A053624 Highly composite odd numbers: odd numbers where d(n) increases to a record.

1, 3, 9, 15, 45, 105, 225, 315, 945, 1575, 2835, 3465, 10395, 17325, 31185, 45045, 121275, 135135, 225225, 405405, 675675, 1576575, 2027025, 2297295, 3828825, 6891885, 11486475, 26801775, 34459425, 43648605, 72747675, 130945815

Offset: 1

Author

Stefano Lanfranco (lastefano(AT)yahoo.it), Mar 21 2000

Keywords

• nonn
• nice

Comments

Also numbers k such that the number of partitions of k into consecutive integers is a record. For example, 45 = 22+23 = 14+15+16 = 7+8+9+10+11 = 5+6+7+8+9+10 = 1+2+3+4+5+6+7+8+9, six such partitions, but all smaller terms have fewer such partitions (15 has four). See A000005 comments and A038547 formula. - Rick L. Shepherd, Apr 20 2008

From Hartmut F. W. Hoft, Mar 29 2022: (Start)

Also the odd parts of the numbers in A340506, see also comments in A250071.

A140864 is a subsequence. (End)

Positions of records in A001227, i.e., integers whose number of odd divisors sets a new record. - Bernard Schott, Jul 18 2022

Conjecture: all terms after the first three terms are congruent to 5 mod 10. - Harvey P. Dale, Jul 05 2023

From Keith F. Lynch, Jan 12 2024: (Start)

Dale's conjecture is correct. a(n) can't be even, since then a(n)/2 would be a smaller number with the same number of odd divisors. The respective powers of the successive odd primes can't increase, since if they did, swapping them would give a smaller number with the same number of divisors, e.g., 3^2 * 5^4 has the same number of divisors as 3^4 * 5^2, and the latter is smaller. As such, every a(n) must be an odd multiple of 5, hence congruent to 5 mod 10, unless it's simply a power of 3. But multiplying a power of 3 by 3 gives just one more divisor while multiplying a power of 3 by 5 doubles the number of divisors, so after a(n) = 9 all a(n) must be congruent to 5 mod 10, i.e., have a rightmost decimal digit of 5.

This has three equivalent definitions:

* Odd numbers with more divisors than any smaller odd number.

* Numbers with more odd divisors than any smaller number, i.e., record high values of A001227.

* Numbers with a greater excess of odd divisors over even divisors than any smaller number, i.e., record high values of A048272. (End)

Examples

			9 is in the sequence because 9 has 3 divisors {1, 3, 9}, which is more than any previous odd number.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..363 (terms 1..170 from Ray Chandler)
• David Ryan, Mathematical Harmony Analysis, arXiv preprint arXiv:1603.08904 [cs.SD], 2016.

Crossrefs

Cf. A002182, A053640, A000005. Subsequence of A147516.

Cf. A140864, A237593, A250071, A279387, A340506.

Cf. A005408, A001227, A038547.

Programs

Mathematica

nn = 10^6; maxd = 0;
Reap[For[n = 1, n <= nn, n += 2, If[(nd = DivisorSigma[0, n]) > maxd, Print[n]; Sow[n]; maxd = nd]]][[2, 1]] (* Jean-François Alcover, Sep 20 2018, from PARI *)
next[n_] := Module[{k=n, r=DivisorSigma[0, n]}, While[DivisorSigma[0, k]<=r, k+=2]; k]
a053624[n_] := NestList[next, 1, n-1]/; n>=1 (* returns n numbers *)
a053624[31] (* Hartmut F. W. Hoft, Mar 29 2022 *)
DeleteDuplicates[Table[{n,DivisorSigma[0,n]},{n,1,131*10^6,2}],GreaterEqual[ #1[[2]],#2[[2]]]&][[;;,1]] (* Harvey P. Dale, Jul 05 2023 *)

PARI

lista(nn) = {maxd = 0; forstep (n=1, nn, 2, if ((nd = numdiv(n)) > maxd, print1(n, ", "); maxd = nd;););} \\ Michel Marcus, Apr 21 2014