A364547 Odd numbers k such that k is a multiple of A005940(k).
1, 3, 5, 1035, 524295, 16777217
Offset: 1
Examples
1035 is included because 1034 in binary is "10000001010", which Doudna isomorphism maps to 345 = 3*5*23, which thus divides 1035 (= 3^2 * 5 * 23). Note that there are six 0's in the binary representation between its most significant bit and the trailing "1010", thus we get the prime factors A000040(1+1) = 3, A000040(1+1+1) = 5 and A000040(1+1+1+6) = 23. 524295 is included because 524294 in binary is "10000000000000000110", which Doudna isomorphism maps to 549 = 3^2 * 61, which thus divides 524295 (= 3^2 * 5 * 61 * 191). Note that there are sixteen 0's in the binary representation between its most significant bit and the trailing "110", thus we get the prime factors A000040(2) = 3 and A000040(2+16) = 61. 16777217 = 2^24 + 1 is included because A000040(1+24) = 97, and 16777217 = 97*257*673.
Crossrefs
Programs
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Mathematica
nn = 2^20 + 2; Array[Set[a[#], #] &, 2]; {1}~Join~Reap[Do[If[EvenQ[n], Set[a[n], 2 a[n/2]], a[n] = k = Times @@ Power @@@ Map[{Prime[PrimePi[#1] + 1], #2} & @@ # &, FactorInteger[a[(n + 1)/2]]]; If[Divisible[n, a[n]], Sow[n]]], {n, 3, nn}] ][[-1, 1]] (* Michael De Vlieger, Jul 28 2023 *) -
PARI
A005940(n) = { my(p=2, t=1); n--; until(!n\=2, if((n%2), (t*=p), p=nextprime(p+1))); t }; isA364547(n) = ((n%2)&&!(n%A005940(n)));
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