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a(n) = [x^n] Product_{k=1..n} 1/(1 - x^k)^(n-k+1).
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%I #16 Aug 21 2018 08:58:55

%S 1,1,4,17,80,384,1887,9385,47139,238488,1213588,6204547,31844710,

%T 163978344,846741721,4382945317,22735196277,118151632006,615032941924,

%U 3206257881171,16736910271178,87472908459696,457662760258109,2396899780970552,12564645719730297

%N a(n) = [x^n] Product_{k=1..n} 1/(1 - x^k)^(n-k+1).

%C Number of partitions of n with 1 kind of n's, 2 kinds of (n-1)'s, ..., n kinds of 1's, see example. [_Joerg Arndt_, May 17 2013]

%H Vaclav Kotesovec, <a href="/A206228/b206228.txt">Table of n, a(n) for n = 0..1000</a>

%F a(n) ~ c * d^n / sqrt(n), where d = A270915 = 5.3527013334866426877724158141653278798514832712869470973196907560641... and c = 0.2030089852709942695768237484498370155967795685257713505678384193773498... - _Vaclav Kotesovec_, Aug 21 2018

%e Let [x^n] F(x) denote the coefficient of x^n in F(x); then

%e a(0) = 1;

%e a(1) = [x] 1/(1-x) = 1;

%e a(2) = [x^2] 1/((1-x)^2*(1-x^2)) = 4;

%e a(3) = [x^3] 1/((1-x)^3*(1-x^2)^2*(1-x^3)) = 17;

%e a(4) = [x^4] 1/((1-x)^4*(1-x^2)^3*(1-x^3)^2*(1-x^4)) = 80; ...

%e as illustrated below.

%e The coefficients in Product_{k=1..n} 1/(1-x^k)^(n-k+1) for n=0..9 begin:

%e n=0: [(1), 0, 0, 0, 0, 0, 0, ...];

%e n=1: [1,(1), 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...];

%e n=2: [1, 2,(4), 6, 9, 12, 16, 20, 25, 30, 36, 42, ...]; (A002620)

%e n=3: [1, 3, 8, (17), 33, 58, 97, 153, 233, 342, 489, 681, ...]; (A002625)

%e n=4: [1, 4, 13, 34, (80), 170, 339, 636, 1141, 1964, 3270, ...];

%e n=5: [1, 5, 19, 58, 157,(384), 874, 1869, 3803, 7408, 13907, ...];

%e n=6: [1, 6, 26, 90, 273, 746, (1887), 4474, 10062, 21620, ...];

%e n=7: [1, 7, 34, 131, 438, 1314, 3632, (9385), 22940, 53466, ...];

%e n=8: [1, 8, 43, 182, 663, 2158, 6445, 17944, (47139), 117842, ...];

%e n=9: [1, 9, 53, 244, 960, 3361, 10757, 32008, 89651, (238488), ...]; ...

%e where the coefficients in parenthesis start this sequence.

%e Incidentally, the antidiagonal sums in the above table form A206119.

%e From _Joerg Arndt_, May 17 2013: (Start)

%e There are a(3)=17 partitions of 3 into 1 kind of 3's, 2 kinds of 2's, and 3 kinds of 1's:

%e 01: [ 1:0 1:0 1:0 ]

%e 02: [ 1:0 1:0 1:1 ]

%e 03: [ 1:0 1:0 1:2 ]

%e 04: [ 1:0 1:1 1:1 ]

%e 05: [ 1:0 1:1 1:2 ]

%e 06: [ 1:0 1:2 1:2 ]

%e 07: [ 1:0 2:0 ]

%e 08: [ 1:0 2:1 ]

%e 09: [ 1:1 1:1 1:1 ]

%e 10: [ 1:1 1:1 1:2 ]

%e 11: [ 1:1 1:2 1:2 ]

%e 12: [ 1:1 2:0 ]

%e 13: [ 1:1 2:1 ]

%e 14: [ 1:2 1:2 1:2 ]

%e 15: [ 1:2 2:0 ]

%e 16: [ 1:2 2:1 ]

%e 17: [ 3:0 ]

%e (End)

%t Table[SeriesCoefficient[Product[1/(1 - x^k)^(n-k+1), {k, 1, n}], {x, 0, n}], {n, 0, 40}] (* _Vaclav Kotesovec_, Aug 21 2018 *)

%o (PARI) {a(n)=polcoeff(prod(k=1,n,1/(1-x^k+x*O(x^n))^(n-k+1)),n)}

%o for(n=0,41,print1(a(n),", "))

%Y Cf. A206119, A206229.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Feb 05 2012