A360247 Numbers for which the prime indices have the same mean 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, 127, 128, 129, 130
Offset: 1
Keywords
Examples
The prime indices of 900 are {3,3,2,2,1,1} with mean 2, and the distinct prime indices are {1,2,3} also with mean 2, so 900 is in the sequence.
Crossrefs
These partitions are counted by A360243.
Programs
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Maple
isA360247 := proc(n) local ifs,pidx,pe,meanAll,meanDist ; if n = 1 then return true ; end if ; ifs := ifactors(n)[2] ; # list of prime indices with multiplicity pidx := [] ; for pe in ifs do [numtheory[pi](op(1,pe)),op(2,pe)] ; pidx := [op(pidx),%] ; end do: meanAll := add(op(1,pe)*op(2,pe),pe=pidx) / add(op(2,pe),pe=pidx) ; meanDist := add(op(1,pe),pe=pidx) / nops(pidx) ; if meanAll = meanDist then true; else false; end if; end proc: for n from 1 to 130 do if isA360247(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],Mean[prix[#]]==Mean[Union[prix[#]]]&]
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