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A369653
a(n) = 1 if A327860(n) is a multiple of 3, otherwise 0, where A327860 is the arithmetic derivative of the primorial base exp-function.
9
1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1
OFFSET
0
COMMENTS
Conjecture: asymptotic mean is 4/9: One third of the primorial base representations (A049345 from which the intermediate value x = A276086(n) is built from) have "2" as their second rightmost digit (out of 0, 1, 2). That 2 is converted by A276086 to 3^2, so as x is a multiple of nine, it is guaranteed that x' [= A003415(x)] gives a multiple of three. If the second rightmost digit is "1", then x = A276086(n) is a multiple of 3 (but not of 9), and its arithmetic derivative, x' = 3*(x/3)' + (x/3), certainly is not a multiple of 3. This leaves the case where the second rightmost digit is "0", and conjecturing now in this case that the modulo 3 remainders are eventually evenly distributed for x' (see A359430 and A373256), at least among the number of the form A048103, it gives in this case 1/3 * 1/3 = 1/9 probability that x' is a multiple of 3, so the total probability would be 1/3 + 0 + 1/9 = 4/9. Note that Sum_{i=1..10^n} a(i), for n = 1..8 gives: 3, 43, 443, 4443, 44443, 444443, 4444444, 44444444, so empirically the value 4/9 seems very sharp. - revised Jun 18 2024.
Compare also to A369034 and A373601.
FORMULA
a(n) = A079978(A327860(n)) = A359430(A276086(n)).
If n == 0 or 1 (mod 6), a(n) = A373601(n), if n == 2 or 3 (mod 6), a(n) = 0, and if n == 4 or 5 (mod 6), a(n) = 1. - Antti Karttunen, Jun 18 2024
PROG
(PARI)
A327860(n) = { my(s=0, m=1, p=2, e); while(n, e = (n%p); m *= (p^e); s += (e/p); n = n\p; p = nextprime(1+p)); (s*m); };
A369653(n) = !(A327860(n)%3);
(PARI) A369653(n) = { my(p=2, c1=0, c2=0); while(n, if(3==p, if((n%p), return(2==(n%p))), if(1==(p%3), c1 += (n%p), if(2==(p%3), c2 += (n%p)))); n = n\p; p = nextprime(1+p)); 0==((c1-c2)%3); }; \\ Antti Karttunen, Jun 18 2024
CROSSREFS
Characteristic function of A369654.
Sequence in context: A166253 A159638 A187615 * A120528 A287663 A195376
KEYWORD
nonn
AUTHOR
Antti Karttunen, Jan 28 2024
STATUS
approved