A299111 -id:A299111 - cp's OEIS frontend

A338869 Shortest most frequent distance among first n primes.

1, 1, 2, 2, 2, 2, 2, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 30, 30, 30, 30, 6, 30, 6, 6, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30

Offset: 2

Andres Cicuttin, Nov 13 2020

nonn

Conjecture: Shortest most frequent distance among first n primes is a primorial number (A002110) for n>1.
This sequence is quite related to A338238 with which it shares many terms.
The corresponding frequencies of the most frequent distances among n first primes are in A283371.

			For n = 2, the distance between the first two primes 2 and 3 is 1, so the only possible distance is also the most frequent one, then a(2) = 1.
For n = 3, the distances between the first three primes 2, 3 and 5 are 1 = 3 - 2, 3 = 5 - 2, and 2 = 5 - 3, so all three distances are different, have the same frequency, and the shortest among them is 1, then a(3) = 1.
For n = 4, the five different distances between the first four primes 2, 3, 5 and 7 are 1 = 3 - 2, 2 = 5 - 3 = 7 - 5, 3 = 7 - 4 , 4 = 7 - 3 and 5 = 7 - 2, then a(3) = 2 because 2 is the most common distance (two cases) compared with the other distances which appear only once.
For n = 32, the most frequent distances are 30 and 6, and both appear with the same frequency (19 cases), then a(32) = 6 because 6 is the shortest between 30 and 6.

Andres Cicuttin, Log-log plot of the first 2^12 terms

Cf. A338238, A002110, A299111, A283371.

a[n_]:=Module[{pset, p2s,diffp2s,sd,sdgb,sdgbst},
pset=Prime[Range[n]]; (* First n primes *)
p2s=Subsets[pset,{2}]; (* All possible pairs of primes *)
(* Compute all possible distances and the corresponding frequencies *)
diffp2s=Map[Differences,p2s]//Flatten//Tally ;
(* Sort pairs {distance, frequency} by decreasing frequency *)
sd=Sort[diffp2s,#1[[2]]>#2[[2]]&];
(* Gather pairs {dist, freq} with same maximum frequency *)
sdgb=GatherBy[sd,sd[[1]][[2]]==#[[2]] &];
(* Sort selected pairs {dist, freq} with maximum frequency according to increasing distance *)
sdgbst=Sort[sdgb[[1]],#1[[1]]<#2[[1]]&];
(* Finally select and return the minimum distance among those with same maximum frequency *)
sdgbst[[1]][[1]] //Return];
Table[a[n],{n,2,100}]

A338238 Minimum number of rotations for a second maximum cyclic autocorrelation of the first n terms of the characteristic function of primes.

Original entry on oeis.org

1, 1, 1, 2, 3, 2, 2, 2, 2, 2, 4, 2, 6, 2, 8, 2, 4, 2, 6, 6, 6, 6, 6, 4, 6, 6, 6, 6, 6, 2, 6, 6, 6, 6, 12, 6, 6, 6, 6, 6, 18, 6, 6, 6, 6, 6, 24, 6, 6, 6, 6, 6, 24, 6, 6, 6, 24, 6, 6, 6, 6, 6, 6, 6, 24, 6, 6, 6, 6, 6, 24, 6, 6, 6, 6, 6, 24, 6, 6, 6, 6, 6, 30, 6, 30, 6, 12, 6, 30, 6, 6, 6, 6, 6, 30, 6, 30

Offset: 2

Andres Cicuttin, Oct 17 2020

nonn

It seems that most frequent terms among the first ones assume values 1, 2, 6, 30, 210, 2310, . . . Primorials? Several scatter plots of sequences of different lengths suggest this pattern (See Link).

			The primes among the first 5 positive integers (1,2,3,4,5) are 2, 3, and 5, then the corresponding characteristic function of primes is (0,1,1,0,1) (See A010051) and the corresponding five possible cyclic autocorrelations are the dot products between (0,1,1,0,1) and its rotations as shown here below:
(0,1,1,0,1).(0,1,1,0,1) = 0*0 + 1*1 + 1*1 + 0*0 + 1*1 = 3, (0 rotations)
(0,1,1,0,1).(1,0,1,1,0) = 0*1 + 1*0 + 1*1 + 0*1 + 1*0 = 1, (1 rotation)
(0,1,1,0,1).(0,1,0,1,1) = 0*0 + 1*1 + 1*0 + 0*1 + 1*1 = 2, (2 rotations)
(0,1,1,0,1).(1,0,1,0,1) = 0*1 + 1*0 + 1*1 + 0*0 + 1*1 = 2, (3 rotations)
(0,1,1,0,1).(1,1,0,1,0) = 0*1 + 1*1 + 1*0 + 0*1 + 1*0 = 1, (4 rotations)
The maximum value of the cyclic autocorrelation is always trivially obtained with zero rotations. In this example, the maximum value is 3 and the second maximum is 2, then a(5)=2 because it is needed a minimum of 2 rotations to obtain the second maximum.

Andres Cicuttin, Several scatter plots of sequences of different lengths

Cf. A010051, A002110, A337802, A299111, A338132.

nmax = 2^7;
b = Table[If[PrimeQ[i], 1, 0], {i, 1, nmax}];
tab = Table[Table[b[[1;;n]].RotateRight[b[[1;;n]], j], {j, 1, n-1}], {n, 2, nmax}];
tabmaxs = Table[Max[tab[[n]]], {n, 1, nmax-1}];
a = Table[First@Position[tab[[j]], tabmaxs[[j]]], {j, 1, nmax-1}] // Flatten

A299053 Minimum value of the cyclic autocorrelation of first n primes.

Original entry on oeis.org

4, 12, 31, 62, 133, 224, 377, 558, 865, 1304, 1805, 2462, 3337, 4280, 5389, 6726, 8449, 10264, 12663, 15294, 18061, 21200, 24961, 29166, 34173, 39508, 45017, 50870, 57141, 63788, 72299, 81234, 91365, 101732, 113327, 125166, 138355, 152348, 167179, 182862

Offset: 1

Andres Cicuttin, Feb 01 2018

nonn

Maximum values of the cyclic autocorrelation of first n primes are in A024450.

If we use this definition with integers instead of primes it is obtained A088003.

			For n = 4 the four possible cyclic autocorrelations of first four primes are:
(2,3,5,7).(2,3,5,7) = 2*2 + 3*3 + 5*5 + 7*7 = 4 + 9 + 25 + 49 = 87,
(2,3,5,7).(7,2,3,5) = 2*7 + 3*2 + 5*3 + 7*5 = 14 + 6 + 15 + 35 = 70,
(2,3,5,7).(5,7,2,3) = 2*5 + 3*7 + 5*2 + 7*3 = 10 + 21 + 10 + 21 = 62,
(2,3,5,7).(3,5,7,2) = 2*3 + 3*5 + 5*7 + 7*2 = 6 + 15 + 35 + 14 = 70,
then a(4)=62 because 62 is the minimum among the four values.

Alois P. Heinz, Table of n, a(n) for n = 1..10000

Cf. A024450, A014342, A299111, A088003.

a:= n-> min(seq(add(ithprime(i)*ithprime(irem(i+k, n)+1), i=1..n), k=1..n)):
seq(a(n), n=1..40);  # Alois P. Heinz, Feb 06 2018

p[n_]:=Prime[Range[n]];
Table[Table[p[n].RotateRight[p[n],j],{j,0,n-1}]//Min,{n,1,36}]

a(n) = vecmin(vector(n, k, sum(i=1, n, prime(i)*prime(1+(i+k)%n)))); \\ Michel Marcus, Feb 07 2018

a(n) = Min_{k=1..n} Sum_{i=1..n} prime(i)*prime(1 + (i+k) mod n).

A337327 Maximum value of the cyclic self-convolution of the first n terms of the characteristic function of primes.

Original entry on oeis.org

0, 1, 2, 2, 3, 2, 4, 3, 3, 3, 4, 4, 6, 5, 4, 4, 6, 5, 8, 7, 6, 6, 8, 7, 8, 7, 6, 6, 8, 7, 10, 9, 8, 8, 8, 8, 10, 9, 8, 8, 10, 10, 12, 12, 10, 11, 12, 12, 12, 13, 12, 12, 14, 13, 14, 13, 12, 12, 12, 12, 14, 13, 12, 13, 12, 14, 14, 15, 12, 14, 14, 16, 16, 18

Offset: 1

Andres Cicuttin, Aug 23 2020

			The primes among the first 5 positive integers (1,2,3,4,5) are 2, 3, and 5, then the corresponding characteristic function of primes is (0,1,1,0,1) (see A010051) and the corresponding five possible cyclic self-convolutions are the dot products between (0,1,1,0,1) and the rotations of its mirrored version as shown below:
  (0,1,1,0,1).(1,0,1,1,0) = 0*1 + 1*0 + 1*1 + 0*1 + 1*0 = 1,
  (0,1,1,0,1).(0,1,0,1,1) = 0*0 + 1*1 + 1*0 + 0*1 + 1*1 = 2,
  (0,1,1,0,1).(1,0,1,0,1) = 0*1 + 1*0 + 1*1 + 0*0 + 1*1 = 2,
  (0,1,1,0,1).(1,1,0,1,0) = 0*1 + 1*1 + 1*0 + 0*1 + 1*0 = 1,
  (0,1,1,0,1).(0,1,1,0,1) = 0*0 + 1*1 + 1*1 + 0*0 + 1*1 = 3.
Then a(5)=3 because 3 is the maximum among the five values.

Rémy Sigrist, Table of n, a(n) for n = 1..10000
Andres Cicuttin, Graph of first 2^10 terms

Cf. A010051, A299111, A014342.

b[n_]:=Table[If[PrimeQ[i],1,0],{i,1,n}];
Table[Max@Table[b[n].RotateRight[Reverse[b[n]],j],{j,0,n-1}],{n,1,100}]

a(n) = vecmax(vector(n, k, sum(i=1, n, isprime(n-i+1)*isprime(1+(i+k)%n)))); \\ Michel Marcus, Aug 26 2020

A337802 Minimum value of the cyclic self-convolution of the first n terms of the characteristic function of primes.

Original entry on oeis.org

0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset: 1

Andres Cicuttin, Sep 22 2020

nonn

In the first 1000 terms, a(n) = 1 only for n = 3, 5, 7, 11, 13, 17, 19, 23, 31, 43, 47, 61, 73, 107, 109, 113, 181, 199, and 467.

Is there an index k such that a(n) = 0 for n > k ?

			The primes among the first 5 positive integers (1,2,3,4,5) are 2, 3, and 5, then the corresponding characteristic function of primes is (0,1,1,0,1) (see A010051) and the corresponding five possible cyclic self-convolutions are the dot products between (0,1,1,0,1) and the rotations of its mirrored version as shown below:
  (0,1,1,0,1).(1,0,1,1,0) = 0*1 + 1*0 + 1*1 + 0*1 + 1*0 = 1,
  (0,1,1,0,1).(0,1,0,1,1) = 0*0 + 1*1 + 1*0 + 0*1 + 1*1 = 2,
  (0,1,1,0,1).(1,0,1,0,1) = 0*1 + 1*0 + 1*1 + 0*0 + 1*1 = 2,
  (0,1,1,0,1).(1,1,0,1,0) = 0*1 + 1*1 + 1*0 + 0*1 + 1*0 = 1,
  (0,1,1,0,1).(0,1,1,0,1) = 0*0 + 1*1 + 1*1 + 0*0 + 1*1 = 3.
Then a(5)=1 because 1 is the minimum among the five values.

Cf. A337327, A010051, A299111, A014342.

b[n_] := Table[If[PrimeQ[i], 1, 0], {i, 1, n}];
Table[Min@Table[b[n].RotateRight[Reverse[b[n]], j], {j, 0, n - 1}], {n, 1, 100}]

a(n) = vecmin(vector(n, k, sum(i=1, n, isprime(n-i+1)*isprime(1+(i+k)%n)))); \\ Michel Marcus, Sep 23 2020

A294172 Maximum value of the cyclic convolution of the first n positive integers with themselves.

Original entry on oeis.org

1, 5, 13, 28, 50, 83, 126, 184, 255, 345, 451, 580, 728, 903, 1100, 1328, 1581, 1869, 2185, 2540, 2926, 3355, 3818, 4328, 4875, 5473, 6111, 6804, 7540, 8335, 9176, 10080, 11033, 12053, 13125, 14268, 15466, 16739, 18070, 19480, 20951, 22505, 24123, 25828, 27600

Offset: 1

Andres Cicuttin, Feb 10 2018

nonn

Conjecture: a(n) = (n*(13 + 3*(-1)^n + 24*n + 14*n^2))/48, and then lim_{n -> infinity} a(n)/n^3 = 7/24.

The conjectured formula is true. See links. - Sela Fried, Aug 13 2024.

			For n = 4, the four possible cyclic convolutions of the first four positive integers with themselves are:
(1,2,3,4).(4,3,2,1) = 1*4 + 2*3 + 3*2 + 4*1 = 4 + 6 + 6 + 4 = 20,
(1,2,3,4).(3,2,1,4) = 1*3 + 2*2 + 3*1 + 4*4 = 3 + 4 + 3 + 16 = 26,
(1,2,3,4).(2,1,4,3) = 1*2 + 2*1 + 3*4 + 4*3 = 2 + 2 + 12 + 12 = 28,
(1,2,3,4).(1,4,3,2) = 1*1 + 2*4 + 3*3 + 4*2 = 1 + 8 + 9 + 8 = 26,
then a(4)=28 because 28 is the maximum among the four values.

Sela Fried, On the maximum value of the cyclic convolution of the first n positive integers with themselves (A294172), 2024.
Sela Fried, Proofs of some Conjectures from the OEIS, arXiv:2410.07237 [math.NT], 2024. See p. 8.

Cf. A000292, A299111.

a[n_] := Max[Table[Range[n].RotateRight[Reverse[Range[n]], k], {k, 0, n - 1}]];
Table[a[n], {n, 1, 45}]

a(n) = vecmax(vector(n, k, sum(i=1, n, (n-i+1)*(1+(i+k) % n)))); \\ Michel Marcus, Feb 11 2018

a(n) = Max {x; x=Sum_{i=1..n}(n-i+1)*(1+(i+k) mod n); for k=1..n}.
Conjectures from Colin Barker, Feb 11 2018: (Start)
G.f.: x*(1 + 3*x + 2*x^2 + x^3) / ((1 - x)^4*(1 + x)^2).
a(n) = 2*a(n-1) + a(n-2) - 4*a(n-3) + a(n-4) + 2*a(n-5) - a(n-6) for n>6.
(End)

A338132 List of lengths of subsequences of the characteristic function of those primes for which the minimum cyclic self-convolution is 1. Also n such that A337802(n)=1.

Original entry on oeis.org

3, 5, 7, 11, 13, 17, 19, 23, 31, 43, 47, 61, 73, 107, 109, 113, 181, 199, 467

Offset: 1

Andres Cicuttin, Oct 11 2020

Is this sequence finite?

Cf. A337802, A010051, A299111.

b[n_] := Table[If[PrimeQ[i], 1, 0], {i, 1, n}];
tab = Table[Min@Table[b[n].RotateRight[Reverse[b[n]], j], {j, 0, n - 1}], {n, 1, 2^12}]; (* to explore the first 2^12 subsequences *)
Flatten[Position[tab, 1]]

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A338869 Shortest most frequent distance among first n primes.

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A338238 Minimum number of rotations for a second maximum cyclic autocorrelation of the first n terms of the characteristic function of primes.

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A299053 Minimum value of the cyclic autocorrelation of first n primes.

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A337327 Maximum value of the cyclic self-convolution of the first n terms of the characteristic function of primes.

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A337802 Minimum value of the cyclic self-convolution of the first n terms of the characteristic function of primes.

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A294172 Maximum value of the cyclic convolution of the first n positive integers with themselves.

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A338132 List of lengths of subsequences of the characteristic function of those primes for which the minimum cyclic self-convolution is 1. Also n such that A337802(n)=1.

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