A204999 -id:A204999 - 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

A204998 a(n) = k^2 - j^2, where (k^2,j^2) is the least pair of distinct squares for which n divides their difference.

Original entry on oeis.org

3, 8, 3, 8, 5, 12, 7, 8, 9, 20, 11, 12, 13, 28, 15, 16, 17, 72, 19, 20, 21, 44, 23, 24, 75, 52, 27, 28, 29, 60, 31, 32, 33, 68, 35, 72, 37, 76, 39, 40, 41, 84, 43, 44, 45, 92, 47, 48, 147, 200, 51, 52, 53, 108, 55, 56, 57, 116, 59, 60, 61, 124, 63, 64, 65, 132, 67, 68, 69, 140, 71, 72, 73, 148, 75, 76, 77, 156, 79, 80, 81, 164

Offset: 1

Clark Kimberling, Jan 21 2012

nonn

For a guide to related sequences, see A204892.

Antti Karttunen, Table of n, a(n) for n = 1..23005

Cf. A204994, A204892, A204996, A204997, A204999.

Mathematica
```
(See the program at A204994.)
```

A204998(n) = { my(d); for(k=sqrtint(1+n), oo, for(j=1,k-1,if(!((d=(k^2)-(j^2))%n),return(d),if(dAntti Karttunen, Sep 28 2018

a(n) = A204996(n) - A204997(n).

More terms from Antti Karttunen, Sep 28 2018

A204905 Least k such that n divides k^2-j^2 for some j satisfying 1<=j

Original entry on oeis.org

2, 3, 2, 3, 3, 4, 4, 3, 5, 6, 6, 4, 7, 8, 4, 5, 9, 9, 10, 6, 5, 12, 12, 5, 10, 14, 6, 8, 15, 8, 16, 6, 7, 18, 6, 9, 19, 20, 8, 7, 21, 10, 22, 12, 7, 24, 24, 7, 14, 15, 10, 14, 27, 12, 8, 9, 11, 30, 30, 8

Offset: 1

Clark Kimberling, Jan 21 2012

nonn

See A204892 for a discussion and guide to related sequences.

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

Cf. A000290, A204892.

s[n_] := s[n] = n^2; z1 = 600; z2 = 60;
Table[s[n], {n, 1, 30}]     (* A000290 *)
u[m_] := u[m] = Flatten[Table[s[k] - s[j], {k, 2, z1}, {j, 1, k - 1}]][[m]]
Table[u[m], {m, 1, z1}]     (* A120070 *)
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}]   (* A204994 *)
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}]       (* A204905 *)
Table[j[n], {n, 1, z2}]       (* A204995 *)
Table[s[k[n]], {n, 1, z2}]    (* A204996 *)
Table[s[j[n]], {n, 1, z2}]    (* A204997 *)
Table[s[k[n]] - s[j[n]], {n, 1, z2}]     (* A204998 *)
Table[(s[k[n]] - s[j[n]])/n, {n, 1, z2}] (* A204999 *)

A205032 a(n) = (s(k)-s(j))/n, where (s(k),s(j)) is the least pair of oblong numbers (A002378) for which n divides their difference; a(n) = (1/n)*A205031(n).

Original entry on oeis.org

4, 2, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2

Offset: 1

Clark Kimberling, Jan 22 2012

nonn

For a guide to related sequences, see A204892.

Even n such that a(n) = 2 are: 2, 12, 20, 56, 72, 132, 156, 240, 272, 380, 552, 812, 992, 1056, 1332, 1640, 1892, 2256, 2756, 3540, 3660, 4032, 4160, 4556, 5112, 5256, 6320, 6972, 7656, 7832, ... - Antti Karttunen, Nov 06 2018

Antti Karttunen, Table of n, a(n) for n = 1..14025

Cf. A002378, A205018, A205031, A204892.

Cf. also A204897, A204999, A205007.

Mathematica
```
(See the program at A205018.)
```

A205032(n) = for(k=2,oo,my(sk=k*(k+1)); for(j=1,k-1,if(!((sk-((j+1)*j))%n),return((sk-((j+1)*j))/n)))); \\ Antti Karttunen, Nov 06 2018

A205032(n) = for(k=sqrtint(n)-1,oo,my(sk=k*(k+1), d); for(j=1,k-1,d=(sk-((j+1)*j)); if(0==(d%n),return(d/n),if(dAntti Karttunen, Nov 06 2018

Definition edited and more terms from Antti Karttunen, Nov 06 2018

Showing 1-4 of 4 results.

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

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A204998 a(n) = k^2 - j^2, where (k^2,j^2) is the least pair of distinct squares for which n divides their difference.

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A204905 Least k such that n divides k^2-j^2 for some j satisfying 1<=j

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A205032 a(n) = (s(k)-s(j))/n, where (s(k),s(j)) is the least pair of oblong numbers (A002378) for which n divides their difference; a(n) = (1/n)*A205031(n).

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