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Sum of all parts of the last section of the set of partitions of n.
50

%I #67 Oct 02 2023 20:14:46

%S 1,3,5,11,15,31,39,71,94,150,196,308,389,577,750,1056,1353,1881,2380,

%T 3230,4092,5412,6821,8935,11150,14386,17934,22834,28281,35735,43982,

%U 55066,67551,83821,102365,126267,153397,188001,227645,277305,334383

%N Sum of all parts of the last section of the set of partitions of n.

%C Row sums of the triangles A135010, A138121, A138151 and others related to the section model of partitions (see A135010 and A138121).

%C From _Omar E. Pol_, Jan 20 2021: (Start)

%C Convolution of A000203 and A002865.

%C Convolution of A340793 and A000041.

%C Row sums of triangles A339278, A340426, A340583. (End)

%C a(n) is also the sum of all divisors of all terms of n-th row of A336811. These divisors are also all parts in the last section of the set of partitions of n. - _Omar E. Pol_, Jul 27 2021

%C Row sums of A336812. - _Omar E. Pol_, Aug 03 2021

%H Vaclav Kotesovec, <a href="/A138879/b138879.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = A000041(n)*n - A000041(n-1)*(n-1) = A138880(n) + A000041(n-1).

%F a(n) = A066186(n) - A066186(n-1), for n>=1.

%F a(n) ~ exp(Pi*sqrt(2*n/3)) * Pi/(12*sqrt(2*n)) * (1 - (72 + 13*Pi^2) / (24*Pi*sqrt(6*n)) + (7/12 + 3/(2*Pi^2) + 217*Pi^2/6912)/n - (15*sqrt(3/2)/(16*Pi) + 115*Pi/(288*sqrt(6)) + 4069*Pi^3/(497664*sqrt(6)))/n^(3/2)). - _Vaclav Kotesovec_, Oct 21 2016, extended Jul 06 2019

%F G.f.: x*(1 - x)*f'(x), where f(x) = Product_{k>=1} 1/(1 - x^k). - _Ilya Gutkovskiy_, Apr 13 2017

%e a(6)=31 because the parts of the last section of the set of partitions of 6 are (6), (3,3), (4,2), (2,2,2), (1), (1), (1), (1), (1), (1), (1), so the sum is a(6) = 6 + 3 + 3 + 4 + 2 + 2 + 2 + 2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 31.

%e From _Omar E. Pol_, Aug 13 2013: (Start)

%e Illustration of initial terms:

%e . _ _ _ _ _ _

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%e . |_ _ _ _| |_| |_|

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%e . _ _ |_ _ _| |_| |_| |_|

%e . _ |_ _| |_| |_| |_| |_|

%e . |_| |_| |_| |_| |_| |_|

%e .

%e . 1 3 5 11 15 31

%e .

%e (End)

%e On the other hand for n = 6 the 6th row of triangle A336811 is [6, 4, 3, 2, 2, 1, 1] and the sum of all divisors of these terms is [1 + 2 + 3 + 6] + [1 + 2 + 4] + [1 + 3] + [1 + 2] + [1 + 2] + [1] + [1] = 31, so a(6) = 31. - _Omar E. Pol_, Jul 27 2021

%p A066186 := proc(n) n*combinat[numbpart](n) ; end proc:

%p A138879 := proc(n) A066186(n)-A066186(n-1) ; end proc:

%p seq(A138879(n),n=1..80) ; # _R. J. Mathar_, Jan 27 2011

%t Table[PartitionsP[n]*n - PartitionsP[n-1]*(n-1), {n, 1, 50}] (* _Vaclav Kotesovec_, Oct 21 2016 *)

%o (PARI) for(n=1, 50, print1(numbpart(n)*n - numbpart(n - 1)*(n - 1),", ")) \\ _Indranil Ghosh_, Mar 19 2017

%o (Python)

%o from sympy.ntheory import npartitions

%o print([npartitions(n)*n - npartitions(n - 1)*(n - 1) for n in range(1, 51)]) # _Indranil Ghosh_, Mar 19 2017

%Y Cf. A000041, A000203, A002865, A066186, A133041, A135010, A138121, A138135 - A138138, A138151, A138880, A139100, A237593, A336811, A336812, A338156, A339278, A340035, A340426, A340583, A340793.

%K nonn

%O 1,2

%A _Omar E. Pol_, Apr 30 2008

%E a(34) corrected by _R. J. Mathar_, Jan 27 2011