A140670 -id:A140670 - 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

A135481 a(n) = 2^A007814(n+1) - 1.

Original entry on oeis.org

0, 1, 0, 3, 0, 1, 0, 7, 0, 1, 0, 3, 0, 1, 0, 15, 0, 1, 0, 3, 0, 1, 0, 7, 0, 1, 0, 3, 0, 1, 0, 31, 0, 1, 0, 3, 0, 1, 0, 7, 0, 1, 0, 3, 0, 1, 0, 15, 0, 1, 0, 3, 0, 1, 0, 7, 0, 1, 0, 3, 0, 1, 0, 63, 0, 1, 0, 3, 0, 1, 0, 7, 0, 1, 0, 3, 0, 1, 0, 15, 0, 1, 0, 3, 0, 1, 0, 7, 0, 1, 0, 3, 0, 1, 0, 31, 0, 1, 0, 3, 0, 1, 0

Offset: 0

N. J. A. Sloane, based on a message from Guy Steele and Don Knuth, Mar 01 2008

This is Guy Steele's sequence GS(1, 6) (see A135416).

Antti Karttunen, Table of n, a(n) for n = 0..16384

Cf. A006519, A007814, A135416, A140670, A136013 (partial sums).

using IntegerSequences
A135481List(len) = [Bits("CNIMP", n+1, n) for n in 0:len]
println(A135481List(100))  # Peter Luschny, Sep 25 2021

Maple
```
GS(1,6,200); # see A135416
```

Table[BitAnd[i, BitNot[i+1]], {i, 0, 200}] (* Peter Luschny, Jun 01 2011 *)

a(n) = 2^valuation(n+1, 2)-1; \\ Michel Marcus, Nov 19 2017

a(n) = bitand(bitneg(n+1), n); \\ Ruud H.G. van Tol, Apr 05 2023

def A135481(n): return ~(n+1)&n # Chai Wah Wu, Jul 06 2022

a(n) = A006519(n+1) - 1. - R. J. Mathar, Feb 10 2016

a(0) = 0 prepended by Andrey Zabolotskiy, Oct 08 2019, based on Lothar Esser's contribution

A204987 Least k such that n divides 2^k - 2^j for some j satisfying 1 <= j < k.

Original entry on oeis.org

2, 2, 3, 3, 5, 3, 4, 4, 7, 5, 11, 4, 13, 4, 5, 5, 9, 7, 19, 6, 7, 11, 12, 5, 21, 13, 19, 5, 29, 5, 6, 6, 11, 9, 13, 8, 37, 19, 13, 7, 21, 7, 15, 12, 13, 12, 24, 6, 22, 21, 9, 14, 53, 19, 21, 6, 19, 29, 59, 6, 61, 6, 7, 7, 13, 11, 67, 10, 23, 13, 36, 9, 10, 37, 21, 20, 31, 13, 40, 8, 55, 21, 83, 8, 9, 15, 29, 13

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 2^2 - 2^1, so a(2)=2;
3 divides 2^3 - 2^1, so a(3)=3;
4 divides 2^3 - 2^2, so a(4)=3;
5 divides 2^5 - 2^1, so a(5)=5.

Antti Karttunen, Table of n, a(n) for n = 1..6556

Cf. A000079, A007733, A007814, A054703, A204892, A204988.

s[n_] := s[n] = 2^n; z1 = 1000; 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}]     (* A204985 *)
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}]     (* A204986 *)
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}]     (* A204987 *)
Table[j[n], {n, 1, z2}]     (* A204988 *)
Table[s[k[n]], {n, 1, z2}]  (* A204989 *)
Table[s[j[n]], {n, 1, z2}]  (* A140670 ? *)
Table[s[k[n]] - s[j[n]], {n, 1, z2}]     (* A204991 *)
Table[(s[k[n]] - s[j[n]])/n, {n, 1, z2}] (* A204992 *)
%%/2  (* A204990=(1/2)*A204991 *)

A204987etA204988(n) = { my(k=2); while(1,for(j=1,k-1,if(!(((2^k)-(2^j))%n),return([k,j]))); k++); }; \\ (Computes also A204988 at the same time) - Antti Karttunen, Nov 19 2017

a(n)={my(k=valuation(n,2)); max(k, 1) + znorder(Mod(2, n>>k))} \\ Andrew Howroyd, Aug 08 2018

a(n) = max(1, A007814(n)) + A007733(n). - Andrew Howroyd, Aug 08 2018

More terms from Antti Karttunen, Nov 19 2017

A160467 a(n) = 1 if n is odd; otherwise, a(n) = 2^(k-1) where 2^k is the largest power of 2 that divides n.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 1, 4, 1, 1, 1, 2, 1, 1, 1, 8, 1, 1, 1, 2, 1, 1, 1, 4, 1, 1, 1, 2, 1, 1, 1, 16, 1, 1, 1, 2, 1, 1, 1, 4, 1, 1, 1, 2, 1, 1, 1, 8, 1, 1, 1, 2, 1, 1, 1, 4, 1, 1, 1, 2, 1, 1, 1, 32, 1, 1, 1, 2, 1, 1, 1, 4, 1, 1, 1, 2, 1, 1, 1, 8, 1, 1, 1, 2, 1, 1, 1, 4, 1, 1, 1, 2, 1, 1, 1, 16

Offset: 1

Johannes W. Meijer, May 24 2009, Jun 28 2011

Fifth factor of the row sums A160466 of the Eta triangle A160464.
From Peter Luschny, May 31 2009: (Start)
Let odd(n) be the characteristic function of the odd numbers (A000035) and sigma(n) the number of 1's in binary expansion of n (A000120). Then a(n) = 2^(sigma(n-1) - sigma(n) + odd(n)).
Let B_{n} be the Bernoulli number. Then this sequence is also
a(n) = denominator(4*(4^n-1)*B_{2*n}/n). (End)

Antti Karttunen, Table of n, a(n) for n = 1..16384

Cf. A000265, A001620, A007814, A026741, A140670, A160464, A220466, A006519, A000010.

nmax:=96: p:= floor(log[2](nmax)): for n from 1 to nmax do a(n):=1 end do: for q from 1 to p do for n from 1 to nmax do if n mod 2^q = 0 then a(n):= 2^(q-1) end if: end do: end do: seq(a(n), n=1..nmax);
From Peter Luschny, May 31 2009: (Start)
a := proc(n) local sigma; sigma := proc(n) local i; add(i,i=convert(n,base,2)) end; 2^(sigma(n-1)-sigma(n)+`if`(type(n,odd),1,0)) end: seq(a(n), n=1..96);
a := proc(n) denom(4*(4^n-1)*bernoulli(2*n)/n) end: seq(a(n), n=1..96); (End)

a[n_] := If[OddQ[n], 1, 2^(IntegerExponent[n, 2] - 1)]; Array[a, 100] (* Amiram Eldar, Jul 02 2020 *)

A160467(n) = 2^max(valuation(n,2)-1,0); \\ Antti Karttunen, Nov 18 2017, after Max Alekseyev's Feb 09 2011 formula.

def A160467(n): return max(1,(n&-n)>>1) # Chai Wah Wu, Jul 08 2022

a(n) = A026741(n)/A000265(n). - Paul Curtz, Apr 18 2010
a(n) = 2^max(A007814(n) - 1, 0). - Max Alekseyev, Feb 09 2011
a((2*n-1)*2^p) = A011782(p), p >= 0 and n >= 1. - Johannes W. Meijer, Jan 25 2013
a(n) = (1 + A140670(n))/2. - Antti Karttunen, Nov 18 2017
From Amiram Eldar, Dec 31 2022: (Start)
Dirichlet g.f.: zeta(s)*(2^s-2+1/2^s)/(2^s-2).
Sum_{k=1..n} a(k) ~ (1/(4*log(2)))*n*log(n) + (5/8 + (gamma-1)/(4*log(2)))*n, where gamma is Euler's constant (A001620). (End)
a(n) = A006519(n)/gcd(n,2). - Ridouane Oudra, Feb 08 2025
a(n) = A000010(A006519(n)). - Ridouane Oudra, Jul 27 2025

Keyword mult added by Max Alekseyev, Feb 09 2011

Name changed by Antti Karttunen, Nov 18 2017

A330569 a(n) = 1 if n is odd, otherwise a(n) = 2^(v-1)+1 where v is the 2-adic valuation of n (A007814(n)).

Original entry on oeis.org

1, 2, 1, 3, 1, 2, 1, 5, 1, 2, 1, 3, 1, 2, 1, 9, 1, 2, 1, 3, 1, 2, 1, 5, 1, 2, 1, 3, 1, 2, 1, 17, 1, 2, 1, 3, 1, 2, 1, 5, 1, 2, 1, 3, 1, 2, 1, 9, 1, 2, 1, 3, 1, 2, 1, 5, 1, 2, 1, 3, 1, 2, 1, 33, 1, 2, 1, 3, 1, 2, 1, 5, 1, 2, 1, 3, 1, 2, 1, 9, 1, 2, 1, 3, 1

Offset: 1

N. J. A. Sloane, Jan 07 2020

Conjecture: a(n) is the height of the bottom region in column n of the Conant gasket in A328078.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A007814, A328078, A006519 A040001, A140670.

a[n_] := If[OddQ[n], 1, 1 + 2^(IntegerExponent[n, 2] - 1)]; Array[a, 100] (* Amiram Eldar, Aug 30 2024 *)

a(n) = if(n % 2, 1, 1 + 1 << (valuation(n, 2) - 1)); \\ Amiram Eldar, Aug 30 2024

def A330569(n): return 1+(0 if n&1 else 1<<(~n & n-1).bit_length()-1) # Chai Wah Wu, Jul 01 2022

a(n) = (A006519(n) + A040001(n))/2. - Ridouane Oudra, Jul 30 2025

Showing 1-5 of 5 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).

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A135481 a(n) = 2^A007814(n+1) - 1.

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A204987 Least k such that n divides 2^k - 2^j for some j satisfying 1 <= j < k.

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A160467 a(n) = 1 if n is odd; otherwise, a(n) = 2^(k-1) where 2^k is the largest power of 2 that divides n.

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A330569 a(n) = 1 if n is odd, otherwise a(n) = 2^(v-1)+1 where v is the 2-adic valuation of n (A007814(n)).

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