A190138 Final number of terms obtained with Euler's recurrence formula when computing the sum of divisors of n.

1, 2, 3, 5, 9, 15, 27, 46, 80, 138, 238, 413, 713, 1235, 2136, 3695, 6393, 11057, 19130, 33091, 57246, 99032, 171315, 296365, 512682, 886902, 1534266, 2654154, 4591475, 7942870, 13740526, 23769981, 41120131, 71134474, 123056829, 212878289, 368262059, 637063333

Offset: 1

Author

Michel Marcus, Dec 19 2012

Keywords

• nonn

Comments

It appears that a(n) is the number of compositions of n whose parts are pentagonal numbers. See Neville link. - Michel Marcus, Jul 28 2017

Examples

			For n=5, start with row 5 of A195310: [4, 3, 0]. Then replace 4 by row 4: [3, 2], replace 3 by row 3: [2, 1]. The row is now [3, 2, 2, 1, 0].
Repeat process until all terms are 0: [4, 3, 0], [3, 2, 2, 1, 0], [2, 1, 1, 0, 1, 0, 0, 0], [1, 0, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0, 0, 0].
The final array has 9 items, hence a(5) = 9.
		

Links

• Leonhard Euler, Jordan Bell, A demonstration of a theorem on the order observed in the sums of divisors, arXiv:math/0507201 [math.HO], 2005-2009.
• Leonhard Euler, Jordan Bell, An observation on the sums of divisors, arXiv:math/0411587 [math.HO], 2004-2009.
• N. Robbins, On compositions whose parts are polygonal numbers, Annales Univ. Sci. Budapest., Sect. Comp. 43 (2014) 239-243. See p. 242.

Crossrefs

Cf. A000203, A001318, A195310.

Programs

Mathematica

rows = 30;
gpenta[n_] := If[EvenQ[n], n(3n/2+1)/4, (n+1)(3n+1)/8];
T[n_, k_] := n - gpenta[k];
Do[row[n] = DeleteCases[Table[T[n, k], {k, n}], _?Negative], {n, rows}];
a[n_] := a[n] = row[n] //. j_?Positive :> Sequence @@ row[j] // Length;
Table[Print["a(", n, ") = ", a[n]]; a[n], {n, rows}] (* Jean-François Alcover, Sep 22 2018 *)

PARI

A001318(n) = { return((3*n^2 + 2*n + (n%2) * (2*n + 1)) / 8);}
A195310(n) = {if (n == 0, return ([0])); nb = 1; vec = vector(0); nn = n; while (nn >=0, nn = n - A001318(nb); if (nn >=0, vec = concat(vec, nn)); nb++;); return(vec);}
A190138(m) = { vval = vector(m); for (n=1, m, vec = A195310(n); svec = 0; for (k=1, length(vec), if (vec[k] == 0, svec += 1, svec += vval[vec[k]]);); vval[n] = svec;); for (n=1, m, print1(vval[n], ", "););}