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A206228 a(n) = [x^n] Product_{k=1..n} 1/(1 - x^k)^(n-k+1).

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%I A206228 #16 Aug 21 2018 08:58:55
%S A206228 1,1,4,17,80,384,1887,9385,47139,238488,1213588,6204547,31844710,
%T A206228 163978344,846741721,4382945317,22735196277,118151632006,615032941924,
%U A206228 3206257881171,16736910271178,87472908459696,457662760258109,2396899780970552,12564645719730297
%N A206228 a(n) = [x^n] Product_{k=1..n} 1/(1 - x^k)^(n-k+1).
%C A206228 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 A206228 Vaclav Kotesovec, <a href="/A206228/b206228.txt">Table of n, a(n) for n = 0..1000</a>
%F A206228 a(n) ~ c * d^n / sqrt(n), where d = A270915 = 5.3527013334866426877724158141653278798514832712869470973196907560641... and c = 0.2030089852709942695768237484498370155967795685257713505678384193773498... - _Vaclav Kotesovec_, Aug 21 2018
%e A206228 Let [x^n] F(x) denote the coefficient of x^n in F(x); then
%e A206228 a(0) = 1;
%e A206228 a(1) = [x] 1/(1-x) = 1;
%e A206228 a(2) = [x^2] 1/((1-x)^2*(1-x^2)) = 4;
%e A206228 a(3) = [x^3] 1/((1-x)^3*(1-x^2)^2*(1-x^3)) = 17;
%e A206228 a(4) = [x^4] 1/((1-x)^4*(1-x^2)^3*(1-x^3)^2*(1-x^4)) = 80; ...
%e A206228 as illustrated below.
%e A206228 The coefficients in Product_{k=1..n} 1/(1-x^k)^(n-k+1) for n=0..9 begin:
%e A206228 n=0: [(1), 0, 0, 0, 0, 0, 0, ...];
%e A206228 n=1: [1,(1), 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...];
%e A206228 n=2: [1, 2,(4), 6, 9, 12, 16, 20, 25, 30, 36, 42, ...]; (A002620)
%e A206228 n=3: [1, 3, 8, (17), 33, 58, 97, 153, 233, 342, 489, 681, ...]; (A002625)
%e A206228 n=4: [1, 4, 13, 34, (80), 170, 339, 636, 1141, 1964, 3270, ...];
%e A206228 n=5: [1, 5, 19, 58, 157,(384), 874, 1869, 3803, 7408, 13907, ...];
%e A206228 n=6: [1, 6, 26, 90, 273, 746, (1887), 4474, 10062, 21620, ...];
%e A206228 n=7: [1, 7, 34, 131, 438, 1314, 3632, (9385), 22940, 53466, ...];
%e A206228 n=8: [1, 8, 43, 182, 663, 2158, 6445, 17944, (47139), 117842, ...];
%e A206228 n=9: [1, 9, 53, 244, 960, 3361, 10757, 32008, 89651, (238488), ...]; ...
%e A206228 where the coefficients in parenthesis start this sequence.
%e A206228 Incidentally, the antidiagonal sums in the above table form A206119.
%e A206228 From _Joerg Arndt_, May 17 2013: (Start)
%e A206228 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 A206228 01:  [ 1:0  1:0  1:0  ]
%e A206228 02:  [ 1:0  1:0  1:1  ]
%e A206228 03:  [ 1:0  1:0  1:2  ]
%e A206228 04:  [ 1:0  1:1  1:1  ]
%e A206228 05:  [ 1:0  1:1  1:2  ]
%e A206228 06:  [ 1:0  1:2  1:2  ]
%e A206228 07:  [ 1:0  2:0  ]
%e A206228 08:  [ 1:0  2:1  ]
%e A206228 09:  [ 1:1  1:1  1:1  ]
%e A206228 10:  [ 1:1  1:1  1:2  ]
%e A206228 11:  [ 1:1  1:2  1:2  ]
%e A206228 12:  [ 1:1  2:0  ]
%e A206228 13:  [ 1:1  2:1  ]
%e A206228 14:  [ 1:2  1:2  1:2  ]
%e A206228 15:  [ 1:2  2:0  ]
%e A206228 16:  [ 1:2  2:1  ]
%e A206228 17:  [ 3:0  ]
%e A206228 (End)
%t A206228 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 A206228 (PARI) {a(n)=polcoeff(prod(k=1,n,1/(1-x^k+x*O(x^n))^(n-k+1)),n)}
%o A206228 for(n=0,41,print1(a(n),", "))
%Y A206228 Cf. A206119, A206229.
%K A206228 nonn
%O A206228 0,3
%A A206228 _Paul D. Hanna_, Feb 05 2012