A204984 -id:A204984 - cp's OEIS frontend

A204892 Least k such that n divides s(k)-s(j) for some j in [1,k), where s(k)=prime(k).

2, 3, 3, 4, 4, 5, 7, 5, 5, 6, 6, 7, 10, 7, 7, 8, 8, 9, 13, 9, 9, 10, 16, 10, 16, 10, 10, 11, 11, 12, 19, 12, 20, 12, 12, 13, 22, 13, 13, 14, 14, 15, 24, 15, 15, 16, 25, 16, 26, 16, 16, 17, 29, 17, 30, 17, 17, 18, 18, 19, 31, 19, 32, 19, 19, 20, 33, 20, 20, 21

Offset: 1

Clark Kimberling, Jan 20 2012

nonn

Suppose that (s(i)) is a strictly increasing sequence in the set N of positive integers. For i in N, let r(h) be the residue of s(i+h)-s(i) mod n, for h=1,2,...,n+1. There are at most n distinct residues r(h), so that there must exist numbers h and h' such that r(h)=r(h'), where 0<=h

Corollary: for each n, there are infinitely many pairs (j,k) such that n divides s(k)-s(j), and this result holds if s is assumed unbounded, rather than strictly increasing.

Guide to related sequences:

...

s(n)=prime(n), primes

... k(n), j(n): A204892, A204893

... s(k(n)),s(j(n)): A204894, A204895

... s(k(n))-s(j(n)): A204896, A204897

s(n)=prime(n+1), odd primes

... k(n), j(n): A204900, A204901

... s(k(n)),s(j(n)): A204902, A204903

... s(k(n))-s(j(n)): A109043(?), A000034(?)

s(n)=prime(n+2), primes >=5

... k(n), j(n): A204908, A204909

... s(k(n)),s(j(n)): A204910, A204911

... s(k(n))-s(j(n)): A109043(?), A000034(?)

s(n)=prime(n)*prime(n+1) product of consecutive primes

... k(n), j(n): A205146, A205147

... s(k(n)),s(j(n)): A205148, A205149

... s(k(n))-s(j(n)): A205150, A205151

s(n)=(prime(n+1)+prime(n+2))/2: averages of odd primes

... k(n), j(n): A205153, A205154

... s(k(n)),s(j(n)): A205372, A205373

... s(k(n))-s(j(n)): A205374, A205375

s(n)=2^(n-1), powers of 2

... k(n), j(n): A204979, A001511(?)

... s(k(n)),s(j(n)): A204981, A006519(?)

... s(k(n))-s(j(n)): A204983(?), A204984

s(n)=2^n, powers of 2

... k(n), j(n): A204987, A204988

... s(k(n)),s(j(n)): A204989, A140670(?)

... s(k(n))-s(j(n)): A204991, A204992

s(n)=C(n+1,2), triangular numbers

... k(n), j(n): A205002, A205003

... s(k(n)),s(j(n)): A205004, A205005

... s(k(n))-s(j(n)): A205006, A205007

s(n)=n^2, squares

... k(n), j(n): A204905, A204995

... s(k(n)),s(j(n)): A204996, A204997

... s(k(n))-s(j(n)): A204998, A204999

s(n)=(2n-1)^2, odd squares

... k(n), j(n): A205378, A205379

... s(k(n)),s(j(n)): A205380, A205381

... s(k(n))-s(j(n)): A205382, A205383

s(n)=n(3n-1), pentagonal numbers

... k(n), j(n): A205138, A205139

... s(k(n)),s(j(n)): A205140, A205141

... s(k(n))-s(j(n)): A205142, A205143

s(n)=n(2n-1), hexagonal numbers

... k(n), j(n): A205130, A205131

... s(k(n)),s(j(n)): A205132, A205133

... s(k(n))-s(j(n)): A205134, A205135

s(n)=C(2n-2,n-1), central binomial coefficients

... k(n), j(n): A205010, A205011

... s(k(n)),s(j(n)): A205012, A205013

... s(k(n))-s(j(n)): A205014, A205015

s(n)=(1/2)C(2n,n), (1/2)*(central binomial coefficients)

... k(n), j(n): A205386, A205387

... s(k(n)),s(j(n)): A205388, A205389

... s(k(n))-s(j(n)): A205390, A205391

s(n)=n(n+1), oblong numbers

... k(n), j(n): A205018, A205028

... s(k(n)),s(j(n)): A205029, A205030

... s(k(n))-s(j(n)): A205031, A205032

s(n)=n!, factorials

... k(n), j(n): A204932, A204933

... s(k(n)),s(j(n)): A204934, A204935

... s(k(n))-s(j(n)): A204936, A204937

s(n)=n!!, double factorials

... k(n), j(n): A204982, A205100

... s(k(n)),s(j(n)): A205101, A205102

... s(k(n))-s(j(n)): A205103, A205104

s(n)=3^n-2^n

... k(n), j(n): A205000, A205107

... s(k(n)),s(j(n)): A205108, A205109

... s(k(n))-s(j(n)): A205110, A205111

s(n)=Fibonacci(n+1)

... k(n), j(n): A204924, A204925

... s(k(n)),s(j(n)): A204926, A204927

... s(k(n))-s(j(n)): A204928, A204929

s(n)=Fibonacci(2n-1)

... k(n), j(n): A205442, A205443

... s(k(n)),s(j(n)): A205444, A205445

... s(k(n))-s(j(n)): A205446, A205447

s(n)=Fibonacci(2n)

... k(n), j(n): A205450, A205451

... s(k(n)),s(j(n)): A205452, A205453

... s(k(n))-s(j(n)): A205454, A205455

s(n)=Lucas(n)

... k(n), j(n): A205114, A205115

... s(k(n)),s(j(n)): A205116, A205117

... s(k(n))-s(j(n)): A205118, A205119

s(n)=n*(2^(n-1))

... k(n), j(n): A205122, A205123

... s(k(n)),s(j(n)): A205124, A205125

... s(k(n))-s(j(n)): A205126, A205127

s(n)=ceiling[n^2/2]

... k(n), j(n): A205394, A205395

... s(k(n)),s(j(n)): A205396, A205397

... s(k(n))-s(j(n)): A205398, A205399

s(n)=floor[(n+1)^2/2]

... k(n), j(n): A205402, A205403

... s(k(n)),s(j(n)): A205404, A205405

... s(k(n))-s(j(n)): A205406, A205407

			Let s(k)=prime(k).  As in A204890, the ordering of differences s(k)-s(j), follows from the arrangement shown here:
k...........1..2..3..4..5...6...7...8...9
s(k)........2..3..5..7..11..13..17..19..23
...
s(k)-s(1)......1..3..5..9..11..15..17..21..27
s(k)-s(2).........2..4..8..10..14..16..20..26
s(k)-s(3)............2..6..8...12..14..18..24
s(k)-s(4)...............4..6...10..12..16..22
...
least (k,j) such that 1 divides s(k)-s(j) for some j is (2,1), so a(1)=2.
least (k,j) such that 2 divides s(k)-s(j): (3,2), so a(2)=3.
least (k,j) such that 3 divides s(k)-s(j): (3,1), so a(3)=3.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Cf. A000040, A204890.

s[n_] := s[n] = Prime[n]; z1 = 400; z2 = 50;
Table[s[n], {n, 1, 30}]          (* A000040 *)
u[m_] := u[m] = Flatten[Table[s[k] - s[j],
   {k, 2, z1}, {j, 1, k - 1}]][[m]]
Table[u[m], {m, 1, z1}]          (* A204890 *)
v[n_, h_] := v[n, h] = If[IntegerQ[u[h]/n], h, 0]
w[n_] := w[n] = Table[v[n, h], {h, 1, z1}]
d[n_] := d[n] = First[Delete[w[n],
   Position[w[n], 0]]]
Table[d[n], {n, 1, z2}]          (* A204891 *)
k[n_] := k[n] = Floor[(3 + Sqrt[8 d[n] - 1])/2]
m[n_] := m[n] = Floor[(-1 + Sqrt[8 n - 7])/2]
j[n_] := j[n] = d[n] - m[d[n]] (m[d[n]] + 1)/2
Table[k[n], {n, 1, z2}]          (* A204892 *)
Table[j[n], {n, 1, z2}]          (* A204893 *)
Table[s[k[n]], {n, 1, z2}]       (* A204894 *)
Table[s[j[n]], {n, 1, z2}]       (* A204895 *)
Table[s[k[n]] - s[j[n]], {n, 1, z2}]     (* A204896 *)
Table[(s[k[n]] - s[j[n]])/n, {n, 1, z2}] (* A204897 *)
(* Program 2: generates A204892 and A204893 rapidly *)
s = Array[Prime[#] &, 120];
lk = Table[NestWhile[# + 1 &, 1, Min[Table[Mod[s[[#]] - s[[j]], z], {j, 1, # - 1}]] =!= 0 &], {z, 1, Length[s]}]
Table[NestWhile[# + 1 &, 1, Mod[s[[lk[[j]]]] - s[[#]], j] =!= 0 &], {j, 1, Length[lk]}]
(* Peter J. C. Moses, Jan 27 2012 *)

a(n)=forprime(p=n+2,,forstep(k=p%n,p-1,n,if(isprime(k), return(primepi(p))))) \\ Charles R Greathouse IV, Mar 20 2013

A204979 Least k such that n divides 2^(k-1)-2^(j-1) for some j satisfying 1<=j

Original entry on oeis.org

2, 3, 3, 4, 5, 4, 4, 5, 7, 6, 11, 5, 13, 5, 5, 6, 9, 8, 19, 7, 7, 12, 12, 6, 21, 14, 19, 6, 29, 6, 6, 7, 11, 10, 13, 9, 37, 20, 13, 8, 21, 8, 15, 13, 13, 13, 24, 7, 22, 22

Offset: 1

Clark Kimberling, Jan 21 2012

nonn

See A204892 for a discussion and guide to related sequences.

			1 divides 2^2-2^1, so a(1)=2
2 divides 2^3-2^2, so a(2)=3
3 divides 2^3-2^1, so a(3)=3
4 divides 2^4-2^3, so a(4)=4

Cf. A000079, A204892.

s[n_] := s[n] = 2^(n - 1); z1 = 800; z2 = 50;
Table[s[n], {n, 1, 30}]       (* A000079 *)
u[m_] := u[m] = Flatten[Table[s[k] - s[j], {k, 2, z1}, {j, 1, k - 1}]][[m]]
Table[u[m], {m, 1, z1}]       (* A130328 *)
v[n_, h_] := v[n, h] = If[IntegerQ[u[h]/n], h, 0]
w[n_] := w[n] = Table[v[n, h], {h, 1, z1}]
d[n_] := d[n] = First[Delete[w[n], Position[w[n], 0]]]
Table[d[n], {n, 1, z2}]       (* A204939 *)
k[n_] := k[n] = Floor[(3 + Sqrt[8 d[n] - 1])/2]
m[n_] := m[n] = Floor[(-1 + Sqrt[8 n - 7])/2]
j[n_] := j[n] = d[n] - m[d[n]] (m[d[n]] + 1)/2
Table[k[n], {n, 1, z2}]       (* A204979 *)
Table[j[n], {n, 1, z2}]       (* A001511 ? *)
Table[s[k[n]], {n, 1, z2}]    (* A204981 *)
Table[s[j[n]], {n, 1, z2}]    (* A006519 ? *)
Table[s[k[n]] - s[j[n]], {n, 1, z2}]     (* A204983 *)
Table[(s[k[n]] - s[j[n]])/n, {n, 1, z2}] (* A204984 *)

A204983 a(n) = 2^(k-1)-2^(j-1), where (2^(k-1),2^(j-1)) is the least pair of distinct positive powers of 2 for which n divides 2^(k-1)-2^(j-1).

Original entry on oeis.org

1, 2, 3, 4, 15, 6, 7, 8, 63, 30, 1023, 12, 4095, 14, 15, 16, 255, 126, 262143, 60, 63, 2046, 2047, 24, 1048575, 8190, 262143, 28, 268435455, 30, 31, 32, 1023, 510, 4095, 252, 68719476735, 524286, 4095, 120, 1048575, 126, 16383, 4092, 4095

Offset: 1

Clark Kimberling, Jan 21 2012

nonn

For a guide to related sequences, see A204892.

(Conjecture) Equivalently, the solution set of 2^p * (2^q - 1) = x * y, OR 2^q - 1 = 2^p * x * y, for at most one of the naturals x and y being given; unknown p and q in the integers; then a(n) = 2^p * (2^q - 1) where p and q are directly related to n (see formula). - Andrew T. Porter, Dec 20 2022

Cf. A204979, A204892, A204984.
Cf. A000079, A173787.
Cf. A007733, A007814.

Mathematica
```
(See the program at A204979.)
```

a(n) = for (k=1, oo, for (j=1, k-1, my(d=2^(k-1)-2^(j-1)); if (!(d % n), return(d)););); \\ Michel Marcus, Sep 16 2023

Conjecture: a(n) = 2^A007814(n) * (2^A007733(n) - 1). - Andrew T. Porter, Dec 20 2022

Showing 1-3 of 3 results.

cp's OEIS Frontend

A204892 Least k such that n divides s(k)-s(j) for some j in [1,k), where s(k)=prime(k).

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

A204979 Least k such that n divides 2^(k-1)-2^(j-1) for some j satisfying 1<=j

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

A204983 a(n) = 2^(k-1)-2^(j-1), where (2^(k-1),2^(j-1)) is the least pair of distinct positive powers of 2 for which n divides 2^(k-1)-2^(j-1).

Views

Author

Keywords

Comments

Crossrefs

Programs

Mathematica

PARI

Formula