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.

User: Edward Early

Edward Early's wiki page.

Edward Early has authored 5 sequences.

A058886 Sum of the row of the character table of S_n corresponding to the partition 2,1^{n-2}.

Original entry on oeis.org

2, 1, 2, 3, 4, 5, 5, 7, 9, 11, 12, 15, 18, 20, 23, 28, 33, 37, 42, 48, 56, 63, 70, 80, 92, 102, 114, 129, 145, 161, 178, 199, 223, 246, 271, 302, 335, 368, 404, 447, 493, 540, 591, 649, 713, 779, 848, 929, 1017, 1106, 1203, 1312, 1429, 1551, 1682, 1828, 1986
Offset: 2

Author

Edward Early, Jan 08 2001

Keywords

Comments

Also partial sums of the number of self-conjugate partitions (A000700) with the last term subtracted. - Edward Early, May 30 2012

Extensions

More terms added by Edward Early, May 30 2012

A076269 Size of largest antichain in partition lattice Par(n).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 2, 2, 3, 4, 4, 5, 6, 7, 9, 10, 11, 14, 17, 20, 24, 29, 35, 40, 48, 55
Offset: 0

Author

Edward Early, Nov 05 2002

Keywords

Comments

Par(n) is the set of partitions of n under "dominance order": partition P is <= partition Q iff the sum of the largest k parts of P is <= the corresponding sum for Q for all k.

Examples

			a(10)=4; one antichain consists of 5+1+1+1+1+1, 4+3+1+1+1, 4+2+2+2 and 3+3+3+1.
		

Crossrefs

Programs

  • Mathematica
    leq[p_, q_] := If[Length[p]
    				

Formula

Order of growth is between n^(-5/2)e^(Pi*sqrt(2n/3)) and n^(-1)e^(Pi*sqrt(2n/3)).

Extensions

Edited by Dean Hickerson, Nov 09 2002
a(22)-a(26) by Paul Tabatabai, Dec 05 2018

A076779 Maximum number of disjoint maximum-size antichains in partition lattice Par(n).

Original entry on oeis.org

1, 1, 2, 3, 5, 7, 2, 3, 7, 3, 2, 3, 4, 3, 3, 2, 5, 5, 2, 1
Offset: 0

Author

Edward Early, Nov 15 2002

Keywords

Crossrefs

Extensions

a(17)-a(19) from Sean A. Irvine, Apr 16 2025

A058884 Partial sums of the partition function (A000041), with the last term subtracted. Also the sum of the row of the character table for S_n corresponding to the partition n-1,1 for n>1. Also the sum over all partitions lambda of n of one less than the number of 1's in lambda.

Original entry on oeis.org

-1, 0, 0, 1, 2, 5, 8, 15, 23, 37, 55, 83, 118, 171, 238, 332, 453, 618, 827, 1107, 1460, 1922, 2504, 3253, 4188, 5380, 6860, 8722, 11024, 13895, 17421, 21787, 27122, 33677, 41653, 51390, 63179, 77496, 94755, 115600, 140632, 170725, 206717, 249804, 301151, 362367, 435077, 521439, 623674, 744695
Offset: 0

Author

Edward Early, Jan 08 2001

Keywords

Comments

For n>=1 number of up-steps in all partitions of n (represented as weakly increasing lists), see example. - Joerg Arndt, Sep 03 2014

Examples

			a(6) = 8 because the 11 partitions of 6
01:  [ 1 1 1 1 1 1 ]
02:  [ 1 1 1 1 2 ]
03:  [ 1 1 1 3 ]
04:  [ 1 1 2 2 ]
05:  [ 1 1 4 ]
06:  [ 1 2 3 ]
07:  [ 1 5 ]
08:  [ 2 2 2 ]
09:  [ 2 4 ]
10:  [ 3 3 ]
11:  [ 6 ]
contain 0+1+1+1+1+2+1+0+1+0+0 = 8 up-steps. - _Joerg Arndt_, Sep 03 2014
		

Crossrefs

Cf. A218074 (up-steps in partitions into distinct parts).

Programs

  • Maple
    a:= proc(n) uses combinat; add(numbpart(k), k=0..n-1)-numbpart(n) end:
    seq(a(n), n=0..49);
  • Mathematica
    p[n_] := IntegerPartitions[n]; l[n_] := Length[p[n]]; Table[Count[Flatten[p[n]], 1] - l[n], {n, 0, 30}] (* Clark Kimberling, Mar 08 2012 *)
  • PARI
    a(n) = {sum(k=0, n-1, numbpart(k)) - numbpart(n)} \\ Andrew Howroyd, Apr 21 2023
    
  • PARI
    Vec((2*x - 1)/(1 - x)/eta(x + O(x^51))) \\ Andrew Howroyd, Apr 21 2023

Formula

From Andrew Howroyd, Apr 21 2023: (Start)
a(n) = A000070(n-1) - A000041(n) for n > 0.
G.f.: (2*x - 1)*P(x)/(1 - x) where P(x) is the g.f. of A000041. (End)

Extensions

More terms from James Sellers, Sep 28 2001

A045883 a(n) = ((3*n+1)*2^n - (-1)^n)/9.

Original entry on oeis.org

0, 1, 3, 9, 23, 57, 135, 313, 711, 1593, 3527, 7737, 16839, 36409, 78279, 167481, 356807, 757305, 1601991, 3378745, 7107015, 14913081, 31224263, 65244729, 136081863, 283348537, 589066695, 1222872633, 2535223751, 5249404473, 10856722887, 22429273657, 46290203079
Offset: 0

Author

Keywords

Comments

Without the initial zero, PSumSIGN transform of A001787. - Michael Somos, Jul 10 2003
Number of rises (drops) in the compositions of n+2 with parts in N.
From Michel Lagneau, Jan 13 2012: (Start)
This sequence is connected with the Collatz problem. We consider the array T(i,j) where the i-th row gives the parity trajectory of i, for example for i = 6, the infinite trajectory is 6 -> 3 -> 10 ->5 -> 16 ->8 -> 4 -> 2 -> 1 -> 4 -> 2 -> 1 -> 4->2-> 1 ... and T(6,j) = [0,1,0,1,0,0,0,0,1,0,0,1,...,1,0,0,1,...]. Now, we consider the sum of the digits 1 of each array T(i,j), where
a(1) = sum of the digits "1" of T(i,j), i = 1..2^1 and j = 1;
a(2) = sum of the digits "1" of T(i,j), i = 1..2^2 and j = 1..2;
a(3) = sum of the digits "1" of T(i,j), i = 1..2^3 and j = 1..3;
a(n) = Sum_{i=1..2^n}(Sum_{j=1..n} T(i,j)) = Sum_{i=1..n} A001045(n)*2^(n-i) = convolution of A001045 and A000079 (see the formula below).
The number of digits "0" equals A113861(n) = n*2^n - a(n) because n and 2^n are the dimensions of each array.
An important result is that the ratio r = A113861(n) / A045883(n) tends towards 2 when n tends towards infinity. In other words, when the array tends towards infinity, the ratio r = (number of divisions by 2) / (number of multiplications by 3) tends towards 2, even if there exists divergent trajectories. That is the problem! For each possible divergent infinite trajectory, r < 2 even though the global ratio r is 2.
Conclusion:
1. For each number n with a convergent trajectory T(n,k), k = 1..infinity, or for each row of the array T(i,j), the ratio r tends towards 2 (the proof is easy because the trajectory becomes periodic from a certain index 1001001001...).
2. For each array of dimension n X 2^n, the radio r tends towards 2.
3. If there exists a number n such that the trajectory is divergent, this trajectory is random and r tends towards a real x such that 1 < = r < = x < 2.
4. In order to establish a proof of the Collatz problem from this considerations (if that is possible), it is necessary to prove that a ratio < 2 for an infinite row (or several rows) of an infinite array T(i,j) is incompatible with r = 2, the exact ratio for this array. (End)
a(n) is the distance spectral radius of the dimension-regular generalized recursive circulant graph (commonly known as multiplicative circulant graph) of order 2^n. - John Rafael M. Antalan, Sep 25 2020
Total sum over all compositions of n of the absolute differences between consecutive parts, assuming an initial part 0. - Alois P. Heinz, Apr 30 2025

Crossrefs

Partial sums of A059570, bisection: A014916.
Row sums of triangle A094953.

Programs

  • Magma
    [((3*n+1)*2^n-(-1)^n)/9: n in [0..35]]; // Vincenzo Librandi, Jun 15 2017
  • Maple
    A045883:=n->((3*n+1)*2^n-(-1)^n)/9; seq(A045883(n), n=0..30); # Wesley Ivan Hurt, Mar 21 2014
  • Mathematica
    nn=31;a=x^2(1-x)/(1-x-2x^2)/(1-2x);b=x^2/(1-2x)^2;Drop[CoefficientList[Series[(b-a)/2,{x,0,nn}],x],2] (* Geoffrey Critzer, Mar 21 2014 *)
    CoefficientList[Series[x / ((1 + x) (1 - 2 x)^2), {x, 0, 33}], x] (* Vincenzo Librandi, Jun 15 2017 *)
    LinearRecurrence[{3, 0, -4}, {0, 1, 3}, 33] (* Jean-François Alcover, Sep 27 2017 *)
  • PARI
    {a(n) = if( n<-1, 0, ((3*n + 1)*2^n - (-1)^n) / 9)};
    

Formula

G.f.: x/((1+x)*(1-2*x)^2).
a(n) = 3*a(n-1) - 4*a(n-3).
Convolution of A001045 and A000079. G.f.: x/((1-2*x)(1-x-2*x^2)). - Paul Barry, May 21 2004
Starting with "1" = triangle A049260 * the odd integers as a vector. - Gary W. Adamson, Mar 06 2012
a(n) = A140960(n)/2. - J. M. Bergot, May 21 2013
From Wolfdieter Lang, Jun 14 2017: (Start)
a(n) = f(n)*2^n, where f(n) is a rational Fibonacci type sequence based on fuse(a,b) = (a+b+1)/2 with f(0) = 0, f(1) = 1/2 and f(n) = fuse(f(n-1), f(n-2)), for n >= 2. For fuse(a,b) see the Jeff Erickson link under A188545. Proof with f(n) = (3*n+1 - (-1)^n/2^n)/9, n >= 0, by induction.
a(n) = a(n-1) + 2*a(n-2) + 2^(n-1), n >= 0, with input a(-2) = 1/4 and a(-1) = 0. See also A127984. (End)
E.g.f.: (exp(2*x)*(1 + 6*x) - cosh(x) + sinh(x))/9. - Stefano Spezia, Apr 09 2025
a(n) = Sum_{k=0..n+2} k * A238343(n+2,k). - Alois P. Heinz, Apr 30 2025

Extensions

Simpler description from Vladeta Jovovic, Jul 18 2002