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A024786
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Number of 2's in all partitions of n.
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29
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0, 1, 1, 3, 4, 8, 11, 19, 26, 41, 56, 83, 112, 160, 213, 295, 389, 526, 686, 911, 1176, 1538, 1968, 2540, 3223, 4115, 5181, 6551, 8191, 10269, 12756, 15873, 19598, 24222, 29741, 36532, 44624, 54509, 66261, 80524, 97446, 117862, 142029, 171036, 205290, 246211
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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-1 with a distinguished part different from all the others. [Comment corrected by Emeric Deutsch, Aug 13 2008]
In general the number of times that j appears in the partitions of n equals Sum_{k<n, k = n (mod j)} P(k). In particular this gives a formula for a(n), A024787, ..., A024794, for j = 2,...,10; it generalizes the formula given for A000070 for j=1. - Jose Luis Arregui (arregui(AT)posta.unizar.es), Apr 05 2002
Equals row sums of triangle A173238 [From Gary W. Adamson, Feb 13 2010]
The sums of two successive terms give A000070. - Omar E. Pol, Jul 12 2012
a(n) is also the difference between the sum of second largest and the sum of third largest elements in all partitions of n. More generally, the number of occurrences of k in all partitions of n equals the difference between the sum of k-th largest and the sum of (k+1)st largest elements in all partitions of n. And more generally, the sum of the number of occurrences of k, k+1, k+2..k+m in all partitions of n equals the difference between the sum of k-th largest and the sum of (k+m+1)st largest elements in all partitions of n. - Omar E. Pol, Oct 25 2012
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REFERENCES
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Manosij Ghosh Dastidar and Sourav Sen Gupta, Generalization of a few results in Integer Partitions, Arxiv preprint arXiv:1111.0094, 2011
E. Deutsch et al., Problem 11237, Amer. Math. Monthly, 115 (No. 7, 2008), 666-667. [From Emeric Deutsch, Aug 13 2008]
J. Riordan, Combinatorial Identities, Wiley, 1968, p. 184.
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LINKS
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Alois P. Heinz, Table of n, a(n) for n = 1..1000
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FORMULA
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a(n) = sum{k=1 to floor(n/2)} A000041(n-2k). - Christian G. Bower, Jun 22 2000
a(n) = Sum_{k<n, k = n (mod 2)} P(k), P(k) =number of partitions of k as in A000041, P(0) = 1. - Jose Luis Arregui (arregui(AT)posta.unizar.es), Apr 05 2002
G.f.: x/((1-x)*(1-x^2)^2))*product(1/(1-x^j), j=3..infty) from Riordan reference second term, last eq.
a(n) = A006128(n-1) - A194452(n-1). - Omar E. Pol, Nov 20 2011
a(n) = A181187(n,2) - A181187(n,3). - Omar E. Pol, Oct 25 2012
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EXAMPLE
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Contribution from Omar E. Pol, Oct 25 2012 (Start):
For n = 7 we have:
--------------------------------------
. Number
Partitions of 7 of 2's
--------------------------------------
7 .............................. 0
4 + 3 .......................... 0
5 + 2 .......................... 1
3 + 2 + 2 ...................... 2
6 + 1 .......................... 0
3 + 3 + 1 ...................... 0
4 + 2 + 1 ...................... 1
2 + 2 + 2 + 1 .................. 3
5 + 1 + 1 ...................... 0
3 + 2 + 1 + 1 .................. 1
4 + 1 + 1 + 1 .................. 0
2 + 2 + 1 + 1 + 1 .............. 2
3 + 1 + 1 + 1 + 1 .............. 0
2 + 1 + 1 + 1 + 1 + 1 .......... 1
1 + 1 + 1 + 1 + 1 + 1 + 1 ...... 0
------------------------------------
. 24 - 13 = 11
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The difference between the sum of the second column and the sum of the third column of the set of partitions of 7 is 24 - 13 = 11 and equals the number of 2's in all partitions of 7, so a(7) = 11.
(End)
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MAPLE
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b:= proc(n, i) option remember; local f, g;
if n=0 or i=1 then [1, 0]
else f:= b(n, i-1); g:= `if`(i>n, [0$2], b(n-i, i));
[f[1]+g[1], f[2]+g[2]+`if`(i=2, g[1], 0)]
fi
end:
a:= n-> b(n, n)[2]:
seq (a(n), n=1..50); # Alois P. Heinz, May 18 2012
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MATHEMATICA
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<< DiscreteMath`Combinatorica`; Table[ Count[ Flatten[ Partitions[n]], 2], {n, 1, 50} ]
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CROSSREFS
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Cf. A066633, A024787, A024788, A024789, A024790, A024791, A024792, A024793, A024794.
Column 2 of A060244.
First differences of A000097.
Cf. A173238 [From Gary W. Adamson, Feb 13 2010]
Sequence in context: A212548 A212549 A212550 * A097497 A006167 A137504
Adjacent sequences: A024783 A024784 A024785 * A024787 A024788 A024789
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KEYWORD
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nonn
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AUTHOR
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Clark Kimberling
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STATUS
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approved
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