A286288 -id:A286288 - cp's OEIS frontend

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.

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

Offset: 1

Fred Schneider, Mar 29 2017

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
a(n) > 3.37*10^14 for n > 18. - Robert Gerbicz, May 14 2017

			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.

Robert Israel, Dickson's conjecture implies all a(n)>0

Cf. A000005, A000203, A006558, A013632, A028965, A075046, A285893.

See also A286287, A286288, A286289, A303577, A303578.

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 *)

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

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

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

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

A285893 Least number to start a run of exactly n nondecreasing values of sigma (sum of divisors, A000203).

Original entry on oeis.org

45, 5, 313, 1, 356067536821, 36721681

Offset: 1

M. F. Hasler, May 06 2017

a(6) = 36721681, see also A028965.

The analogous sequence based on tau = A000005 instead of sigma is A284597.

			We have the following values of sigma for n = 1..10:
         n   1   2   3   4   5   6   7   8   9  10  ...
   sigma(n)  0   1   1   2   1   2   1   3   2   2  ...
We see a run of 4 nondecreasing values starting at 1, ending at 4, therefore a(4) = 1. There is a run of 2 nondecreasing values starting at 5, ending at 6, therefore a(2) = 5.
Correspondingly, a run of length 1 corresponds to a number n such that sigma(n-1) > sigma(n) > sigma(n+1). This happens first at a(1) = 45.

Cf. A000005, A000203, A028965, A284597 (analog for sigma_0), A286287 (analog for omega = A001221), A286288 (analog for bigomega = A001222).

Function[s, {45}~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[1, #] &, 10^6]] (* Michael De Vlieger, May 06 2017 *)

alias(A,A285893);A=vector(19);apply(scan(N,s=1,t=sigma(s))=for(k=s+1,N,t>(t=sigma(k))||next;k-s>#A||A[k-s]||printf("a(%d)=%d,",k-s,s)||A[k-s]=s;s=k);done,[10^8]) \\ Search may be extended using scan(END,START).

a(5)-a(6) from Giovanni Resta, May 07 2017

A286287 Least number to start a run of exactly n nondecreasing values of little omega (A001221).

Original entry on oeis.org

106, 11, 13, 7, 512, 1, 1941, 141, 6847, 211, 195031, 82321, 808083, 534077, 3355906, 526093, 526889774, 127890361, 22529949392, 118968284927, 164159173895, 244022049199, 3022058317713, 585927201061

Offset: 1

N. J. A. Sloane, May 16 2017

a(17) > 10^7.
a(18) = 127890361; a(n) > 4*10^8 for n=17 and for n >= 19. - Jon E. Schoenfield, Jul 16 2017
a(n) > 6*10^12 for n >= 25. - Giovanni Resta, Jul 16 2019

			We have omega(10) = 2, omega(11) = 1, omega(12) = 2, omega(13) = 1. Therefore 11 starts a run of exactly 2 consecutive integers (11, 12) which have nondecreasing (here: strictly increasing) values of omega.
The 6 numbers from 1 through 6 yield values (0, 1, ..., 1, 2) for omega, therefore a(6) = 1. The 4 numbers from 7 through 10 yield values (1, 1, 1, 2) for omega, therefore a(4) = 7.
A run of length 1 is a single number n such that omega(n-1) > omega(n) > omega(n+1). (If we had "<=" in one of the cases, it would be part of a run of at least 2 numbers with nondecreasing omega.) This first happens for a(1) = 106.

M. F. Hasler, Posting to Sequence Fans Mailing List, May 06 2017

Cf. A001221, A284597 (analog for tau = A000005), A285893 (analog for sigma = A000203).

A286289 Least number to start a run of exactly n nondecreasing values of the Euler phi function (A000010).

Original entry on oeis.org

314, 6, 315, 14, 1

Offset: 1

N. J. A. Sloane, May 16 2017

a(6) > 10^7. - Michael De Vlieger, May 19 2017

a(6) > 10^13. - Giovanni Resta, Nov 12 2019

			From _Michael De Vlieger_, May 19 2017: (Start)
A run of subsequent numbers with nondecreasing phi is of length 1 if it consists of a single number n with phi(n-1) > phi(n) > phi(n+1) (else n belongs to a run of length >= 2). This happens first for a(1) = 314.
Phi(14..18) = (6, 8, 8, 16, 6), therefore the first run of 4 numbers with nondecreasing phi(= A000010) starts at a(4) = 14. (End)

M. F. Hasler, Posting to Sequence Fans Mailing List, May 06 2017

Cf. A000010, A284597, A285893, A286287, A286288.

Prepend[#, Module[{k = 2}, While[Sign@ Differences@ EulerPhi[k + {-1, 0, 1}] != {-1, -1}, k++];k]] &@ Function[s, Function[r, If[Length@ # > 0, #[[1, 1]], -1] &@ Select[s, Length@ # == r &]] /@ Range@ Max@ Map[Length, s]]@ DeleteCases[SplitBy[MapIndexed[Function[k, (2 Boole[#1 <= #2] - 1) k & @@ #1]@ First@ #2 &, Partition[Array[EulerPhi, 10^7], 2, 1]], Sign], w_ /; First@ w < 0] (* Michael De Vlieger, May 19 2017 *)

A364804 a(n) is the smallest number k such that the number of prime divisors (counted with multiplicity) of the n numbers from k through k+n-1 are in nondescending order.

Original entry on oeis.org

1, 1, 1, 1, 121, 121, 2521, 2521, 162121, 460801, 23553169, 23553169, 244068841, 913535283

Offset: 1

Ilya Gutkovskiy, Aug 08 2023

Smallest initial number k of n consecutive numbers satisfying bigomega(k) <= bigomega(k+1) <= ... <= bigomega(k+n-1).

			a(5) = 121 = a(6) as bigomega(121) = bigomega(122) = bigomega(123) = 2 < bigomega(124) = bigomega(125) = 3 < bigomega(126) = 4.

Cf. A001222, A075046, A286288, A364805.

k = 1; Do[While[t = Table[PrimeOmega[i], {i, k, k + n - 1}]; t != Sort[t], k++]; Print[k], {n, 1, 10}]

a(n) = my(k=1, list=List(vector(n, i, bigomega(i)))); while (vecsort(list) != list, listpop(list, 1); k++; listput(list, bigomega(k+n-1))); k; \\ Michel Marcus, Aug 14 2023

a(11)-a(14) from Michel Marcus, Aug 14 2023

cp's OEIS Frontend

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.

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A285893 Least number to start a run of exactly n nondecreasing values of sigma (sum of divisors, A000203).

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A286287 Least number to start a run of exactly n nondecreasing values of little omega (A001221).

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A286289 Least number to start a run of exactly n nondecreasing values of the Euler phi function (A000010).

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A364804 a(n) is the smallest number k such that the number of prime divisors (counted with multiplicity) of the n numbers from k through k+n-1 are in nondescending order.

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