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%I #103 Jul 19 2023 14:15:46
%S 1,1,2,3,4,4,7,5,7,9,9,7,14,8,11,16,13,10,20,11,17,21,16,13,27,17,18,
%T 26,22,16,35,17,23,31,23,25,41,20,25,36,33,22,46,23,31,48,30,25,52,29,
%U 38,47,36,28,57,37,41,52,37,31,71,32,39,62,44,43,69,35,45,62,57,37,79,38
%N Number of nondecreasing arithmetic progressions of positive integers with sum n.
%C From _Gus Wiseman_, May 03 2019: (Start)
%C a(n) is the number of integer partitions of n with equal differences. The Heinz numbers of these partitions are given by A325328. For example, the a(1) = 1 through a(9) = 9 partitions are:
%C 1 2 3 4 5 6 7 8 9
%C 11 21 22 32 33 43 44 54
%C 111 31 41 42 52 53 63
%C 1111 11111 51 61 62 72
%C 222 1111111 71 81
%C 321 2222 333
%C 111111 11111111 432
%C 531
%C 111111111
%C (End)
%C From _Petros Hadjicostas_, Sep 29 2019: (Start)
%C We show how _Leroy Quet_'s g.f. Sum_{n >= 0} a(n)*x^n = 1/(1-x) + Sum_{k >= 2} x^k/(1-x^(k*(k-1)/2))/(1-x^k) in the Formula section below can be derived from _Graeme McRae_'s g.f. for A049982 (see one of the links below).
%C Let b(n) = A049982(n) for n >= 1. Then _Graeme McRae_ proved that Sum_{n >= 1} b(n)*x^n = Sum_{k >= 2} x^t(k)/(x^t(k) - x^t(k-1) - x^k + 1) = Sum_{k >= 2} x^t(k)/((1 - x^k) * (1 - x^t(k-1))), where t(k) = A000217(k) = k*(k+1)/2.
%C Since a(n) - b(n) = A000005(n) for n >= 1, to finish the proof, we only need to show that K(x) := 1 + Sum_{n >= 1} a(n)*x^n - Sum_{n >= 1} b(n)*x^n is the g.f. of A000005 (= number of divisors). But it is easy to show that K(x) = 1 + Sum_{k >= 1} x^k/(1 - x^k) = 1 + Sum_{n >= 1} A000005(n)*x^n (Lambert series for the number of divisors function). (End)
%H Lars Blomberg, <a href="/A049988/b049988.txt">Table of n, a(n) for n = 0..10000</a> (Corrected by _Gus Wiseman_, May 03 2019)
%H Lars Blomberg, <a href="/A049988/a049988.cs.txt">C# program for calculating b-file</a> (needs to be updated for a(0) = 1 - _Gus Wiseman_, May 07 2019).
%H Sadek Bouroubi and Nesrine Benyahia Tani, <a href="http://ftp.math.uni-rostock.de/pub/romako/heft64/bou64.pdf">Integer partitions into arithmetic progressions</a>, Rostok. Math. Kolloq. 64 (2009), 11-16.
%H Sadek Bouroubi and Nesrine Benyahia Tani, <a href="https://www.emis.de/journals/INTEGERS/papers/j7/j7.Abstract.html">Integer partitions into arithmetic progressions with an odd common difference</a>, Integers 9(1) (2009), 77-81.
%H F. Javier de Vega, <a href="https://arxiv.org/abs/2003.13378">An extension of Furstenberg's theorem of the infinitude of primes</a>, arXiv:2003.13378 [math.NT], 2020.
%H F. Javier de Vega, <a href="https://arxiv.org/abs/2004.09505">A Complete Solution of the Partitions of a Number into Arithmetic Progressions</a>, arXiv:2004.09505 [math.NT], 2020.
%H F. Javier de Vega, <a href="https://doi.org/10.17654/0972555523015">On the parabolic partitions of a number</a>, J. Alg., Num. Theor., and Appl. (2023) Vol. 61, No. 2, 135-169.
%H Graeme McRae, <a href="https://web.archive.org/web/20081122034835/http://2000clicks.com/MathHelp/BasicSequenceA049982.htm">Counting arithmetic sequences whose sum is n</a>.
%H Graeme McRae, <a href="/A049988/a049988.pdf">Counting arithmetic sequences whose sum is n</a> [Cached copy]
%H Augustine O. Munagi, <a href="https://www.emis.de/journals/INTEGERS/papers/k7/k7.Abstract.html">Combinatorics of integer partitions in arithmetic progression</a>, Integers 10(1) (2010), 73-82.
%H Augustine O. Munagi and Temba Shonhiwa, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL11/Shonhiwa/shonhiwa13.html">On the partitions of a number into arithmetic progressions</a>, Journal of Integer Sequences 11 (2008), Article 08.5.4.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Arithmetic_progression">Arithmetic progression</a>.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Lambert_series">Lambert series</a>.
%H Gus Wiseman, <a href="/A325325/a325325.txt">Sequences counting and ranking integer partitions by the differences of their successive parts</a>.
%F G.f.: 1/(1-x) + Sum_{k>=2} x^k/(1-x^(k*(k-1)/2))/(1-x^k). - _Leroy Quet_, Apr 08 2010. [Edited by _Gus Wiseman_, May 03 2019]
%F a(n) = A049982(n) + A000005(n) = A049980(n) + A000005(n) - 1 for n >= 1. - _Petros Hadjicostas_, Sep 28 2019
%t a[n_]:=If[n==0,1,Block[{i,c=Floor[(n-1)/2]+DivisorSigma[0,n]},Do[i=1;While[i*k<n,If[Mod[2*(n-i*k),k*(k-1)]==0,c++];i++],{k,3,(Sqrt[1+8*n]-1)/2}];c]];a/@Range[0,73] (* Giovanni Resta,Feb 16 2013. Edited by _Gus Wiseman_, May 07 2019 *)
%t Table[Length[Select[IntegerPartitions[n],SameQ@@Differences[#]&]],{n,0,30}] (* _Gus Wiseman_, May 03 2019 *)
%o (PARI) seq(n)={Vec(1/(1-x) + sum(k=2, n, x^k/(1 - x^(k*(k-1)/2))/(1-x^k) + O(x*x^n)))} \\ _Andrew Howroyd_, Sep 28 2019
%Y Cf. A000005, A000217, A007862, A047966, A049982, A049983, A049986, A049987, A129654, A240026, A240027, A307824, A320466, A325325, A325328.
%K nonn
%O 0,3
%A _Clark Kimberling_
%E Edited by _Max Alekseyev_, May 03 2010
%E a(0) = 1 prepended by _Gus Wiseman_, May 03 2019
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