A366891 Lexicographically earliest infinite sequence such that a(i) = a(j) => A365425(i) = A365425(j), A206787(A163511(i)) = A206787(A163511(j)) and A336651(A163511(n)) = A336651(A163511(j)) for all i, j >= 0.

1, 1, 1, 2, 1, 3, 2, 4, 1, 5, 3, 6, 2, 7, 4, 8, 1, 9, 5, 10, 3, 11, 6, 12, 2, 13, 7, 14, 4, 15, 8, 16, 1, 17, 9, 18, 5, 19, 10, 20, 3, 21, 11, 22, 6, 23, 12, 24, 2, 25, 13, 26, 7, 27, 14, 28, 4, 29, 15, 30, 8, 14, 16, 31, 1, 32, 17, 33, 9, 34, 18, 35, 5, 36, 19, 37, 10, 38, 20, 39, 3, 40, 21, 41, 11, 42, 22, 43, 6, 44, 23

Offset: 0

Author

Antti Karttunen, Nov 04 2023

Keywords

• nonn

Comments

Restricted growth sequence transform of the triplet [A365425(n), A206787(A163511(n)), A336651(A163511(n))], and also by conjecture, of sequence b(n) = A351040(A163511(n)).

For all i, j >= 0:

a(i) = a(j) => A365395(i) = A365395(j),

a(i) = a(j) => A366874(i) = A366874(j),

a(i) = a(j) => A366881(i) = A366881(j).

Links

• Antti Karttunen, Table of n, a(n) for n = 0..65537

Crossrefs

Cf. A000265, A000593, A163511, A347385, A351040, A351461, A365425.

Differs from A366806 for the first time at n=105, where a(105) = 52, while A366806(105) = 19.

Differs from A366881 for the first time at n=511, where a(511) = 249, while A366881(511) = 7.

Programs

PARI

up_to = 65537;
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; };
A000265(n) = (n>>valuation(n,2));
A046523(n) = { my(f=vecsort(factor(n)[, 2], , 4), p); prod(i=1, #f, (p=nextprime(p+1))^f[i]); }; \\ From A046523
A163511(n) = if(!n, 1, my(p=2, t=1); while(n>1, if(!(n%2), (t*=p), p=nextprime(1+p)); n >>= 1); (t*p));
A365425(n) = A046523(A000265(A163511(n)));
A206787(n) = sumdiv(n, d, d*issquarefree(2*d));
A336651(n) = { my(f=factor(n>>valuation(n,2))); prod(i=1, #f~, f[i,1]^(f[i,2]-1)); };
A366891aux(n) = [A365425(n), A206787(A163511(n)), A336651(A163511(n))];
v366891 = rgs_transform(vector(1+up_to,n,A366891aux(n-1)));
A366891(n) = v366891[1+n];

Formula

For all n >= 1, a(n) = a(2*n) = a(A000265(n)).