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.

Showing 1-9 of 9 results.

A369240 Irregular triangle read by rows, where row n lists in ascending order all numbers k whose arithmetic derivative k' is equal to the n-th partial sum of primorials, A143293(n). Rows of length zero are simply omitted, i.e., when A369239(n) = 0.

Original entry on oeis.org

14, 45, 74, 198, 5114, 10295, 65174, 1086194, 40354813, 20485574, 465779078, 12101385979, 15237604243, 18046312939, 29501083259, 52467636437, 65794608773, 86725630997, 87741700037, 131833085077, 168380217557, 176203950283, 177332276971, 226152989747, 292546582253, 307379277253, 321317084917, 342666536237, 348440115979
Offset: 1

Views

Author

Antti Karttunen, Jan 19 2024

Keywords

Comments

Only two nonsquarefree terms are currently known: 45, 198.
See comments in A369239 for an explanation why rows with an odd n generally have more terms than those with an even n.

Examples

			Row 1 has no terms because there are no numbers whose arithmetic derivative is equal to 3 = A143293(1).
Row 2 has just one term: 14 (= 2 * 7), with A003415(14) = 2+7 = 9 = A143293(2).
Row 3 has two terms: 45 (= 3^2 * 5) and 74 (= 2 * 37), with A003415(3*3*5) = (3*3) + (3*5) + (3*5) = 39, and A003415(2*37) = 2+37 = 39 = A143293(3).
Row 4 has one term: 198 (= 2 * 3^2 * 11).
Row 5 has two terms: 5114 (= 2 * 2557) and 10295 (= 5 * 29 * 71).
Row 6 has one term: 65174 (= 2 * 32587).
Row 7 has two terms: 1086194 (= 2 * 543097) and 40354813 (= 97 * 541 * 769).
Row 8 has one term: 20485574 (= 2 * 10242787).
Row 9 has 27 terms:
  465779078 (= 2 * 1049 * 222011),
  12101385979 (= 79 * 151 * 1014451),
  15237604243 (= 67 * 2659 * 85531),
  18046312939 (= 79 * 3931 * 58111),
  29501083259 (= 179 * 431 * 382391),
  52467636437 (= 233 * 8501 * 26489),
  65794608773 (= 449 * 761 * 192557),
  86725630997 (= 449 * 2213 * 87281),
  87741700037 (= 449 * 2381 * 82073),
  131833085077 (= 613 * 12241 * 17569),
  etc., up to the last one of them:
  680909375411 (= 8171 * 8219 * 10139).
Row 10 has no terms.
Row 11 has 319 terms, beginning as:
  293420849770 (= 2 * 5 * 157 * 186892261),
  414527038034 (= 2 * 207263519017),
  12092143168139 (= 59 * 5231 * 39180191),
  16359091676491 (= 79 * 91291 * 2268319),
  20784361649963 (= 167 * 251 * 495845639),
  etc., up to the last one of them:
  17866904665985941 (= 224869 * 248041 * 320329).
Row 12 has just one term: 318745032938881 (= 71 * 173 * 307 * 1259 * 67139).
Row 13 probably has thousands of terms. Interestingly, many of them appear in clusters that share a smallest prime factor. For example the following five:
  390120053091860677 (= 1321 * 23563 * 12533283799),
  407566547631686353 (= 1321 * 121687 * 2535429439),
  410999481465461617 (= 1321 * 547999 * 567752023),
  411668623600396429 (= 1321 * 1701571 * 183144919),
  411913933485848977 (= 1321 * 8787799 * 35483263),
  and also these:
  3846842704473466739 (= 20231 * 31601 * 6017086469),
  4300947161911032233 (= 20231 * 43319 * 4907590697),
  4437898843097002379 (= 20231 * 47969 * 4572980861),
  6130224093530040341 (= 20231 * 692459 * 437587529),
  6210584908378844243 (= 20231 * 1275569 * 240664037).
		

Crossrefs

Cf. A328243 (same sequence sorted into ascending order).
Cf. A369239 (number of terms on row n), A369243 (the first element of each row), A369244 (the last element of each row).
Cf. also A366890.

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.

A351029 Number of integers whose arithmetic derivative is equal to the n-th primorial.

Original entry on oeis.org

0, 1, 3, 19, 114, 905, 9494, 124181, 2044847, 43755729, 1043468388, 30309948250
Offset: 1

Views

Author

Antti Karttunen, Feb 01 2022

Keywords

Comments

Number of integers k such that A003415(k) = A002110(n).
a(7) = A116979(7) + 1 since 1547371'=510510 and 1547371=7^2*23*1373 and every other example has only two prime factors. a(8) > A116979(8) because there is at least one term k in A327978 for which A003415(k) = 9699690 = A002110(8), which is not semiprime, that k being 79332523 = 17^2 * 277 * 991. - Edited by Craig J. Beisel, Sep 13 2022 and Antti Karttunen, Jan 05 2023
Most such k are semiprimes, i.e., are "Goldbachian solutions", counted by A116979. The non-semiprime solutions (A366890) form a very tiny minority, and are counted by A369000. - Antti Karttunen, Jan 19 2024

Examples

			a(1) = 0 because there are no such k that A003415(k) = 2 = A002110(1).
a(2) = 1 because there is only one number, 9, such that A003415(9) = A002110(2) = 6.
a(3) = 3 because there are exactly three numbers, k = 161, 209, 221, for which A003415(k) = A002110(3) = 30. (See A327978). These are all semiprime solutions, generated by the partitions of 30 into 2 primes: 30 = 7 + 23 = 11 + 19 = 13 + 17, and we have 7*23 = 161; 11*19 = 209; 13*17 = 221.
		

Crossrefs

Cf. A002110, A002620, A003415, A099302, A099303, A116979, A327978, A366890 (nonsemiprime solutions), A368703 (the least of solutions), A368704 (the largest of solutions), A369000.
Cf. also A369239.

Programs

  • PARI
    A002110(n) = prod(i=1,n,prime(i));
    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]));
    A351029(n) = { my(g=A002110(n)); sum(k=1,A002620(g),A003415(k)==g); }; \\ Very naive and slow. See comments in A327978.
    
  • PARI
    A351029(n) = {v=prod(j=1,n,prime(j)); c=0; for(k=2, v^2/4, d=0; m=factor(k); for(i=1, matsize(m)[1], d+=(m[i,2]/m[i,1])*k; if(d>v, break;); ); if(d==v, c=c+1; ); ); c;} \\ Craig J. Beisel, Sep 13 2022

Formula

a(n) = Sum_{k=1..A002620(A002110(n))} [A003415(k) = A002110(n)], where [ ] is the Iverson bracket.
a(n) = A116979(n) + A369000(n). - Antti Karttunen, Jan 19 2024

Extensions

a(7) from Craig J. Beisel, Sep 13 2022
a(8)..a(12) [the last based on the value of A116979(12)] from Antti Karttunen, Jan 09 2024

A328243 Numbers whose arithmetic derivative (A003415) is larger than 1 and one of the terms of A143293 (partial sums of primorials).

Original entry on oeis.org

14, 45, 74, 198, 5114, 10295, 65174, 1086194, 20485574, 40354813, 465779078, 12101385979, 15237604243, 18046312939, 29501083259, 52467636437, 65794608773, 86725630997, 87741700037, 131833085077, 168380217557, 176203950283, 177332276971, 226152989747, 292546582253
Offset: 1

Views

Author

Antti Karttunen, Oct 10 2019

Keywords

Comments

From David A. Corneth, Oct 12 2019: (Start)
Let k' be the arithmetic derivative of k. Then to find terms of the form k = p * q where p, q are prime, we could see that k' = p + q. Then as one of them needs to be two, say p, needs to be 2, we have q = A143293(m) - 2 a prime. This would give terms 2 * q.
If terms are of the form k = p * q * r where p, q, r are distinct primes then k' = p*q + p*r + q*r. For m we like, we could solve p*q + p*r + q*r = A143293(m). checking p * q below some bound, we can solve for r and get r = (A143293(m) - p*q) / (p + q). With some extra constraints and searching different prime signatures, one might confirm terms found are all below some chosen upper bound. (End)
See sequences A369239 and A369240 for more observations and insights about the terms of this sequence. - Antti Karttunen, Jan 22 2024

Crossrefs

Sequence A369240 sorted into ascending order.

Programs

  • 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]));
    A143293(n) = if(n==0, 1, my(P=1, s=1); forprime(p=2, prime(n), s+=P*=p); (s)); \\ From A143293.
    A276150(n) = { my(s=0, p=2, d); while(n, d = (n%p); s += d; n = (n-d)/p; p = nextprime(1+p)); (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; };
    isA328243(n) = { my(u=A003415(n)); ((u>1)&&(1==A276150(A276086(u)))); }; \\ This is very slow program!
    k=0; for(n=1,A002620(A143293(6)),if(isA328243(n), k++; print1(n,", ")));

Formula

A327969(a(n)) <= 5 for all n.

Extensions

a(12)-a(25) from David A. Corneth and Giovanni Resta, Oct 12 2019

A369245 Number of representations of the n-th Euclid number, A002110(n) + 1, as a sum (p*q + p*r + q*r) with three odd primes p <= q <= r. (Definition implies that p=3 and q > 3).

Original entry on oeis.org

0, 0, 0, 0, 1, 2, 2, 1, 0, 0, 1, 1, 0, 1, 2, 0, 1
Offset: 0

Views

Author

Antti Karttunen, Jan 22 2024

Keywords

Comments

Number of representations of the n-th Euclid number, A002110(n) + 1, as a sum of the form 3*(p+q) + p*q, where p and q are odd primes.
Question: Will there be an eventual growth spurt for this sequence? Even though all solutions must be multiples of 3 (but not of 9), because A006862(n) == 1 (mod 3), for n > 1, and the solutions belong to a set listed by A369461.
Similar sequence A369242 grows more vigorously because A033312(n) == -1 (mod 3) for n >= 3, thus allowing non-multiples of 3 as solutions. See comments in A369252.

Examples

			a(4) = 1 as there exists a natural number 399 = 3 * 7 * 19, whose arithmetic derivative (indicated with 399', see A003415) is computed as ((3*7) + (3*19) + (7*19)) = 211 = 1 + prime(4)# = A006862(4), and because 399 is the unique term in A046316 that satisfies the condition.
a(17) >= 1 because there exists (at least one) solution k = 4903038892893242229501 = 3 * 17 * 96138017507710631951 with A003415(k) = 1+A002110(17).
For other cases, see examples in A369246.
		

Crossrefs

Cf. also A116979, A369000, A369239 for similar counts, also A369241, A369242 and A369247.

Programs

  • PARI
    \\ Needs also program from A369054.
    A002110(n) = prod(i=1,n,prime(i));
    A369245(n) = A369054(A002110(n)+1);
    
  • PARI
    \\ Optimized version of above, employs the fact that solutions must all be multiples of 3. Outputs also terms for A369246.
    search_for_3k1_cases(n) = if(3!=(n%4),0, my(p = 5, q, c=0); while(1, q = (n-(3*p)) / (3+p); if(q < p, return(c), if(1==denominator(q) && isprime(q),c++; write("b369246_by_search_order_to.txt", n, " ", 3*p*q))); p = nextprime(1+p)));
    A002110(n) = prod(i=1,n,prime(i));
    A369245(n) = search_for_3k1_cases(A002110(n)+1);

Formula

a(n) = A369054(A006862(n)).

A369243 a(n) is the least integer k whose arithmetic derivative is equal to the n-th partial sum of primorials, and 0 if no such k exists.

Original entry on oeis.org

2, 0, 14, 45, 198, 5114, 65174, 1086194, 20485574, 465779078, 0, 293420849770, 318745032938881
Offset: 0

Views

Author

Antti Karttunen, Jan 19 2024

Keywords

Comments

a(n) = the smallest integer k for which A003415(k) = A143293(n), and 0 if no such k exists.

Examples

			a(0) = 2 as the least number k such that A003415(k) = A143293(0) = 1 is 2.
a(1) = 0 as there is no number k such that A003415(k) = A143293(1) = 3.
		

Crossrefs

Cf also A368703.

Formula

a(n) <= A369244(n).

A369244 a(n) is the greatest integer k whose arithmetic derivative is equal to the n-th partial sum of primorials, and 0 if no such k exists.

Original entry on oeis.org

0, 14, 74, 198, 10295, 65174, 40354813, 20485574, 680909375411, 0, 17866904665985941, 318745032938881
Offset: 1

Views

Author

Antti Karttunen, Jan 19 2024

Keywords

Comments

a(n) = the largest integer k for which A003415(k) = A143293(n), and 0 if no such k exists.

Crossrefs

Cf. also A368704.

Formula

a(n) >= A369243(n).

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).

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

Views

Author

Antti Karttunen, Nov 20 2024

Keywords

Comments

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.

Crossrefs

Row 9 of irregular triangle A377992.
Subsequence of A099308, and after the initial term, subsequence of A046389.
Showing 1-9 of 9 results.