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 1-7 of 7 results.

A048249 Number of distinct values produced from sums and products of n unity arguments.

Original entry on oeis.org

1, 2, 3, 4, 6, 9, 11, 17, 23, 30, 44, 60, 80, 114, 156, 212, 296, 404, 556, 770, 1065, 1463, 2032, 2795, 3889, 5364, 7422, 10300, 14229, 19722, 27391, 37892, 52599, 73075, 101301, 140588, 195405, 271024, 376608, 523518, 726812, 1010576, 1405013, 1952498
Offset: 1

Views

Author

Keywords

Comments

Values listed calculated by exhaustive search algorithm.
For n+1 operands (n operations) there are (2n)!/((n!)((n+1)!)) possible postfix forms over a single operator. For each such form, there are 2^n ways to assign 2 operators (here, sum and product). Calculate results and eliminate duplicates.
Number of distinct positive integers that can be obtained by iteratively adding or multiplying together parts of an integer partition until only one part remains, starting with 1^n. - Gus Wiseman, Sep 29 2018

Examples

			a(3)=3 since (in postfix): 111** = 11*1* = 1, 111*+ = 11*1+ = 111+* = 11+1* = 2 and 111++ = 11+1+ = 3. Note that at n=7, the 11 possible values produced are the set {1,2,3,4,5,6,7,8,9,10,12}. This is the first n for which there are "skipped" values in the set.
		

Crossrefs

Programs

  • Maple
    b:= proc(n) option remember; `if`(n=1, {1}, {seq(seq(seq(
         [f+g, f*g][], g=b(n-i)), f=b(i)), i=1..iquo(n, 2))})
        end:
    a:= n-> nops(b(n)):
    seq(a(n), n=1..35);  # Alois P. Heinz, May 05 2019
  • Mathematica
    ReplaceListRepeated[forms_,rerules_]:=Union[Flatten[FixedPointList[Function[pre,Union[Flatten[ReplaceList[#,rerules]&/@pre,1]]],forms],1]];
    Table[Length[Select[ReplaceListRepeated[{Array[1&,n]},{{foe___,x_,mie___,y_,afe___}:>Sort[Append[{foe,mie,afe},x+y]],{foe___,x_,mie___,y_,afe___}:>Sort[Append[{foe,mie,afe},x*y]]}],Length[#]==1&]],{n,10}] (* Gus Wiseman, Sep 29 2018 *)
  • Python
    from functools import cache
    @cache
    def f(m):
        if m == 1: return {1}
        out = set()
        for j in range(1, m//2+1):
            for x in f(j):
                for y in f(m-j):
                    out.update([x + y, x * y])
        return out
    def a(n): return len(f(n))
    print([a(n) for n in range(1, 40)]) # Michael S. Branicky, Aug 03 2022

Formula

Equals partial sum of "number of numbers of complexity n" (A005421). - Jonathan Vos Post, Apr 07 2006

Extensions

More terms from David W. Wilson, Oct 10 2001
a(43)-a(44) from Alois P. Heinz, May 05 2019

A319850 Number of distinct positive integers that can be obtained, starting with the initial interval partition (1, ..., n), by iteratively adding or multiplying together parts until only one part remains.

Original entry on oeis.org

1, 2, 5, 21, 94, 446, 2287, 12568, 78509
Offset: 1

Views

Author

Gus Wiseman, Sep 29 2018

Keywords

Examples

			The n-th row lists all integers that can be obtained starting with (1, ..., n):
  1
  2 3
  5 6 7 8 9
  9 10 11 12 13 14 15 16 17 18 19 20 21 24 25 26 27 28 30 32 36
		

Crossrefs

Programs

  • Mathematica
    ReplaceListRepeated[forms_,rerules_]:=Union[Flatten[FixedPointList[Function[pre,Union[Flatten[ReplaceList[#,rerules]&/@pre,1]]],forms],1]];
    Table[Length[Select[ReplaceListRepeated[{Range[n]},{{foe___,x_,mie___,y_,afe___}:>Sort[Append[{foe,mie,afe},x+y]],{foe___,x_,mie___,y_,afe___}:>Sort[Append[{foe,mie,afe},x*y]]}],Length[#]==1&]],{n,6}]

A319909 Number of distinct positive integers that can be obtained by iteratively adding any two or multiplying any two non-1 parts of an integer partition until only one part remains, starting with 1^n.

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 4, 5, 10, 15, 21, 34, 49, 68, 101, 142, 197, 280, 387, 538, 751, 1045, 1442, 2010, 2772, 3865, 5339, 7396, 10273, 14201, 19693
Offset: 0

Views

Author

Gus Wiseman, Oct 01 2018

Keywords

Examples

			We have
   7 = 1+1+1+1+1+1+1,
   8 = (1+1)*(1+1+1)+1+1,
   9 = (1+1)*(1+1)*(1+1)+1,
  10 = (1+1+1+1+1)*(1+1),
  12 = (1+1+1)*(1+1+1+1),
so a(7) = 5.
		

Crossrefs

Programs

  • Mathematica
    ReplaceListRepeated[forms_,rerules_]:=Union[Flatten[FixedPointList[Function[pre,Union[Flatten[ReplaceList[#,rerules]&/@pre,1]]],forms],1]];
    mexos[ptn_]:=If[Length[ptn]==0,{0},Union@@Select[ReplaceListRepeated[{Sort[ptn]},{{foe___,x_,mie___,y_,afe___}:>Sort[Append[{foe,mie,afe},x+y]],{foe___,x_?(#>1&),mie___,y_?(#>1&),afe___}:>Sort[Append[{foe,mie,afe},x*y]]}],Length[#]==1&]];
    Table[Length[mexos[Table[1,{n}]]],{n,30}]

A319855 Minimum number that can be obtained by iteratively adding or multiplying together parts of the integer partition with Heinz number n until only one part remains.

Original entry on oeis.org

0, 1, 2, 1, 3, 2, 4, 1, 4, 3, 5, 2, 6, 4, 5, 1, 7, 4, 8, 3, 6, 5, 9, 2, 6, 6, 6, 4, 10, 5, 11, 1, 7, 7, 7, 4, 12, 8, 8, 3, 13, 6, 14, 5, 7, 9, 15, 2, 8, 6, 9, 6, 16, 6, 8, 4, 10, 10, 17, 5, 18, 11, 8, 1, 9, 7, 19, 7, 11, 7, 20, 4, 21, 12, 8, 8, 9, 8, 22, 3, 8
Offset: 1

Views

Author

Gus Wiseman, Sep 29 2018

Keywords

Comments

The Heinz number of an integer partition (y_1, ..., y_k) is prime(y_1) * ... * prime(y_k).

Examples

			a(30) = 5 because the minimum number that can be obtained starting with (3,2,1) is 3+2*1 = 5.
		

Crossrefs

Programs

  • Mathematica
    ReplaceListRepeated[forms_,rerules_]:=Union[Flatten[FixedPointList[Function[pre,Union[Flatten[ReplaceList[#,rerules]&/@pre,1]]],forms],1]];
    nexos[ptn_]:=If[Length[ptn]==0,{0},Union@@Select[ReplaceListRepeated[{Sort[ptn]},{{foe___,x_,mie___,y_,afe___}:>Sort[Append[{foe,mie,afe},x+y]],{foe___,x_,mie___,y_,afe___}:>Sort[Append[{foe,mie,afe},x*y]]}],Length[#]==1&]];
    Table[Min[nexos[If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]]]],{n,100}]

Formula

a(1) = 0, a(n) = max(A056239(n) - A007814(n), 1). - Charlie Neder, Oct 03 2018

A319856 Maximum number that can be obtained by iteratively adding or multiplying together parts of the integer partition with Heinz number n until only one part remains.

Original entry on oeis.org

0, 1, 2, 2, 3, 3, 4, 3, 4, 4, 5, 4, 6, 5, 6, 4, 7, 6, 8, 6, 8, 6, 9, 6, 9, 7, 8, 8, 10, 9, 11, 6, 10, 8, 12, 9, 12, 9, 12, 9, 13, 12, 14, 10, 12, 10, 15, 9, 16, 12, 14, 12, 16, 12, 15, 12, 16, 11, 17, 12, 18, 12, 16, 9, 18, 15, 19, 14, 18, 16, 20, 12, 21, 13
Offset: 1

Views

Author

Gus Wiseman, Sep 29 2018

Keywords

Comments

The Heinz number of an integer partition (y_1, ..., y_k) is prime(y_1) * ... * prime(y_k).

Examples

			a(30) = 9 because the maximum number that can be obtained starting with (3,2,1) is 3*(2+1) = 9.
		

Crossrefs

Programs

  • Mathematica
    ReplaceListRepeated[forms_,rerules_]:=Union[Flatten[FixedPointList[Function[pre,Union[Flatten[ReplaceList[#,rerules]&/@pre,1]]],forms],1]];
    nexos[ptn_]:=If[Length[ptn]==0,{0},Union@@Select[ReplaceListRepeated[{Sort[ptn]},{{foe___,x_,mie___,y_,afe___}:>Sort[Append[{foe,mie,afe},x+y]],{foe___,x_,mie___,y_,afe___}:>Sort[Append[{foe,mie,afe},x*y]]}],Length[#]==1&]];
    Table[Max[nexos[If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]]]],{n,100}]

A212342 Sequence of coefficients of x^0 in marked mesh pattern generating function Q_{n,132}^(0,3,0,0)(x).

Original entry on oeis.org

1, 1, 2, 5, 9, 14, 20, 27, 35, 44, 54, 65, 77, 90, 104, 119, 135, 152, 170, 189, 209, 230, 252, 275, 299, 324, 350, 377, 405, 434, 464, 495, 527, 560, 594, 629, 665, 702, 740, 779, 819, 860, 902, 945, 989, 1034, 1080, 1127, 1175, 1224, 1274, 1325, 1377, 1430, 1484, 1539, 1595, 1652, 1710, 1769, 1829, 1890, 1952, 2015, 2079
Offset: 0

Views

Author

N. J. A. Sloane, May 09 2012

Keywords

Crossrefs

A201163 is similar. - Robert Price, Jun 02 2012

Programs

  • Mathematica
    QQ0[t, x] = (1 - (1-4*x*t)^(1/2)) / (2*x*t); QQ1[t, x] = 1/(1 - t*QQ0[t, x]); QQ2[t, x] = (1 + t*(QQ1[t, x] - QQ0[t, x]))/(1 - t*QQ0[t, x]); QQ3[t, x] = (1 + t*(QQ2[t, x] - QQ0[t, x] + t*(QQ1[t, x] - QQ0[t,  x])))/(1 - t*QQ0[t, x]); q=Simplify[Series[QQ3[t, x], {t, 0, 35}]]; CoefficientList[q /. x -> 0, t] (* Robert Price, Jun 04 2012 *)

Formula

For n>=2, a(n)=(n^2+n-2)/2. - Robert Price, Jun 02 2012
For n>=5, a(n)=3*a(n-1)-3*a(n-2)+a(n-3). G.f.: (1-2*x+2*x^2+x^3-x^4)/(1-x)^3. - Colin Barker, Jul 06 2012

Extensions

a(10)-a(35) from Robert Price, Jun 02 2012
Added a(0) to correspond to given offset and to be consistent with A212346, Robert Price, Jun 02 2012

A319907 Number of distinct integers that can be obtained by iteratively adding any two or multiplying any two non-1 parts of an integer partition until only one part remains, starting with the integer partition with Heinz number n.

Original entry on oeis.org

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

Views

Author

Gus Wiseman, Oct 01 2018

Keywords

Comments

The Heinz number of an integer partition (y_1, ..., y_k) is prime(y_1) * ... * prime(y_k).

Examples

			The Heinz number of (3,3,2) is 75 and we have
    3+3+2 = 8,
    3+3*2 = 9,
    3*3+2 = 11,
  (3+3)*2 = 12,
  3*(3+2) = 15,
    3*3*2 = 18,
so a(75) = 6.
		

Crossrefs

Programs

  • Mathematica
    ReplaceListRepeated[forms_,rerules_]:=Union[Flatten[FixedPointList[Function[pre,Union[Flatten[ReplaceList[#,rerules]&/@pre,1]]],forms],1]];
    mexos[ptn_]:=If[Length[ptn]==0,{0},Union@@Select[ReplaceListRepeated[{Sort[ptn]},{{foe___,x_,mie___,y_,afe___}:>Sort[Append[{foe,mie,afe},x+y]],{foe___,x_?(#>1&),mie___,y_?(#>1&),afe___}:>Sort[Append[{foe,mie,afe},x*y]]}],Length[#]==1&]];
    Table[Length[mexos[If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]]]],{n,100}]
Showing 1-7 of 7 results.