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A096373
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Number of partitions of n such that the least part occurs exactly twice.
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42
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0, 1, 0, 2, 1, 3, 3, 6, 5, 11, 11, 17, 20, 30, 33, 49, 56, 77, 92, 122, 143, 190, 225, 287, 344, 435, 516, 648, 770, 951, 1134, 1388, 1646, 2007, 2376, 2868, 3395, 4078, 4807, 5749, 6764, 8042, 9449, 11187, 13101, 15463, 18070, 21236, 24772, 29021, 33764
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OFFSET
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1,4
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COMMENTS
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Also number of partitions of n such that the difference between the two largest distinct parts is 2 (it is assumed that 0 is a part in each partition). Example: a(6)=3 because we have [4,2], [3,1,1,1] and [2,2,2]. - Emeric Deutsch, Apr 08 2006
Number of partitions p of n+2 such that min(p) + (number of parts of p) is a part of p. - Clark Kimberling, Feb 27 2014
Number of partitions of n+1 such that the two smallest parts differ by one. - Giovanni Resta, Mar 07 2014
Also the number of integer partitions of n with an even number of parts that cannot be grouped into pairs of distinct parts. These are also integer partitions of n with an even number of parts whose greatest multiplicity is greater than half the number of parts. - Gus Wiseman, Oct 26 2018
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LINKS
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FORMULA
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G.f.: Sum_{m>0} (x^(2*m) / Product_{i>m} (1-x^i)}. More generally, g.f. for number of partitions of n such that the least part occurs exactly k times is Sum_{m>0} (x^(k*m)/Product_{i>m} (1-x^i)}.
G.f.: sum(x^(2k-2)*(1-x^(k-1))/product(1-x^j, j=1..k), k=1..infinity). - Emeric Deutsch, Apr 08 2006
a(n) ~ exp(Pi*sqrt(2*n/3)) * Pi / (12*sqrt(2)*n^(3/2)). - Vaclav Kotesovec, Jun 02 2018
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EXAMPLE
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a(6)=3 because we have [4,1,1], [3,3] and [2,2,1,1].
G.f. = x^2 + 2*x^4 + x^5 + 3*x^6 + 3*x^7 + 6*x^8 + 5*x^9 + 11*x^10 + 11*x^11 + ...
The a(2) = 1 through a(10) = 11 partitions where the least part occurs exactly twice (zero terms not shown):
(11) (22) (311) (33) (322) (44) (522) (55)
(211) (411) (511) (422) (711) (433)
(2211) (3211) (611) (4311) (622)
(3311) (5211) (811)
(4211) (32211) (3322)
(22211) (4411)
(5311)
(6211)
(33211)
(42211)
(222211)
The a(2) = 1 through a(10) = 11 partitions that cannot be grouped into pairs of distinct parts (zero terms not shown):
(11) (22) (2111) (33) (2221) (44) (3222) (55)
(1111) (3111) (4111) (2222) (6111) (3331)
(111111) (211111) (5111) (321111) (4222)
(221111) (411111) (7111)
(311111) (21111111) (222211)
(11111111) (331111)
(421111)
(511111)
(22111111)
(31111111)
(1111111111)
(End)
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MAPLE
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g:=sum(x^(2*k)/product(1-x^j, j=k+1..80), k=1..70): gser:=series(g, x=0, 55): seq(coeff(gser, x, n), n=1..51); # Emeric Deutsch, Apr 08 2006
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MATHEMATICA
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(* do first *) Needs["DiscreteMath`Combinatorica`"] (* then *) f[n_] := Block[{p = Partitions[n], l = PartitionsP[n], c = 0, k = 1}, While[k < l + 1, q = PadLeft[ p[[k]], 3]; If[ q[[1]] != q[[3]] && q[[2]] == q[[3]], c++ ]; k++ ]; c]; Table[ f[n], {n, 51}] (* Robert G. Wilson v, Jul 23 2004 *)
Table[Count[IntegerPartitions[n+2], p_ /; MemberQ[p, Length[p] + Min[p]]], {n, 50}] (* Clark Kimberling, Feb 27 2014 *)
p[n_, m_] := If[m == n, 1, If[m > n, 0, p[n, m] = Sum[p[n-m, k], {k, m, n}]]];
a[n_] := Sum[p[n+1-k, k+1], {k, n/2}]; Array[a, 100] (* Giovanni Resta, Mar 07 2014 *)
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PROG
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(PARI) {q=sum(m=1, 100, x^(2*m)/prod(i=m+1, 100, 1-x^i, 1+O(x^60)), 1+O(x^60)); for(n=1, 51, print1(polcoeff(q, n), ", "))} - Klaus Brockhaus, Jul 21 2004
(PARI) {a(n) = if( n<0, 0, polcoeff( ( 1 - (1 - x - x^2) / eta(x + x^4 * O(x^n)) ) * (1 - x) / x^3, n))} /* Michael Somos, Feb 28 2014 */
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CROSSREFS
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Cf. A000070, A000569, A007717, A025065, A141268, A209816, A261049, A320328, A320891, A320921, A320922.
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KEYWORD
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easy,nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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