OFFSET
1,1
COMMENTS
Can 5 consecutive numbers be divisible by the total binary weight of their divisors? If they exist, then they are larger than 10^11.
EXAMPLE
1377595575 is a term since the 4 consecutive numbers from 1377595575 to 1377595578 are all terms of A093705.
MATHEMATICA
divQ[n_] := Divisible[n, DivisorSum[n, DigitCount[#, 2, 1] &]]; div = divQ /@ Range[4]; Reap[Do[If[And @@ div, Sow[k - 4]]; div = Join[Rest[div], {divQ[k]}], {k, 5, 5*10^9}]][[2, 1]]
SequencePosition[Table[If[Mod[n, Total[Flatten[IntegerDigits[#, 2]&/@Divisors[n]]]]==0, 1, 0], {n, 526*10^8}], {1, 1, 1, 1}][[;; , 1]] (* The program will take a long time to run. *) (* Harvey P. Dale, May 28 2023 *)
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Amiram Eldar, Oct 31 2020
STATUS
approved