A173239 -id:A173239 - cp's OEIS frontend

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

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

Clark Kimberling

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

			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)

Alois P. Heinz, Table of n, a(n) for n = 1..5000
David Benson, Radha Kessar, and Markus Linckelmann, Hochschild cohomology of symmetric groups in low degrees, arXiv:2204.09970 [math.GR], 2022.
Philip Cuthbertson, Fixed hooks in arbitrary columns of partitions, Integers (2025) Vol. 25, Art. No. A28. See p. 3.
Eric Weisstein's World of Mathematics, q-Pochhammer Symbol

Cf. A066633, A024786, A024788, A024789, A024790, A024791, A024792, A024793, A024794, A263232.

Cf. A173239. - Gary W. Adamson, Feb 13 2010

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

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 *)

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^2/((1 - x^3)*(x)inf), where (q)_inf is the q-Pochhammer symbol (the Euler function). - _Vladimir Reshetnikov, Nov 22 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

A173241 Euler transform of A051064, the ruler function sequence for k=3.

Original entry on oeis.org

1, 1, 2, 4, 6, 9, 16, 22, 33, 51, 71, 100, 147, 199, 275, 384, 515, 692, 944, 1242, 1645, 2186, 2847, 3706, 4848, 6231, 8019, 10330, 13153, 16729, 21305, 26864, 33858, 42658, 53366, 66668, 83277, 103378, 128200, 158846, 195895, 241237, 296860, 363796, 445285, 544465, 663520

Offset: 0

Gary W. Adamson, Feb 13 2010

nonn

Let P(x) = polcoeff A000041: (1 + x + 2x^2 + 3x^3 + 5x^4 + 7x^5 + ...) and
A(x) = polcoeff A173241: (1 + x + 2x^2 + 4x^3 + 6x^4 + 9x^5 + ...); then
P(x) = A(x) / A(x^3).
A092119 = Euler transform of the ruler function for k=2: A001511.

			Equals 1/((1-x)*(1-x^2)*(1-x^3)^2*(1-x^4)*(1-x^5)*(1-x^6)^2*(1-x^7)*...); where in (1-x)^k, k = A051064: (1, 1, 2, 1, 1, 2, 1, 1, 3, ...).

Seiichi Manyama, Table of n, a(n) for n = 0..10000

Cf. A000041, A001511, A051064, A092119, A173238, A173239.

N=66; x='x+O('x^N); /* that many terms */
gf=1/prod(e=0, ceil(log(N)/log(3)), eta(x^(3^e)));
Vec(gf) /* show terms */ /* Joerg Arndt, Jun 21 2011 */

G.f.: 1/Product_{k>=0} P(x^(3^k)) where P(x)=Product_{k>=1} (1-x^k). - Joerg Arndt, Jun 21 2011
Euler transform of A051064, where A051064 = the ruler function for k=3:
(1, 1, 2, 1, 1, 2, 1, 1, 3, 1, 1, 2, ...).

More terms from Joerg Arndt, Jun 21 2011

A173238 Triangle by columns, A000041 in every column shifted down twice for columns > 0.

Original entry on oeis.org

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

Offset: 0

Gary W. Adamson, Feb 13 2010

Row sums = A024786 starting with A024786(2): (1, 1, 3, 4, 8, 11, 19, 26, ...) = number of 2's in all partitions of n.
A173238 as an infinite lower triangular matrix * [1, 2, 3, ...] = A103650.
Let the triangle = M. Then Lim_{n->inf} M^n = (1, 1, 3, 4, 10, 13, 26, ...), the left-shifted vector considered as a sequence = A092119, the Euler transform of the ruler sequence, A001511.
Given P(x) = polcoeff A000041 = (1 + x + 2x^2 + 3x^3 + 5x^4 + 7x^5 + ...), then P(x) = A(x)/A(x^2), where A(x) = polcoeff A092119: (1 + x + 3x^2 + 4x^3 + ...).
Conjectures: Given the infinite set of triangles with A000041 in every column shifted down 0, 1, 2, ... n times, row sums of n-th triangle (where A173238 = 2nd in the set) = the numbers of n's in all partitions of n. E.g., row sums = of A173238 = A024786, the numbers of 2's in all partitions of n.
Similarly, row sums of triangle A173239 with columns >0 shifted down thrice = numbers of 3's in all partitions of n, and so on. Refer to comments in A000041 regarding the numbers of 1's in partitions of n.
...
Second conjecture: Given the infinite set of analogous triangles with columns shifted down 2, 3, 4, ..., k times, we let such triangles = T(k) and perform lim_{n->inf} T^n(k), obtaining the left-shifted vectors considered as sequences. The conjecture states that the infinite set of such left-shifted vectors = the Euler transform of the infinite set of Ruler functions starting with the ruler function for k=2 = A001511: (1, 2, 1, 3, 1, 2, 1, ...)
To obtain the k-th ruler functions, begin with the natural numbers, 1,...2,...3,...4,...5,...6,...7,...8,...9,...; then for k = 2 we get A001511: 1,...2,...1,...3,...1,...2,...1,...4,...1,...; by finding the highest exponent of k dividing n, then adding 1. Similarly, for k = 2, we obtain A051064: 1,...1,...2,...1,...1,...2,...1,...1,...3,...
Next, we obtain the Euler transforms of the ruler functions (e.g., Euler transform of A001511 = A092119: (1, 1, 3, 4, 10, 13, 26, ...), noting that A092119 is lim_{n->inf} A173238^n, the left-shifted vector.
...
Third conjecture: Let P(x) = polcoeff A000041 = (1 + x + 2x^2 + 3x^3 + ...), and A(k)(x) = the Euler transform of k-th ruler sequence, (k=2,3,...). Then P(x) = A(k)(x) / A(k)(x^k).
Examples: for k=2, A(x) = A092119: (1, 1, 3, 4, 10, 13, ...), then P(x) = (1 + x + 2x^2 + 3x^3 + ...) = (1 + x + 3x^2 + 4x^3 + ...) / (1 + x^2 + 3x^4 + 4x^6 + 10x^8 + ...). For k=3 relating to triangle A173238, the left-shifted vector = the Euler transform of A051064 = A(x) for k=3, then P(x) = A(x) / A(x^3).
The conjecture extends the analogous conclusions to all k.
From Gary W. Adamson, Feb 25 2010: (Start)
Proof of second conjecture received from Helmut Prodinger 02/28/10 with a summary by R. J. Mathar:
Consider Product_{n>=0} z^(t^n)/(1-z^(t^n)) = Sum_{k>=1} (1+v_t(k))z^k where v_t(n) is the number of trailing zeros in the t-ary expansion of n, and its Euler transform A(z) = product_{k >= 1} 1/(1-z^k)^{1+v_t(k)}, then A(z)/A(z^t) = product_{k >= 1} 1/(1-z^k) is the partition generating function.
Here is the proof: A(z)/A(z^t) = Product_{k>=1} (1-z^(tk))^(1+v_t(k))/(1-z^k)^(1+v_t(k))
= Product_{k>=1} (1-z^(tk))^(v_t(tk))/(1-z^k)^(1+v_t(k))
= Product_{k>=1} (1-z^k)^(v_t(k))/(1-z^k)^(1+v_t(k)) (*)
= Product_{k>=1} 1/(1-z^k)
as desired. Notice that for (*), that v_t(n)=0 if n is not divisible by t. [Helmut Prodinger, hproding(AT)sun.ac.za, Feb 28 2010] (End)

			First few rows of the triangle:
    1;
    1;
    2,   1;
    3,   1;
    5,   2,  1;
    7,   3,  1;
   11,   5,  2,  1;
   15,   7,  3,  1;
   22,  11,  5,  2,  1;
   30,  15,  7,  3,  1;
   42,  22, 11,  5,  2, 1;
   56,  30, 15,  7,  3, 1;
   77,  42, 22, 11,  5, 2, 1;
  101,  56, 30, 15,  7, 3, 1;
  135,  77, 42, 22, 11, 5, 2, 1;
  176, 101, 56, 30, 15, 7, 3, 1;
  ...

Michael De Vlieger, Table of n, a(n) for n = 0..9999 (rows 0 <= n <= 199, flattened)

Cf. A000041, A001511, A092119, A173239.

Table[PartitionsP[n - 2 k], {n, 17}, {k, Floor[n/2]}] // Flatten (* Michael De Vlieger, Nov 23 2021 *)

T(n,k) = A000041(n-2*k) for k=0..floor(n/2).

a(24), a(25) corrected by Georg Fischer, Nov 23 2021

cp's OEIS Frontend

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

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A173241 Euler transform of A051064, the ruler function sequence for k=3.

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A173238 Triangle by columns, A000041 in every column shifted down twice for columns > 0.

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