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.

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A235383 Fibonacci numbers that are the product of other Fibonacci numbers.

Original entry on oeis.org

8, 144
Offset: 1

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Author

Robert C. Lyons, Jan 08 2014

Keywords

Comments

This sequence and A229037 and A235265 are winners in the contest held at the 2014 AMS/MAA Joint Mathematics Meetings. - T. D. Noe, Jan 20 2014
Carmichael's theorem implies that 8 and 144 are the only terms of this sequence.
First two terms of A061899, A111687, A172150, A212703, and A231851. - Omar E. Pol, Jan 21 2014
Saha and Karthik conjectured (without reference to Carmichael's theorem) that the only positive integers k for which A001175(k^2) = A001175(k) are 6 and 12. (A000045(6) = 8 and A000045(12) = 144.) - L. Edson Jeffery, Feb 13 2014
Y. Bugeaud, M. Mignotte, and S. Siksek proved that 8 and 144 are the only nontrivial perfect power Fibonacci numbers. - Robert C. Lyons, Dec 23 2015

Examples

			The Fibonacci number 8 is in the sequence because 8=2*2*2, and 2 is a Fibonacci number that is not equal to 8. The Fibonacci number 144 is in the sequence because 144=3*3*2*2*2*2, and both 2 and 3 are Fibonacci numbers that are not equal to 144.
		

Crossrefs

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

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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

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.

A371632 Lexicographically earliest sequence of positive integers such that no three terms a(j), a(j+k), a(j+2k) (for any j and k) form a progression of the form p, p+q, p-q, where q >= 0.

Original entry on oeis.org

1, 1, 2, 1, 1, 2, 2, 3, 3, 2, 3, 2, 4, 1, 3, 2, 1, 3, 2, 3, 4, 1, 2, 3, 4, 3, 4, 4, 5, 4, 1, 5, 5, 4, 1, 4, 2, 5, 5, 6, 5, 5, 6, 6, 7, 7, 3, 7, 6, 6, 8, 6, 6, 5, 7, 7, 8, 7, 1, 8, 8, 9, 9, 8, 5, 3, 9, 9, 10, 9, 6, 8, 8, 5, 9, 9, 5, 8, 6, 10, 1, 7, 10, 6, 6, 4, 4, 8, 3, 10
Offset: 1

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Author

Neal Gersh Tolunsky, May 24 2024

Keywords

Comments

This sequence avoids one of the six permutations of a set of three integers in arithmetic progression. For example, the set {1,2,3} can be ordered as tuples (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), and (3, 2, 1). In this sequence, we avoid (2,3,1) and other progressions of the form p, p+q, p-q, for all q >= 0.

Crossrefs

A373010 Lexicographically earliest sequence of positive integers such that no three terms a(j), a(j+k), a(j+2k) (for any j and k) form a progression of the form p, p-2*q, p-q, where q >= 0.

Original entry on oeis.org

1, 1, 2, 1, 1, 2, 2, 3, 3, 1, 1, 3, 1, 1, 2, 4, 3, 3, 4, 3, 2, 2, 4, 4, 2, 2, 5, 1, 1, 3, 1, 1, 2, 4, 4, 5, 1, 1, 3, 1, 1, 5, 4, 5, 3, 6, 5, 6, 5, 4, 6, 6, 4, 3, 4, 3, 3, 4, 3, 6, 2, 6, 5, 7, 3, 6, 6, 3, 2, 7, 6, 7, 5, 5, 2, 2, 6, 2, 2, 4, 5, 1, 1, 2, 1, 1, 5, 2, 6, 7
Offset: 1

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Author

Neal Gersh Tolunsky, May 22 2024

Keywords

Comments

This sequence avoids one of the six permutations of a set of three integers in arithmetic progression. For example, the set {1,2,3} can be ordered as tuples (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), and (3, 2, 1). In this sequence, we avoid (3,1,2) and other progressions of the form p, p-2*q, p-q, for all q >= 0.

Crossrefs

Formula

a(n)=1 iff n in A003278.

A373052 Lexicographically earliest sequence of positive integers such that no three terms a(j), a(j+k), a(j+2k) (for any j and k) form a weakly decreasing arithmetic progression.

Original entry on oeis.org

1, 1, 2, 1, 1, 2, 2, 3, 3, 1, 1, 2, 1, 1, 2, 2, 3, 3, 2, 3, 3, 4, 4, 5, 4, 4, 5, 1, 3, 2, 4, 1, 1, 2, 1, 3, 2, 4, 2, 5, 1, 2, 2, 1, 3, 3, 4, 3, 4, 5, 2, 4, 3, 5, 5, 6, 3, 4, 3, 6, 4, 4, 5, 5, 4, 5, 5, 6, 6, 7, 6, 6, 7, 7, 5, 8, 6, 8, 6, 7, 7, 2, 7, 7, 2, 8
Offset: 1

Views

Author

Neal Gersh Tolunsky, May 20 2024

Keywords

Crossrefs

Programs

  • PARI
    \\ See Links section.

A373111 Lexicographically earliest sequence of positive integers such that no three terms a(j), a(j+k), a(j+2k) (for any j and k) form a progression of the form c, c+2d, c+d, where d >= 0.

Original entry on oeis.org

1, 1, 2, 1, 1, 2, 2, 3, 3, 1, 1, 3, 1, 1, 4, 4, 3, 2, 3, 3, 4, 4, 5, 4, 4, 5, 5, 1, 1, 4, 1, 1, 5, 5, 6, 5, 1, 1, 6, 1, 1, 2, 2, 5, 2, 2, 5, 6, 6, 7, 2, 2, 7, 2, 2, 8, 7, 6, 5, 6, 8, 8, 9, 5, 6, 9, 8, 9, 2, 2, 7, 3, 2, 8, 2, 3, 8, 7, 3, 7, 4, 1, 1, 6, 1, 1, 9, 8
Offset: 1

Views

Author

Neal Gersh Tolunsky, May 25 2024

Keywords

Comments

This sequence avoids one of the six permutations of a set of three integers in arithmetic progression. For example, the set {1,2,3} can be ordered as tuples (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), and (3, 2, 1). In this sequence, we avoid (1,3,2) and other progressions of the form c, c+2d, c+d, for all d >= 0.

Crossrefs

Formula

a(n)=1 iff n in A003278.

A381597 Lexicographically earliest sequence of positive integers such that for any t and k, with k>=1, where t = a(n) = a(n+k) = a(n+2*k), only one occurrence of k, for a given t, appears anywhere in the sequence.

Original entry on oeis.org

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

Views

Author

Scott R. Shannon, Mar 01 2025

Keywords

Comments

See A381599 for the index where n first appears, and A381598 for the index where three consecutive n's appears.

Examples

			a(1) = a(2) = a(3) = 1, which is the first appearance of three 1's separated by one term.
a(4) = 2 as 1 cannot be chosen as that would form a(2) = a(3) = a(4) = 1, but three 1's separated by one term has already appeared.
a(5) = 1, which also forms three 1's separated by two terms, a(1) = a(3) = a(5) = 1.
a(17) = 3 as 1 cannot be chosen as that would form a(15) = a(16) = a(17) = 1, but three 1's separated by one term has already appeared, while choosing 2 would form a(11) = a(14) = a(17) = 2, but three 2's separated by three terms has already appeared at a(4) = a(7) = a(10) = 2.
		

Crossrefs

Cf. A381598 (triplets), A381599 (where n first appears), A370708 (indices where 1's appear), A281511, A229037.

A364057 Lexicographically earliest infinite sequence of positive integers such that every subsequence {a(j), a(j+k), a(j+2k)} (j, k >= 1) is unique.

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 2, 4, 5, 6, 7, 8, 5, 1, 2, 9, 3, 4, 6, 7, 1, 10, 11, 12, 13, 8, 14, 15, 16, 3, 17, 9, 18, 4, 7, 19, 5, 2, 11, 12, 20, 6, 1, 8, 21, 22, 9, 23, 24, 13, 14, 3, 10, 16, 17, 25, 26, 19, 27, 6, 28, 11, 15, 20, 22, 29, 12, 21, 16, 23, 30, 18, 31, 32
Offset: 1

Views

Author

Samuel Harkness, Oct 19 2023

Keywords

Comments

To find a(n), two criteria must be satisfied:
1. Every subsequence {a(n-2k), a(n-k) a(n)} created by a(n) must be unique.
2. a(n) cannot create the scenario where a future a(m) will create multiple {a(m-2k), a(m-k), a(m)} regardless of choice for a(m). The first time this is the sole reason a candidate is denied is at a(10), see Example below.
Will every subsequence of 3 positive integers appear in arithmetic progression in this sequence?
Will every positive integer occur infinitely many times?
For n >= 3, a(n) != a(n+1).
In the 74 initially published terms, numbers on average seem to reoccur at (very) roughly twice the index of their previous occurrence. This seems worthy of better quantification when further terms are established. - Peter Munn, Nov 03 2023

Examples

			For a(9), we first try 1. If a(9) were 1, {a(3), a(6), a(9)} would be {1, 1, 1}, but this already occurred at {a(1), a(2), a(3)}.
Next, try 2. If a(9) were 2, {a(3), a(6), a(9)} would be {1, 1, 2}, but this already occurred at {a(2), a(3), a(4)}.
Next, try 3. If a(9) were 3, {a(3), a(6), a(9)} would be {1, 1, 3}, but this already occurred at {a(1), a(3), a(5)}.
Next, try 4. If a(9) were 4, {a(1), a(5), a(9)} would be {1, 3, 4}, but this already occurred at {a(2), a(5), a(8)}.
Then, try 5. New subsequences at indices {a(1), a(5), a(9)} = {1, 3, 5}, {a(3), a(6), a(9)} = {1, 1, 5}, {a(5), a(7), a(9)} = {3, 2, 5}, and {a(7), a(8), a(9)} = {2, 4, 5} are formed, none of which have occurred at any {a(j), a(j+k), a(j+2k)} (for any j and k) previously. No 5 has occurred previously, so criteria (2) in Comments must be satisfied. Thus a(9) = 5.
a(10) is the first time a candidate is denied solely because it would create a guaranteed future duplicate. Note that no subsequences prevent a(10) from being 4.
n    = 1  2  3  4  5  6  7  8  9 10 11 12 13 14
a(n) = 1  1  1  2  3  1  2  4  5 [4]          X
                      |           |           |
          |                 |                 |
If a(10) were 4, {a(2), a(8), a(14)} = {a(6), a(10), a(14)} = {1, 4, X}, making a subsequence {a(j), a(j+k), a(j+2k)} which is not unique. Therefore a(10) != 4.
		

Crossrefs

Programs

  • MATLAB
    See Links section.
    (C++) See Links section.

A248627 Index at which the first n appears in A248625 = least nonnegative sequence with no AP (a(n), a(n+d), a(n+2d)).

Original entry on oeis.org

0, 2, 94, 7, 21, 120, 143, 23, 26, 59, 66, 72, 65, 78, 162, 195, 147, 149, 219, 79, 180, 172, 177, 196, 212, 202, 193, 201, 260, 373, 303, 386, 644, 294, 446, 378, 289, 419, 361, 505, 514, 877, 519, 835, 940, 494, 593, 753, 883, 957, 500, 484, 560, 406, 466
Offset: 0

Views

Author

M. F. Hasler, Oct 10 2014

Keywords

Crossrefs

Formula

a(n) = A241752(n)-1.

A276205 a(0) = a(2) = a(3) = 0. For n>2 a(n) is the smallest nonnegative integer such that there is no arithmetic progression j,k,m,n (of length 4) such that a(j)+a(k)+a(m) = a(n).

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 2, 0, 0, 0, 1, 2, 1, 3, 0, 0, 0, 4, 0, 1, 2, 2, 3, 1, 4, 0, 0, 1, 0, 0, 0, 5, 3, 0, 7, 1, 0, 4, 2, 4, 2, 3, 5, 1, 1, 4, 1, 3, 0, 0, 0, 1, 0, 0, 2, 0, 0, 0, 9, 2, 8, 10, 0, 4, 0, 0, 0, 2, 1, 7, 13, 4, 12, 4, 6, 7, 4, 4, 2, 0, 10, 2, 2, 1, 3, 1, 0, 0, 0, 12, 0, 9, 1, 0, 5, 2, 1, 17, 0, 3, 5, 0, 1, 1, 0, 0, 8, 3, 0, 0, 0, 15, 12, 9, 10, 11, 1, 5
Offset: 0

Views

Author

Michal Urbanski, Aug 24 2016

Keywords

Comments

This sequence, unlike A276204 (defined similarly) is seemingly irregular.
a(n) <= n/3. - Robert Israel, Aug 24 2016
The graph (and the definition) are reminiscent of A229037. - N. J. A. Sloane, Aug 29 2016

Examples

			For n = 6 we have that:
a(6)>0, because a(0)+a(2)+a(4)=0 and 0,2,4,6 is an arithmetic progression.
a(6)>1, because a(3)+a(4)+a(5)=1 and 3,4,5,6 is an arithmetic progression.
there is no such arithmetic progression j,k,m,6 that a(j)+a(k)+a(m)=2, so a(6) = 2.
		

Crossrefs

Cf. A276204 (length 3), A276206 (length 5), A276207 (any length).
Cf. also A229037.

Programs

  • Maple
    for i from 0 to 2 do A[i]:= 0 od:
    for n from 3 to 200 do
      Forbid:= {seq(A[n-d]+A[n-2*d]+A[n-3*d],d=1..floor(n/3))};
      A[n]:= min({$0..max(Forbid)+1} minus Forbid)
    od:
    seq(A[i],i=0..200); # Robert Israel, Aug 24 2016
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