%I #56 Apr 27 2022 18:39:03

%S 0,0,1,1,2,4,6,9,15,21,31,45,63,87,122,164,222,298,395,519,683,885,

%T 1146,1475,1887,2401,3050,3845,4837,6060,7563,9402,11664,14405,17751,

%U 21807,26715,32634,39784,48352,58649,70969,85690,103232,124143,148951,178407,213277,254509

%N Number of 3's in all partitions of n.

%C Starting with the first 1 = row sums of triangle A173239. - _Gary W. Adamson_, Feb 13 2010

%C The sums of three successive terms give A000070. - _Omar E. Pol_, Jul 12 2012

%C 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

%H Alois P. Heinz, <a href="/A024787/b024787.txt">Table of n, a(n) for n = 1..1000</a>

%H David Benson, Radha Kessar, and Markus Linckelmann, <a href="https://arxiv.org/abs/2204.09970">Hochschild cohomology of symmetric groups in low degrees</a>, arXiv:2204.09970 [math.GR], 2022.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/q-PochhammerSymbol.html">q-Pochhammer Symbol</a>

%F a(n) = A181187(n,3) - A181187(n,4). - _Omar E. Pol_, Oct 25 2012

%F a(n) = Sum_{k=1..floor(n/3)} A263232(n,k). - _Alois P. Heinz_, Nov 01 2015

%F 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

%F 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

%F 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

%e From _Omar E. Pol_, Oct 25 2012: (Start)

%e For n = 7 we have:

%e --------------------------------------

%e . Number

%e Partitions of 7 of 3's

%e --------------------------------------

%e 7 .............................. 0

%e 4 + 3 .......................... 1

%e 5 + 2 .......................... 0

%e 3 + 2 + 2 ...................... 1

%e 6 + 1 .......................... 0

%e 3 + 3 + 1 ...................... 2

%e 4 + 2 + 1 ...................... 0

%e 2 + 2 + 2 + 1 .................. 0

%e 5 + 1 + 1 ...................... 0

%e 3 + 2 + 1 + 1 .................. 1

%e 4 + 1 + 1 + 1 .................. 0

%e 2 + 2 + 1 + 1 + 1 .............. 0

%e 3 + 1 + 1 + 1 + 1 .............. 1

%e 2 + 1 + 1 + 1 + 1 + 1 .......... 0

%e 1 + 1 + 1 + 1 + 1 + 1 + 1 ...... 0

%e ------------------------------------

%e . 13 - 7 = 6

%e 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.

%e (End)

%p b:= proc(n, i) option remember; local g;

%p if n=0 or i=1 then [1, 0]

%p else g:= `if`(i>n, [0$2], b(n-i, i));

%p b(n, i-1) +g +[0, `if`(i=3, g[1], 0)]

%p fi

%p end:

%p a:= n-> b(n, n)[2]:

%p seq(a(n), n=1..100); # _Alois P. Heinz_, Oct 27 2012

%t Table[ Count[ Flatten[ IntegerPartitions[n]], 3], {n, 1, 50} ]

%t 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_ *)

%t Join[{0, 0}, (1/((1 - x^3) QPochhammer[x]) + O[x]^50)[[3]]] (* _Vladimir Reshetnikov_, Nov 22 2016 *)

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

%Y Cf. A173239. - _Gary W. Adamson_, Feb 13 2010

%K nonn,easy

%O 1,5

%A _Clark Kimberling_