A327975 -id:A327975 - cp's OEIS frontend

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.

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)));

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

A351259 First noncomposite number reached when iterating the map x -> x', when starting from x = A351255(n). Here x' is the arithmetic derivative of x, A003415.

Original entry on oeis.org

1, 2, 3, 5, 5, 7, 5, 7, 31, 7, 41, 71, 191, 2711, 7, 5, 7, 41, 103, 59, 71, 271, 71, 1031, 2887, 439, 5, 5, 7, 631, 251, 401, 3491, 1031, 1319, 17747, 9733, 1931, 16319, 77351, 131, 5, 419, 7079, 22343, 971, 5981, 6861581, 419, 18731, 11903, 33937, 7079, 15287, 15287, 6143, 6944111, 1415651, 11, 13, 5, 61, 103, 401, 631

Offset: 1

Antti Karttunen, Feb 11 2022

For the initial 105367 19-smooth terms of A351255, the last 7 occurs here at a(54796), with A351255(54796) = 289993286583 = 3^2 * 7 * 11 * 13^2 * 19^5, and the last 5 occurs here at a(65777), with A351255(65777) = 391899820830375516750 = 2 * 3^2 * 5^3 * 7^3 * 13^3 * 17^3 * 19^6, already a moderately high starting value, in whose vicinity most ending primes for successful iterations are much larger. This observation motivates a conjecture: Even from large numbers with high exponents in their prime factorization it is sometimes possible to reach a small prime. Compare to the conjecture 8 in Ufnarovski & Åhlander paper.

			From A351255(27) = 2625 it takes 12 iterations of the map x -> A003415(x) to reach zero: 2625 -> 2825 -> 1155 -> 886 -> 445 -> 94 -> 49 -> 14 -> 9 -> 6 -> 5 -> 1 -> 0. Two steps before the final zero is the first and only prime on the path, 5, therefore a(27) = 5.

Antti Karttunen, Table of n, a(n) for n = 1..12878 (computed for all 17-smooth terms of A351255)
Antti Karttunen, 105368 initial terms, without indices (computed for all 19-smooth terms of A351255, and also for A276086(9699690) = 23)
Victor Ufnarovski and Bo Åhlander, How to Differentiate a Number, J. Integer Seqs., Vol. 6, 2003, #03.3.4.

Cf. A000040, A003415, A276086, A327975, A327977, A351078, A351255, A351257, A351261.

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(m=1, p=2); while(n, m *= (p^(n%p)); n = n\p; p = nextprime(1+p)); (m); };
A351078(n) = { while(n>1&&!isprime(n), n = A003415checked(n)); (n); };
for(n=0, 2^9, u=A276086(n); p = A351078(u); if(p>0,print1(p, ", ")));

a(n) = A351078(A351255(n)).

a(1) = 1, and for n > 1, a(n) = A003415^[A351257(n)-2](A351255(n)). [This means: take the (A351257(n)-2)-th arithmetic derivative of A351255(n)].

A328115 Numbers n from which zero or more applications of A003415 will lead to 5.

Original entry on oeis.org

5, 6, 9, 14, 33, 49, 62, 94, 177, 445, 817, 886, 913, 961, 1146, 1155, 1633, 1719, 1822, 2006, 2173, 2209, 2306, 2625, 2649, 2825, 3255, 3430, 3450, 3871, 4018, 4405, 4414, 4575, 4638, 4869, 5294, 5687, 6009, 6162, 6506, 6849, 8110, 8113, 8126, 8334, 8570, 9741, 9770, 10222, 10635, 11221, 12014, 13377, 13458, 13767, 13993, 15351

Offset: 1

Antti Karttunen, Oct 07 2019

nonn

Union of {5} with those terms of k of A099308 for which A327968(k) = 5.

Antti Karttunen, Table of n, a(n) for n = 1..10000

Cf. A003415, A099308, A327968, A327975, A328117.

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));
isA328115(n) = { while((n>5), n = A003415checked(n)); (5==n); };
k=0; n=0; while(k<100, n++; if(isA328115(n), k++; print1(n, ", ")));

A328385 If n is of the form p^p, a(n) = n, otherwise a(n) is the first number found by iterating the map x -> A003415(x) that is different from n and either a prime, or whose degree (A051903) differs from the degree of n.

Original entry on oeis.org

0, 1, 1, 4, 1, 5, 1, 12, 6, 7, 1, 16, 1, 9, 8, 32, 1, 21, 1, 24, 7, 13, 1, 44, 10, 8, 27, 32, 1, 31, 1, 80, 9, 19, 12, 96, 1, 7, 16, 68, 1, 41, 1, 48, 39, 25, 1, 608, 14, 39, 20, 56, 1, 81, 16, 92, 13, 31, 1, 96, 1, 9, 51, 640, 18, 61, 1, 72, 8, 59, 1, 156, 1, 16, 55, 80, 18, 71, 1, 3424, 108, 43, 1, 128, 13, 45, 32, 140, 1, 123, 20, 96, 19

Offset: 1

Antti Karttunen, Oct 14 2019

nonn

			For n = 3, 3 is a prime, thus a(3) = 1.
For n = 4, A003415(4) = 4, thus as it is among the fixed points of A003415 and a(4) = 4.
For n = 8 = 2^3, its "degree" is A051903(33) = 3, but A003415(8) = 12 = 2^2 * 3, with degree 2, thus a(8) = 12.
For n = 21 = 3*7, A051903(21) = 1, the first derivative A003415(21) = 10 = 2*5 is of the same degree as A051903(10) = 1, but then continuing, we have A003415(10) = 7, which is a prime, thus a(21) = 7.
For n = 33 = 3*11, A051903(33) = 1, A003415(33) = 14 = 2*7, is of the same degree, but on the second iteration, A003415(14) = 9 = 3^2, with A051903(9) = 2, different from the initial degree, thus a(33) = 9.

Antti Karttunen, Table of n, a(n) for n = 1..16384
Antti Karttunen, Data supplement: n, a(n) computed for n = 1..65537

Cf. A003415, A051674, A051903, A157037.
Cf. A328384 (the number of iterations needed to reach such a number).
Cf. also A327968, A327975, A327977, A328252, A328305, A328310, A328311, A328320, A328321.

A003415(n) = if(n<=1, 0, my(f=factor(n)); n*sum(i=1, #f~, f[i, 2]/f[i, 1]));
A051903(n) = if((1==n),0,vecmax(factor(n)[, 2]));
A328385(n) = { my(d=A051903(n), u=A003415(n)); while(u && (u!=n) && !isprime(u) && A051903(u)==d, n = u; u = A003415(u)); (u); };

a(1) = 0 [as here the degrees of 0 and 1 are considered different].
a(p) = 1 for all primes.
a(A051674(n)) = A051674(n).
a(A157037(n)) = A003415(A157037(n)), a prime.
a(A328252(n)) = A003415(A328252(n)), a squarefree number.
a(n) = A003415^(k)(n), when k = abs(A328384(n)). [Taking the abs(A328384(n))-th arithmetic derivative of n gives a(n)]

cp's OEIS Frontend

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

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A351259 First noncomposite number reached when iterating the map x -> x', when starting from x = A351255(n). Here x' is the arithmetic derivative of x, A003415.

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A328115 Numbers n from which zero or more applications of A003415 will lead to 5.

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A328385 If n is of the form p^p, a(n) = n, otherwise a(n) is the first number found by iterating the map x -> A003415(x) that is different from n and either a prime, or whose degree (A051903) differs from the degree of n.

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