Displaying 1-10 of 15 results found.
Inverse Möbius transform of A328308, which is the characteristic function of numbers that eventually reach zero when iterated with the arithmetic derivative.
+20
4
1, 2, 2, 2, 2, 4, 2, 2, 3, 4, 2, 4, 2, 4, 3, 2, 2, 6, 2, 4, 4, 4, 2, 4, 3, 3, 3, 4, 2, 7, 2, 2, 4, 4, 3, 6, 2, 4, 3, 4, 2, 8, 2, 4, 4, 4, 2, 4, 3, 5, 3, 3, 2, 6, 3, 4, 4, 4, 2, 7, 2, 4, 5, 2, 4, 8, 2, 4, 3, 7, 2, 6, 2, 3, 4, 4, 4, 6, 2, 4, 3, 4, 2, 8, 4, 3, 3, 4, 2, 9, 3, 4, 4, 4, 3, 4, 2, 6, 5, 5, 2, 6, 2, 3, 6
PROG
(PARI)
A003415checked(n) = if(n<=1, 0, my(f=factor(n), s=0); for(i=1, #f~, if(f[i, 2]>=f[i, 1], return(0), s += f[i, 2]/f[i, 1])); (n*s));
A328308(n) = if(!n, 1, while(n>1, n = A003415checked(n)); (n));
Numbers m whose k-th arithmetic derivative is zero for some k. Complement of A099309.
+10
28
0, 1, 2, 3, 5, 6, 7, 9, 10, 11, 13, 14, 17, 18, 19, 21, 22, 23, 25, 29, 30, 31, 33, 34, 37, 38, 41, 42, 43, 46, 47, 49, 53, 57, 58, 59, 61, 62, 65, 66, 67, 70, 71, 73, 77, 78, 79, 82, 83, 85, 89, 93, 94, 97, 98, 101, 103, 105, 107, 109, 113, 114, 118, 121, 126, 127, 129, 130
COMMENTS
The first derivative of 0 and 1 is 0. The second derivative of a prime number is 0.
EXAMPLE
18 is on this list because the first through fifth derivatives are 21, 10, 7, 1, 0.
MATHEMATICA
dn[0]=0; dn[1]=0; dn[n_]:=Module[{f=Transpose[FactorInteger[n]]}, If[PrimeQ[n], 1, Plus@@(n*f[[2]]/f[[1]])]]; d1=Table[dn[n], {n, 40000}]; nLim=200; lst={1}; i=1; While[i<=Length[lst], currN=lst[[i]]; pre=Intersection[Flatten[Position[d1, currN]], Range[nLim]]; pre=Complement[pre, lst]; lst=Join[lst, pre]; i++ ]; Union[lst]
PROG
(PARI)
\\ The following program would get stuck in nontrivial loops. However, we assume that the conjecture 3 in Ufnarovski & Åhlander paper holds ("The differential equation n^(k) = n has only trivial solutions p^p for primes p").
A003415checked(n) = if(n<=1, 0, my(f=factor(n), s=0); for(i=1, #f~, if(f[i, 2]>=f[i, 1], return(0), s += f[i, 2]/f[i, 1])); (n*s));
isA099308(n) = if(!n, 1, while(n>1, n = A003415checked(n)); (n)); \\ Antti Karttunen, Feb 14 2022
CROSSREFS
Cf. A003415 (arithmetic derivative of n), A099307 (least k such that the k-th arithmetic derivative of n is zero), A099309 (complement, numbers whose k-th arithmetic derivative is nonzero for all k), A351078 (first noncomposite reached when iterating the derivative from these numbers), A351079 (the largest term on such paths).
Cf. also A256750 (number of iterations needed to reach either 0 or a number with a factor of the form p^p), A327969, A351088.
Multiplicative with a(p^e) = 1 if p > e, otherwise 0.
+10
28
1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1
FORMULA
Multiplicative with a(p^e) = [e < p], where [ ] is the Iverson bracket.
Sum_{k=1..n} a(k) ~ c * n, where c = Product_{p prime} (1 - 1/p^p) = 0.7219902344... . - Amiram Eldar, Jan 07 2023
MATHEMATICA
f[p_, e_] := If[e < p, 1, 0]; a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 100] (* Amiram Eldar, Jan 06 2023 *)
PROG
(PARI) A359550(n) = { my(f = factor(n)); prod(k=1, #f~, (f[k, 2]<f[k, 1])); };
(Python)
from sympy import factorint
def A359550(n): return int(all(map(lambda d:d[0]>d[1], factorint(n).items()))) # Chai Wah Wu, Jan 06 2023
CROSSREFS
Characteristic function of A048103.
Numbers n whose k-th arithmetic derivative is nonzero for all k. Complement of A099308.
+10
20
4, 8, 12, 15, 16, 20, 24, 26, 27, 28, 32, 35, 36, 39, 40, 44, 45, 48, 50, 51, 52, 54, 55, 56, 60, 63, 64, 68, 69, 72, 74, 75, 76, 80, 81, 84, 86, 87, 88, 90, 91, 92, 95, 96, 99, 100, 102, 104, 106, 108, 110, 111, 112, 115, 116, 117, 119, 120, 122, 123, 124, 125, 128, 132
COMMENTS
Numbers of the form n = m*p^p (where p is prime), i.e., multiples of some term in A051674, have n' = (m + m')*p^p, which is again of the same form, but strictly larger iff m > 1. Therefore successive derivatives grow to infinity in this case, and they are constant when m = 1. There are other terms in this sequence, but I conjecture that they all eventually lead to a term of this form, e.g., 26 -> 15 -> 8 etc. - M. F. Hasler, Apr 09 2015
PROG
(PARI) is(n)=until(4>n=factorback(n~)*sum(i=1, #n, n[2, i]/n[1, i]), for(i=1, #n=factor(n)~, n[1, i]>n[2, i]||return(1))) \\ M. F. Hasler, Apr 09 2015
CROSSREFS
Cf. A003415 (arithmetic derivative of n), A099307 (least k such that the k-th arithmetic derivative of n is zero), A099308 (numbers whose k-th arithmetic derivative is zero for some k).
Cf. A341999 (characteristic function),
a(n) = 1 if the k-th arithmetic derivative is nonzero for all k >= 0, otherwise 0.
+10
13
0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0
COMMENTS
Characteristic function of A099309.
FORMULA
a(n) = 1 if n is in A100716 or ends there by repeated applications of A003415, otherwise a(n) = 0 (when n instead reaches 0 by such iteration).
For all n > 0, a(n) = [ A256750(n) < 1].
For all n > 0, a(n) >= [ A129251(n)>0], i.e., if A129251(n) is nonzero, then certainly a(n) = 1.
For all n > 1, a(n) >= [ A341997(n) > 1].
PROG
(PARI)
A003415checked(n) = if(n<=1, 0, my(f=factor(n), s=0); for(i=1, #f~, if(f[i, 2]>=f[i, 1], return(0), s += f[i, 2]/f[i, 1])); (n*s));
A341999(n) = if(!n, n, while(n>1, n = A003415checked(n)); (!n));
a(n) tells how many numbers m there are in range 0..n such that the k-th arithmetic derivative of A276086(m) is zero for some k >= 0.
+10
9
1, 2, 3, 4, 5, 6, 7, 8, 8, 9, 9, 9, 10, 10, 10, 11, 11, 11, 11, 11, 12, 13, 13, 13, 13, 13, 13, 13, 14, 14, 15, 16, 17, 18, 18, 19, 19, 20, 21, 21, 22, 22, 22, 23, 23, 23, 24, 25, 25, 26, 27, 27, 27, 27, 27, 27, 27, 27, 27, 27, 28, 29, 29, 29, 29, 30, 30, 31, 32, 32, 32, 33, 33, 34, 35, 35, 36, 36, 36, 37, 37, 37, 37, 37, 38
COMMENTS
a(n) = the number of such terms m in A328116 that m <= n.
Although in principle A276086 moves any n out of the "all hope lost" zone A100716 (where A328308 is always zero), back to its complement A048103, by comparing the ratio of this and A328309 it can be seen (see the Plot2-link in the Links-section) that such a transfer actually lessens the chances that by just iterating A003415 one could reach zero from there. Note also how the effect of the primorial base representation can be clearly seen in the folds and warps of that plot.
FORMULA
a(0) = 1; for n > 0, a(n) = a(n-1) + A328306(n).
PROG
(PARI)
up_to = 65537;
A003415checked(n) = if(n<=1, 0, my(f=factor(n), s=0); for(i=1, #f~, if(f[i, 2]>=f[i, 1], return(0), s += f[i, 2]/f[i, 1])); (n*s));
A276086(n) = { my(i=0, m=1, pr=1, nextpr); while((n>0), i=i+1; nextpr = prime(i)*pr; if((n%nextpr), m*=(prime(i)^((n%nextpr)/pr)); n-=(n%nextpr)); pr=nextpr); m; };
A328308(n) = if(!n, 1, while(n>1, n = A003415checked(n)); (n));
A328307list(up_to) = { my(v=vector(up_to), s= A328306(0)); for(i=1, up_to, s += A328306(i); v[i] = s); (v); };
v328307 = A328307list(up_to);
A328307(n) = if(!n, 1, v328307[n]);
a(n) = 1 if there is no prime p such that p^p divides the arithmetic derivative of n, and 0 otherwise.
+10
9
0, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1
COMMENTS
Question: What is the asymptotic mean of this sequence (and its complement A341996)? Knowing the value for A360111 would solve this. See also related sequences like A354874 and A368916.
FORMULA
For all n >= 1, a(n) >= A354874(n).
PROG
(PARI)
A003415(n) = if(n<=1, 0, my(f=factor(n)); n*sum(i=1, #f~, f[i, 2]/f[i, 1]));
A359550(n) = { my(f = factor(n)); prod(k=1, #f~, (f[k, 2]<f[k, 1])); };
CROSSREFS
Characteristic function of A358215.
a(n) = 1 if k-th arithmetic derivative of A276086(n) is zero for some k, otherwise 0.
+10
7
1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0
PROG
(PARI)
A003415checked(n) = if(n<=1, 0, my(f=factor(n), s=0); for(i=1, #f~, if(f[i, 2]>=f[i, 1], return(0), s += f[i, 2]/f[i, 1])); (n*s));
A276086(n) = { my(i=0, m=1, pr=1, nextpr); while((n>0), i=i+1; nextpr = prime(i)*pr; if((n%nextpr), m*=(prime(i)^((n%nextpr)/pr)); n-=(n%nextpr)); pr=nextpr); m; };
A328308(n) = if(!n, 1, while(n>1, n = A003415checked(n)); (n));
CROSSREFS
Characteristic function of A328116.
a(n) = 1 if there is no factor of the form p^p in n, but for some k-th arithmetic derivative (k >= 1) of n such a factor exists; otherwise 0.
+10
7
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 0, 1
COMMENTS
Question: What can be said about the distribution of 0's and 1's in this sequence? Compare also to A328308, A341996 and A359543.
FORMULA
a(n) = [ A256750(n) < 0], where [ ] is the Iverson bracket.
EXAMPLE
a(15) = 1, because although 15 itself is not in A100716, its arithmetic derivative 15' = 8 is there.
a(26) = 1, as although neither 26 nor 26' = 15 are in A100716, the second derivative of 26, 26'' = 15' = 8 is there.
PROG
(PARI)
A003415checked(n) = if(n<=1, 0, my(f=factor(n), s=0); for(i=1, #f~, if(f[i, 2]>=f[i, 1], return(0), s += f[i, 2]/f[i, 1])); (n*s));
A327936(n) = { my(f = factor(n)); for(k=1, #f~, f[k, 2] = (f[k, 2]>=f[k, 1])); factorback(f); };
A341999(n) = if(!n, n, while(n>1, n = A003415checked(n)); (!n));
CROSSREFS
Characteristic function of A359547.
Number of integers x in range A002110(n) .. A002110(1+n)-1 such that the k-th arithmetic derivative of A276086(x) is zero for some k, where A002110(n) is the n-th primorial.
+10
6
1, 4, 8, 44, 216, 1474, 11130, 92489
COMMENTS
Ratio a(n) / A061720(n) develops as:
0: 1 / 1 = 1.0
1: 4 / 4 = 1.0
2: 8 / 24 = 0.333...
3: 44 / 180 = 0.244...
4: 216 / 2100 = 0.1029...
5: 1474 / 27720 = 0.05317...
6: 11130 / 480480 = 0.02316...
7: 92489 / 9189180 = 0.01006...
Computing term a(8) would need processing over 213393180 integers whose greatest prime factor is 23, from single A351255(105368) = 23 at start to product (2^1)*(3^2)*(5*4)*(7^6)*(11^10)*(13^12)*(17^16)*(19^18)*(23^22) at the end of the batch [number whose size in binary is 346 bits], and would required factoring integers of comparable size and more (see A351261), that might not all be easily factorable.
EXAMPLE
There are eight terms [6, 7, 9, 12, 15, 20, 21, 28] that are >= A002110(2) and < A002110(3) in A328116 for which the corresponding terms [5, 10, 30, 25, 150, 375, 750, 5625] in A276086 (and A351255) are all in A099308, therefore a(2) = 8.
PROG
(PARI)
\\ Memoization would work quite badly here. (See comments in A351255. In practice sequence A328306 was computed first, up to its term a(9699690). Same data is available in A328116.)
A002110(n) = prod(i=1, n, prime(i));
A003415checked(n) = if(n<=1, 0, my(f=factor(n), s=0); for(i=1, #f~, if(f[i, 2]>=f[i, 1], return(0), s += f[i, 2]/f[i, 1])); (n*s));
A328308(n) = if(!n, 1, while(n>1, n = A003415checked(n)); (n));
A276086(n) = { my(m=1, p=2); while(n, m *= (p^(n%p)); n = n\p; p = nextprime(1+p)); (m); };
CROSSREFS
Cf. A002110, A003415, A061720, A099308, A328306, A328307, A328308, A328116, A351067, A351069, A351255, A351261, A351072 (partial sums).
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