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 12 results. Next

A254436 A component sequence of A254296.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 1, 2, 1, 2, 1, 4, 3, 6, 3, 6, 5, 8, 7, 10, 7, 12, 9, 14, 11, 16, 14, 19, 17, 22, 20, 28, 23, 31, 26, 34, 32, 40, 35, 43, 38, 51, 46, 59, 51, 64, 61, 74, 71, 84, 76, 94, 86, 104, 96, 114, 108, 126, 120, 138, 132, 157, 146, 171
Offset: 1

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Author

Md. Towhidul Islam, Feb 28 2015

Keywords

Comments

This sequence is a component of the formula for counting A254296.
If m=ceiling(log_3(2k)), define n=(3^(m-1)+1)/2+(3^(m-2))-k for k in the range (3^(m-1)+1)/2<=k<=(3^(m-1)-1)/2+(3^(m-2)). Then this sequence gives the first 3^(m-2) terms.

Crossrefs

Formula

If m=ceiling(log_3(2k)), define n=(3^(m-1)+1)/2+(3^(m-2))-k for k in the range (3^(m-1)+1)/2<=k<=(3^(m-1)-1)/2+(3^(m-2)).
Then a(n)=Sum_{d=ceiling((3k+2)/5)..(3^(m-1)-1)/2} Sum_{p=ceiling((d-1)/3..2d-k-1} A254296(p).

A254439 Median of terms of A254296 in the range (3^(n-1)+1)/2 to (3^n-1)/2.

Original entry on oeis.org

1, 1, 2, 7, 47, 682, 23132, 1913821, 397731998, 212521309666, 297464368728296
Offset: 1

Views

Author

Md. Towhidul Islam, Mar 01 2015

Keywords

Comments

As described in A254296, all the 'feasible' partitions of natural numbers (3^(n-1)+1)/2 to (3^n-1)/2 has n parts. A254439 lists the "median of the range ((3^(n-1)+1)/2)-th to ((3^n-1)/2)-th terms of A254296".
From conjectured formula, it appears that next terms are 1107102779611719118, 11090084422457163934046, 302002529294596303158583642. - Benedict W. J. Irwin, Nov 16 2016

Examples

			As described in sequence A254296, "feasible" partitions of the integers 41 through 121 consist of 5 parts. The number 3^(5-1) = 81 has 47 "feasible" partitions, which is the median of the range from the 41st to the 121st term of A254296.
		

Crossrefs

Programs

  • C
    /* C Code to make Mathematica Code for conjectured n-th term n>3 */
    #include 
    int main(int argc, char* argv[]){
    int i, n=atoi(argv[1])-3;
    printf("F[a_,x_,k_]:=Sum[x,{a,1,k}]\n");
    for(i=1; i<=n; i++)printf("F[i%d,",i);
    printf("3i%d-1,",n);
    for(i=n-1; i>0; i--)printf("3i%d-1],",i);
    printf("2]\n");
    return 0;
    }
    /* Benedict Irwin, Nov 16 2016 */
  • Mathematica
    F[a_, x_, k_] := Sum[x, {a, 1, k}]
    F[i1, 3*i1 - 1, 2]
    F[i1, F[i2, 3*i2 - 1, 3*i1 - 1], 2]
    F[i1, F[i2, F[i3, 3*i3 - 1, 3*i2 - 1], 3*i1 - 1], 2]
    F[i1, F[i2, F[i3, F[i4, 3*i4 - 1, 3*i3 - 1], 3*i2 - 1], 3*i1 - 1], 2] (* Examples of how to get first few terms, use the C code to generate the n-th term of the conjectured formula, Benedict W. J. Irwin, Nov 16 2016 *)

Formula

a(n) = A254296(3^(n-1)).
Conjecture: for n>3, a(n+3) = Sum_{i_1=1..2} Sum_{i_2=1..3*i_1-1} ... Sum_{i_n..3*i_(n-1)-1} (3*i_n - 1). - Benedict W. J. Irwin, Nov 16 2016

A254440 Palindromes in A254296.

Original entry on oeis.org

1, 2, 3, 5, 7, 9, 11, 33, 101, 111, 131, 262, 313, 23132, 86068
Offset: 1

Views

Author

Md. Towhidul Islam, Mar 19 2015

Keywords

A254435 Squares in A254296.

Original entry on oeis.org

1, 9, 81, 729, 1296, 23532201
Offset: 1

Views

Author

Md. Towhidul Islam, Feb 03 2015

Keywords

Comments

They seem to be in the ((3^(n-1) + 1)/2)-th to ((3^(n-1)-1)/2 + 3^(n-2))-th values of A254296, though this is not proved. The terms are the squares of: 1, 3, 9, 27, 36, 4851.

Crossrefs

A254442 Triangle read by rows: T(n,k) is the total number of parts of denomination k used in all n-part feasible partitions described in A254296.

Original entry on oeis.org

1, 4, 1, 1, 19, 7, 10, 3, 3, 2, 2, 1, 1, 201, 62, 124, 27, 37, 35, 42, 31, 35, 16, 16, 14, 14, 12, 12, 9, 9, 7, 7, 5, 5, 3, 3, 2, 2, 1, 1, 5020, 1271, 3551, 431, 719, 840, 1128, 851, 1051, 255, 303, 327, 369, 370, 408, 358, 387, 340, 366, 309, 330, 262, 280, 248, 264, 226, 238, 183, 183, 173, 173, 162, 162, 150, 150
Offset: 1

Views

Author

Md. Towhidul Islam, May 12 2015

Keywords

Comments

Row n contains 3^(n-1) terms.
Sum of row n equals n*A254430(n).

Examples

			Triangle begins:
1;
4, 1, 1;
19, 7, 10, 3, 3, 2, 2, 1, 1;
201, 62, 124, 27, 37, 35, 42, 31, 35, 16, 16, 14, 14, 12, 12, 9, 9, 7, 7, 5, 5, 3, 3, 2, 2, 1, 1;
5020, 1271, 3551, 431, 719, 840, 1128, 851, 1051, 255, 303, 327, 369, 370, 408, 358, 387, 340, 366, 309, 330, 262, 280, 248, 264, 226, 238, 183, 183, 173, 173, 162, 162, 150, 150, 139, 139, 127, 127, 115, 115, 104, 104, 93, 93, 81, 81, 72, 72, 63, 63, 54, 54, 47, 47, 40, 40, 33, 33, 28, 28, 23, 23, 18, 18, 15, 15, 12, 12, 9, 9, 7, 7, 5, 5 ,3, 3, 2, 2, 1, 1;
		

Crossrefs

A062051 Number of partitions of n into powers of 3.

Original entry on oeis.org

1, 1, 1, 2, 2, 2, 3, 3, 3, 5, 5, 5, 7, 7, 7, 9, 9, 9, 12, 12, 12, 15, 15, 15, 18, 18, 18, 23, 23, 23, 28, 28, 28, 33, 33, 33, 40, 40, 40, 47, 47, 47, 54, 54, 54, 63, 63, 63, 72, 72, 72, 81, 81, 81, 93, 93, 93, 105, 105, 105, 117, 117, 117, 132, 132, 132, 147, 147, 147, 162
Offset: 0

Views

Author

Amarnath Murthy, Jun 06 2001

Keywords

Comments

Number of different partial sums of 1+[1,*3]+[1,*3]+..., where [1,*3] means we can either add 1 or multiply by 3. E.g., a(6)=3 because we have 6=1+1+1+1+1+1=(1+1)*3=1*3+1+1+1. - Jon Perry, Jan 01 2004
Also number of partitions of n into distinct 3-smooth parts. E.g., a(10) = #{9+1, 8+2, 6+4, 6+3+1, 4+3+2+1} = #{9+1, 3+3+3+1, 3+3+1+1+1+1, 3+1+1+1+1+1+1+1, 1+1+1+1+1+1+1+1+1+1} = 5. - Reinhard Zumkeller, Apr 07 2005
Starts to differ from A008650 at a(81). - R. J. Mathar, Jul 31 2010
If m=ceiling(log_3(2k)) and n=(3^m+1)/2-k for k in the range (3^(m-1)+1)/2+(3^(m-2))<=k<=(3^m-1)/2, this sequence gives the number of "feasible" partitions described in the sequence A254296. For instance, the terms starting at 121st term of A254296 backwards to 68th term of A254296 provide the first 54 terms of this sequence. - Md. Towhidul Islam, Mar 01 2015
From Gary W. Adamson, Sep 03 2016: (Start)
Let M =
1, 0, 0, 0, 0, ...
1, 0, 0, 0, 0, ...
1, 0, 0, 0, 0, ...
1, 1, 0, 0, 0, ...
1, 1, 0, 0, 0, ...
1, 1, 0, 0, 0, ...
1, 1, 1, 0, 0, ...
1, 1, 1, 0, 0, ...
..., where the leftmost column is all 1's, and all other columns are 1's shifted down thrice. Lim_{k=1..inf} M^k has a single nonzero column, which gives the sequence. (End)

Examples

			a(4) = 2 and the partitions are 3+1, 1+1+1+1;
a(9) = 5 and the partitions are 9; 3+3+3; 3+3+1+1+1; 3+1+1+1+1+1+1; 1+1+1+1+1+1+1+1+1.
		

Crossrefs

Programs

  • Mathematica
    nn=70;a=Product[1/(1-x^(3^i)),{i,0,4}];CoefficientList[Series[a,{x,0,nn}],x] (* Geoffrey Critzer, Oct 30 2012 *)
  • PARI
    { n=15; v=vector(n); for (i=1,n,v[i]=vector(2^(i-1))); v[1][1]=1; for (i=2,n, k=length(v[i-1]); for (j=1,k, v[i][j]=v[i-1][j]+1; v[i][j+k]=v[i-1][j]*3)); c=vector(n); for (i=1,n, for (j=1,2^(i-1), if (v[i][j]<=n, c[v[i][j]]++))); c } \\ Jon Perry
    
  • Python
    from functools import lru_cache
    @lru_cache(maxsize=None)
    def A062051(n): return A062051(n-1)+(0 if n%3 else A062051(n//3)) if n>2 else 1 # Chai Wah Wu, Sep 21 2022

Formula

a(n) = A005704([n/3]).
G.f.: Product_{k>=0} 1/(1-x^(3^k)). - R. J. Mathar, Jul 31 2010
If m = ceiling(log_3(2k)), define n = (3^m + 1)/2 - k for k in the range (3^(m-1)+1)/2 + (3^(m-2)) <= k <= (3^m-1)/2. Then, a(n) = Sum_{s=ceiling((k-1)/3)..(3^(m-1)-1)/2} a(s). This gives the first 2(3^(m-1))/3 terms. - Md. Towhidul Islam, Mar 01 2015
G.f.: 1 + Sum_{i>=0} x^(3^i) / Product_{j=0..i} (1 - x^(3^j)). - Ilya Gutkovskiy, May 07 2017

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), Jun 11 2001

A254438 Natural numbers k such that k is a multiple of its number of "feasible" partitions.

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 10, 11, 12, 13, 28, 30, 33, 36, 38, 39, 40, 72, 92, 110, 114, 116, 118, 119, 120, 121, 330, 350, 355, 357, 360, 362, 363, 364, 1086, 1088, 1090, 1091, 1092, 1093, 3248, 3270, 3273, 3276, 3278, 3279, 3280, 9792, 9828, 9830, 9834, 9836, 9838, 9839, 9840, 9841, 29376, 29512, 29515, 29517, 29520, 29522, 29523, 29524
Offset: 1

Views

Author

Md. Towhidul Islam, Mar 01 2015

Keywords

Comments

This sequence lists the natural numbers k that are divisible by A254296(k).

Examples

			For n=1,2,3, A254296(n)=1, so they are in the sequence.
For n=4,6,8,10, A254296(n)=2, so they are in the sequence.
For n=5,9, A254296(n)=2, so they are not in the sequence.
		

Crossrefs

Programs

  • Mathematica
    (* This program is not suitable to compute a large number of terms. *)
    okQ[v_] := Module[{s=0}, For[i=1, i <= Length[v], i++, If[v[[i]] > 2s+1, Return[False], s += v[[i]]]]; Return[True]];
    b[n_] := b[n] = With[{k = Ceiling[Log[3, 2 n]]}, Select[Reverse /@ IntegerPartitions[n, {k}], okQ] // Length];
    Reap[Do[If[Divisible[k, b[k]], Print[k]; Sow[k]], {k, 1, 120}]][[2, 1]] (* Jean-François Alcover, Nov 03 2018 *)

Extensions

a(48)-a(64) added by Md. Towhidul Islam, Apr 18 2015

A254430 Number of "feasible" partitions with n parts.

Original entry on oeis.org

1, 3, 16, 183, 4804, 299558, 45834625, 17696744699, 17644374475261, 46279884666882734, 324101360547203133793
Offset: 1

Views

Author

Md. Towhidul Islam, Jan 30 2015

Keywords

Comments

This sequence answers the question: "How many sellers can each be provided with a distinct set of n-part 'feasible' weights described in A254296?" It counts all the n-part "feasible" partitions of all the natural numbers from (3^(n-1)+1)/2 to (3^n-1)/2. Here n resembles m in A254296.

Examples

			For n=2, we count 2nd through 4th values of A254296. So a(2)=1+1+1=3.
For n=3, we count 5th through 13th values from A254296. So a(3)= 2+2+3+2+2+2+1+1+1 = 16.
For n=4, a(4)= Sum of 14th through 40th terms of A254296, that is, 183.
		

Crossrefs

Programs

  • Mathematica
    okQ[v_] := Module[{s = 0}, For[i = 1, i <= Length[v], i++, If[v[[i]] > 2s + 1, Return[False], s += v[[i]]]]; Return[True]];
    a254296[n_] := With[{k = Ceiling[Log[3, 2n]]}, Select[Reverse /@ IntegerPartitions[n, {k}], okQ] // Length];
    a[n_] := Sum[a254296[p], {p, (3^(n-1) + 1)/2, (3^n - 1)/2}];
    Array[a, 5] (* Jean-François Alcover, Nov 04 2018, after Charles R Greathouse IV in A254296 *)

Formula

a(n) = Sum_{p=(3^(n-1)+1)/2..(3^n-1)/2} A254296(p).

Extensions

a(9)-a(11) from Md. Towhidul Islam, Apr 18 2015

A254431 Number of "feasible" partitions of the smallest natural number of length n.

Original entry on oeis.org

1, 1, 2, 10, 131, 3887, 262555, 42240104, 16821037273, 17094916187012, 45374905859155948
Offset: 1

Views

Author

Md. Towhidul Islam, Jan 30 2015

Keywords

Comments

The sequence lists the number of "feasible" partitions of the first natural number (3^(n-1)+1)/2 of length n. Here n resembles m in A254296 which describes "feasible" partitions.

Examples

			The smallest natural numbers "feasibly" partitionable into 1, 2, 3, 4 and 5 parts respectively are 1,2,5,14 and 41. From A254296, the number of "feasible" partitions of them are 1,1,2,10 and 131.
		

Crossrefs

Formula

a(n) = A254296((3^(n-1)+1)/2).

Extensions

a(10)-a(11) from Md. Towhidul Islam, Apr 18 2015

A254433 Maximum number of "feasible" partitions of length n.

Original entry on oeis.org

1, 1, 3, 12, 140, 3950, 263707, 42285095, 16825391023, 17095967464466, 45375565948693336
Offset: 1

Views

Author

Md. Towhidul Islam, Feb 03 2015

Keywords

Comments

a(n) gives the highest value in the (3^(n-1)+1)/2-th through the (3^n-1)/2-th terms of the sequence A254296. It lists the highest possible number of "feasible" partitions into n parts.

Examples

			The numbers 2, 3 and 4 are "feasibly" partitionable into 2 parts. Each of them has 1 feasible partitions. So a(2)=1.
The numbers 14 to 40 are "feasibly" partitionable into 4 parts. Among them 16, 18, 19 and 22 each has the highest 12 "feasible" partitions. So a(4)=12.
The numbers 122 to 364 are "feasibly" partitionable into 6 parts. Among them 124 has the highest 3950 "feasible" partitions. So a(6)=3950.
		

Crossrefs

Formula

The first term is 1. For n>=2, a(n) = A254296((3^(n-1)+5)/2).

Extensions

a(9) corrected and a(10)-a(11) added by Md. Towhidul Islam, Apr 18 2015
Showing 1-10 of 12 results. Next