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 1-6 of 6 results.

A005282 Mian-Chowla sequence (a B_2 sequence): a(1) = 1; for n>1, a(n) = smallest number > a(n-1) such that the pairwise sums of elements are all distinct.

Original entry on oeis.org

1, 2, 4, 8, 13, 21, 31, 45, 66, 81, 97, 123, 148, 182, 204, 252, 290, 361, 401, 475, 565, 593, 662, 775, 822, 916, 970, 1016, 1159, 1312, 1395, 1523, 1572, 1821, 1896, 2029, 2254, 2379, 2510, 2780, 2925, 3155, 3354, 3591, 3797, 3998, 4297, 4433, 4779, 4851
Offset: 1

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Author

Keywords

Comments

An alternative definition is to start with 1 and then continue with the least number such that all pairwise differences of distinct elements are all distinct. - Jens Voß, Feb 04 2003. [However, compare A003022 and A227590. - N. J. A. Sloane, Apr 08 2016]
Rachel Lewis points out [see link] that S, the sum of the reciprocals of this sequence, satisfies 2.158435 <= S <= 2.158677. Similarly, the sum of the squares of reciprocals of this sequence converges to approximately 1.33853369 and the sum of the cube of reciprocals of this sequence converges to approximately 1.14319352. - Jonathan Vos Post, Nov 21 2004; comment changed by N. J. A. Sloane, Jan 02 2020
Let S denote the reciprocal sum of a(n). Then 2.158452685 <= S <= 2.158532684. - Raffaele Salvia, Jul 19 2014
From Thomas Ordowski, Sep 19 2014: (Start)
Known estimate: n^2/2 + O(n) < a(n) < n^3/6 + O(n^2).
Conjecture: a(n) ~ n^3 / log(n)^2. (End)

Examples

			The second term is 2 because the 3 pairwise sums 1+1=2, 1+2=3, 2+2=4 are all distinct.
The third term cannot be 3 because 1+3 = 2+2. But it can be 4, since 1+4=5, 2+4=6, 4+4=8 are distinct and distinct from the earlier sums 1+1=2, 1+2=3, 2+2=4.
		

References

  • P. Erdős and R. Graham, Old and new problems and results in combinatorial number theory. Monographies de L'Enseignement Mathématique (1980).
  • S. R. Finch, Mathematical Constants, Cambridge, 2003, Section 2.20.2.
  • R. K. Guy, Unsolved Problems in Number Theory, E28.
  • A. M. Mian and S. D. Chowla, On the B_2-sequences of Sidon, Proc. Nat. Acad. Sci. India, A14 (1944), 3-4.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Row 2 of A347570.
Cf. A051788, A080200 (for differences between terms).
Different from A046185. Cf. A011185.
See also A003022, A227590.
A259964 has a greater sum of reciprocals.
Cf. A002858.

Programs

  • Haskell
    import Data.Set (Set, empty, insert, member)
    a005282 n = a005282_list !! (n-1)
    a005282_list = sMianChowla [] 1 empty where
       sMianChowla :: [Integer] -> Integer -> Set Integer -> [Integer]
       sMianChowla sums z s | s' == empty = sMianChowla sums (z+1) s
                            | otherwise   = z : sMianChowla (z:sums) (z+1) s
          where s' = try (z:sums) s
                try :: [Integer] -> Set Integer -> Set Integer
                try []     s                      = s
                try (x:sums) s | (z+x) `member` s = empty
                               | otherwise        = try sums $ insert (z+x) s
    -- Reinhard Zumkeller, Mar 02 2011
    
  • Maple
    a[1]:= 1: P:= {2}: A:= {1}:
    for n from 2 to 100 do
      for t from a[n-1]+1 do
        Pt:= map(`+`,A union {t},t);
        if Pt intersect P = {} then break fi
      od:
      a[n]:= t;
      A:= A union {t};
      P:= P union Pt;
    od:
    seq(a[n],n=1..100); # Robert Israel, Sep 21 2014
  • Mathematica
    t = {1}; sms = {2}; k = 1; Do[k++; While[Intersection[sms, k + t] != {}, k++]; sms = Join[sms, t + k, {2 k}]; AppendTo[t, k], {49}]; t (* T. D. Noe, Mar 02 2011 *)
  • PARI
    A005282_vec(N, A=[1], U=[0], D(A, n=#A)=vector(n-1, k, A[n]-A[n-k]))={ while(#A2 && U=U[k-1..-1]);A} \\ M. F. Hasler, Oct 09 2019
    
  • PARI
    aupto(L)= my(S=vector(L), A=[1]); for(i=2, L, for(j=1, #A, if(S[i-A[j]], next(2))); for(j=1, #A, S[i-A[j]]=1); A=concat(A, i)); A \\ Ruud H.G. van Tol, Jun 30 2025
    
  • Python
    from itertools import count, islice
    def A005282_gen(): # generator of terms
        aset1, aset2, alist = set(), set(), []
        for k in count(1):
            bset2 = {k<<1}
            if (k<<1) not in aset2:
                for d in aset1:
                    if (m:=d+k) in aset2:
                        break
                    bset2.add(m)
                else:
                    yield k
                    alist.append(k)
                    aset1.add(k)
                    aset2 |= bset2
    A005282_list = list(islice(A005282_gen(),30)) # Chai Wah Wu, Sep 05 2023

Formula

a(n) = A025582(n) + 1.
a(n) = (A034757(n)+1)/2.

Extensions

Examples added by N. J. A. Sloane, Jun 01 2008

A080201 Numbers that do not occur as differences between terms of the Mian-Chowla variant A051788.

Original entry on oeis.org

49, 50, 71, 72, 76, 82, 90, 93, 95, 96, 119, 128, 139, 143, 152, 162, 172, 173, 180, 182, 185, 188
Offset: 1

Views

Author

Hugo Pfoertner, Feb 05 2003

Keywords

Comments

Terms are only conjectures. Elements might be eliminated by higher terms of A051788, which was checked up to 4*10^6.

References

Crossrefs

A080222 Record-setting differences between adjacent elements of the Mian-Chowla sequence A005282.

Original entry on oeis.org

1, 2, 4, 5, 8, 10, 14, 21, 26, 34, 48, 71, 74, 90, 113, 143, 153, 249, 270, 299, 346, 453, 535, 940, 1052, 1226, 1347, 1365, 2443, 2511, 4253, 4254, 6116, 7339, 8898, 13621, 15567, 17940, 21061, 21307, 25558, 35749, 39437, 46664, 62709
Offset: 1

Views

Author

Hugo Pfoertner, Feb 07 2003

Keywords

Examples

			a(12)=71 because A005282(17)-A005282(16)=361-290=71 is greater than all previous differences. a(45)=A005282(619)-A005282(618)=3738616-3675907=62709
		

References

Crossrefs

A080932 Non-occurring pairwise differences between the elements of the Mian-Chowla sequence variant A058335.

Original entry on oeis.org

26, 33, 50, 55, 68, 93, 94, 98, 107, 108, 109, 115, 122, 138, 144, 150, 155, 163, 178, 181, 182, 183, 185, 186, 193, 196, 200, 202, 204, 208, 210, 212, 223, 227, 232, 235, 239, 242, 245, 250, 253, 257, 263, 264, 268
Offset: 1

Views

Author

Hugo Pfoertner, Feb 24 2003

Keywords

References

Crossrefs

A247556 Exact differential base (a B_2 sequence) constructed as follows: Start with a(0)=0. For n>=1, let S be the set of all differences a(j)-a(i) for 0 <= i < j <= n-1, and let d be the smallest positive integer not in S. If, for every i in 1..n-1, a(n-1) + d - a(i) is not in S, then a(n) = a(n-1) + d. Otherwise, let r be the smallest positive integer such that, for every i in 1..n-1, neither a(n-1) + r - a(i) nor a(n-1) + r + d - a(i) is in S; then a(n) = a(n-1) + r and a(n+1) = a(n) + d.

Original entry on oeis.org

0, 1, 3, 7, 12, 20, 30, 44, 65, 80, 96, 143, 165, 199, 224, 306, 332, 415, 443, 591, 624, 678, 716, 934, 973, 1134, 1174, 1449, 1491, 1674, 1720, 2113, 2161, 2468, 2517, 2855, 2906, 2961, 3245, 3302, 3711, 3772, 4081, 4148, 4603, 4673, 5557, 5628, 5917, 5989
Offset: 0

Views

Author

Thomas Ordowski, Sep 19 2014

Keywords

Comments

Every positive integer is uniquely represented as a difference of two distinct elements of the base set. This is a B_2 sequence.
By the definition of this sequence, with d as the smallest unused difference among terms a(0)..a(n-1), we assign a(n) = a(n-1) + d, provided that this would not cause any difference to be repeated; otherwise, we assign a(n) = a(n-1) + r and a(n+1) = a(n) + d, where r is the smallest integer that allows this assignment of a(n) and a(n+1) without causing any difference to be repeated. Thus, at each step, the smallest unused difference d is either used immediately (as a(n) - a(n-1)) or delayed by one step (and used as a(n+1) - a(n)). In this way, the sequence includes every positive integer as a difference (unlike the Mian-Chowla sequence A005282, which omits differences 33, 88, 98, 99, ...; see A080200).
The set is an optimization of Browkin's base, where r = a(n-1) + 1.
The series Sum_{n>=0} 1/(a(n+1) - a(n)) is divergent.
Conjecture: lim inf_{n->oo} (a(n+1) - a(n))/n = 1/2.

Examples

			Given a(0)=0, a(1)=1, a(2)=3, a(3)=7, the differences used are 1,2,3,4,6,7, so d=5, and we can use a(4) = a(3)+d = 7+5 = 12 because appending a(4)=12 to the sequence will result in the differences 12-0=12, 12-1=11, 12-3=9, 12-7=5, none of which had already been used.
Similarly, given a(0)..a(4) = 0,1,3,7,12, the differences used are 1..7,9,11,12, so d=8, and we can use a(5) = a(4)+d = 12+8 = 20 because the resulting differences will be 20, 19, 17, 13, 8, none of which had already been used.
Proceeding as above, we get a(6)=30 and a(7)=44.
Given a(0)..a(7) = 0,1,3,7,12,20,30,44, the differences used are 1..14,17..20,23..24,27,29..30,32,37,41,43..44, so d=15, but we cannot use a(8) = a(7)+d = 44+15 = 59 because the difference 29 would be repeated: 59-30 = 30-1. Thus, we must find the smallest r such that using both a(8) = a(7)+r and a(9) = a(8)+d will not repeat any differences. The smallest such r is 21, so a(8) = a(7)+r = 44+21 = 65 and a(9) = a(8)+d = 65+15 = 80.
		

References

  • Jerzy Browkin, Rozwiązanie pewnego zagadnienia A. Schinzla (Polish) [The solution of a certain problem of A. Schinzel], Roczniki Polskiego Towarzystwa Matematycznego [Annals Polish Mathematical Society], Seria I, Prace Matematyczne III (1959).

Crossrefs

Cf. A001856, where a(1)=1, a(2)=2, a(2n+1)=2*a(2n), a(2n+2) = a(2n+1) + d.
Cf. A005282 (Mian-Chowla sequence), A025582.
Cf. A080200.

Formula

a(n) >= A025582(n+1) and for n <= 10 is here equality.
Conjecture: a(n) ~ log(log(n))*A025582(n+1), where A025582(m)+1 = A005282(m) is the Mian-Chowla sequence.

Extensions

More terms from Jon E. Schoenfield, Jan 18 2015
Edited by Jon E. Schoenfield, Jan 22 2015

A080933 Smallest non-occurring pairwise difference between the elements of a Mian-Chowla sequence (A005282) variant starting with (1,n).

Original entry on oeis.org

33, 49, 26, 15, 30, 19, 44, 13, 38, 50, 54, 58, 44, 46, 25, 20, 45, 10, 13, 84, 38, 15, 71, 33, 35, 54, 31, 16, 57, 10, 42, 26, 15, 14, 33, 14, 15, 32, 34, 16, 25, 28, 25, 16, 36, 16, 16, 25, 28, 40, 16, 31, 33, 28, 15, 31, 15, 22, 31, 33, 15, 21, 49, 51, 28
Offset: 2

Views

Author

Hugo Pfoertner, Feb 24 2003

Keywords

Comments

For large n the sequence consists of increasingly long runs of "33", which is the first term of A080200 (the smallest non-occurring difference in A005282), with some interspersed other terms. E.g. for n=100000..100100 we have a(n)=88 for n=100005,100021,100041,100087,100099, a(100072)=98 and otherwise a(n)=33.

Examples

			a(2)=33 because the smallest non-occurring pairwise difference between the terms of A005282 (starting with 1,2) is A080200(1)=33. a(3)=49 because the smallest non-occurring pairwise difference between the terms of A051788 (starting with 1,3) is A080201(1)=49. a(4)=26 because the smallest non-occurring pairwise difference between the terms of A058335 (starting with 1,4) is A080932(1)=26.
		

References

Crossrefs

Showing 1-6 of 6 results.