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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A034757 a(1)=1, a(n) = smallest odd number such that all sums of pairs of (not necessarily distinct) terms in the sequence are distinct.

Original entry on oeis.org

1, 3, 7, 15, 25, 41, 61, 89, 131, 161, 193, 245, 295, 363, 407, 503, 579, 721, 801, 949, 1129, 1185, 1323, 1549, 1643, 1831, 1939, 2031, 2317, 2623, 2789, 3045, 3143, 3641, 3791, 4057, 4507, 4757, 5019, 5559, 5849, 6309, 6707, 7181, 7593
Offset: 1

Views

Author

Wouter Meeussen, Jun 01 2000

Keywords

Comments

a(1) = 1, a(n) = least number such that every difference a(i)-a(j) is a distinct even number. - Amarnath Murthy, Apr 07 2004

Examples

			5 is not in the sequence since 5+1 is already obtainable from 3+3, 9 is excluded since 1, 3 and 7 are in the sequence and would collide with 1+9
		

Crossrefs

Partial sums of A287178.

Programs

  • Haskell
    a034757 = (subtract 1) . (* 2) . a005282  -- Reinhard Zumkeller, Dec 18 2012
    
  • Mathematica
    seq2={1, 3}; Do[le=Length[seq2]; t=Last[seq2]+2; While[Length[Expand[(Plus @@ (x^seq2) + x^t)^2]] < Pochhammer[3, le]/le!, t=t+2]; AppendTo[seq2, t], {20}]; Print@seq2
  • Python
    from itertools import count, islice
    def A034757_gen(): # generator of terms
        aset1, aset2, alist = set(), set(), []
        for k in count(1,2):
            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.update(bset2)
    A034757_list = list(islice(A034757_gen(),30)) # Chai Wah Wu, Sep 05 2023

Formula

a(n) = 2*A005282(n)-1. (David Wasserman)

Extensions

An incorrect comment from Amarnath Murthy, also dated Apr 07 2004, has been deleted.
Offset fixed by Reinhard Zumkeller, Dec 18 2012

A062065 a(1) = 1; for n >= 1, a(n+1) is smallest number such that the sums of any one, two or three of a(1), ..., a(n) are distinct (repetitions not allowed).

Original entry on oeis.org

1, 2, 4, 8, 15, 28, 52, 96, 165, 278, 460, 663, 980, 1332, 1864, 2609, 3375, 4769, 5600, 6776, 9141, 11505, 14453, 17404, 21904, 25023, 31159, 35006, 42780, 51792, 55799, 68834, 75036, 87163, 96746, 116231, 128924, 144085, 172606, 193507, 207826
Offset: 1

Views

Author

Olivier Gérard, Jun 26 2001

Keywords

Examples

			1,2,1+2 are different so a(2) = 2; 1,2,3,1+2,1+3,2+3,1+2+3 are not all different (3 = 1+2) so a(3) is not 3; 1,2,4,1+2,1+4,2+4,1+2+4 are all different so a(3) = 4.
		

Crossrefs

Programs

  • PARI
    {unique(v)=local(b); b=1; for(j=2,length(v),if(v[j-1]==v[j],b=0)); b}
    {news(v,q)=local(s); s=[]; for(i=1,length(v),s=concat(s,v[i]+q)); s}
    {m=210000; print1(p=1,","); w1=[p]; w2=[]; w3=[]; q=p+1; while(qKlaus Brockhaus, May 17 2003
    
  • Python
    from itertools import count, islice
    def A062065_gen(): # generator of terms
        aset2, aset3, alist = set(), set(), [1]
        yield 1
        for k in count(2):
            bset2, bset3 = set(), set()
            if not (k in aset2 or k in aset3):
                for a in alist:
                    if (b2:=a+k) in aset2 or b2 in aset3:
                        break
                    bset2.add(b2)
                else:
                    for a2 in aset2:
                        if (b3:=a2+k) in aset2 or b3 in aset3:
                            break
                        bset3.add(b3)
                    else:
                        yield k
                        alist.append(k)
                        aset2.update(bset2)
                        aset3.update(bset3)
    A062065_list = list(islice(A062065_gen(),20)) # Chai Wah Wu, Sep 10 2023

Extensions

More terms from Naohiro Nomoto, Oct 07 2001
Terms a(27) to a(41) from Klaus Brockhaus, May 17 2003

A060276 a(1) = 2; a(n) = smallest prime > a(n-1) such that the sum of any three nondecreasing terms, chosen from a(1), ..., a(n-1) and a(n), is unique.

Original entry on oeis.org

2, 3, 7, 19, 59, 73, 211, 257, 631, 919, 1291, 1979, 3229, 4397, 5557, 7151, 10657, 12049, 17827, 19577, 25919, 32143, 35951, 46141, 54499, 64433, 81199, 92507, 116009, 132511, 145303, 171763, 193679, 232417, 260549, 289573, 302009, 340111, 424967, 465151, 506507
Offset: 1

Views

Author

Naohiro Nomoto, Mar 23 2001

Keywords

Examples

			For {2,3,5} the sums are not unique: 2+2+5 = 3+3+3. Three terms chosen from {2,3,7} can be 2+2+2; 2+2+3; 2+3+3; 3+3+3; 2+2+7; 2+3+7; 3+3+7; 2+7+7; 3+7+7; 7+7+7; the sums are all distinct, so a(3) = 7.
		

Crossrefs

Cf. A051912.

Programs

  • PARI
    {unique(v)=local(b); b=1; for(j=2,length(v),if(v[j-1]==v[j],b=0)); b}
    {news(v,q)=local(s); s=[]; for(i=1,length(v),s=concat(s,v[i]+q)); s}
    {m=310000; print1(p=2,","); w1=[p]; w2=[p+p]; w3=[p+p+p]; q=nextprime(p+1); while(q
    				
  • Python
    from itertools import count, islice
    from sympy import nextprime
    def A060276_gen(): # generator of terms
        aset1, aset2, aset3, alist, k = set(), set(), set(), [], 2
        while True:
            bset2, bset3 = {k<<1}, {3*k}
            if 3*k not in aset3:
                for d in aset1:
                    if (m:=d+(k<<1)) in aset3:
                        break
                    bset2.add(d+k)
                    bset3.add(m)
                else:
                    for d in aset2:
                        if (m:=d+k) in aset3:
                            break
                        bset3.add(m)
                    else:
                        yield k
                        alist.append(k)
                        aset1.add(k)
                        aset2.update(bset2)
                        aset3.update(bset3)
            k = nextprime(k)
    A060276_list = list(islice(A060276_gen(),40)) # Chai Wah Wu, Sep 05 2023

Extensions

Edited and extended by Klaus Brockhaus, May 16 2003
More terms from Chai Wah Wu, Sep 05 2023

A096772 A B3-sequence: a(1) = 1; for n>1, a(n) = smallest number > a(n-1) such that the sums of any three terms are all distinct.

Original entry on oeis.org

1, 2, 5, 14, 33, 72, 125, 219, 376, 573, 745, 1209, 1557, 2442, 3098, 4048, 5298, 6704, 7839, 10987, 12332, 15465, 19144, 24546, 28974, 34406, 37769, 45864, 50877, 61372, 68303, 77918, 88545, 101917, 122032, 131625, 148575, 171237, 197815, 201454
Offset: 1

Views

Author

Rick L. Shepherd, Aug 15 2004

Keywords

Comments

This is the B3-sequence analog of the Mian-Chowla B2-sequence (A005282): Let a(1)=1; then use the greedy algorithm to choose the smallest a(n) > a(n-1) such that all sums a(i) + a(j) + a(k) are distinct for 1 <= i <= j <= k <= n. The reciprocal sum of the sequence for the first forty terms is 1.837412....

Crossrefs

Row 3 of A347570.
Cf. A005282 (Mian-Chowla B2-sequence). A051912.

Programs

  • Python
    from itertools import count, islice
    def A096772_gen(): # generator of terms
        aset1, aset2, aset3, alist = set(), set(), set(), []
        for k in count(1):
            bset2, bset3 = {k<<1}, {3*k}
            if 3*k not in aset3:
                for d in aset1:
                    if (m:=d+(k<<1)) in aset3:
                        break
                    bset2.add(d+k)
                    bset3.add(m)
                else:
                    for d in aset2:
                        if (m:=d+k) in aset3:
                            break
                        bset3.add(m)
                    else:
                        yield k
                        alist.append(k)
                        aset1.add(k)
                        aset2 |= bset2
                        aset3 |= bset3
    A096772_list = list(islice(A096772_gen(),30)) # Chai Wah Wu, Sep 05 2023

Formula

a(n) = A051912(n-1) + 1. - Peter Kagey, Oct 20 2021
Previous Showing 11-14 of 14 results.