cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

User: Eugen J. Ionascu

Eugen J. Ionascu's wiki page.

Eugen J. Ionascu has authored 4 sequences.

A178797 Number of regular octahedra that can be formed using the points in an (n+1)X(n+1)X(n+1) lattice cube.

Original entry on oeis.org

0, 1, 8, 32, 104, 261, 544, 1000, 1696, 2759, 4296, 6434, 9352, 13243, 18304, 24774, 32960, 43223, 55976, 71752, 90936, 113973, 141312, 173436, 210960, 254587, 305000, 364406, 432824, 511421, 600992, 702556, 817200, 946131, 1090392, 1251238
Offset: 1

Author

Eugen J. Ionascu, Jun 15 2010

Keywords

Examples

			a(2)=1 because there is 1 way to form a regular octahedron using points of a {0,1,2}^3 lattice cube.
		

Crossrefs

Extensions

Edited by Ray Chandler, Jul 27 2010

A103139 Woolbright sequence: the maximum number of kings on an n X n chessboard such that every single king is attacking a number of other kings that is smaller or equal to the number of empty spaces around it.

Original entry on oeis.org

1, 2, 6, 9, 15, 22, 28, 39, 49, 59, 73
Offset: 1

Author

Eugen J. Ionascu, Mar 17 2005

Keywords

Comments

Lower bounds for terms following 59 are as follows: 73, 86, 102, 117, 136, 153, 173, 195, 216, 239, 266, 289, 318, 345, 375, 405, 438, 471, 504, 540, 576, 614, 654, 693, 735, 777, ...

Examples

			a(3)=6. Indeed, on a 3 X 3 chessboard one can arrange six kings on two side columns to satisfy the requirement. It is not possible to arrange seven kings since the center has to be empty and then at least one of the squares in the middle of the sides must have a king on it, which requires at least three empty spaces around, and that is impossible.
		

References

  • Bernardo Recamán, The Bogotá Puzzles, Dover Publications, 2020, p. 19.

Formula

a(n) = n^2 - gamma_{1/2}(n) = approx floor(3*(n^2+1)/5). (I assume this is a lower bound? - N. J. A. Sloane)

Extensions

One more term [from the Ionascu et al. paper] from Vladeta Jovovic, Sep 17 2008

A102032 a(n) is the smallest positive integer k such that, if kn is written in base 2, it requires exactly n ones.

Original entry on oeis.org

1, 3, 7, 15, 11, 21, 89, 255, 167, 307, 349, 1365, 1259, 6729, 6417, 65535, 28431, 29127, 54757, 209715, 274627, 750685, 706649, 5592405, 2663383, 9679163, 14913005, 186946121, 37025579, 353440017, 1175487521, 4294967295
Offset: 1

Author

Eugen J. Ionascu, Aug 03 2007

Keywords

Comments

This sequence can be considered for any base b. If one calculates the arithmetic mean of the sequence d(n):=a(n)/2^n, i.e. (d(1)+d(2)+...+d(n))/n, one obtains a sequence converging to zero.
From Robert Israel, Aug 26 2015: (Start)
a(2*n) <= (2^A070939(a(n)) + 1)*a(n).
if n is odd, a(n) <= (2^(n*r)-1)/(n*(2^r-1)) where r = A002326((n-1)/2). (End)

Examples

			Example: If n=7 then 7(89)=623 which written in base 2 is 1001101111 using exactly 7 ones and 89 is the smallest positive integer with this property. Hence a(7)=89. The number 1001101111 is usually known as Niven number in base 2. We called 623 a minimal Niven number.
		

References

  • I. Vardi, Niven numbers, Computational Recreations in Mathematics, Addison-Wesley, 1991, pp. 19 and 28--31.

Crossrefs

Programs

  • Maple
    with(numtheory):
    fjv6:=proc(n,m)
    local i,j,k,l,x,x1,y,y1,z,z1,w,stopp,s,t,u,v,A,F,G,out;
    i:=n;stopp:=0;
    x1:=2^(m*i+6)-1; x:=x1 mod i;j:=0;
    while stopp=0 and j<=m*i+5 do
    l:=j;
    while stopp=0 and l<=m*i+4 do
    k:=l;
    while stopp=0 and k<=m*i+3 do
    s:=k;
    while stopp=0 and s<=m*i+2 do
    t:=s;
    while stopp=0 and t<=m*i+1 do
    v:=t;
    while stopp=0 and v<=m*i do y1:=2^(m*i+5-j)+2^(m*i+4-l)+2^(m*i+3-k)+2^(m*i+2-s)+2^(m*i+1-t)+2^(m*i-v); y:=y1 mod i;
    if y=x then z:=(x1-y1)/i;out:=[m*i, z];
    stopp:=1;
    fi;
    v:=v+1;od;t:=t+1;od;s:=s+1;od; k:=k+1; od;l:=l+1;od;j:=j+1;od;
    if stopp=0 then out:=[m*i,0];fi;
    out;
    end:
    formula:=proc(n)
    local x,y,B,expon,outputis, theOddFactor;
    x:=n+1;B:=ifactors(x);expon:=B[2][1][2];theOddFactor:=(n+1)/2^expon;
    y:=isprime(n);
    if theOddFactor=1 and y=true then outputis:=[n,(2^(n+expon-1)+2^n-2^(n-expon)-1)/n];fi;
    if theOddFactor>1 or y=false then outputis:=fjv6(n,1);fi;
    lprint(outputis[1],outputis[2]);
    end:
    fjfromis6:=proc(n,m)
    local k,B,expon, theoddfac,par,stopp,av,sub;
    av:=0;for k from n to m do
    par:=k mod 2;
    if par=0 then B:=ifactors(k);expon:=B[2][1][2];theoddfac:=k/2^expon;
    sub:=fjv6(theoddfac,2^expon);
    lprint(sub[1], sub[2]); fi;
    stopp:=0;
    if par=1 then formula(k); fi;
    od;
    end:
    fjfromis6(1,185);
    # Alternative:
    F:= proc(k,x,n,dmax)
       option remember;
       local d,z,v;
       if k = 0 then
          if x = 0 then return 0 else return infinity fi
       end;
       for d from k-1 to dmax do
          v:= procname(k-1,(x - 2^d) mod n,n, d-1) ;
          if v < 2^d then return v + 2^d fi
       od;
       infinity;
    end proc:
    seq(F(n,0,n,infinity)/n, n=1..100); # Robert Israel, Aug 26 2015
  • Mathematica
    F[k_, x_, n_, dMax_] := F[k, x, n, dMax] = Module[{d, z, v}, If[k == 0, If[x == 0, Return[0], Return[Infinity]]]; For[d = k - 1, d <= dMax, d++, v = F[k - 1, Mod[x - 2^d, n], n, d - 1]; If[v < 2^d, Return[v + 2^d]]]; Infinity];
    Table[F[n, 0, n, Infinity]/n, {n, 1, 32}] (* Jean-François Alcover, Jun 22 2020, after Robert Israel *)
  • PARI
    a(n)=my(K=n);while(hammingweight(K)!=n,K+=n);K/n \\ Charles R Greathouse IV, Feb 04 2013

Formula

a(n) = 2^n-1 if n=2^k or a(n) = (2^(n+k-1)+2^n-2^(n-k)-1)/n if n=2^k-1 is a prime number; unknown for other values of n.

Extensions

Edited by Ray Chandler, Nov 16 2008

A101201 Maximal number of kings in the toroidal king's graph on an n X n board such that each king is attacking no more than four other kings.

Original entry on oeis.org

0, 2, 5, 9, 15, 21, 28, 37, 47, 60, 71, 84
Offset: 1

Author

Eugen J. Ionascu, Aug 12 2008

Keywords

Comments

All the numbers listed so far were calculated using LpSolveIDE. It can be shown that the sequence of densities a(n)/n^2 has a limit as n goes to infinity, which is the supremum of all the elements in the sequence. With the help of the computer, it was shown that a(n) is not more than 0.608956n^2.

Examples

			a(2)=2 because one can check that any arrangement of two kings will satisfy the requirement but any arrangement of three kings will not.
		

Crossrefs

Formula

a(n) is approximately 3n^2/5 (conjecture).