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.

Showing 1-4 of 4 results.

A248625 Lexicographically earliest sequence of nonnegative integers such that no triple (a(n),a(n+d),a(n+2d)) is in arithmetic progression, for any d>0, n>=0.

Original entry on oeis.org

0, 0, 1, 0, 0, 1, 1, 3, 3, 0, 0, 1, 0, 0, 1, 1, 3, 3, 1, 3, 3, 4, 4, 7, 4, 4, 8, 0, 0, 1, 0, 0, 1, 1, 3, 3, 0, 0, 1, 0, 0, 1, 1, 3, 3, 1, 3, 3, 4, 4, 7, 4, 4, 8, 8, 3, 3, 4, 4, 9, 4, 4, 9, 1, 9, 12, 10, 9, 7, 10, 12, 9, 11, 9, 9, 11, 9, 10, 13, 19, 12, 0, 0, 1, 0, 0, 1, 1, 3, 3
Offset: 0

Views

Author

M. F. Hasler, Oct 10 2014

Keywords

Comments

The sequence is constructed in the greedy way, appending at each step the least nonnegative integer such that no subsequence of equidistant terms contains an AP.
Every nonnegative integer seems to appear in this sequence - see A248627 for the corresponding indices.
Sequence A229037 is the analog for positive integers (and indices).

Examples

			Start with a(0)=a(1)=0, smallest possible choice and trivially satisfying the constraint since no 3-term subsequence is possible.
Then one must take a(2)=1 since otherwise [0,0,0] would be an AP.
Then one can take again a(3)=a(4)=0, and a(5)=1.
Now appending 0 would yield the AP (0,0,0) by extracting terms with indices 0,3,6; therefore a(6)=1.
Now a(7) cannot be 0 not 1 nor 2 since else a(3)=0, a(5)=1, a(7)=2 would be an AP, therefore a(7)=3 is the least possible choice.

Crossrefs

Cf. A005836, A005839, A023717, ..., A020664.
Cf. A248627, A229037, A241752.

Programs

  • PARI
    [DD(v)=vecextract(v,"^1")-vecextract(v,"^-1"), hasAP(a,m=3)=u=vector(m,i,1);v=vector(m,i,i-1);for(i=1,#a-m+1,for(s=1,(#a-i)\(m-1),#Set(DD(t=vecextract(a,(i)*u+s*v)))==1&&return
([i,s,t])))]; a=[]; for(n=1,90,a=concat(a,0);while(hasAP(a),a[#a]++);print1(a[#a]","));a

Formula

a(n) = A229037(n+1)+1.

A248641 Lexicographically earliest positive sequence which does not contain a 4-term equidistant subsequence (a(n+k*d); k=0,1,2,3) in arithmetic progression.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 2, 2, 2, 3, 1, 1, 1, 2, 1, 2, 2, 2, 3, 3, 3, 1, 1, 3, 1, 1, 1, 2, 2, 1, 2, 2, 1, 2, 2, 2, 3, 3, 2, 3, 2, 3, 3, 5, 1, 1, 1, 3, 1, 1, 3, 1, 1, 1, 2, 2, 2, 3, 1, 2, 1, 1, 1, 2, 2, 2, 3, 4, 2, 3, 2, 2, 2, 3, 3, 1, 3, 3, 3, 5, 5, 4, 1, 1, 1, 3, 1, 2, 3, 1, 5, 3, 2, 6, 1, 3, 2, 2, 3, 2, 1, 1, 3, 3, 1, 1, 1
Offset: 0

Views

Author

M. F. Hasler, Oct 10 2014

Keywords

Comments

See A248625 for more information, links and examples.
It is a variation of A229037 where 3-term is replaced by 4-term (and with “lead index” 0 instead of 1)

Links

Crossrefs

Cf. A248625, A248639, A248640, A248627, A229037, A241752.

Programs

  • PARI
    a=[];for(n=1,190,a=concat(a,1);while(hasAP(a,4),a[#a]++));a \\ See A248625 for hasAP().
  • SageMath
    cpdef FourFree(int n):
   cdef int i, r, k, s, L1, L2, L3
   cdef list L, Lb
   cdef set b
   L=[1, 1, 1]
   for k in range(3, n):
      b=set()
      for i in range(k):
         if 3*((k-i)/3)==k-i:
            r=(k-i)/3
            L1, L2, L3=L[i], L[i+r], L[i+2*r]
            s=3*(L2-L1)+L1
            if s>0 and L3==2*(L2-L1)+L1:
               b.add(s)
      if 1 not in b:
         L.append(1)
      else:
         Lb=list(b)
         Lb.sort()
         for t in Lb:
            if t+1 not in b:
               L.append(t+1)
               break
   return L
# Sébastien Palcoux, Aug 28 2019

A248639 Least nonnegative sequence which does not contain a 4-term equidistant subsequence (a(n+k*d); k=0,1,2,3) in arithmetic progression.

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 2, 0, 0, 0, 1, 0, 1, 1, 1, 2, 2, 2, 0, 0, 2, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1, 2, 2, 1, 2, 1, 2, 2, 4, 0, 0, 0, 2, 0, 0, 2, 0, 0, 0, 1, 1, 1, 2, 0, 1, 0, 0, 0, 1, 1, 1, 2, 3, 1, 2, 1, 1, 1, 2, 2, 0, 2, 2, 2, 4, 4, 3, 0, 0, 0, 2, 0, 1, 2, 0, 4, 2, 1, 5, 0, 2, 1, 1, 2, 1, 0, 0, 2, 2, 0, 0, 0, 3, 0, 0, 1, 1, 1, 4, 1, 2, 3, 0, 1, 2, 1, 0, 3, 3, 4, 1, 1, 3
Offset: 0

Views

Author

M. F. Hasler, Oct 10 2014

Keywords

Comments

See A248625 for more information, links and examples.
See A248641 for the "positive integers" variant.

Crossrefs

Cf. A248625, A248640, A248641, A248627, A229037, A241752.

Programs

  • PARI
    a=[];for(n=1,190,a=concat(a,0);while(hasAP(a,4),a[#a]++));a \\ See A248625 for hasAP().

A248640 Least nonnegative sequence which does not contain a 5-term equidistant subsequence (a(n+k*d); k=0,1,2,3,4) in arithmetic progression.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0
Offset: 0

Views

Author

M. F. Hasler, Oct 10 2014

Keywords

Comments

See A248625 for more information, links and examples.

Crossrefs

Cf. A248625, A248639, A248627, A229037, A241752.

Programs

  • PARI
    a=[];for(n=1,190,a=concat(a,0);while(hasAP(a,5),a[#a]++));a \\ See A248625 for hasAP(). Use concat(a,1) for the "positive integer" variant.
Showing 1-4 of 4 results.

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