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A053624
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Highly composite odd numbers (1): where d(n) increases to a record.
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24
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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
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OFFSET
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1,2
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COMMENTS
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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
Also the odd parts of the numbers in A340506, see also comments in A250071.
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
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)
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LINKS
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EXAMPLE
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9 is in the sequence because 9 has 3 divisors {1, 3, 9}, which is more than any previous odd number.
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MATHEMATICA
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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 *)
DeleteDuplicates[Table[{n, DivisorSigma[0, n]}, {n, 1, 131*10^6, 2}], GreaterEqual[ #1[[2]], #2[[2]]]&][[;; , 1]] (* Harvey P. Dale, Jul 05 2023 *)
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PROG
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(PARI) lista(nn) = {maxd = 0; forstep (n=1, nn, 2, if ((nd = numdiv(n)) > maxd, print1(n, ", "); maxd = nd; ); ); } \\ Michel Marcus, Apr 21 2014
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CROSSREFS
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KEYWORD
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nonn,nice
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AUTHOR
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Stefano Lanfranco (lastefano(AT)yahoo.it), Mar 21 2000
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STATUS
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approved
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