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A284597
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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.
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8
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46, 5, 43, 1, 1613, 241, 17011, 12853, 234613, 376741, 78312721, 125938261, 4019167441, 16586155153, 35237422882, 1296230533473, 42301168491121, 61118966262061
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OFFSET
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1,1
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COMMENTS
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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
Conjecture follows from Dickson's conjecture (see link). - Robert Israel, Mar 30 2020
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
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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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 *)
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PROG
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(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
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
(PARI) A284597=vector(19); apply(scan(N, s=1, t=numdiv(s))=for(k=s+1, N, t>(t=numdiv(k))||next; k-s>#A284597||A284597[k-s]||printf(" a(%d)=%d, ", k-s, s)||A284597[k-s]=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
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CROSSREFS
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KEYWORD
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nonn,hard,more
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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