1, max}]*Sum[x^k/(1+x^k), {k, 1, max}] + (1/2)*Product[1-x^k, {k, 1, max}]*Sum[x^k/(1-x^k), {k, 1, max}] + O[x]^(max+1); CoefficientList[s, x] (* Jean-François Alcover, Dec 27 2015 *)

Formula

a(n) = (1/2)*A015723(n)+(1/2)*sum{k=0..A235963(n)-1, (-1)^A110654(k)*A000005(n-A001318(k))}.
G.f.: (1/2)*prod(k>=1, 1+x^k ) * sum(k>=1, x^k/(1+x^k) ) + (1/2)*prod(k>=1, 1-x^k) * sum(k>=1, x^k/(1-x^k) ).
a(n) ~ 3^(1/4) * log(2) * exp(Pi*sqrt(n/3)) / (4*Pi*n^(1/4)). - Vaclav Kotesovec, May 27 2018

A238133 Difference between A238131(n) and A238132(n).

Original entry on oeis.org

0, 1, 1, -1, -1, -3, 0, -2, 1, 2, 1, 2, 4, 1, -1, 4, -2, -1, -3, -1, -2, -2, -6, 0, -1, 1, -4, 0, 3, 2, 2, 2, 3, 0, 4, 7, 0, 0, 2, -3, 7, -2, -1, -3, -2, -4, 0, -3, -3, -2, -1, -10, -1, 0, 1, -1, 0, -6, 2, 2, 0, 4, 3, 4, 0, 2, 4, 3, 0, 5, 8, 2, 0, 1, -1, 1, -3
Offset: 0

Views

Author

Mircea Merca, Feb 18 2014

Keywords

Comments

Difference between the number of parts in all partitions of n into odd number of distinct parts and the number of parts in all partitions of n into even number of distinct parts.
The convolution of A000005 and A010815.

Crossrefs

Programs

  • Maple
    A238133 := proc(n)
        add( numtheory[tau](k)*A010815(n-k),k=0..n) ;
    end proc: # R. J. Mathar, Jun 18 2016
    # second Maple program:
    b:= proc(n, i) option remember; `if`(i*(i+1)/2n, 0, (p->
          [p[2], p[1], p[4]+p[2], p[3]+p[1]])(b(n-i, i-1)))))
        end:
    a:= n-> (p-> p[4]-p[3])(b(n$2)):
    seq(a(n), n=0..100);  # Alois P. Heinz, Jun 18 2016
  • Mathematica
    Table[SeriesCoefficient[QPochhammer[x] (Log[1 - x] + QPolyGamma[1, x])/Log[x], {x, 0, n}], {n, 0, 80}] (* Vladimir Reshetnikov, Nov 20 2016 *)

Formula

a(n) = Sum_{k=0..A235963(n)-1} (-1)^A110654(k) * A000005(n-A001318(k)).
G.f.: Product_{k>=1} (1-x^k) * Sum_{k>=1} x^k/(1-x^k).
G.f.: (x)_inf * (log(1-x) + psi_x(1))/log(x), where psi_q(z) is the q-digamma function, (q)_inf is the q-Pochhammer symbol (the Euler function).
Showing 1-5 of 5 results.