A319716 Filter sequence combining the largest proper divisor of n (A032742) with modulo 6 residue of the smallest prime factor, A010875(A020639(n)).
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 5, 11, 7, 12, 13, 14, 5, 15, 7, 16, 17, 18, 5, 19, 20, 21, 22, 23, 5, 24, 7, 25, 26, 27, 28, 29, 7, 30, 31, 32, 5, 33, 7, 34, 35, 36, 5, 37, 38, 39, 40, 41, 5, 42, 43, 44, 45, 46, 5, 47, 7, 48, 49, 50, 51, 52, 7, 53, 54, 55, 5, 56, 7, 57, 58, 59, 60, 61, 7, 62, 63, 64, 5, 65, 66, 67, 68, 69, 5, 70, 71, 72, 73, 74, 75, 76, 7, 77, 78, 79, 5, 80, 7, 81, 82, 83, 5, 84, 7, 85, 86, 87, 5, 88, 89, 90, 91, 92, 93, 94, 43
Offset: 1
Keywords
Examples
For n = 55 = 5*11 and 121 = 11*11, 55 = 121 = 1 mod 6 and 11 is their common largest proper divisor, thus they are allotted the same number by the restricted growth sequence transform, that is a(55) = a(121) = 43 (which is the number allotted). Note that such nontrivial equivalence classes may only contain numbers that are 5-rough, A007310, with no prime factors 2 or 3.
Links
- Antti Karttunen, Table of n, a(n) for n = 1..100000
Crossrefs
Programs
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PARI
up_to = 100000; rgs_transform(invec) = { my(om = Map(), outvec = vector(length(invec)), u=1); for(i=1, length(invec), if(mapisdefined(om,invec[i]), my(pp = mapget(om, invec[i])); outvec[i] = outvec[pp] , mapput(om,invec[i],i); outvec[i] = u; u++ )); outvec; }; A032742(n) = if(1==n,n,n/vecmin(factor(n)[,1])); A286476(n) = if(1==n,n,(6*A032742(n) + (n % 6))); v319716 = rgs_transform(vector(up_to,n,A286476(n))); A319716(n) = v319716[n];
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