A327966 -id:A327966 - cp's OEIS frontend

A327967 Positions of records in A327966.

0, 1, 2, 6, 9, 14, 33, 62, 177, 414, 1155, 1719, 2625, 4018, 6849, 9770, 17675, 30206, 90609, 164088, 336006, 757995, 1290874, 2029875, 4059746, 7037655, 17594075, 50850483, 68589598, 186888243, 373659254

Offset: 0

Antti Karttunen, Oct 01 2019

nonn

Starting offset is zero to align with the indexing used in A189760, as this is conjecturally also the least k such that A327966(k) = n.
For at least n = 1, 3, 4, 5, 6, 7, 10, 14, 15, 17, 18, 21, 23, 24, 25, 26, 27, 28, 29, a(n) = A327965(a(1+n)). For example, 30206 = A327965(90609) and 90609 = A327965(164088).
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, ...
See also comments in A189760.

Cf. A327965, A327966.

Differs from A189760 for the first time at n=19, as a(19) = 164088, while A189760(19) = 260343.

A003415(n) = {my(fac); if(n<1, 0, fac=factor(n); sum(i=1, matsize(fac)[1], n*fac[i, 2]/fac[i, 1]))}; \\ From A003415
A327938(n) = { my(f = factor(n)); for(k=1, #f~, f[k,2] = (f[k,2]%f[k,1])); factorback(f); };
A327965(n) = if(n<=1,0,A327938(A003415(n)));
A327966(n) = if(!n,0,1+A327966(A327965(n)));
k=0; n=0; m=-1; while(k<32, if(!(n%2^27),print1("(",n,"),")); if((t=A327966(n))>m, write("b327967.txt", k, " ", n); m = t; k++); n++);

A327969 The length of a shortest path from n to zero when using the transitions x -> A003415(x) and x -> A276086(x), or -1 if no zero can ever be reached from n.

Original entry on oeis.org

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

Offset: 0

Antti Karttunen, Oct 07 2019

The terms of this sequence are currently known only up to n=23, with the value of a(24) still being uncertain. For the tentative values of the later terms, see sequence A328324 which gives upper bounds for these terms, many of which are very likely also exact values for them.
As A051903(A003415(n)) >= A051903(n)-1, it means that it takes always at least A051903(n) steps to a prime if iterating solely with A003415.
Some known values and upper bounds from n=24 onward:
a(24) <= 11.
a(25) = 4.
a(26) = 7.
a(27) <= 22.
a(33) = 4.
a(39) = 4.
a(40) = 5.
a(42) = 3.
a(44) <= 10.
a(45) = 5.
a(46) = 5.
a(48) = 9.
a(49) = 6.
a(50) = 6.
a(55) = 7.
a(74) = 5.
a(77) = 6.
a(80) <= 18.
a(111) = 6.
a(112) = 8.
a(125) <= 9.
a(240) = 7.
a(625) <= 10.
a(875) = 8.
From Antti Karttunen, Feb 20 2022: (Start)
a(2556) <= 20.
a(5005) <= 19.
What is the value of a(128), and is A328324(128) well-defined?
When I created this sequence, I conjectured that by applying two simple arithmetic operations "arithmetic derivative" (A003415) and "primorial base exp-function" (A276086) in some combination, and starting from any positive integer, we could always reach zero (via a prime and 1).
At the first sight it seems almost certain that the conjecture holds, as it is always possible at every step to choose from two options (which very rarely meet, see A351088), leading to an exponentially growing search tree, and also because A276086 always jumps out of any dead-end path with p^p-factors (dead-end from the arithmetic derivative's point of view). However, it should be realized that one can reach the terms of either A157037 or A327978 with a single step of A003415 only from squarefree numbers (or respectively, cubefree numbers that are not multiples of 4, see A328234), and in general, because A003415 decreases the maximal exponent of the prime factorization (A051903) at most by one, if the maximal exponent in the prime factorization of n is large, there is a correspondingly long path to traverse if we take only A003415-steps in the iteration, and any step could always lead with certain probability to a p^p-number. Note that the antiderivatives of primorials with a square factor seem quite rare, see A351029.
And although taking a A276086-step will always land us to a p^p-free number (which a priori is not in the obvious dead-end path of A003415, although of course it might eventually lead to one), it (in most cases) also increases the magnitude of number considerably, that tends to make the escape even harder. Particularly, in the majority of cases A276086 increases the maximal exponent (which in the preimage is A328114, "maximal digit value used when n is written in primorial base"), so there will be even a longer journey down to squarefree numbers when using A003415. See the sequences A351067 and A351071 for the diminishing ratios suggesting rapidly diminishing chances of successfully reaching zero from larger terms of A276086. Also, the asymptotic density of A276156 is zero, even though A351073 may contain a few larger values.
On the other hand, if we could prove that by (for example) continuing upwards with any p^p-path of A003415 we could eventually reach with a near certainty a region of numbers with low values of A328114 (i.e., numbers with smallish digits in primorial base, like A276156), then the situation might change (see also A351089). However, a few empirical runs seemed to indicate otherwise.
For all of the above reasons, I now conjecture that there are natural numbers from which it is not possible to reach zero with any combination of steps. For example 128 or 5^5 = 3125.
(End)

			Let -A> stand for an application of A003415 and -B> for an application of A276086, then, we have for example:
a(8) = 6 as we have 8 -A>  12 -B>  25 -A> 10 -A>  7 -A> 1 -A> 0, six transitions in total (and there are no shorter paths).
a(15) = 6 as we have 15 -B> 150 -A> 185 -A> 42 -A> 41 -A> 1 -A> 0, six transitions in total (and there are no shorter paths).
a(20) = 7, as 20 -B> 375 -A> 350 -A> 365 -A> 78 -A> 71 -A> 1 -A> 0, and there are no shorter paths.
For n=112, we know that a(112) cannot be larger than eight, as A328099^(8)(112) = 0, so we have a path of length 8 as 112 -A> 240 -B> 77 -A> 18 -A> 21 -A> 10 -A> 7 -A> 1 -A> 0. Checking all 32 combinations of the paths of lengths of 5 starting from 112 shows that none of them or their prefixes ends with a prime, thus there cannot be any shorter path, and indeed a(112) = 8.
a(24) <= 11 as A328099^(11)(24) = 0, i.e., we have 24 -A> 44 -A> 48 -A> 112 -A> 240 -B> 77 -A> 18 -A> 21 -A> 10 -A> 7 -A> 1 -A> 0. On the other hand, 24 -B> 625 -B> 17794411250 -A> 41620434625 -A> 58507928150 -A> 86090357185 -A> 54113940517 -A> 19982203325 -A> 12038411230 -A> 8426887871 -A> 1 -A> 0, thus offering another path of length 11.

Cf. A003415, A046099, A051674, A051903, A068346, A276086, A276087, A327859, A327860, A328099, A328112, A328114, A328116, A328307.
Cf. A328324 (a sequence giving upper bounds, computed with restricted search space).
Sequences for whose terms k, value a(k) has a guaranteed constant upper bound: A000040, A002110, A143293, A157037, A192192, A327978, A328232, A328233, A328239, A328240, A328243, A328249, A328313.
Sequences for whose terms k, it is guaranteed that a(k) has finite value > 0, even if not bound by a constant: A099308, A328116.
Cf. also A256750, A327966, A328110, A351029, A351088, A351067, A351071, A351073, A351089 and A351255, A351256, A351257, A351258, A351261.

A003415(n) = if(n<=1, 0, my(f=factor(n)); n*sum(i=1, #f~, f[i, 2]/f[i, 1]));
A276086(n) = { my(m=1, p=2); while(n, m *= (p^(n%p)); n = n\p; p = nextprime(1+p)); (m); };
A327969(n,searchlim=0) = if(!n,n,my(xs=Set([n]),newxs,a,b,u); for(k=1,oo, print("n=", n, " k=", k, " xs=", xs); newxs=Set([]); for(i=1,#xs,u = xs[i]; a = A003415(u); if(0==a, return(k)); if(isprime(a), return(k+2)); b = A276086(u); if(isprime(b), return(k+1+(u>2))); newxs = setunion([a],newxs); if(!searchlim || (b<=searchlim),newxs = setunion([b],newxs))); xs = newxs));

a(0) = 0, a(p^p) = 1 + a(A276086(p^p)) for primes p, and for other numbers, a(n) = 1+min(a(A003415(n)), a(A276086(n))).
a(p) = 2 for all primes p.
For all n, a(n) <= A328324(n).
Let A stand the transition x -> A003415(x), and B stand for x -> A276086(x). The following sequences give some constant upper limits, because it is guaranteed that the combination given in brackets (the leftmost A or B is applied first) will always lead to a prime:
For all n, a(A157037(n)) = 3.  [A]
For n > 1, a(A002110(n)) = 3.  [B]
For all n, a(A192192(n)) <= 4. [AA]
For all n, a(A327978(n)) = 4.  [AB]
For all n, a(A328233(n)) <= 4. [BA]
For all n, a(A143293(n)) <= 4. [BB]
For all n, a(A328239(n)) <= 5. [AAA]
For all n, a(A328240(n)) <= 5. [BAA]
For all n, a(A328243(n)) <= 5. [ABB]
For all n, a(A328313(n)) <= 5. [BBB]
For all n, a(A328249(n)) <= 6. [BAAA]
For all k in A046099, a(k) >= 4, and if A328114(k) > 1, then certainly a(k) > 4.

A256750 Start with n, and repeatedly apply the arithmetic derivative A003415. |a(n)| = the number of iterations to reach 0 (then a(n) is taken nonnegative) or a number having a factor of the form p^p with prime p, in which case a(n) = -|a(n)|.

Original entry on oeis.org

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

Offset: 0

M. F. Hasler, Apr 09 2015

sign

Under iterations of the arithmetic derivative, the orbit of some numbers ends in zero, and the orbit of all others (I conjecture) reaches a number of the form m*p^p with prime p, from where on it keeps this form and grows to infinity iff m>1, or remains at this fixed point if m=1.

This is an extension of the sequence A099307 which counts the steps to reach 0 or yields 0 if this never happens.

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

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, positions of terms > 0 after the initial 0), A099309 (numbers whose k-th arithmetic derivative is nonzero for all k, positions of terms <= 0 after the initial 0), A359547 (positions of negative terms), A327934 (positions of -1's).

Cf. also A327966, A327969 (A328324).

w = {}; nn = 2^16; k = 1; While[Set[m, #^#] <= nn &[Prime[k]], AppendTo[w, m]; k++]; a3415[n_] := a3415[n] = Which[n < 2, 0, PrimeQ[n], 1, True, n Total[#2/#1 & @@@ FactorInteger[n]]]{0, 1}~Join~Reap[Do[Which[PrimeQ[n], Sow[2], MemberQ[w, n], Sow[0], True, Sow@ If[#[[-1]] == 0, Length[#] - 1, -Length[#] + 1] &[NestWhileList[a3415, n, And[! Divisible[#, 4], FreeQ[w, #]] &, 1]]], {n, 2, nn}] ][[-1, -1]] (* Michael De Vlieger, Jan 04 2023 *)

a(n,c=0)={n&&until(!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(-c));c++);c}

a(n) = 0 <=> n = 0 or n = m*p^p for some prime p and some m >= 1 (which is a fixed point iff m = 1).
a(n) = 1 <=> n = 1.
a(n) = 2 <=> n is prime.
a(n) <= 0 <=> n is in A099309 U {0}. If n > 0, the iterations of A003415 applied to n end in a nonzero fixed point or grow to infinity.
a(n) > 0 <=> n is in A099308 \ {0}.
A099307(n) = min { 0, a(n) }.

A327965 "Tamed variant" of arithmetic derivative: a(0) = a(1) = 0; for n > 1, a(n) = A327938(A003415(n)).

Original entry on oeis.org

0, 0, 1, 1, 1, 1, 5, 1, 3, 6, 7, 1, 1, 1, 9, 2, 2, 1, 21, 1, 6, 10, 13, 1, 11, 10, 15, 1, 2, 1, 31, 1, 5, 14, 19, 3, 15, 1, 21, 1, 17, 1, 41, 1, 3, 39, 25, 1, 7, 14, 45, 5, 14, 1, 3, 1, 23, 22, 31, 1, 23, 1, 33, 51, 3, 18, 61, 1, 18, 26, 59, 1, 39, 1, 39, 55, 5, 18, 71, 1, 11, 1, 43, 1, 31, 22, 45, 2, 35, 1, 123, 5, 6

Offset: 0

Antti Karttunen, Oct 01 2019

nonn

Applying A327938 to the result of A003415(n) ensures that all terms stay in A048103, and that all iteration paths will (hopefully) terminate in zero. See A327966.

Antti Karttunen, Table of n, a(n) for n = 0..20000
Antti Karttunen, Data supplement: n, a(n) computed for n = 0..100000

Cf. A003415, A048103, A327938, A327966, A327967.

Cf. also A327859.

A003415(n) = {my(fac); if(n<1, 0, fac=factor(n); sum(i=1, matsize(fac)[1], n*fac[i, 2]/fac[i, 1]))}; \\ From A003415
A327938(n) = { my(f = factor(n)); for(k=1, #f~, f[k,2] = (f[k,2]%f[k,1])); factorback(f); };
A327965(n) = if(n<=1,0,A327938(A003415(n)));

a(0) = a(1) = 0; for n > 1, a(n) = A327938(A003415(n)).

A327968 a(0) = a(1) = 0, a(prime) = 1, and for all other numbers, a(n) = the first noncomposite reached when iterating A327965, or -1 if no noncomposite is ever reached.

Original entry on oeis.org

0, 0, 1, 1, 1, 1, 5, 1, 3, 5, 7, 1, 1, 1, 5, 2, 2, 1, 7, 1, 5, 7, 13, 1, 11, 7, 2, 1, 2, 1, 31, 1, 5, 5, 19, 3, 2, 1, 7, 1, 17, 1, 41, 1, 3, 1, 7, 1, 7, 5, 1, 5, 5, 1, 3, 1, 23, 13, 31, 1, 23, 1, 5, 5, 3, 7, 61, 1, 7, 2, 59, 1, 1, 1, 1, 1, 5, 7, 71, 1, 11, 1, 43, 1, 31, 13, 1, 2, 3, 1, 11, 5, 5, 19, 5, 5, 17, 1, 7, 1, 3, 1, 5, 1, 41, 71

Offset: 0

Antti Karttunen, Oct 02 2019

nonn

Of the prime terms, 5 seems to be the most common (8886 occurrences among the first 100001 terms). See also A327975.

Antti Karttunen, Table of n, a(n) for n = 0..10000
Antti Karttunen, Data supplement: n, a(n) computed for n = 0..100000

Cf. A003415, A189760, A327938, A327965, A327966, A327967, A327975, A327977.

A003415(n) = {my(fac); if(n<1, 0, fac=factor(n); sum(i=1, matsize(fac)[1], n*fac[i, 2]/fac[i, 1]))}; \\ From A003415
A327938(n) = { my(f = factor(n)); for(k=1, #f~, f[k,2] = (f[k,2]%f[k,1])); factorback(f); };
A327965(n) = if(n<=1,0,A327938(A003415(n)));
A327968(n) = if(n<=1,0,if(isprime(n),1, while((n>1)&&!isprime(n), n = A327965(n)); (n)));

A327975 Breadth-first reading of the subtree rooted at 5 of the tree where each parent node is the arithmetic derivative (A003415) of all its children.

Original entry on oeis.org

5, 6, 9, 14, 33, 49, 62, 94, 177, 817, 961, 445, 913, 1633, 2173, 2209, 1146, 886, 1822, 4414, 19193, 25829, 32393, 41033, 47429, 57929, 64133, 88229, 101753, 111173, 116729, 129413, 138233, 148553, 160229, 173093, 183929, 188453, 208613, 216773, 232229, 235913, 244229, 249929, 257573, 262793, 272633, 278153, 282533, 288329, 294473, 304613, 316229, 320933, 322853, 323429, 327653, 328313, 1155, 2649

Offset: 1

Antti Karttunen, Oct 02 2019

Permutation of A328115.
The branching degree of vertex v is given by A099302(v).
Leaves form a subsequence of A098700.
Most terms of A189760 (apart from 0, 1, 2, 414, ...) seem to be located in this tree, in positions where they have no smaller siblings.
For any number k at level n (where 5 is at level 2), we have A256750(k) = A327966(k) = n.
Question: Does this subtree contain infinitely long paths? How many? Cf. conjecture number 8 in Ufnarovski and Ahlander paper, and a similar tree starting from 7, A327977.

			Because we have A003415(5) = 1, A003415(6) = 5, A003415(9) = 6, A003415(14) = 9, A003415(33) = A003415(49) = 14, A003415(62) = 33, etc, this subtree is laid out as below. The terms of this sequence are obtained by scanning each successive level of the tree from left to right, from the node 5 onward:
   (0)
    |
   (1)
    |
    5
    |
    6
    |
    9
    |
    14________________
    |                 |
    33               49
    |                 |
    62________       94_____________________________
    |    |    |       |       |      |      |       |
    |    |    |       |       |      |      |       |
   177  817  961     445     913   1633   2173    2209
              |       |       |                     |
              |       |       |                     |
            1146     886    1822                  4414
              |       |       |                     |
              |       |       |                     |
            (19193,  (1155,  (19921, ..., 829921)  (22045, ..., 4870849)
             25829,   2649,                        [49 children for 4414]
               ...,  ...,    [27 children for 1822]
            328313)  196249)
                     [19 children for 886]
        [38 children
         for 1146]
The row lengths thus start as: 1, 1, 1, 1, 2, 2, 8, 4, 133 (= 38+19+27+49), ...

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

Cf. A003415, A098699, A098700, A099302, A099303, A099307, A099308, A189760, A256750, A327966, A327968, A328115.

Cf. A327977 for the subtree starting from 7, and also A263267 for another similar tree.

A002620(n) = ((n^2)>>2);
A003415(n) = {my(fac); if(n<1, 0, fac=factor(n); sum(i=1, matsize(fac)[1], n*fac[i, 2]/fac[i, 1]))}; \\ From A003415
A327975list(e) = { my(lista=List([5]), f); for(n=1, e, f = lista[n]; for(k=1,1+A002620(f),if(A003415(k)==f, listput(lista,k)))); Vec(lista); };
v328975 = A327975list(21);
A327975(n) = v328975[n];

# uses[A003415]
def A327975():
  '''Breadth-first reading of irregular subtree rooted at 5, defined by the edge-relation A003415(child) = parent.'''
  yield 5
  for x in A327975():
    for k in [1 .. 1+(x*x)//2]:
      if A003415(k) == x: yield k
def take(n, g):
  '''Returns a list composed of the next n elements returned by generator g.'''
  z = []
  if 0 == n: return z
  for x in g:
    z.append(x)
    if n > 1: n = n-1
    else: return(z)
take(60, A327975())

A327977 Breadth-first reading of the subtree rooted at 7 of the tree where each parent node is the arithmetic derivative (A003415) of all its children.

Original entry on oeis.org

7, 10, 21, 25, 18, 38, 46, 65, 77, 217, 361, 129, 205, 493, 529, 98, 426, 718, 170, 254, 462, 982, 1501, 2077, 2257, 2105, 2933, 6953, 11513, 14393, 16469, 17813, 19769, 21653, 24053, 25769, 27413, 29993, 34553, 35369, 41273, 42233, 42869, 44969, 45113, 45173, 11917, 27757, 38881, 45937, 62317, 76897, 84781, 102637, 111457, 114481, 117217, 118477, 120781, 127117, 128881, 501, 1141

Offset: 1

Antti Karttunen, Oct 02 2019

Permutation of A328117.
The branching degree of vertex v is given by A099302(v).
Leaves form a subsequence of A098700.
For any number k at level n (where 7 is at level 2), we have A256750(k) = A327966(k) = n.
Question: Does this subtree contain infinitely long paths? How many? Cf. conjecture number 8 in Ufnarovski and Ahlander paper. As an example of possible beginning of such a sequence they give: 1 ← 7 ← 10 ← 25 ← 46 ← 129 ← 170 ← 501 ← 414 ← 2045.

			The subtree is laid out as below. The terms of this sequence are obtained by scanning each successive level of the tree from left to right, from the node 7 onward:
   (0)
    |
   (1)
    |
    7
    |
    10______________________________
    |                               |
    21________                     25
    |         |                     |
    18___    38_____               46_________________________________
    |    |    |     |               |            |      |             |
    65   77  217   361____         129____      205    493_____      529
         |          |     |         |     |             |      |
         98        426   718       170   254           462    982
         |          |     |         |     |             |      |
        [3]       [21]   [15]      [9]   [9]           [28]   [17]
On the last level illustrated above, the numbers in brackets [ ] tell how many children the node has. E.g, there are three for 98: 1501, 2077, 2257, as A003415(1501) = A003415(2077) = A003415(2257) = 98, and nine for 170: 501, 1141, 2041, 2869, 4309, 5461, 6649, 6901, 7081.

Antti Karttunen, Table of n, a(n) for n = 1..204
Victor Ufnarovski and Bo Ahlander, How to Differentiate a Number, J. Integer Seqs., Vol. 6, 2003.

Cf. A003415, A098699, A098700, A099302, A099303, A099307, A099308, A189760, A256750, A327966, A327968, A328117.

Cf. A327975 for the subtree starting from 5, and also A263267 for another similar tree.

A002620(n) = ((n^2)>>2);
A003415(n) = {my(fac); if(n<1, 0, fac=factor(n); sum(i=1, matsize(fac)[1], n*fac[i, 2]/fac[i, 1]))}; \\ From A003415
A327977list(e) = { my(lista=List([7]), f); for(i=1, e, f = lista[i]; for(k=1,1+A002620(f),if(A003415(k)==f, listput(lista,k)))); Vec(lista); };

\\ With precomputed large A328117, use this:
v328117 = readvec("a328117.txt");
A327977list(e) = { my(lista=List([7]), f, i); for(n=1, e, f = lista[n]; print("n=",n," #lista=", #lista, " A002620(",f,")=",A002620(f)); my(u=1+A002620(f)); if(u>=v328117[#v328117],print("Not enough precomputed terms of A328117 as search upper limit ", u, " > ", v328117[#v328117], " (the last item in v328117). Number of expansions so far=", n); return(1/0)); i=1; while(v328117[i]A003415(v328117[i])==f, listput(lista,v328117[i])); i++)); Vec(lista); };
v327977 = A327977list(114);
A327977(n) = v327977[n];
for(n=1,#v327977,write("b327977.txt", n, " ", A327977(n)));

# uses[A003415]
def A327977():
  '''Breadth-first reading of irregular subtree rooted at 7, defined by the edge-relation A003415(child) = parent. Starts giving terms from 7 onward, after a(0) = 0 and a(1) = 1.'''
  yield 7
  for x in A327977():
    for k in [1 .. 1+floor((x*x)/2)]:
      if(A003415(k) == x): yield k
def take(n, g):
  '''Returns a list composed of the next n elements returned by generator g.'''
  z = []
  if 0 == n: return(z)
  for x in g:
    z.append(x)
    if n > 1: n = n-1
    else: return(z)
take(52, A327977())

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

Original entry on oeis.org

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

A328117 Numbers such that zero or more applications of A003415 (arithmetic derivative) will yield 7.

Original entry on oeis.org

7, 10, 18, 21, 25, 38, 46, 65, 77, 98, 129, 170, 205, 217, 254, 361, 414, 426, 462, 493, 501, 529, 718, 753, 982, 998, 1141, 1362, 1501, 1502, 2041, 2045, 2077, 2105, 2257, 2285, 2869, 2933, 2998, 3102, 3133, 3706, 4066, 4078, 4309, 4497, 4885, 5213, 5214, 5461, 5837, 6363, 6410, 6546, 6649, 6749, 6901, 6913, 6937, 6953, 7011

Offset: 1

Antti Karttunen, Oct 06 2019

nonn

Union of {7} with those terms of k of A099308 for which A327968(k) = 7.

			a(4) = 21 is included as A003415(A003415(21)) = 7.
a(627664) = 4294966334 (= 2^32 - 962) is included as A003415^(14)(4294966334) = 7. Note that A327966(4294966334) = 16.

Antti Karttunen, Table of n, a(n) for n = 1..10000
Antti Karttunen, Terms computed up to a(627664) = 4294966334; (all terms in range 1 .. 2^32)

Cf. A003415, A099308, A327966, A327968, A327977, A328115.

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));
isA328117(n) = { while((n>7), n = A003415checked(n)); (7==n); };

cp's OEIS Frontend

A327967 Positions of records in A327966.

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A327969 The length of a shortest path from n to zero when using the transitions x -> A003415(x) and x -> A276086(x), or -1 if no zero can ever be reached from n.

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A256750 Start with n, and repeatedly apply the arithmetic derivative A003415. |a(n)| = the number of iterations to reach 0 (then a(n) is taken nonnegative) or a number having a factor of the form p^p with prime p, in which case a(n) = -|a(n)|.

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A327965 "Tamed variant" of arithmetic derivative: a(0) = a(1) = 0; for n > 1, a(n) = A327938(A003415(n)).

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A327968 a(0) = a(1) = 0, a(prime) = 1, and for all other numbers, a(n) = the first noncomposite reached when iterating A327965, or -1 if no noncomposite is ever reached.

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A327975 Breadth-first reading of the subtree rooted at 5 of the tree where each parent node is the arithmetic derivative (A003415) of all its children.

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Sage

A327977 Breadth-first reading of the subtree rooted at 7 of the tree where each parent node is the arithmetic derivative (A003415) of all its children.

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Sage

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

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