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 A323253 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). 2
 1, 1, 1, 844, 74849, 671346, 8870025 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,4 COMMENTS Smallest number k such that n or more consecutive integers starting at k have the same number of proper prime power divisors. a(8) > 10^9. - Vaclav Kotesovec, Sep 01 2019 a(8) <= 254023231417746. - David A. Corneth, Sep 01 2019 a(8) > 10^13. - Giovanni Resta, Sep 05 2019 LINKS EXAMPLE 671346 = 2 * 3^2 * 13 * 19 * 151, 671347 = 17^2 * 23 * 101, 671348 = 2^2 * 47 * 3571, 671349 = 3 * 7^2 * 4567, 671350 = 2 * 5^2 * 29 * 463, 671351 = 53^2 * 239. These the first 6 consecutive numbers with the same number of proper prime power divisors, so a(6) = 671346. MATHEMATICA 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 *) (* faster program *) fak = Table[f = FactorInteger[j]; Total[Transpose[f][]] - 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 *) PROG (PARI) excess(n) = bigomega(n) - omega(n); score(n) = my(t=excess(n)); for(k=1, oo, if(excess(n+k) != t, return(k))); upto(nn) = my(n=1); for(k=1, nn, while(score(k) >= n, print1(k, ", "); n++)); \\ Daniel Suteu, Sep 01 2019 CROSSREFS Cf. A001221, A001222, A006558, A045983, A045984, A046660, A246547. Sequence in context: A078144 A071320 A338628 * A160212 A188296 A187861 Adjacent sequences:  A323250 A323251 A323252 * A323254 A323255 A323256 KEYWORD nonn,more AUTHOR Ilya Gutkovskiy, Aug 30 2019 EXTENSIONS a(7) from Daniel Suteu and Vaclav Kotesovec, Sep 01 2019 STATUS approved

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Last modified January 27 15:42 EST 2022. Contains 350607 sequences. (Running on oeis4.)