%I #68 Mar 30 2020 20:56:55
%S 46,5,43,1,1613,241,17011,12853,234613,376741,78312721,125938261,
%T 4019167441,16586155153,35237422882,1296230533473,42301168491121,
%U 61118966262061
%N 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.
%C 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
%C Initial values computed with a brute force C++ program.
%C 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
%C Conjecture follows from Dickson's conjecture (see link). - _Robert Israel_, Mar 30 2020
%C 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
%C 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
%C a(n) > 3.37*10^14 for n > 18. - _Robert Gerbicz_, May 14 2017
%H Robert Israel, <a href="/A284597/a284597.pdf">Dickson's conjecture implies all a(n)>0</a>
%e 241 = 241^1 => 2 divisors
%e 242 = 2^1 * 11^2 => 6 divisors
%e 243 = 3^5 => 6 divisors
%e 244 = 2^2 * 61^1 => 6 divisors
%e 245 = 5^1 * 7^2 => 6 divisors
%e 246 = 2^1 * 3^1 * 41^1 => 8 divisors
%e 247 = 13^1 * 19^1 => 4 divisors
%e 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.
%e 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.
%t 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 *)
%o (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
%o (Python)
%o from sympy import divisor_count
%o def A284597(n):
%o count, starti, s, i = 0,1,0,1
%o while True:
%o d = divisor_count(i)
%o if d < s:
%o if count == n:
%o return starti
%o starti = i
%o count = 0
%o s = d
%o i += 1
%o count += 1 # _Chai Wah Wu_, May 04 2017
%o (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
%Y Cf. A000005, A000203, A006558, A013632, A028965, A075046, A285893.
%Y See also A286287, A286288, A286289, A303577, A303578.
%K nonn,hard,more
%O 1,1
%A _Fred Schneider_, Mar 29 2017
%E a(1), a(2), a(4) corrected by _Bill McEachen_ and _Giovanni Resta_, Apr 26 2017
%E a(17)-a(18) from _Robert Gerbicz_, May 14 2017