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 1-10 of 10 results.

A167809 Number of admissible bases in the postage stamp problem for n denominations and h = 2 stamps.

Original entry on oeis.org

1, 2, 5, 17, 65, 292, 1434, 7875, 47098, 305226, 2122983, 15752080, 124015310, 1031857395, 9041908204, 83186138212, 801235247145, 8059220936672, 84463182889321
Offset: 1

Views

Author

Yogy Namara (yogy.namara(AT)gmail.com), Nov 12 2009

Keywords

Comments

A basis 1 = b_1 < b_2 ... < b_n is admissible if all the values 1 <= x <= b_n are obtainable as a sum of at most h (not necessarily distinct) numbers in the basis.
Conjecture: a(n) >= A000108(n). - Michael Chu, May 16 2022

References

  • R. K. Guy, Unsolved Problems in Number Theory, C12.

Links

Crossrefs

Other enumerations with different parameters: A167809 (h = 2), A167810 (h = 3), A167811 (h = 4), A167812 (h = 5), A167813 (h = 6), A167814 (h = 7).
For h = 2, cf. A008932.

Extensions

a(17) from simple depth-first search by Jukka Kohonen, Jun 16 2016
a(18)-a(19) from depth-first search by Jukka Kohonen, Jul 30 2016

A167810 Number of admissible basis in the postage stamp problem for n denominations and h = 3 stamps.

Original entry on oeis.org

1, 3, 13, 86, 760, 8518, 116278, 1911198, 37063964, 835779524, 21626042510, 635611172160, 21033034941826, 777710150809009
Offset: 1

Views

Author

Yogy Namara (yogy.namara(AT)gmail.com), Nov 12 2009

Keywords

Comments

A basis 1 = b_1 < b_2 ... < b_n is admissible if all the values 1 <= x <= b_n is obtainable as a sum of at most h (not necessarily distinct) numbers in the basis.

References

  • R. K. Guy, Unsolved Problems in Number Theory, C12.

Links

Crossrefs

Other enumerations with different parameters: A167809 (h = 2), A167810 (h = 3), A167811 (h = 4), A167812 (h = 5), A167813 (h = 6), A167814 (h = 7).
For h = 2, cf. A008932.
A152112 is essentially the same sequence by definition. [From Herbert Kociemba, Jul 14 2010]

Extensions

Terms a(1) to a(12) verified and new terms a(13) and a(14) added by Herbert Kociemba, Jul 14 2010

A167811 Number of admissible basis in the postage stamp problem for n denominations and h = 4 stamps.

Original entry on oeis.org

1, 4, 26, 291, 4752, 109640, 3380466, 136053274, 6963328612, 444765731559
Offset: 1

Views

Author

Yogy Namara (yogy.namara(AT)gmail.com), Nov 12 2009

Keywords

Comments

A basis 1 = b_1 < b_2 ... < b_n is admissible if all the values 1 <= x <= b_n is obtainable as a sum of at most h (not necessarily distinct) numbers in the basis.

References

  • R. K. Guy, Unsolved Problems in Number Theory, C12.

Links

Crossrefs

Other enumerations with different parameters: A167809 (h = 2), A167810 (h = 3), A167811 (h = 4), A167812 (h = 5), A167813 (h = 6), A167814 (h = 7).
For h = 2, cf. A008932.

A167812 Number of admissible basis in the postage stamp problem for n denominations and h = 5 stamps.

Original entry on oeis.org

1, 5, 45, 750, 20881, 880325, 54329413, 4727396109, 563302698378
Offset: 1

Views

Author

Yogy Namara (yogy.namara(AT)gmail.com), Nov 12 2009

Keywords

Comments

A basis 1 = b_1 < b_2 ... < b_n is admissible if all the values 1 <= x <= b_n is obtainable as a sum of at most h (not necessarily distinct) numbers in the basis.

References

  • R. K. Guy, Unsolved Problems in Number Theory, C12.

Links

Crossrefs

Other enumerations with different parameters: A167809 (h = 2), A167810 (h = 3), A167811 (h = 4), A167812 (h = 5), A167813 (h = 6), A167814 (h = 7).
For h = 2, cf. A008932.

A167813 Number of admissible basis in the postage stamp problem for n denominations and h = 6 stamps.

Original entry on oeis.org

1, 6, 71, 1694, 73126, 5235791, 593539539, 102141195784
Offset: 1

Views

Author

Yogy Namara (yogy.namara(AT)gmail.com), Nov 12 2009

Keywords

Comments

A basis 1 = b_1 < b_2 ... < b_n is admissible if all the values 1 <= x <= b_n is obtainable as a sum of at most h (not necessarily distinct) numbers in the basis.

References

  • R. K. Guy, Unsolved Problems in Number Theory, C12.

Links

Crossrefs

Other enumerations with different parameters: A167809 (h = 2), A167810 (h = 3), A167811 (h = 4), A167812 (h = 5), A167813 (h = 6), A167814 (h = 7).
For h = 2, cf. A008932.

A167814 Number of admissible basis in the postage stamp problem for n denominations and h = 7 stamps.

Original entry on oeis.org

1, 7, 105, 3407, 217997, 24929035, 4863045067
Offset: 1

Views

Author

Yogy Namara (yogy.namara(AT)gmail.com), Nov 12 2009

Keywords

Comments

A basis 1 = b_1 < b_2 ... < b_n is admissible if all the values 1 <= x <= b_n is obtainable as a sum of at most h (not necessarily distinct) numbers in the basis.

References

  • R. K. Guy, Unsolved Problems in Number Theory, C12.

Links

Crossrefs

Other enumerations with different parameters: A167809 (h = 2), A167810 (h = 3), A167811 (h = 4), A167812 (h = 5), A167813 (h = 6), A167814 (h = 7).
For h = 2, cf. A008932.

A123509 Rohrbach's problem: a(n) is the largest integer such that there exists a set of n integers that is a basis of order 2 for (0, 1, ..., a(n)-1).

Original entry on oeis.org

1, 3, 5, 9, 13, 17, 21, 27, 33, 41, 47, 55, 65, 73, 81, 93, 105, 117, 129, 141, 153, 165, 181, 197, 213
Offset: 1

Views

Author

Warren D. Smith, Oct 02 2006

Keywords

Comments

Notation: N[q] = the set of q+1 elements inside {0,1,...,N-1}
Length of the longest sequence of consecutive integers that can be obtained from a set of n distinct integers by summing any two integers in the set or doubling any one. - Jon E. Schoenfield, Jul 16 2017
According to Zhining Yang, Jul 08 2017, a(13) to a(20) are 65, 70, 79, 90, 101, 112, 123, 134, but there is some doubt about these terms, and they should be confirmed before they are accepted. They do not agree with the conjecture, so perhaps the VBA program is not correct.
The definition of Rohrbach's Problem in the paper of S. Gunturk and M. B. Nathanson in the links is different from the one here. In the paper, the set should contain n nonnegative integers instead of integers. The result should be equal to A001212(n-1)+1 according to the definition in the paper since adding one 0 before any set for A001212(n-1) provides a set of the problem. The data provided by Zhining Yang is obviously wrong since a(n) >= A001212(n-1)+1. And A302648 provides another lower bound of this array since a(n) >= 2*A302648(n)+1. - Zhao Hui Du, Apr 13 2018

Examples

			Example: 8[3]: 0,1,3,4 means {0,1,2,...,8} is covered thus: 0=0+0, 1=0+1, 2=1+1, 3=0+3, 4=0+4=1+3, 5=1+4, 6=3+3, 7=3+4, 8=4+4.
N[q]: set
------------------------------
3[2]: 0,1,
4[3]: 0,1,2,
5[3]: 0,1,2,
6[3]: 0,2,3,
7[4]: 0,1,2,3,
8[4]: 0,1,3,4,
9[4]: 0,1,3,4,
10[5]: 0,1,2,4,5,
11[5]: 0,1,2,4,5,
12[5]: 0,1,3,5,6,
13[5]: 0,1,3,5,6,
14[6]: 0,1,2,4,6,7,
15[6]: 0,1,2,4,6,7,
16[6]: 0,1,3,5,7,8,
17[6]: 0,1,3,5,7,8,
18[6]: 0,2,3,7,8,10,
19[7]: 0,1,2,4,6,8,9,
20[7]: 0,1,3,5,7,9,10,
21[7]: 0,1,3,5,7,9,10,
22[7]: 0,2,3,7,8,10,11,
23[8]: 0,1,2,4,6,8,10,11,
24[8]: 0,1,3,5,7,9,11,12,
25[8]: 0,1,3,5,7,9,11,12,
26[8]: 0,2,3,7,8,10,12,13,
27[8]: 0,1,3,4,9,10,12,13,
28[8]: 0,2,3,7,8,12,13,15,
29[9]: 0,1,3,5,7,9,11,13,14,
30[9]: 0,2,3,7,8,10,12,14,15,
31[9]: 0,1,3,4,9,10,12,14,15,
32[9]: 0,2,3,7,8,12,13,15,16,
a(5)=13 because we can obtain at most a total of 13 consecutive integers from a set of 5 integers by summing any two integers in the set or doubling any one; from the 5-integer set {1,2,4,6,7}, we can obtain all 13 integers in the interval [2..14] as follows: 2=1+1, 3=1+2, 4=2+2, 5=1+4, 6=2+4, 7=1+6, 8=2+6, 9=2+7, 10=4+6, 11=4+7, 12=6+6, 13=6+7, 14=7+7.
a(16)=90 because we can obtain at most a total of 90 consecutive integers from a set of 16 integers by summing any two integers in the set or doubling any one: from the 16-integer set {1,2,4,5,8,9,10,17,18,22,25,36,47,58,69,80}, we can obtain all 90 integers in the interval [2..91]. - _Jon E. Schoenfield_, Jul 16 2017

Links

Crossrefs

Cf. A001212, A008932.

Formula

a(n) = A001212(n-1)+1 (conjecture). - R. J. Mathar, Oct 08 2006. Comment from Martin Fuller, Mar 18 2009: I agree with this conjecture.
lim inf a(n) / n^2 > 0.2857 lim sup a(n) / n^2 < 0.4789 - Charles R Greathouse IV, Aug 11 2007

Extensions

More terms (from Smith's web site) from R. J. Mathar, Oct 08 2006
Entry revised by N. J. A. Sloane, Aug 06 2017
a(13)-a(25) from Herzog et al. added by Stefano Spezia, Jul 05 2024

A152111 An increasing basis of order 3. See Comments for full definition.

Original entry on oeis.org

0, 1, 2, 4, 8, 9, 16, 18, 32, 36, 64, 65, 72, 73, 128, 130, 144, 146, 256, 260, 288, 292, 512, 513, 520, 521, 576, 577, 584, 585, 1024, 1026, 1040, 1042, 1152, 1154, 1168, 1170, 2048, 2052, 2080, 2084, 2304, 2308, 2336, 2340, 4096, 4097, 4104, 4105, 4160
Offset: 1

Views

Author

David S. Newman, Mar 22 2009

Keywords

Comments

Using the terminology of A008932, call a set A a basis of order h if every number can be written as the sum of h (not necessarily distinct) elements of A. Call a basis an increasing basis of order h if its elements are arranged in increasing order, a0 < a1 < a2 < ...
This sequence is constructed as follows: Take the union of the following three sets: (1) the set of all nonnegative numbers which can be written in base two as sums of powers, k, of 2, where k is congruent to 0 mod 3; (2) the set of all nonnegative numbers which can be written in base two as sums of powers, k, of 2, where k is congruent to 1 mod 3; (3) the set of all nonnegative numbers which can be written in base two as sums of powers, k, of 2, where k is congruent to 2 mod 3.
Numbers of the form A033045(k), or 2*A033045(k), or 4*A033045(k). - R. J. Mathar, Sep 21 2009
There are 3*2^i - 1 terms up to 8^i. - David A. Corneth, Aug 02 2017

Links

Crossrefs

Cf. A008932, A033045, A152112, A284116.

Programs

  • Maple
    ismod3 := proc(n,m) b := convert(n,base,2) ; for i from 1+((m+1) mod 3) to nops(b) by 3 do if op(i,b) <> 0 then RETURN(false) ; fi; od: for i from 1 + ((m+2) mod 3) to nops(b) by 3 do if op(i,b) <> 0 then RETURN(false) ; fi; od: true ; end: for n from 0 to 20700 do if ismod3(n,0) or ismod3(n,1) or ismod3(n,2) then printf("%d,",n); fi; od: # R. J. Mathar, Sep 21 2009
  • PARI
    upto(n) = {my(i = 1, r, res = List()); while(1, b = binary(i); r = sum(i=1, #b, 8^i*b[#b+1-i])>>3; if(r > n, break); listput(res, r); i+=2); q = #res; for(i=1,  q, e = res[i] << 1; while(e <= n, listput(res, e); e=e<<1)); listput(res, 0); listsort(res); res} \\ David A. Corneth, Aug 02 2017

Extensions

More terms from R. J. Mathar, Sep 21 2009

A152112 Number of increasing initial sequences of bases of order 3.

Original entry on oeis.org

1, 1, 3, 13, 86, 760, 8518
Offset: 1

Views

Author

David S. Newman, Mar 22 2009

Keywords

Comments

Using the terminology of A008932, call a set A a basis of order h if every number can be written as the sum of h (not necessarily distinct) elements of A. Call a basis an increasing basis of order h if its elements are arranged in increasing order, a0
Consider the set of all initial subsequences of any length {a0, a1, a2, ..., an} of all the increasing bases. These can be ordered in lexicographic order, giving, for h = 3:
0
0,1
0,1,2
0,1,3
0,1,4

Crossrefs

Cf. A008932, A152111.

Programs

  • Mathematica
    f[A_]:=
(AAA={};
For [ii=1,ii<=Length[A],ii++,
For[jj=1,jj<=Length[A],jj++,
For [kk=1,kk<=Length[A],kk++,
AAA=Union[AAA,{A[[ii]]+A[[jj]]+A[[kk]]}]]]];
For[ii=1,ii<=Length[AAA],ii++,
If[ii==Length[AAA],max=ii-1];
If[AAA[[ii]]>ii-1,max=ii-2;Break[]]]);
index=1;
seq[1]={0,1};
rindex=1;
newindex=1;
For[k=1,k<=5,k++,
jbegin=rindex;jend=index;
For[j=jbegin,j<=jend,j++,
f[seq[j]];
For[i=Max[seq[j]]+1,i<=max+1,i++,index++;seq[index]=Append[seq[rindex],i]
];rindex=rindex+1;
]]
For[i=1,i<=index,i++,Print[i," ",seq[i]]] (* David S. Newman, Dec 29 2014 *)

Extensions

a(6)-a(7) from David S. Newman, Dec 29 2014

A193258 Sprimes: A sparse prime-like set of numbers that are constructed recursively to satisfy a Goldbach-type conjecture.

Original entry on oeis.org

1, 3, 7, 11, 13, 27, 31, 35, 49, 61, 77, 79, 93, 101, 115, 117, 133, 163, 183, 187, 193, 235, 245, 257, 271, 279, 323, 335, 343, 381, 399, 439, 481, 497, 507, 535, 549, 569, 619, 669, 681, 693, 713, 739, 815, 833, 863, 905, 941, 973, 1033, 1053, 1089, 1119
Offset: 1

Views

Author

Ian R Harris, Aug 26 2011

Keywords

Comments

Closely related to A123509, and the concept of a "basis". A key difference is that the set described here can only generate even numbers, and 0 is not allowed in the set.

Examples

			a(1)=1.
Note that S1={1}, so A={1}.
Now m=min{N\A}=2.
Thus C1={3} (amongst the natural numbers only 3 can be added to 1 to give 4).
Since 3 is the only candidate, a(2)=3.
To get a(3), we repeat steps 2) to 6).
So, S2={1,3}, A={1,2,3}, m=min{N\A}=4.
Thus the candidate set is C2={5,7} (we can add 5 to 3 to get 8, or 7 to 1 to get 8).
Then w5=|({(5+1)/2, (5+3)/2} union{5}) intersect {4,5,6,7,8,9,10,...}|=|{4,5}|=2.
And w7=|({(7+1)/2, (7+3)/2} union {7}) intersect {4,5,6,7,8,9,10,11,12,13,14,...}|=|{4,5,7}|=3.
Since of the two candidates, 7 has the higher worth, then a(3)=7.

Links

Crossrefs

Cf. A000040, A008578, A000959, A123509, A126684, A008932.

Programs

  • Sage
    @cached_function
def A193258(n):
    if n == 1: return 1
    S = set(A193258(i) for i in [1..n-1])
    A = set((i+j)/2 for i, j in cartesian_product([S, S]))
    m = next(i for i in PositiveIntegers() if i not in A)
    C = set(2*m-i for i in S if 2*m-i > A193258(n-1))
    worthfn = lambda c: len(set((c+i)/2 for i in S).difference(A))
    wc = sorted(list((worthfn(c), c) for c in C)) # sort by worth and by c
    return wc[-1][1]
# D. S. McNeil, Aug 29 2011

Formula

The set of numbers S is chosen to satisfy the Goldbach conjecture. That is any even positive number must be able to be written as the sum of exactly two members of S (typically there are multiple ways to do this). The members of the set are generated by a deterministic recursive algorithm as follows (N is the set of positive integers):
1) a(1)=1
2) Given Sk={a(1),...,a(k)}, form the set A={n in N | exists a(i), a(j) in Sk, (a(i)+a(j))=2n}.
3) Let m=min{N\A}. (Then 2m is the smallest positive even number which cannot be formed from sums of the current finite list of sprimes.)
4) Define candidate set Ck={n in \N| n > a(k), exists a(i) in Sk such that a(i)+n=2m}. (This is a set of possible choices for a(k+1).)
5) To each member of Ck, assign "worth" wi=|({(i+a(j))/2| a(j) in Sk} union {i}) intersect {N\A}|. (This assigns to each candidate a worth equal to the number of new values that will be added to set A if the candidate is added to Sk.)
6) a(k+1)=max{i in Ck | wi=maximum (over {j in Ck}) (wj)}.
Repeat steps 2) to 6).
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