A360249 Numbers for which the prime indices have the same median as the distinct prime indices.
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16, 17, 19, 21, 22, 23, 25, 26, 27, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 41, 42, 43, 46, 47, 49, 51, 53, 55, 57, 58, 59, 61, 62, 64, 65, 66, 67, 69, 70, 71, 73, 74, 77, 78, 79, 81, 82, 83, 85, 86, 87, 89, 90, 91, 93, 94, 95, 97, 100, 101, 102, 103, 105, 106, 107, 109, 110, 111, 113, 114, 115, 118, 119, 121, 122, 123, 125, 126, 127, 128, 129, 130
Offset: 1
Keywords
Examples
The prime indices of 126 are {1,2,2,4} with median 2 and distinct prime indices {1,2,4} with median 2, so 126 is in the sequence. The prime indices of 180 are {1,1,2,2,3} with median 2 and distinct prime indices {1,2,3} with median 2, so 180 is in the sequence.
Crossrefs
Programs
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Maple
isA360249 := proc(n) local ifs,pidx,pe,medAll,medDist ; if n = 1 then return true ; end if ; ifs := ifactors(n)[2] ; pidx := [] ; for pe in ifs do numtheory[pi](op(1,pe)) ; pidx := [op(pidx),seq(%,i=1..op(2,pe))] ; end do: medAll := stats[describe,median](sort(pidx)) ; pidx := convert(convert(pidx,set),list) ; medDist := stats[describe,median](sort(pidx)) ; if medAll = medDist then true; else false; end if; end proc: for n from 1 to 130 do if isA360249(n) then printf("%d,",n) ; end if; end do: # R. J. Mathar, May 22 2023
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Mathematica
prix[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]]; Select[Range[100],Median[prix[#]]==Median[Union[prix[#]]]&]
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