cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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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.

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

Views

Author

Antti Karttunen, Oct 07 2019

Keywords

Comments

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)

Examples

			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.
		

Crossrefs

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.

Programs

  • PARI
    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));

Formula

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.

A327978 Numbers whose arithmetic derivative (A003415) is a primorial number (A002110) > 1.

Original entry on oeis.org

9, 161, 209, 221, 2189, 2561, 3281, 3629, 5249, 5549, 6401, 7181, 7661, 8321, 8909, 9089, 9869, 10001, 10349, 10541, 10961, 11009, 11021, 29861, 38981, 52601, 66149, 84101, 93029, 97481, 132809, 150281, 158969, 163301, 197669, 214661, 227321, 235721, 285449, 321989, 338021, 357881, 369701, 381449, 385349, 416261, 420089, 442889
Offset: 1

Views

Author

Antti Karttunen, Oct 09 2019

Keywords

Comments

Numbers n such that A327859(n) = A276086(A003415(n)) is an odd prime.
Composite terms in A328232.
Although it first might seem that the numbers whose arithmetic derivative is A002110(k) all appear before any of those whose arithmetic derivative is A002110(k+1), that is not true, as for example, we have a(56) = 570149, and A003415(570149) = 2310, a(57) = 570209, and A003415(570209) = 30030, but then a(58) = 573641 with A003415(573641) = 2310 again.
Because this is a subsequence of A327862 (all primorials > 1 are of the form 4k+2), only odd numbers are present.
Conjecture: No multiples of 5 occur in this sequence, and no multiples of 3 after the initial 9.
Of the first 10000 terms, all others are semiprimes (with 9 the only square one), except 1547371 = 7^2 * 23 * 1373 and 79332523 = 17^2 * 277 * 991, the latter being the only known term whose decimal expansion ends with 3. If all solutions were semiprimes p*q such that p+q = A002110(k) for some k > 1 (see A002375), it would be a sufficient reason for the above conjecture to hold. - David A. Corneth and Antti Karttunen, Oct 11 2019
In any case, the solutions have to be of the form "odd numbers with an even number of prime factors with multiplicity" (see A235992), and terms must also be cubefree (A004709), as otherwise the arithmetic derivative would not be squarefree.
Sequence A366890 gives the non-Goldbachian solutions, i.e., numbers that are not semiprimes. See also A368702. - Antti Karttunen, Jan 17 2024

Crossrefs

Cf. A351029 (number of k for which k' = A002110(n)).
Cf. A368703, A368704 (the least and the greatest k for which k' = A002110(n)).
Cf. A366890 (terms that are not semiprimes), A368702 (numbers k such that k' is one of the terms of this sequence).
Subsequence of following sequences: A004709, A189553, A327862, A328232, A328234.

Programs

  • Mathematica
    ad[n_] := n * Total @ (Last[#]/First[#] & /@ FactorInteger[n]); primQ[n_] := Max[(f = FactorInteger[n])[[;;,2]]] == 1 && PrimePi[f[[-1,1]]] == Length[f]; Select[Range[10^4], primQ[ad[#]] &] (* Amiram Eldar, Oct 11 2019 *)
  • PARI
    A002620(n) = ((n^2)>>2);
    A003415(n) = if(n<=1, 0, my(f=factor(n)); n*sum(i=1, #f~, f[i, 2]/f[i, 1]));
    A276150(n) = { my(s=0, p=2, d); while(n, d = (n%p); s += d; n = (n-d)/p; p = nextprime(1+p)); (s); };
    isA327978flat(n) = { my(u=A003415(n)); ((u>1)&&(1==A276150(u))); }; \\ Slow!
    k=0; for(n=1,A002620(30030),if(isA327978flat(n), k++; write("b327978.txt", k, " ", n)));

Formula

A327969(a(n)) = 4 for all n.

A369055 Number of representations of 4n-1 as a sum (p*q + p*r + q*r) with three odd primes p <= q <= r.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 1, 2, 0, 0, 1, 1, 1, 0, 1, 0, 1, 0, 2, 1, 0, 1, 1, 0, 0, 2, 2, 1, 0, 0, 2, 0, 0, 0, 1, 1, 2, 2, 1, 1, 0, 1, 2, 0, 0, 0, 3, 0, 1, 0, 2, 1, 2, 0, 3, 1, 1, 1, 0, 0, 1, 1, 0, 2, 0, 0, 5, 2, 0, 0, 2, 1, 1, 0, 2, 0, 1, 1, 2, 2, 0, 2, 1, 0, 2, 0, 3, 1, 0, 0, 4, 1, 0, 1
Offset: 1

Views

Author

Antti Karttunen, Jan 20 2024

Keywords

Comments

Number of solutions to 4n-1 = x', where x' is the arithmetic derivative of x (A003415), and x is a product of three odd primes, A046316.
The number of 0's in range [1..10^n], for n=1..7 are: 8, 46, 288, 2348, 21330, 206355, 2079925, etc.
Goldbach's conjecture can be expressed by claiming that each even number > 4 is an arithmetic derivative of an odd semiprime, as (p*q)' = p+q, where p and q are odd primes. One way to extend Goldbach's conjecture to three primes involves applying the arithmetic derivative to all possible products of three odd primes (A046316) as: (p*q*r)' = (p*q) + (p*r) + (q*r), and asking, "Onto which subset of natural numbers does this map surjectively?" Clearly, the above formula can only produce numbers of the form 4m+3, and furthermore, an analysis at A369252 shows that the trisections of this sequence have quite different expected values, being on average the highest in the trisection A369462, which gives the number of representations for the numbers of the form 12m+11. This motivates a new kind of Goldbach-3 conjecture: "All numbers of the form 12*m-1, with m large enough, have at least one representation as a sum (p*q + p*r + q*r) with three odd primes p <= q <= r." Furthermore, empirical data for sequence A369463 suggests that "large enough" in this case might well be 4224080, as 1+(12*4224079) = 50688949 = A369463(285), with the next term of A369463 so far unknown. Similar conjectures can be envisaged for the arithmetic derivatives of products of four or more primes. - Antti Karttunen, Jan 25 2024

Examples

			a(7) = 1 because 4*7 - 1 = 27, which can be represented as a sum of the form (p*q) + (p*r) + (q*r), with all three primes p, q and r = 3.
a(19) = 2 because 4*19 - 1 = 75, which can be represented as a sum of the form (p*q) + (p*r) + (q*r) in two ways, with p=3, q=3 and r=11, or with p = q = r = 5.
a(9999995) = 0 because (4*9999995)-1 = 39999979, which cannot be expressed as a sum (p*q) + (p*r) + (q*r) for any three odd primes p, q and r, whether distinct or not.
		

Crossrefs

Cf. A369460, A369461, A369462 (trisections), A369450, A369451, A369452 (and their partial sums).
Cf. also A351029, A369239.

Programs

  • PARI
    \\ We iterate over weakly increasing triplets of odd primes:
    A369055list(up_to) = { my(v = [3,3,3], ip = #v, d, u = vector(up_to), lim = -1+(4*up_to)); while(1, d = ((v[1]*v[2]) + (v[1]*v[3]) + (v[2]*v[3])); if(d > lim, ip--, ip = #v; u[(d+1)/4]++); if(!ip, return(u)); v[ip] = nextprime(1+v[ip]); for(i=1+ip,#v,v[i]=v[i-1])); };
    v369055 = A369055list(100001);
    A369055(n) = v369055[n];

Formula

a(n) = A369054(4*n-1).
a(n) = Sum_{i=1..A002620(4*n-1)} A369058(i)*[A003415(i)==4*n-1], where [ ] is the Iverson bracket.

A116979 Number of distinct representations of primorials as the sum of two primes.

Original entry on oeis.org

0, 0, 1, 3, 19, 114, 905, 9493, 124180, 2044847, 43755729, 1043468386, 30309948241
Offset: 0

Views

Author

Jonathan Vos Post, Apr 01 2006

Keywords

Comments

Related to Goldbach's conjecture. Let g(2n) = A002375(n). The primorials produce maximal values of the function g in the following sense: the basic shape of the function g is k*x/log(x)^2 and each primorial requires a larger value of k than the previous one. - T. D. Noe, Apr 28 2006
Relates also to a more generic problem of how many numbers there are such that their arithmetic derivative is equal to the n-th primorial number. See A351029. - Antti Karttunen, Jan 17 2024

Examples

			a(2) = 1 because 2nd primorial = 6 = 3 + 3 uniquely.
a(3) = 3 because 3rd primorial = 30 = 7 + 23 = 11 + 19 = 13 + 17.
a(4) = 19 because 4th primorial = 210 = 11 + 199 = 13 + 197 = 17 + 193 = 19 + 191 = 29 + 181 = 31 + 179 = 37 + 173 = 43 + 167 = 47 + 163 = 53 + 157 = 59 + 151 = 61 + 149 = 71 + 139 = 73 + 137 = 79 + 131 = 83 + 127 = 97 + 113 = 101 + 109 = 103 + 107.
		

Crossrefs

Cf. A002375 (number of decompositions of 2n into unordered sums of two odd primes).

Programs

  • Mathematica
    n=1; Join[{0,0}, Table[n=n*Prime[k]; cnt=0; Do[If[PrimeQ[2n-Prime[i]],cnt++ ], {i,2,PrimePi[n]}]; cnt, {k,2,10}]] (* T. D. Noe, Apr 28 2006 *)

Formula

a(n) = #{p(i) + p(j) = A002110(n) for p(k) = A000040(k) and i >= j}.
a(n) = A351029(n) - A369000(n). - Antti Karttunen, Jan 17 2024

Extensions

More terms from T. D. Noe, Apr 28 2006
a(11)-a(12) from Donovan Johnson, Dec 19 2009

A366890 Irregular triangle, wherein row n lists in ascending order all numbers k whose arithmetic derivative k' is equal to the n-th primorial, A002110(n), and that have more than two prime factors with multiplicity. Rows of length zero are simply omitted, i.e., when A369000(n) = 0.

Original entry on oeis.org

1547371, 79332523, 1102527599503, 25336943536819, 25962012375103, 25970380120783, 66702554987143, 526285951027003, 927949814519899, 7777707036642079, 9584173681667203, 13082430772438171, 22101822021783739, 4958985803436403, 32006922970429003, 32076018550175863, 49806227168831659, 84682266449971639, 97995266657958403
Offset: 1

Views

Author

Antti Karttunen, Jan 09 2024

Keywords

Comments

For n > 0, numbers k such that A003415(k) = A002110(n) and A001222(k) > 2.
Sequence as a whole is not listed in ascending order, even though each batch of solutions for each n for which A369000(n) > 0 are. For example, we have a(14) < a(13) because A003415(22101822021783739) = A002110(12), while A003415(4958985803436403) = A002110(13). See the examples.
Question: Are there any common terms with A036785, that is, with A368697?

Examples

			For rows n=1..6, 9 & 10 nothing is listed, as those rows are empty.
Row for n=7 has just one term: 1547371 (= 7^2 * 23 * 1373). Note that A003415(1547371) = 510510 = A002110(7).
Row for n=8 has just one term: 79332523 (= 17^2 * 277 * 991).
Row for n=11 has two terms:
  1102527599503 (= 11^2 * 11071 * 823033),
  25336943536819 (= 157 * 743 * 5749 * 37781).
Row for n=12 has nine terms:
  25962012375103 (= 7^2 * 8597 * 61630451),
  25970380120783 (= 7^2 * 41387 * 12806141),
  66702554987143 (= 19^2 * 167 * 1106416889),
  526285951027003 (= 73 * 3919 * 7013 * 262313),
  927949814519899 (= 269 * 271 * 1697 * 7501033),
  7777707036642079 (= 2203 * 2791 * 7349 * 172127),
  9584173681667203 (= 2131 * 5953 * 7901 * 95621),
  13082430772438171 (= 3109 * 5861 * 24421 * 29399),
  22101822021783739 (= 8783 * 11777 * 13921 * 15349).
Row for n=13 has 18 terms, and begins with:
  4958985803436403 (= 37^2 * 137 * 26440450451),
and ends with:
  3206697143570677543 (= 36899 * 41983 * 45233 * 45763).
Note that A003415(3206697143570677543) = 304250263527210 = A002110(13).
		

Crossrefs

When the whole sequence is sorted into ascending order, equal to A327978 without any semiprime solutions (solutions in A001358), and also a subsequence of following sequences: A004709, A327862, A328234.

Programs

  • PARI
    \\ See the attached PARI-program

A369239 Number of integers whose arithmetic derivative is larger than 1 and equal to the n-th partial sum of primorial numbers.

Original entry on oeis.org

0, 1, 2, 1, 2, 1, 2, 1, 27, 0, 319, 1
Offset: 1

Views

Author

Antti Karttunen, Jan 18 2024

Keywords

Comments

Note how there are generally less solutions for even n than for odd n. This is explained by the fact that A143293(2n) == 1 (mod 4) and A143293(2n+1) == 3 (mod 4) and the arithmetic derivative A003415 of a product of any three odd primes (A046316) is always of the form 4k+3, therefore the solution set counted by a(2n) does not have any solutions from A046316 that contribute the majority of the solutions counted by a(2n+1). See also A369055.
a(13) >= 1 as there are solutions like 5744093403180469, 12538540924097819, etc., probably thousands or even more in total.
a(14) >= 1 [see examples].

Examples

			a(12) = 1 as there is a unique solution k such that k' = A143293(12) = 7628001653829, that k being 318745032938881 = 71*173*307*1259*67139. It's also the first solution with more than four prime factors.
a(14) >= 1, because as A143293(14)-2 = 13394639596851069-2 = 13394639596851067 is a prime, we have at least one solution, with A003415(2*13394639596851067) = A003415(26789279193702134) = 2+13394639596851067 = A143293(14).
For more examples, see A369240.
		

Crossrefs

Programs

  • PARI
    \\ See the attached program.

A368703 a(n) is the least integer k whose arithmetic derivative is equal to the n-th primorial, or 0 if no such k exists.

Original entry on oeis.org

2, 0, 9, 161, 2189, 29861, 510221, 1547371, 79332523, 9592991561, 265257420749, 1102527599503
Offset: 0

Views

Author

Antti Karttunen, Jan 16 2024

Keywords

Comments

a(n) = the smallest integer k for which A003415(k) = A002110(n), and 0 if no such k exists.
If there are non-Goldbachian solutions (A366890) for some n, i.e., if A369000(n) > 0, then the smallest of them appears here as a value of a(n).
a(12) <= 25962012375103, a(13) <= 4958985803436403, a(14) <= 32442711864461575, a(15) <= 11758779158543465383. - David A. Corneth, Jan 17 2024

Examples

			a(0) = 2 as the least number k such that A003415(k) = A002110(0) = 1 is 2.
a(1) = 0 as there is no number k such that A003415(k) = A002110(1) = 2.
a(7) = 1547371 as it is the least number k such that A003415(k) = A002110(7) = 510510. See also A366890.
		

Crossrefs

Formula

a(n) <= A368704(n).
For n<>1, A003415(a(n)) = A002110(n).

A369000 a(n) is the number of integers, not semiprimes, whose arithmetic derivative is equal to the n-th primorial.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 2, 9, 18
Offset: 1

Views

Author

Antti Karttunen, Jan 16 2024

Keywords

Comments

a(n) = number of solutions k to equation A003415(k) = A002110(n) with A001222(k) > 2.

Crossrefs

Cf. A001222, A002110, A003415, A116979, A351029, A366890 (solutions themselves).

Programs

  • PARI
    \\ See the attached PARI-program.

Formula

a(n) = A351029(n) - A116979(n).

A368704 a(n) is the greatest integer k whose arithmetic derivative is equal to the n-th primorial, and 0 if no such k exists.

Original entry on oeis.org

0, 9, 221, 11021, 1333349, 225450221, 65155115009, 23520996509141, 12442607161209161, 10464232622576957201, 10056127550296456854221, 13766838616355849433396389, 23142055714094182897602595769, 42789661015360144177667200022669, 94522361182930558488466844910827309, 265513312562851938794103367354849976069
Offset: 1

Views

Author

Antti Karttunen, Jan 16 2024

Keywords

Comments

a(n) is the greatest integer k for which A003415(k) = A002110(n), and 0 if no such k exists.
See also comments in A116979.

Examples

			a(1) = 0 as there are no number k such that A003415(k) = A002110(1) = 2.
a(3) = 221 as A003415(221) = A003415(13*17) = A003415(13)*17 + 13*A003415(17) = 1*17 + 13*1 = 30 which is A002110(3) and no k>221 has arithmetic derivative 30. - _David A. Corneth_, Jan 17 2024
		

Crossrefs

Cf. A002110, A003415, A116979, A351029, A368703, A369059 (an upper bound).
Cf. also A369244.

Programs

  • PARI
    a(n) = {if(n==1,return(0)); pr = vecprod(primes(n)); prover2 = pr/2; forprime(p = prover2, oo, if(isprime(pr - p), return(p*(pr-p))))} \\ David A. Corneth, Jan 17 2024

Formula

For n >= 1, A368703(n) <= a(n) <= A369059(n).
For n > 1, A003415(a(n)) = A002110(n).
For n > 1, a(n) = p*q, where p, q are primes, p+q = A002110(n) and q >= p and q - p is minimal. - David A. Corneth, Jan 17 2024 [This depends on Goldbach's conjecture being valid, at least on primorials, for which there is strong empirical evidence though.] - Antti Karttunen, Jan 19 2024

Extensions

More terms from David A. Corneth, Jan 17 2024

A376410 Number of integers whose arithmetic derivative (A003415) is equal to n!, the n-th factorial.

Original entry on oeis.org

0, 1, 4, 13, 40, 186, 952, 5533, 38719, 346207, 3130816, 34444968, 382437431, 4637235152
Offset: 2

Views

Author

Antti Karttunen, Nov 06 2024

Keywords

Comments

For 1! = 1, there are an infinite number of integers k for which A003415(k) = 1 (namely, all the primes), therefore the starting offset is 2.
Like with A351029, also here most of the solutions seem to be squarefree semiprimes, counted by A062311.
Terms a(12)..a(15) were obtained by summing the corresponding terms of A062311 and A377986.

Crossrefs

Programs

  • PARI
    \\ Slow program, for computing just a few terms:
    A002620(n) = ((n^2)>>2);
    A003415(n) = if(n<=1, 0, my(f=factor(n)); n*sum(i=1, #f~, f[i, 2]/f[i, 1]));
    A376410(n) = { my(g=n!); sum(k=1,A002620(g),A003415(k)==g); };
    
  • PARI
    A376410(n) = AntiDeriv(n!);
    AntiDeriv(n,startvlen=1,solsfilename="") = { my(v = vector(startvlen,i,2), ip = #v, r, c=0); while(1, r = A003415vrl(v,n); if(0==r, ip--, if(r > 1, c++; if(solsfilename!="", write(solsfilename, r*factorback(v)))); ip = #v); if(0==ip, v = vector(1+#v,i,2); ip = #v; if(A003415vec(v) > n, return(c)), v[ip] = nextprime(1+v[ip]); for(i=1+ip, #v, v[i]=v[i-1]))); };
    A003415vec(tv) = { my(n=factorback(tv), s=0, m=1, spf); for(i=1,#tv,spf = tv[i]; n /= spf; s += m*n; m *= spf); (s); }; \\ Compute Arithmetic derivative from the vector of primes.
    A003415vrl(pv,lim) = { my(n=factorback(pv), x=lim-n, s=0, m=1, spf, u=n); for(i=1,#pv,spf = pv[i]; u /= spf; s += m*u; m *= spf); if(((x/s)
    				

Formula

a(n) = A099302(A000142(n)).
a(n) = Sum_{k=1..A002620(n!)} [A003415(k) = n!], where [ ] is the Iverson bracket.
a(n) = A062311(n) + A377986(n).
Showing 1-10 of 10 results.