A374674 a(n) is the start of the least run of exactly n consecutive positive numbers with strictly decreasing values of A177329, or -1 if no such run exists.

2, 5, 68, 33, 709, 2313, 13251, 17961, 231881, 525323, 4172904, 7163595

Offset: 1

Author

Amiram Eldar, Jul 16 2024

Keywords

• nonn
• hard
• more

Comments

For n > 1, a(n)! is the start the least run of successive factorials with strictly decreasing number of infinitary divisors (A037445).

a(9) > 170000, if it exists.

Examples

			  n |  a(n) | A177329(k), k = a(n), a(n)+1, ..., a(n)+n-1
  --|-------|------------------------------------------------------
  1 |     2 | 1
  2 |     5 | 4 > 3
  3 |    68 | 31 > 28 > 27
  4 |    33 | 21 > 17 > 16 > 15
  5 |   709 | 199 > 197 > 195 > 193 > 190
  6 |  2313 | 528 > 523 > 519 > 518 > 513 > 508
  7 | 13251 | 2355 > 2354 > 2353 > 2351 > 2350 > 2345 > 2343
  8 | 17961 | 3060 > 3056 > 3051 > 3049 > 3048 > 3047 > 3044 > 3041
		

Crossrefs

Cf. A037445, A177329, A374671, A374672, A374673.

Programs

Mathematica

s[n_] := Module[{e = FactorInteger[n!][[;; , 2]]}, Sum[DigitCount[e[[k]], 2, 1], {k, 1, Length[e]}]]; seq[len_] := Module[{v = Table[0, {len}], w = {s[2]}, c = 0, k = 3, m, s1}, While[c < len, s1 = s[k]; m = Length[w]; If[s1 < w[[m]], AppendTo[w, s1], If[m <= len && v[[m]] == 0, v[[m]] = k-m; c++]; w = {s1}]; k++]; v]; seq[5]

PARI

s(n) = {my(e = factor(n!)[, 2]); sum(k=1, #e, hammingweight(e[k]));}
lista(len) = {my(v = vector(len), w = [s(2)], c = 0, k = 3, m, s1); while(c < len, s1 = s(k); m = #w; if(s1 < w[m], w = concat(w, s1), if(m < = len && v[m] == 0, v[m] = k-m; c++); w = [s1]); k++); v;}

Python

from itertools import count
from collections import Counter
from sympy import factorint
def A374674(n):
    if n==1: return 2
    c, a, l = Counter(), 0, 0
    for m in count(2):
        c += Counter(factorint(m))
        b = sum(map(int.bit_count,c.values()))
        if bChai Wah Wu, Jul 18 2024

Extensions

a(9)-a(12) from Chai Wah Wu, Jul 18 2024