A095986 -id:A095986 - cp's OEIS frontend

A006063 A card-arranging problem: values of n such that there exists a permutation p_1, ..., p_n of 1, ..., n such that i + p_i is a cube for every i.

7, 19, 26, 37, 44, 56, 63, 66, 68, 80, 82, 85, 87, 98, 100, 103, 105, 110, 112, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 128, 129, 130, 131, 132, 133, 135, 147, 149, 150, 151, 152, 155, 156, 159, 171, 173, 174, 175, 176, 177, 178, 179

Offset: 1

N. J. A. Sloane

nonn

Apparently Gardner (1975) quotes Papaikonomou as showing that there can be at most one solution for a given n. However, this is incorrect: see A096680 for n values with more than one such permutation. - Ray Chandler

For any n, the number of permutations is permanent(m), where the n X n matrix m is defined m(i,j) = 1 or 0, depending on whether i+j is a cube or not. Hence, n is in this sequence if permanent(m) > 0.

M. Gardner, Mathematical Games column, Scientific American, Mar 1975.
M. Gardner, Time Travel and Other Mathematical Bewilderments. Freeman, NY, 1988, p. 81.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Cf. A095986 (for squares), A096680.

Conjecture: a(n) = n + 124 for n >= 173, i.e. there is such a permutation for every n >= 173. Verified for 173 <= n <= 1000. - Robert Israel, Aug 28 2018

Entry revised Jul 18 2004 based on comments from Franklin T. Adams-Watters

a(8) and later terms from Ray Chandler, Jul 26 2004

A097082 Number of permutations p of (1,2,3,...,n) such that k+p(k) is a Fibonacci number for 1 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 2, 4, 2, 1, 4, 4, 20, 4, 5, 1, 20, 24, 8, 96, 200, 24, 4, 25, 1, 3, 200, 48, 288, 48, 64, 2304, 1600, 10800, 288, 432, 8, 675, 650, 1, 26, 9, 10400, 1600, 576, 2304, 23040, 576, 2560, 1024, 368640, 516096, 128000, 2240000, 5832000, 32256, 2304, 46656, 64, 96, 91125, 3750, 84500, 6, 1, 676, 9, 261

Offset: 0

John W. Layman, Jul 23 2004

nonn

See A097083 for the positive values of n such that a(n) = 1.

Cf. A010056, A073364, A095986, A097083.

nmax=67; A010056[n_]:=With[{fibs=Fibonacci[Range[nmax]]}, If[MemberQ[fibs, n], 1, 0]]; a[n_]:=Permanent[Table[If[A010056[i+j]==1,1,0],{i,n},{j,n}]]; Join[{1},Array[a,nmax]] (* Stefano Spezia, Mar 03 2024 *)

a(n) = permanent(m), where the n X n matrix m is defined by m(i,j) = 1 or 0 depending on whether i+j is a Fibonacci number or not.

a(30)-a(50) from Ray Chandler, Jul 26 2004
More terms from David Wasserman, Dec 19 2007
a(0)=1 prepended by Stefano Spezia, Mar 04 2024

A096680 A card-arranging problem: values of n such that there exists more than one permutation p_1, ..., p_n of 1, ..., n such that i + p_i is a cube for every i.

Original entry on oeis.org

112, 115, 116, 117, 119, 124, 125, 126, 127, 128, 129, 130, 133, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200, 201, 202, 203, 204, 205, 206, 207, 208, 209, 210, 211, 212

Offset: 1

Ray Chandler, Jul 25 2004

nonn

			117 is in the sequence with permutations
(7,6,...,2,1,117,116,...,9,8) and
(26,25,...,2,1,98,97,...28,27,117,116,...,100,99)

Cf. A006063, A073364, A095986, A097082, A097083.

A096901 Number of permutations p of (1,2,3,...,n) such that k+p(k) is a triangular number for 1<=k<=n.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 1, 1, 2, 4, 3, 9, 14, 13, 52, 124, 161, 181, 715, 2338, 7073, 8624, 15466, 52858, 150365, 316543, 691771, 1681604, 5324919, 15407311, 37417775, 69725286, 155786456, 579599171, 2600274145, 10530031625, 22971756045, 47057778714, 112946192928

Offset: 0

Ray Chandler, Jul 26 2004

nonn

Cf. A006063, A073364, A095986, A097082, A097083, A096680.

A276735 Number of permutations p of (1..n) such that k+p(k) is a semiprime for 1 <= k <= n.

Original entry on oeis.org

1, 0, 0, 1, 1, 3, 5, 12, 87, 261, 324, 1433, 5881, 37444, 196797, 708901, 2020836, 12375966, 105896734, 955344450, 11136621319, 95274505723, 590283352231, 4285001635230, 36417581252044, 272699023606314

Offset: 0

Gary E. Davis, Sep 24 2016

nonn

			a(4) = 1 because the permutation (3,4,1,2) added termwise to (1,2,3,4) yields (4,6,4,6) - both semiprimes - and only that permutation does so. a(5) = 3 because exactly 3 permutations - (3,2,1,5,4), (3,4,1,2,5) & (5,4,3,2,1) - added termwise to (1,2,3,4,5) yield semiprime entries.
From _David A. Corneth_, Sep 28 2016 (Start):
The semiprimes up to 10 are 4, 6, 9 and 10. To find a(5), we add (1,2,3,4,5) to some p. Therefore, p(1) in {3, 5}, p(2) in {2, 4}, p(3) in {1, 3}, p(4) in {2, 5} and p(5) in {1, 4, 5}.
If p(1) = 3 then p(3) must be 1. Then {p(2), p(4), p(5)} = {2, 4, 5} for which there are two possibilities.
If p(1) = 5 then p(3) = 3 and p(4) = 2. Then p(2) = 4 and p(5) = 1. So there's one permutation for which p(1) = 5.
This exhausts the options for p(1) and we found 3 permutations. Therefore, a(5) = 3. (End)

Cf. A073364, A095986, A096904, A097082.

with(LinearAlgebra): with(numtheory):
a:= n-> `if`(n=0, 1, Permanent(Matrix(n, (i, j)->
        `if`((s-> bigomega(s)=2)(i+j), 1, 0)))):
seq(a(n), n=0..16);  # Alois P. Heinz, Sep 28 2016

perms[n0_] :=
Module[{n = n0, S, func, T, T2},
  S = Select[Range[2, 2*n], PrimeOmega[#] == 2 &];
  func[k_] := Cases[S, x_ /; 1 <= x - k <= n] - k;
  T = Tuples[Table[func[k], {k, 1, n}]];
  T2 = Cases[T, x_ /; Length[Union[x]] == Length[x]];
  Length[T2]]
Table[perms[n], {n, 0, 12}]
(* Second program (version >= 10): *)
a[0] = 1; a[n_] := Permanent[Table[Boole[PrimeOmega[i + j] == 2], {i, 1, n}, {j, 1, n}]]; Table[an = a[n]; Print[an]; an, {n, 0, 20}] (* Jean-François Alcover, Jul 25 2017 *)

isok(va, vb)=my(v = vector(#va, j, va[j]+vb[j])); #select(x->(bigomega(x) == 2), v) == #v;
a(n) = my(vpo = numtoperm(n,1)); sum(k=1, n!, vp = numtoperm(n,k); isok(vp, vpo)); \\ Michel Marcus, Sep 24 2016

listA001358(lim)=my(v=List()); forprime(p=2, sqrtint(lim\1), forprime(q=p, lim\p, listput(v, p*q))); Set(v)
has(v)=for(k=1,#v, if(!setsearch(semi, v[k]+k), return(0))); 1
a(n)=local(semi=listA001358(2*n)); sum(k=1,n!, has(numtoperm(n,k))) \\ Charles R Greathouse IV, Sep 28 2016

matperm(M)=my(n=matsize(M)[1], innerSums=vectorv(n)); if(n==0, return(1)); sum(x=1,2^n-1, my(k=valuation(x,2), s=M[,k+1], gray=bitxor(x,x>>1)); if(bittest(gray,k), innerSums+=s,innerSums-=s); (-1)^hammingweight(gray)*factorback(innerSums))*(-1)^n
a(n)=matperm(matrix(n,n,x,y,bigomega(x+y)==2)) \\ Charles R Greathouse IV, Oct 03 2016

More terms from Alois P. Heinz, Sep 28 2016

cp's OEIS Frontend

A006063 A card-arranging problem: values of n such that there exists a permutation p_1, ..., p_n of 1, ..., n such that i + p_i is a cube for every i.

Views

Author

Keywords

Comments

References

Crossrefs

Formula

Extensions

A097082 Number of permutations p of (1,2,3,...,n) such that k+p(k) is a Fibonacci number for 1 <= k <= n.

Views

Author

Keywords

Comments

Crossrefs

Programs

Mathematica

Formula

Extensions

A096680 A card-arranging problem: values of n such that there exists more than one permutation p_1, ..., p_n of 1, ..., n such that i + p_i is a cube for every i.

Views

Author

Keywords

Examples

Crossrefs

A096901 Number of permutations p of (1,2,3,...,n) such that k+p(k) is a triangular number for 1<=k<=n.

Views

Author

Keywords

Crossrefs

Formula

Extensions

A276735 Number of permutations p of (1..n) such that k+p(k) is a semiprime for 1 <= k <= n.

Views

Author

Keywords

Examples

Crossrefs

Programs

Maple

Mathematica

PARI

PARI

PARI

Extensions