A099307 -id:A099307 - cp's OEIS frontend

A189760 Least nonnegative number whose n-th arithmetic derivative (A003415) is zero and lower derivatives are nonzero.

0, 1, 2, 6, 9, 14, 33, 62, 177, 414, 1155, 1719, 2625, 4018, 6849, 9770, 17675, 30206, 90609, 260343, 336006, 757995, 1290874, 2029875, 4059746, 7037655, 17594075, 50850483, 68589598, 186888243, 373659254, 1884639669

Offset: 0

T. D. Noe, Apr 27 2011

a(32) <= 9519378185. - Donovan Johnson, Apr 30 2011
From Antti Karttunen, Oct 02 2019: (Start)
For at least n =  1, 3, 4, 5, 6, 7, 10, 14, 15, 17, 21, 23, 24, 25, 26, 27, 28, 29, we have = a(n) = A003415(a(1+n)), thus we have subsequences like 6, 9, 14, 33, 62, 177 that are obtained by iterating A098699 starting from 6, but as A098699(177) = 0, that run ends there. From a(14) to a(16) we have a run of three such terms: 6849, 9770, 17675. A yet longer such run is from a(23) to a(30): 2029875, 4059746, 7037655, 17594075, 50850483, 68589598, 186888243, 373659254.
Applying A327968 to these terms yields: 0, 0, 1, 5, 5, 5, 5, 5, 5, 7, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, ...
Question: Are there indefinitely long sequences of iterations of A003415 that end with steps ... -> p -> 1 -> 0, with p=5? Are there such sequences for any other prime p? Can we construct a such sequence that is guaranteed to be infinite? See the subtree depicted in A327975 and conjecture #8 in Ufnarovski and Ahlander paper.
(End)

Victor Ufnarovski and Bo Ahlander, How to Differentiate a Number, J. Integer Seqs., Vol. 6, 2003.

Cf. A003415, A098699, A099307, A327965, A327968, A327975, A327977.
Cf. A256750, A327966 (left inverses for this sequence).
Subsequence of A048103. Differs from A327967 for the first time at n=19.

nn = 15; t = Table[0, {nn}]; n = 0; cnt = 0; While[cnt < nn, n++; k = 0; d = n; While[f = Transpose[FactorInteger[d]]; d > 1 && And @@ MapThread[Greater, f], k++; d = Plus @@ (d*f[[2]]/f[[1]])]; If[d == 1, k++; If[k <= nn && t[[k]] == 0, t[[k]] = n; cnt++]]]; Join[{0},t]

Least k such that A099307(k) = n.

For all n >= 0, A256750(a(n)) = A327966(a(n)) = n, A327965(a(n)) = A003415(a(n)). - Antti Karttunen, Oct 02 2019

a(26)-a(31) from Donovan Johnson, Apr 29 2011

A351079 a(n) is the largest term encountered on the path from n to 0 when iterating the map x -> x', or 0 if 0 cannot be reached from n (or if n is 0). Here x' is the arithmetic derivative of x, A003415.

Original entry on oeis.org

0, 1, 2, 3, 0, 5, 6, 7, 0, 9, 10, 11, 0, 13, 14, 0, 0, 17, 21, 19, 0, 21, 22, 23, 0, 25, 0, 0, 0, 29, 31, 31, 0, 33, 34, 0, 0, 37, 38, 0, 0, 41, 42, 43, 0, 0, 46, 47, 0, 49, 0, 0, 0, 53, 0, 0, 0, 57, 58, 59, 0, 61, 62, 0, 0, 65, 66, 67, 0, 0, 70, 71, 0, 73, 0, 0, 0, 77, 78, 79, 0, 0, 82, 83, 0, 85, 0, 0, 0, 89

Offset: 0

Antti Karttunen, Feb 11 2022

nonn

Question: Is there any good upper bound for ratio a(n)/n? See also comments in A351261.

			For n = 15, if we iterate with A003415, we get a path 15 -> 8 -> 12 -> 16 -> 32 -> 80 -> 176 -> 368 -> ..., where the terms just keep on growing without ever reaching zero, therefore a(15) = 0.
For n = 18, its path down to zero, when iterating A003415 is: 18 -> 21 -> 10 -> 7 -> 1 -> 0, and the largest term is 21, therefore a(18) = 21.

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

Cf. A003415, A099307, A099309, A351078, A351261 [= a(A351255(n))].

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));
A351079(n) = { my(m=n); while(n>1, n = A003415checked(n); m = max(m,n)); if(n,m); };

For n > 0, a(n) = 0 if A099307(n) = 0, otherwise a(n) = max(n, a(A003415(n))).

a(0) = 0 and a(A099309(n)) = 0 for all n.

A351256 Maximal exponent in the prime factorization of A351255(n).

Original entry on oeis.org

0, 1, 1, 1, 2, 2, 1, 1, 1, 2, 2, 3, 3, 4, 1, 1, 1, 1, 2, 1, 1, 2, 2, 2, 2, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 4, 4, 4, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 5, 5, 6, 6, 6, 6, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 4, 2, 2, 2, 2, 3, 3, 4, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 6, 6, 6, 6

Offset: 1

Antti Karttunen, Feb 11 2022

nonn

The largest digit in the primorial base representation of A328116(n).

The scatter plot gives some sense of how it is harder to eventually reach zero by iterating A003415, when starting from a number with large exponent(s) in its prime factorization. See also A099308.

			A328116(27) = 50, and A049345(50) = "1310", where the largest primorial base digit is 3, therefore a(27) = 3. Equally, A351255(27) = 2625 = 3 * 5^3 * 7, thus A051903(2625) = 3 and a(27) = 3.

Antti Karttunen, Table of n, a(n) for n = 1..105368 (computed for all 19-smooth terms of A351255, and also for A276086(9699690) = 23)

Cf. A003415, A049345, A051903, A099308, A276086, A328114, A328116, A351255, A351257, A351258.

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));
A051903(n) = if((1==n),0,vecmax(factor(n)[, 2]));
A276086(n) = { my(m=1, p=2); while(n, m *= (p^(n%p)); n = n\p; p = nextprime(1+p)); (m); };
A099307(n) = { my(s=1); while(n>1, n = A003415checked(n); s++); if(n,s,0); };
for(n=0, 2^9, u=A276086(n); c = A099307(u); if(c>0,print1(A051903(u), ", ")));

a(n) = A328114(A328116(n)) = A051903(A351255(n)).

For all n, a(n) < A351257(n). [See A351258 for the differences].

A353515 The length of the shortest path from n to 1 when using the transitions x -> A003415(x) and x -> A003961(x), or -1 if no 1 can ever be reached from n.

Original entry on oeis.org

0, 1, 1, 4, 1, 2, 1, 7, 3, 2, 1, 6, 1, 4, 7, 8, 1, 4, 1, 6, 3, 2, 1, 7, 3, 6, 7, 8, 1, 2, 1, 10, 5, 2, 6, 5, 1, 4, 4, 6, 1, 2, 1, 6, 5, 4, 1, 9, 4, 5, 5, 8, 1, 6, 8, 8, 3, 2, 1, 4, 1, 6, 5, 10, 5, 2, 1, 5, 4, 2, 1, 8, 1, 5, 6, 6, 5, 2, 1, 9, 7, 2, 1, 4, 3, 5, 7, 8, 1, 5, 7, 7, 3, 5, 4, 9, 1, 6, 7, 6, 1, 3, 1, 7, 2

Offset: 1

Antti Karttunen, Apr 23 2022

nonn

This is a variant of A327969 that seems to be less in need of an escape clause. Note that enough prime shifts with A003961 will eventually transform every term of A100716 (which is a subsequence of A099309) to a term of A048103, and that A051903(A003961(n)) = A051903(n). See also the array A344027.

Records 0, 1, 4, 7, 8, 10, 12, 13, 14, 15, 16, 19, ... occur at 1, 2, 4, 8, 16, 32, 128, 256, 768, 1024, 2048, 4096, ..., etc.

			From n = 4, we can reach 1 with just four steps as A003961(4) = 9, A003415(9) = 6, A003415(6) = 5 and A003415(5) = 1, and because there are no shorter paths we have a(4) = 4.
From n = 8, we can reach 1 with seven steps, as A003415(8) = 12, A003961(12) = 45, A003415(45) = 39, A003961(39) = 85, A003415(85) = 22, A003415(22) = 13, A003415(13) = 1, and because there are no shorter paths we have a(8) = 7.
For n = 15, as A003415(15) = 8, we know that a(15) is at most a(8)+1, i.e., 8. But we can do better, as A003961(15) = 35, A003961(35) = 77, A003415(77) = 18, A003415(18) = 21, A003415(21) = 10, A003415(10) = 7, A003415(7) = 1, and because there are no shorter paths we have a(15) = 7.
From n = 49, we can reach 1 in four steps, as A003961(49) = 121, A003415(121) = 22, A003415(22) = 13, A003415(13) = 1. Note that this is less than A099307(49)-1, as it would take 5 steps to reach 1 if using the arithmetic derivative only, 49 -> 14 -> 9 -> 6 -> 5 -> 1.

Antti Karttunen, Table of n, a(n) for n = 1..16383
Index entries for sequences computed from indices in prime factorization

Cf. A003415, A003961, A048103, A051903, A099307, A099308, A099309, A100716, A349905.

Cf. also A327969 and A343221, A344027, A353484.

A003415(n) = if(n<=1, 0, my(f=factor(n)); n*sum(i=1, #f~, f[i, 2]/f[i, 1]));
A003961(n) = { my(f = factor(n)); for(i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); }; \\ From A003961
A353515(n) = if(1==n,0,my(xs=Set([n]),newxs,a,b,u); for(k=1,oo, newxs=Set([]); for(i=1,#xs,u = xs[i]; a = A003415(u); if(1==a, return(k)); if(isprime(a), return(k+1)); b = A003961(u); newxs = setunion([a],newxs); newxs = setunion([b],newxs)); xs = newxs));

a(1) = 0, a(p^p) = 1 + a(A003961(p^p)) for primes p, and for other numbers, a(n) = 1 + min(a(A003415(n)), a(A003961(n))).
a(p) = 1 for all primes p.
a(n) < A099307(n), unless A099307(n) = 0. [I.e., for all n in A099308]

A361869 Let x_0, x_1, x_2, ... be the iterations of the arithmetic derivative A003415 starting with x_0 = n. a(n) is the greatest k such that x_0 > x_1 > ... > x_k.

Original entry on oeis.org

0, 1, 2, 2, 0, 2, 3, 2, 0, 4, 3, 2, 0, 2, 5, 1, 0, 2, 0, 2, 0, 4, 3, 2, 0, 4, 2, 0, 0, 2, 0, 2, 0, 6, 3, 1, 0, 2, 5, 1, 0, 2, 3, 2, 0, 2, 5, 2, 0, 6, 3, 1, 0, 2, 0, 1, 0, 4, 3, 2, 0, 2, 7, 2, 0, 1, 3, 2, 0, 3, 3, 2, 0, 2, 2, 2, 0, 1, 3, 2, 0, 0, 3, 2, 0, 4, 3, 1, 0, 2, 0, 1, 0, 4, 7, 1, 0, 2, 2, 3

Offset: 0

Robert Israel, May 28 2023

nonn

a(n) is the number of iterations of A003415 starting at n until the sequence of iterates stops decreasing.
a(n) = 0 if and only if A003415(n) >= n.
First differs from A099307 at n=15, where a(15) = 1 while A099307(15) = 0.

			a(5) = 2 because x_0 = 5 > x_1 = A003415(5) = 1 > x_2 = A003415(1) = 0, but  x_3 = A003415(0) = 0.
a(6) = 3 because x_0 = 6 > x_1 = A003415(6) = 5 > ... > x_3 = 0 but x_4 = 0.

Robert Israel, Table of n, a(n) for n = 0..10000

Cf. A003415, A099307.

ader:= proc(n) local t;
  n * add(t[2]/t[1], t = ifactors(n)[2])
end proc:
f:= proc(n) option remember; local t;
   t:= ader(n);
   if t < n then procname(t)+1 else 0 fi
end proc:
map(f, [$0..1000]);

A378209 Antiderivatives of 334406399, numbers k for which A003415(k) = A024451(9) = A003415(A002110(9)).

Original entry on oeis.org

223092870, 975351895, 1527890095, 1885679383, 2189118743, 2329696457, 2338611863, 3485765789, 4586671213, 5453593183, 5472849253, 5674340053, 8071055747, 8931775397, 9332889127, 9453996491, 9601098443, 10293819917, 12717530039, 17343441881, 18636581773, 19498393573, 20167656703, 23244839627, 23515890737, 23556538969

Offset: 1

Antti Karttunen, Nov 20 2024

Apart from the initial term A002110(9), all other terms are products of three distinct odd primes, A046389. Compare to the comments in A369239.
Note that A024451(9) = 334406399 = 43 * 163 * 47711 == -1 (mod 12). Compare the sequences A369450, A369451 and A369452 to see why there is such a sudden peak in A377993 at n=9, when compared to the nearby terms before and after.
For all n=1..330, A327969(a(n)) <= 7 = A099307(a(n)), because, when we apply A003415 successively, we get: A003415(334406399) -> 9835475 [= A369651(9)] -> 4893565 -> 978718 -> 564671 (which is a prime) -> 1 -> 0.

Antti Karttunen, Table of n, a(n) for n = 1..330
Index entries for sequences related to primorial numbers

Row 9 of irregular triangle A377992.
Subsequence of A099308, and after the initial term, subsequence of A046389.
Cf. A002110, A003415, A024451, A099307, A327969, A369239, A369651.
Cf. A369450, A369451, A369452.

cp's OEIS Frontend

A189760 Least nonnegative number whose n-th arithmetic derivative (A003415) is zero and lower derivatives are nonzero.

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A351079 a(n) is the largest term encountered on the path from n to 0 when iterating the map x -> x', or 0 if 0 cannot be reached from n (or if n is 0). Here x' is the arithmetic derivative of x, A003415.

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A351256 Maximal exponent in the prime factorization of A351255(n).

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A353515 The length of the shortest path from n to 1 when using the transitions x -> A003415(x) and x -> A003961(x), or -1 if no 1 can ever be reached from n.

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A361869 Let x_0, x_1, x_2, ... be the iterations of the arithmetic derivative A003415 starting with x_0 = n. a(n) is the greatest k such that x_0 > x_1 > ... > x_k.

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A378209 Antiderivatives of 334406399, numbers k for which A003415(k) = A024451(9) = A003415(A002110(9)).

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