

A284597


a(n) is the least number that begins a run of exactly n consecutive numbers with a nondecreasing number of divisors, or 1 if no such number exists.


7



46, 5, 43, 1, 1613, 241, 17011, 12853, 234613, 376741, 78312721, 125938261, 4019167441, 16586155153, 35237422882, 1296230533473, 42301168491121, 61118966262061
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OFFSET

1,1


COMMENTS

The words "begins" and "exactly" in the definition are crucial. The initial values of tau (number of divisors function, A000005) can be partitioned into nondecreasing runs as follows: {1, 2, 2, 3}, {2, 4}, {2, 4}, {3, 4}, {2, 6}, {2, 4, 4, 5}, {2, 6}, {2, 6}, {4, 4}, {2, 8}, {3, 4, 4, 6}, {2, 8}, {2, 6}, {4, 4, 4, 9}, {2, 4, 4, 8}, {2, 8}, {2, 6, 6}, {4}, {2, 10}, ... From this we can see that a(1) = 46 (the first singleton), a(2)=5 (the first pair), a(3)=43 (the first triple), a(4)=1, etc.  Bill McEachen and Giovanni Resta, Apr 26 2017. (see also A303577 and A303578  N. J. A. Sloane, Apr 29 2018
Initial values computed with a brute force C++ program.
It seems very likely that one can always find a(n) and that we never need to take a(n) = 1. But this is at present only a conjecture.  N. J. A. Sloane, May 04 2017
If a(n) > 1, then A013632(a(n)) >= n. Might be useful to help speed up brute force search.  Chai Wah Wu, May 04 2017
The analog sequence for sigma (sum of divisors) instead of tau (number of divisors) is A285893 (see also A028965).  M. F. Hasler, May 06 2017
a(n) > 3.37*10^14 for n > 18.  Robert Gerbicz, May 14 2017


LINKS

Table of n, a(n) for n=1..18.


EXAMPLE

241 = 241^1 => 2 divisors
242 = 2^1 * 11^2 => 6 divisors
243 = 3^5 => 6 divisors
244 = 2^2 * 61^1 => 6 divisors
245 = 5^1 * 7^2 => 6 divisors
246 = 2^1 * 3^1 * 41^1 => 8 divisors
247 = 13^1 * 19^1 => 4 divisors
So, 247 breaks the chain. 241 is the lowest number that is the beginning of exactly 6 consecutive numbers with a nondecreasing number of divisors. So it is the 6th term in the sequence.
Note also that a(5) is not 242, even though tau evaluated at 242, 243,..., 246 gives 5 nondecreasing values, because here we deal with full runs and 242 belongs to the run of 6 values starting at 241.


MATHEMATICA

Function[s, {46}~Join~Map[Function[r, Select[s, Last@ # == r &][[1, 1]]], Range[2, Max[s[[All, 1]] ] ]]]@ Map[{#[[1, 1]], Length@ # + 1} &, DeleteCases[SplitBy[#, #[[1]] >= 0 &], k_ /; k[[1, 1]] < 0]] &@ MapIndexed[{First@ #2, #1} &, Differences@ Array[DivisorSigma[0, #] &, 10^6]] (* Michael De Vlieger, May 06 2017 *)


PROG

(PARI) genit()={for(n=1, 20, q=0; ibgn=1; for(m=ibgn, 9E99, mark1=q; q=numdiv(m); if(mark1==0, summ=0; dun=0; mark2=m); if(q>=mark1, summ+=1, dun=1); if(dun>0&&summ==n, print(n, " ", mark2); break); if(dun>0&&summ!=n, q=0; m=1))); } \\ Bill McEachen, Apr 25 2017
(Python)
from sympy import divisor_count
def A284597(n):
count, starti, s, i = 0, 1, 0, 1
while True:
d = divisor_count(i)
if d < s:
if count == n:
return starti
starti = i
count = 0
s = d
i += 1
count += 1 # Chai Wah Wu, May 04 2017
(PARI) A284597=vector(19); apply(scan(N, s=1, t=numdiv(s))=for(k=s+1, N, t>(t=numdiv(k))next; ks>#A284597A284597[ks]printf(" a(%d)=%d, ", ks, s)A284597[ks]=s; s=k); done, [10^6]) \\ Finds a(1..10) in ~ 1 sec, but would take 100 times longer to get one more term with scan(10^8). You may extend the search using scan(END, START).  M. F. Hasler, May 06 2017


CROSSREFS

Cf. A000005, A000203, A006558, A013632, A028965, A075046, A285893.
See also A286287, A286288, A286289, A303577, A303578.
Sequence in context: A261513 A036204 A270814 * A286288 A051161 A260512
Adjacent sequences: A284594 A284595 A284596 * A284598 A284599 A284600


KEYWORD

nonn,hard,more


AUTHOR

Fred Schneider, Mar 29 2017


EXTENSIONS

a(1), a(2), a(4) corrected by Bill McEachen and Giovanni Resta, Apr 26 2017
a(17)a(18) from Robert Gerbicz, May 14 2017


STATUS

approved



