A128759 Least k such that the Jacobsthal function A048669(k) = n.
1, 2, 15, 6, 105, 30, 1155, 770, 36465, 210, 15015, 6006, 255255, 2310, 8580495, 102102, 4849845, 72930, 20056049013, 74364290, 5898837945, 30030, 3234846615, 881790, 195282582495, 510510, 218257003965, 20281170, 100280245065, 17160990, 934482952262145, 6614136163635
Offset: 1
Keywords
A144311 The length of the longest sequence of consecutive integers, each equal to 1 or -1 modulo at least one of the first n primes.
1, 5, 11, 29, 41, 65, 107, 149, 203, 257, 347, 527, 545, 617, 707, 869, 965, 1079, 1283, 1397, 1529, 1709
Offset: 1
Comments
For n > 1, a(n) == 5 (mod 6).
Examples
For the first, second and fourth terms the sequences are {1}, {1,2,3,4,5}, and {73,74,...,101}. [corrected by _Felix A. Pahl_, May 10 2016]
Links
- StackExchange, OEIS A144311 Generating function, May 10 2016.
- Jinyuan Wang, C++ program.
Extensions
a(8)-a(16) from Max Alekseyev, Nov 18 2009
a(17)-a(22) from Jinyuan Wang, Nov 26 2024
A288815 Paired Jacobsthal function applied to the product of the first n primes.
2, 6, 18, 30, 66, 150, 192, 258, 366, 450, 570, 708, 894, 1044, 1284, 1422, 1656, 1902, 2190, 2460, 2622
Offset: 1
Comments
There is a conjecture about an upper bound on this sequence. Let p_n be the n-th prime. If a(n) < p_n^2 - p_n holds for n>=3 then Goldbach's conjecture and the twin prime conjecture hold as well.
Links
- Mario Ziller, John F. Morack, Divisibility in paired progressions, Goldbach's conjecture, and the infinitude of prime pairs, arXiv:1706.00317 [math.NT], 2017.
- Mario Ziller, John F. Morack, On the computation of the generalised Jacobsthal function for paired progressions, arXiv:1706.03668 [math.NT], 2017.
Formula
a(n) = 6*A072753(n) + 6, for n>=3.
A319148 Irregular triangle T(n,m) where row n lists differences m = j*p - r - 1, with iterator 1 <= j <= A002110(n), p = prime(n+1), and r is the smallest number that exceeds j*p that is coprime to A002110(n+1).
0, 1, 0, 1, 0, 1, 2, 3, 0, 3, 2, 1, 0, 1, 0, 3, 2, 3, 0, 1, 4, 5, 2, 1, 0, 1, 0, 3, 2, 1, 2, 1, 0, 3, 4, 1, 0, 5, 0, 1, 0, 3, 2, 3, 0, 1, 0, 1, 2, 5, 4, 5, 2, 1, 2, 3, 0, 1, 0, 1, 4, 3, 4, 1, 2, 1, 2, 3, 0, 5, 0, 3, 2, 3, 0, 1, 0, 1, 2, 5, 0, 5, 2, 3, 2, 3, 0, 1, 0, 1, 4, 3, 4, 1, 0, 1
Offset: 1
Comments
Let p(i) be primes with p(1)=2, p(n)# the n-th primorial number, and h(n) the Jacobsthal function for primorial p(n)#. Conjecture: gcd(h(n), p(n+1)) = 1.
For a multiple m of a prime n, terms in this sequence give the number of contiguous numbers starting at m+1 which have at least one prime factor < n.
Consider a range s of the first n + 1 primes. Let p be the largest of these primes, i.e., A000040(n+1). Let P be the product of the first n primes, i.e., the primorial A002110(n), and let Q be the product of all the primes in s, i.e., the primorial A002110(n+1). Consider the reduced residue system R of primorial P, that is, those numbers 1 <= r < P such that gcd(r, P) = 1; therefore R = row n of A286941. For each n, we generate the multiples k = j*p, with 1 <= j <= P. For each k, we find the smallest residue r in R that exceeds k and take the difference m = r - k - 1. If no value in R exceeds k, then we use Q + 1 (which is also coprime to Q). Row n is thus a list of these m.
Alternatively, consider a multiple k = j*p, with 1 <= j <= P. We can compute m by iterating i such that the sum (i + k) is coprime to Q and subtracting 1. This technique is more efficient in terms of memory, as it does not require storing the reduced residue system of Q.
For n > 1: The penultimate value m on row n = A040976(n). The number of values m on row n is given by the sequence: 1,1,2,2,10,22,500,...
For n > 3: For any even x = m in row n, the number of x in row n is equal to the count of y in row n where y = x + 1. If x = 0, the count of x and y in row n = A000010(A002110(n-1)). For example, on row 4, A000010(A002110(4-1)) = 8, as 0 and 1 each occur 8 times on row 4. The sequence of counts of x and x+1 pairs on consecutive rows is given by the sequence A059861. For example, for x=0 and y=1 occurring 8 times on row 4, x=2 and y=3 occur 8-3=5 times on row 4 given by the value 3 in A059861. For example, for row 8, x=0 and y=1 occur A000010(A002110(8-1)) = 92160 times on row 8, and x=2 and y=3 occur 92160-22275=69885 times on row 8 given by the value 22275 in A059861.
For 3 < n < 9: The largest value on row n occurs twice, the pattern of occurrence is shown in table 1 of Ziller & Morack in the Links section.
Examples
Triangle begins:
0;
1,0;
1,0,1,2,3,0;
3,2,1,0,1,0,3,2,3,0,1,4,5,2,1,0,1,0,3,2,1,2,1,0,3,4,1,0,5,0;
...
For n = 2, we have s = {2,3,5}, with p = prime(n+1) = 5, P = A002110(2) = 6, and Q = A002110(3) = 30. Then R = row n of A286941 = {1, 7, 11, 13, 17, 19, 23, 29} (we add 31 to this list since we are concerned with the residue that is larger than the largest k and since 31 is the ensuing number coprime to Q). The series of multiples k = j*p are the multiples 5j with 1 <= j <= P, thus {5, 10, 15, 20, 25, 30}. In R, the smallest residues that exceed the multiples k in the immediately aforementioned list are {7, 11, 17, 23, 29, 31}. The differences are {7 - 5, 11 - 10, 17 - 15, 23 - 20, 29 - 25, 31 - 30} or {2, 1, 2, 3, 4, 1}; subtracting one from each we have row 2 = {1, 0, 1, 2, 3, 0}.
For example, the third value on row n=20000 is 15, so all values in the range (3 * prime(20000) + i) to (3 * prime(20000) + i) for 1 <= i <= 15 have at least one prime factor <= prime(n).
Links
- Mario Ziller and John F. Morack, Algorithmic concepts for the computation of Jacobsthal's function, arXiv:1611.03310 [math.NT], 2016.
- Mario Ziller, New computational results on a conjecture of Jacobsthal, arXiv:1903.11973 [math.NT], 2019.
Programs
-
Mathematica
rowToCreate = 3; (* create row n *) redundantDistanceToCheck = 1; (* set to 2 or higher to see n repeating patterns of length primorial[rowToCreate] *) Primorial[n_] := Times @@ Prime[Range[n]] rowValue = 0; primeToUse = Prime[rowToCreate]; distanceToCheck1 = redundantDistanceToCheck*Primorial[rowToCreate]; (* distanceToCheck1=rowToCreate*10000; *)(* uncomment this second option to create the first few values in very large rows up to rowToCreate=7000000000000 *) For[i = primeToUse, i < distanceToCheck1 + 1, i = i + primeToUse, For[x = i + 1, x < distanceToCheck1 + 2, x++, If[FactorInteger[x][[1, 1]] < primeToUse, rowValue++; , x = distanceToCheck1 + 2; Print[rowValue]; rowValue = 0; ]]] (* Jamie Morken, Sep 11 2018 *) (* Program to check the number of composites referenced to row values: *) Row = 100; ColumnOnTheRow = 12; Print["composites>", ColumnOnTheRow*Prime[Row], "=", (NextPrime[ColumnOnTheRow*Prime[Row]]) - (ColumnOnTheRow*Prime[Row]) - 1]; (* Second program: *) Table[Block[{s = Prime@ Range[n + 1], p, P, Q}, p = Last@ s; P = Times @@ Most@ s; Q = Times @@ s; Array[Block[{k = 1}, While[! CoprimeQ[k + p #, Q], k++]; k - 1] &, P]], {n, 4}] // Flatten (* Michael De Vlieger, Sep 11 2018 *)
Comments
Crossrefs
Extensions