Search: a373256 -id:a373256
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A359430
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a(n) = 1 if the arithmetic derivative of n is a multiple of 3, otherwise 0.
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+10
13
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1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1
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OFFSET
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0
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LINKS
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FORMULA
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a(n) = [A003415(n) == 0 (mod 3)], where [ ] is the Iverson bracket.
For n > 0, a(n) = [n == 0 (mod 9)] + [n != 0 (mod 3)]*[A373591(n) == A373592(n) (mod 3)].
(End)
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PROG
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(PARI)
A003415(n) = if(n<=1, 0, my(f=factor(n)); n*sum(i=1, #f~, f[i, 2]/f[i, 1]));
(PARI) A359430(n) = if(!n, 1, if(!(n%3), !(n%9), my(f = factor(n), m1=0, m2=0); for(i=1, #f~, if(1==(f[i, 1]%3), m1 += f[i, 2], m2 += f[i, 2])); 0==((m1-m2)%3))); \\ Antti Karttunen, Jun 13 2024
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CROSSREFS
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Characteristic function of A327863.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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A369653
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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.
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+10
9
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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
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OFFSET
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0
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COMMENTS
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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.
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LINKS
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FORMULA
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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
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PROG
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(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); };
(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
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CROSSREFS
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Characteristic function of A369654.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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6, 15, 16, 24, 28, 32, 33, 40, 42, 49, 51, 52, 56, 60, 69, 70, 76, 78, 80, 87, 88, 91, 96, 98, 100, 104, 105, 114, 123, 124, 130, 132, 133, 136, 140, 141, 148, 150, 152, 154, 159, 168, 169, 172, 175, 176, 177, 182, 184, 186, 190, 195, 200, 204, 213, 217, 220, 222, 231, 232, 238, 240, 244, 245, 247, 248, 249, 250, 258
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OFFSET
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1,1
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LINKS
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PROG
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CROSSREFS
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Nonnegative integers are partitioned between A327863, A373255, and this sequence.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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