A323253 - OEIS

%I #28 Sep 07 2019 18:02:54

%S 1,1,1,844,74849,671346,8870025

%N a(n) is the smallest number k such that factorizations of n consecutive integers starting at k have the same excess of number of primes counted with multiplicity over number of primes counted without multiplicity (A046660).

%C Smallest number k such that n or more consecutive integers starting at k have the same number of proper prime power divisors.

%C a(8) > 10^9. - _Vaclav Kotesovec_, Sep 01 2019

%C a(8) <= 254023231417746. - _David A. Corneth_, Sep 01 2019

%C a(8) > 10^13. - _Giovanni Resta_, Sep 05 2019

%e 671346 = 2 * 3^2 * 13 * 19 * 151,

%e 671347 = 17^2 * 23 * 101,

%e 671348 = 2^2 * 47 * 3571,

%e 671349 = 3 * 7^2 * 4567,

%e 671350 = 2 * 5^2 * 29 * 463,

%e 671351 = 53^2 * 239.

%e These the first 6 consecutive numbers with the same number of proper prime power divisors, so a(6) = 671346.

%t Do[find = 0; k = 0; While[find == 0, k++; If[Length[Union[Table[PrimeOmega[j] - PrimeNu[j], {j, k, k + n - 1}]]] == 1, find = 1; Print[k]]], {n, 1, 5}] (* _Vaclav Kotesovec_, Sep 01 2019 *)

%t (* faster program *) fak = Table[f = FactorInteger[j]; Total[Transpose[f][[2]]] - Length[f], {j, 1, 10000000}]; m = Max[fak]; Table[Min[Table[SequencePosition[fak, ConstantArray[j, n]], {j, 0, m}]], {n, 1, 7}] (* _Vaclav Kotesovec_, Sep 01 2019 *)

%o (PARI)

%o excess(n) = bigomega(n) - omega(n);

%o score(n) = my(t=excess(n)); for(k=1, oo, if(excess(n+k) != t, return(k)));

%o upto(nn) = my(n=1); for(k=1, nn, while(score(k) >= n, print1(k, ", "); n++)); \\ _Daniel Suteu_, Sep 01 2019

%Y Cf. A001221, A001222, A006558, A045983, A045984, A046660, A246547.

%K nonn,more

%O 1,4

%A _Ilya Gutkovskiy_, Aug 30 2019

%E a(7) from _Daniel Suteu_ and _Vaclav Kotesovec_, Sep 01 2019