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.

A024787 Number of 3's in all partitions of n.

Original entry on oeis.org

0, 0, 1, 1, 2, 4, 6, 9, 15, 21, 31, 45, 63, 87, 122, 164, 222, 298, 395, 519, 683, 885, 1146, 1475, 1887, 2401, 3050, 3845, 4837, 6060, 7563, 9402, 11664, 14405, 17751, 21807, 26715, 32634, 39784, 48352, 58649, 70969, 85690, 103232, 124143, 148951, 178407, 213277, 254509
Offset: 1

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Comments

Starting with the first 1 = row sums of triangle A173239. - Gary W. Adamson, Feb 13 2010
The sums of three successive terms give A000070. - Omar E. Pol, Jul 12 2012
a(n) is also the difference between the sum of 3rd largest and the sum of 4th largest elements in all partitions of n. - Omar E. Pol, Oct 25 2012

Examples

			From _Omar E. Pol_, Oct 25 2012: (Start)
For n = 7 we have:
--------------------------------------
.                             Number
Partitions of 7               of 3's
--------------------------------------
7 .............................. 0
4 + 3 .......................... 1
5 + 2 .......................... 0
3 + 2 + 2 ...................... 1
6 + 1 .......................... 0
3 + 3 + 1 ...................... 2
4 + 2 + 1 ...................... 0
2 + 2 + 2 + 1 .................. 0
5 + 1 + 1 ...................... 0
3 + 2 + 1 + 1 .................. 1
4 + 1 + 1 + 1 .................. 0
2 + 2 + 1 + 1 + 1 .............. 0
3 + 1 + 1 + 1 + 1 .............. 1
2 + 1 + 1 + 1 + 1 + 1 .......... 0
1 + 1 + 1 + 1 + 1 + 1 + 1 ...... 0
------------------------------------
.      13 - 7 =                  6
The difference between the sum of the third column and the sum of the fourth column of the set of partitions of 7 is 13 - 7 = 6 and equals the number of 3's in all partitions of 7, so a(7) = 6.
(End)
		

Crossrefs

Programs

  • Maple
    b:= proc(n, i) option remember; local g;
          if n=0 or i=1 then [1, 0]
        else g:= `if`(i>n, [0$2], b(n-i, i));
             b(n, i-1) +g +[0, `if`(i=3, g[1], 0)]
          fi
        end:
    a:= n-> b(n, n)[2]:
    seq(a(n), n=1..100);  # Alois P. Heinz, Oct 27 2012
  • Mathematica
    Table[ Count[ Flatten[ IntegerPartitions[n]], 3], {n, 1, 50} ]
    b[n_, i_] := b[n, i] = Module[{g}, If[n==0 || i==1, {1, 0}, g = If[i>n, {0, 0}, b[n-i, i]]; b[n, i-1] + g + {0, If[i==3, g[[1]], 0]}]]; a[n_] := b[n, n][[2]]; Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Oct 09 2015, after Alois P. Heinz *)
    Join[{0, 0}, (1/((1 - x^3) QPochhammer[x]) + O[x]^50)[[3]]] (* Vladimir Reshetnikov, Nov 22 2016 *)

Formula

a(n) = A181187(n,3) - A181187(n,4). - Omar E. Pol, Oct 25 2012
a(n) = Sum_{k=1..floor(n/3)} A263232(n,k). - Alois P. Heinz, Nov 01 2015
a(n) ~ exp(Pi*sqrt(2*n/3)) / (6*Pi*sqrt(2*n)) * (1 - 37*Pi/(24*sqrt(6*n)) + (37/48 + 937*Pi^2/6912)/n). - Vaclav Kotesovec, Nov 05 2016
G.f.: x^3/((1 - x)*(1 - x^2)*(1 - x^3)) * Sum_{n >= 0} x^(3*n)/( Product_{k = 1..n} 1 - x^k ); that is, convolution of A069905 (partitions into 3 parts, or, modulo offset differences, partitions into parts <= 3) and A008483 (partitions into parts >= 3). - Peter Bala, Jan 17 2021