A320004 Filter sequence combining the largest proper divisor of n (A032742) with n's residue modulo 4 (A010873), and a single bit (A319710) telling whether the smallest prime factor is unitary.
1, 2, 3, 4, 5, 6, 3, 7, 8, 9, 3, 10, 5, 11, 12, 13, 5, 14, 3, 15, 16, 17, 3, 18, 19, 20, 21, 22, 5, 23, 3, 24, 25, 26, 27, 28, 5, 29, 30, 31, 5, 32, 3, 33, 34, 35, 3, 36, 37, 38, 39, 40, 5, 41, 42, 43, 44, 45, 3, 46, 5, 47, 48, 49, 50, 51, 3, 52, 53, 54, 3, 55, 5, 56, 57, 58, 25, 59, 3, 60, 61, 62, 3, 63, 64, 65, 66, 67, 5, 68, 30, 69, 70, 71, 72, 73, 5, 74, 75, 76, 5, 77, 3
Offset: 1
Keywords
Examples
For n = 33 (3*11) and n = 77 (7*11), the modulo 4 residue of the smallest prime factor is 3, and the largest proper divisors (A032742) is also equal 11, and the smallest prime factor is unitary. Thus a(33) = a(77) (= 25, a running count value allotted by rgs-transform).
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])); A286474(n) = if(1==n,n,(4*A032742(n) + (n % 4))); A319710(n) = ((n>1)&&(factor(n)[1,2]>1)); v320004 = rgs_transform(vector(up_to,n,[A286474(n),A319710(n)])); A320004(n) = v320004[n];
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