%I #91 Oct 23 2024 08:54:23
%S 1,1,3,1,3,4,3,1,6,5,3,4,3,5,11,1,3,8,3,6,12,5,3,4,8,5,12,8,3,13,3,1,
%T 12,5,15,12,3,5,12,6,3,15,3,9,26,5,3,4,10,10,12,9,3,17,18,8,12,5,3,17,
%U 3,5,28,1,18,19,3,9,12,17,3,13,3,5,27,9,21,20,3,6,21,5,3,19,18,5,12,12,3
%N G.f.: Sum_{n>=1} n * x^(n*(n+1)/2) / (1 - x^n).
%C Conjecture: a(n) is the total number of parts in all partitions of n into consecutive parts. - _Omar E. Pol_, Apr 23 2017
%C Conjecture: row sums of A285914. - _Omar E. Pol_, Apr 30 2017
%C (The conjectures are true. See _Joerg Arndt_'s proof in the Links section.) - _Omar E. Pol_, Jun 14 2017
%C Row lengths of A299765. - _Omar E. Pol_, Jul 23 2018
%C From _Omar E. Pol_, Oct 20 2019: (Start)
%C Row sums of A328361.
%C a(n) = 3 iff n is an odd prime. (End)
%H David A. Corneth, <a href="/A204217/b204217.txt">Table of n, a(n) for n = 1..10000</a> (First 1024 terms from Paul D. Hanna)
%H Joerg Arndt, <a href="http://list.seqfan.eu/oldermail/seqfan/2017-June/017633.html">Proof of the conjectures of A204217</a>, SeqFan Mailing List, Jun 03 2017.
%H Michael H. Mertens, Ken Ono and Larry Rolen, <a href="https://arxiv.org/abs/1906.07410">Mock modular Eisenstein series with Nebentypus</a>, arXiv:1906.07410 [math.NT], 2019, p. 4, 9.
%F a(k) = 1 iff k = 2^n for n>=0.
%F G.f.: (1/Theta4(x)^2) * Sum_{n>=1} (-1)^(n-1)* n*x^(n*(n+1)/2) * (1 - x^n)/(1 + x^n)^2 where Theta4(x) = 1 + 2*Sum_{n>=1} (-x)^(n^2), due to an identity of Ramanujan.
%F Conjecture: a(n) = Sum_{k=1..n} k*A285898(n,k). - _R. J. Mathar_, Apr 30 2017
%F Conjecture: a(n) = Sum_{k=1..A003056(n)} k*A237048(n,k). - _Omar E. Pol_, Apr 30 2017
%F (The conjectures are true. See _Joerg Arndt_'s proof in the Links section.) - _Omar E. Pol_, Jun 14 2017
%F Sum_{k=1..n} a(k) ~ 2^(3/2) * n^(3/2) / 3. - _Vaclav Kotesovec_, Oct 23 2024
%e G.f.: A(x) = x + x^2 + 3*x^3 + x^4 + 3*x^5 + 4*x^6 + 3*x^7 + x^8 + ...
%e follows by expanding A(x) = x/(1-x) + 2*x^3/(1-x^2) + 3*x^6/(1-x^3) + 4*x^10/(1-x^4) + ...
%e Also, by a Ramanujan identity:
%e A(x)*Theta4(x)^2 = x*(1-x)/(1+x)^2 - 2*x^3*(1-x^2)/(1+x^2)^2 + 3*x^6*(1-x^3)/(1+x^3)^2 - 4*x^10*(1-x^4)/(1+x^4)^2 + 5*x^15*(1-x^5)/(1+x^5)^2 + ...
%e For n = 15 there are four partitions of 15 into consecutive parts: [15], [8, 7], [6, 5, 4] and [5, 4, 3, 2, 1]. The total number of parts is 11, so a(15) = 11. - _Omar E. Pol_, Apr 23 2017
%e From _Omar E. Pol_, Nov 30 2020: (Start)
%e Illustration of initial terms:
%e Diagram
%e n a(n) _
%e 1 1 _|1
%e 2 1 _|1 _
%e 3 3 _|1 |2
%e 4 1 _|1 _|
%e 5 3 _|1 |2 _
%e 6 4 _|1 _| |3
%e 7 3 _|1 |2 |
%e 8 1 _|1 _| _|
%e 9 6 _|1 |2 |3 _
%e 10 5 |1 | | |4
%e ...
%e a(n) is the total length of all vertical line segments that are below and that share one vertex with the horizontal line segments that are in the n-th level of the diagram. For more information about the diagram see A286001 and A237593. (End)
%t terms = 1024; Sum[n*x^(n*(n+1)/2)/(1-x^n), {n, 1, Ceiling[Sqrt[2*terms]]}] + O[x]^(terms+1) // CoefficientList[#, x]& // Rest (* _Jean-François Alcover_, Jun 04 2017 *)
%t Table[Sum[If[n > k*(k-1)/2 && IntegerQ[n/k - (k-1)/2], k, 0], {k, Divisors[2*n]}], {n, 1, 100}] (* _Vaclav Kotesovec_, Oct 23 2024 *)
%o (PARI) {a(n)=polcoeff(sum(m=1,n,m*x^(m*(m+1)/2)/(1-x^m+x*O(x^n))),n)}
%o (PARI) {a(n)=local(Theta4=1+2*sum(m=1,sqrtint(n+1),(-x)^(m^2))+x*O(x^n));polcoeff(1/Theta4^2*sum(m=1,sqrtint(2*n+1),(-1)^(m-1)*m*x^(m*(m+1)/2)*(1-x^m)/(1+x^m+x*O(x^n))^2),n)}
%o (PARI) a(n) = {nb = 0; forpart(v=n, nbp = #v; if ((#Set(v)==#v) && (v[nbp] - v[1] == #v-1), nb += #v); ); nb; } \\ _Michel Marcus_, Apr 23 2017
%o (PARI) a(n) = {my(i=2, t=1); n--; while(n>0, t += (i*(n%i==0)); n-=i; i++); t} \\ _David A. Corneth_, Apr 28 2017
%Y Cf. A204218, A237593, A285914, A286001, A299765, A328361, A357618.
%K nonn,easy,changed
%O 1,3
%A _Paul D. Hanna_, Jan 12 2012