A372430 Positive integers k such that the distinct prime indices of k are a subset of the binary indices of k.
1, 3, 5, 15, 27, 39, 55, 63, 85, 121, 125, 135, 169, 171, 175, 209, 243, 247, 255, 299, 375, 399, 437, 459, 507, 539, 605, 637, 725, 735, 783, 841, 867, 891, 1085, 1215, 1323, 1331, 1375, 1519, 1767, 1815, 1863, 2079, 2125, 2187, 2223, 2295, 2299, 2331, 2405
Offset: 1
Examples
The prime indices of 135 are {2,2,2,3}, and the binary indices are {1,2,3,8}. Since {2,3} is a subset of {1,2,3,8}, 135 is in the sequence. The terms together with their prime indices begin: 1: {} 3: {2} 5: {3} 15: {2,3} 27: {2,2,2} 39: {2,6} 55: {3,5} 63: {2,2,4} 85: {3,7} 121: {5,5} 125: {3,3,3} The terms together with their binary expansions and binary indices begin: 1: 1 ~ {1} 3: 11 ~ {1,2} 5: 101 ~ {1,3} 15: 1111 ~ {1,2,3,4} 27: 11011 ~ {1,2,4,5} 39: 100111 ~ {1,2,3,6} 55: 110111 ~ {1,2,3,5,6} 63: 111111 ~ {1,2,3,4,5,6} 85: 1010101 ~ {1,3,5,7} 121: 1111001 ~ {1,4,5,6,7} 125: 1111101 ~ {1,3,4,5,6,7}
Crossrefs
A070939 gives length of binary expansion.
Programs
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Mathematica
prix[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]]; bix[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1]; Select[Range[1000],SubsetQ[bix[#],prix[#]]&]
Comments