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.

A355576 Number A(n,k) of n-tuples (p_1, p_2, ..., p_n) of positive integers such that p_{i-1} <= p_i <= k^(i-1); square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 2, 1, 0, 1, 1, 3, 7, 1, 0, 1, 1, 4, 24, 44, 1, 0, 1, 1, 5, 58, 541, 516, 1, 0, 1, 1, 6, 115, 3236, 35649, 11622, 1, 0, 1, 1, 7, 201, 12885, 713727, 6979689, 512022, 1, 0, 1, 1, 8, 322, 39656, 7173370, 627642640, 4085743032, 44588536, 1, 0
Offset: 0

Views

Author

Alois P. Heinz, Jul 07 2022

Keywords

Examples

			A(2,3) = 3: (1,1), (1,2), (1,3).
A(3,2) = 7: (1,1,1), (1,1,2), (1,1,3), (1,1,4), (1,2,2), (1,2,3), (1,2,4).
A(3,3) = 24: (1,1,1), (1,1,2), (1,1,3), (1,1,4), (1,1,5), (1,1,6), (1,1,7), (1,1,8), (1,1,9), (1,2,2), (1,2,3), (1,2,4), (1,2,5), (1,2,6), (1,2,7), (1,2,8), (1,2,9), (1,3,3), (1,3,4), (1,3,5), (1,3,6), (1,3,7), (1,3,8), (1,3,9).
Square array A(n,k) begins:
  1, 1,     1,       1,         1,           1,            1, ...
  1, 1,     1,       1,         1,           1,            1, ...
  0, 1,     2,       3,         4,           5,            6, ...
  0, 1,     7,      24,        58,         115,          201, ...
  0, 1,    44,     541,      3236,       12885,        39656, ...
  0, 1,   516,   35649,    713727,     7173370,     46769781, ...
  0, 1, 11622, 6979689, 627642640, 19940684251, 330736663032, ...
		

Crossrefs

Rows n=1-4 give: A000012, A001477, A081436(k-1) for k>0, A354608.
Main diagonal gives A355561.

Programs

  • Maple
    A:= proc(n, k) option remember; `if`(n=0, 1, -add(
          A(j, k)*(-1)^(n-j)*binomial(k^j, n-j), j=0..n-1))
        end:
    seq(seq(A(n, d-n), n=0..d), d=0..12);
  • Mathematica
    A[n_, k_] := A[n, k] = If[n==0, 1, -Sum[A[j, k]*(-1)^(n-j)*Binomial[If[j==0, 1, k^j], n-j], {j, 0, n-1}]];
    Table[Table[A[n, d-n], {n, 0, d}], {d, 0, 12}] // Flatten (* Jean-François Alcover, Sep 21 2022, after Alois P. Heinz *)

A109055 To compute a(n) we first write down 3^n 1's in a row. Each row takes the rightmost 3rd part of the previous row and each element in it equals sum of the elements of the previous row starting with the first of the rightmost 3rd part. The single element in the last row is a(n).

Original entry on oeis.org

1, 1, 3, 24, 541, 35649, 6979689, 4085743032, 7166723910237, 37698139930450365, 594816080266215640710, 28154472624850002001979592, 3997853576535778666975681355079, 1703042427700923785323670557504832751, 2176429411666209822350337722381643148477248
Offset: 0

Views

Author

Augustine O. Munagi, Jun 17 2005

Keywords

Comments

Comment from Franklin T. Adams-Watters, Jul 13 2006: This is the number of subpartitions of the sequence 3^n-1. As such it can also be computed adding forward, with 3^n terms in the n-th line:
1...........................................................................
1.1 1.......................................................................
1.2.3.3..3..3..3..3..3......................................................
1.3.6.9.12.15.18.21.24.24.24.24.24.24.24.24.24.24.24.24.24.24.24.24.24.24.24

Examples

			For example, for n=3 the array looks like this:
1..1..1..1..1........1..1..1..1..1..1..1..1..1..1
........................1..2..3..4..5..6..7..8..9
..........................................7.15.24
...............................................24
Therefore a(3)=24.
		

Crossrefs

Programs

  • Maple
    proc(n::nonnegint) local f,a; if n=0 or n=1 then return 1; end if; f:=L->[seq(add(L[i],i=2*nops(L)/3+1..j),j=2*nops(L)/3+1..nops(L))]; a:=f([seq(1,j=1..3^n)]); while nops(a)>3 do a:=f(a) end do; a[3]; end proc;
  • Mathematica
    A[n_, k_] := A[n, k] = If[n == 0, 1, -Sum[A[j, k]*(-1)^(n - j)*Binomial[If[j == 0, 1, k^j], n - j], {j, 0, n - 1}]];
    a[n_] := A[n, 3];
    Table[a[n], {n, 0, 14}] (* Jean-François Alcover, Apr 01 2024, after Alois P. Heinz in A355576 *)

Extensions

More terms from Paul D. Hanna

A109061 To compute a(n) we first write down 9^n 1's in a row. Each row takes the rightmost 9th part of the previous row and each element in it equals sum of the elements of the previous row starting with the first of the rightmost 9th part. The single element in the last row is a(n).

Original entry on oeis.org

1, 1, 9, 693, 476121, 2940705927, 163444130390781, 81756588582353417271, 368059416198072536171078649, 14912674110246473369128526689667934, 5437955149300119215042866669813503145575607, 17846712348533391270843269203829434120473501691723788
Offset: 0

Views

Author

Augustine O. Munagi, Jun 17 2005

Keywords

Examples

			For example, for n=3 the array, from 2nd row, follows:
1..2..3.....70..71..72..73..74..75..76..77..78..79..80..81
........................73.147.222.298.375.453.532.612.693
.......................................................693
Therefore a(3)=693.
		

Crossrefs

Programs

  • Maple
    proc(n::nonnegint) local f,a; if n=0 or n=1 then return 1; end if; f:=L->[seq(add(L[i],i=8*nops(L)/9+1..j),j=8*nops(L)/9+1..nops(L))]; a:=f([seq(1,j=1..9^n)]); while nops(a)>9 do a:=f(a) end do; a[9]; end proc;
  • Mathematica
    A[n_, k_] := A[n, k] = If[n == 0, 1, -Sum[A[j, k]*(-1)^(n - j)* Binomial[If[j == 0, 1, k^j], n - j], {j, 0, n - 1}]];
    a[n_] := A[n, 9];
    Table[a[n], {n, 0, 12}] (* Jean-François Alcover, Apr 01 2024, after Alois P. Heinz in A355576 *)

Extensions

More terms from Alois P. Heinz, Jul 06 2022

A109056 To compute a(n) we first write down 4^n 1's in a row. Each row takes the rightmost 4th part of the previous row and each element in it equals sum of the elements of the previous row starting with the first of the rightmost 4th part. The single element in the last row is a(n).

Original entry on oeis.org

1, 1, 4, 58, 3236, 713727, 627642640, 2205897096672, 31004442653082720, 1743005531132374350208, 391947224244531572312436328, 352545281714327012273215572739472, 1268416358395092955994185170741834144224, 18254446075150458724007419019753847268167282688
Offset: 0

Views

Author

Augustine O. Munagi, Jun 17 2005

Keywords

Examples

			For example, for n=3 the array looks like this:
1..1.....1..1..1..1..1..1..1..1..1..1..1..1..1..1..1..1..1
............1..2..3..4..5..6..7..8..9.10.11.12.13.14.15.16
...............................................13.27.42.58
........................................................58
Therefore a(4)=58.
		

Crossrefs

Programs

  • Maple
    proc(n::nonnegint) local f,a; if n=0 or n=1 then return 1; end if; f:=L->[seq(add(L[i],i=3*nops(L)/4+1..j),j=3*nops(L)/4+1..nops(L))]; a:=f([seq(1,j=1..4^n)]); while nops(a)>4 do a:=f(a) end do; a[4]; end proc;
  • Mathematica
    A[n_, k_] := A[n, k] = If[n == 0, 1, -Sum[A[j, k]*(-1)^(n - j)* Binomial[If[j == 0, 1, k^j], n - j], {j, 0, n - 1}]];
    a[n_] := A[n, 4];
    Table[a[n], {n, 0, 13}] (* Jean-François Alcover, Apr 01 2024, after Alois P. Heinz in A355576 *)

Extensions

More terms from Alois P. Heinz, Jul 06 2022

A109058 To compute a(n) we first write down 6^n 1's in a row. Each row takes the rightmost 6th part of the previous row and each element in it equals sum of the elements of the previous row starting with the first of the rightmost 6th part. The single element in the last row is a(n).

Original entry on oeis.org

1, 1, 6, 201, 39656, 46769781, 330736663032, 14031372754200653, 3571582237574150514024, 5454701025672508908169570740, 49984143782624329482858175943128416, 2748177454593265010973723857947479180947553, 906585004703475512437226615670665677815744239819376
Offset: 0

Views

Author

Augustine O. Munagi, Jun 17 2005

Keywords

Examples

			For example, for n=3 the array, from 2nd row, follows:
1..2..3.....25..26..27..28..29..30..31..32..33..34..35..36
....................................31..63..96.130.165.201
.......................................................201
Therefore a(3)=201.
		

Crossrefs

Programs

  • Maple
    proc(n::nonnegint) local f,a; if n=0 or n=1 then return 1; end if; f:=L->[seq(add(L[i],i=5*nops(L)/6+1..j),j=5*nops(L)/6+1..nops(L))]; a:=f([seq(1,j=1..6^n)]); while nops(a)>6 do a:=f(a) end do; a[6]; end proc;
  • Mathematica
    A[n_, k_] := A[n, k] = If[n == 0, 1, -Sum[A[j, k]*(-1)^(n - j)* Binomial[If[j == 0, 1, k^j], n - j], {j, 0, n - 1}]];
    a[n_] := A[n, 6];
    Table[a[n], {n, 0, 12}] (* Jean-François Alcover, Apr 01 2024, after Alois P. Heinz in A355576 *)

Extensions

More terms from Alois P. Heinz, Jul 06 2022

A109059 To compute a(n) we first write down 7^n 1's in a row. Each row takes the rightmost 7th part of the previous row and each element in it equals sum of the elements of the previous row starting with the first of the rightmost 7th part. The single element in the last row is a(n).

Original entry on oeis.org

1, 1, 7, 322, 102249, 226742516, 3518406903403, 382149784071841422, 290546585470549214822793, 1546306129153609960601346281449, 57606719909341067627899562630623352149, 15022729501707009545842655841005666468590455864, 27423481304702360472157221630747597794702587610760693525
Offset: 0

Views

Author

Augustine O. Munagi, Jun 17 2005

Keywords

Examples

			For example, for n=3 the array, from 2nd row, follows:
1..2..3.....38..39..40..41..42..43..44..45..46..47..48..49
................................43..87.132.178.225.273.322
.......................................................322
Therefore a(3)=322.
		

Crossrefs

Programs

  • Maple
    proc(n::nonnegint) local f,a; if n=0 or n=1 then return 1; end if; f:=L->[seq(add(L[i],i=6*nops(L)/7+1..j),j=6*nops(L)/7+1..nops(L))]; a:=f([seq(1,j=1..7^n)]); while nops(a)>7 do a:=f(a) end do; a[7]; end proc;
  • Mathematica
    A[n_, k_] := A[n, k] = If[n == 0, 1, -Sum[A[j, k]*(-1)^(n - j)* Binomial[If[j == 0, 1, k^j], n - j], {j, 0, n - 1}]];
    a[n_] := A[n, 7];
    Table[a[n], {n, 0, 12}] (* Jean-François Alcover, Apr 01 2024, after_Alois P. Heinz_ in A355576 *)

Extensions

More terms from Alois P. Heinz, Jul 06 2022

A109060 To compute a(n) we first write down 8^n 1's in a row. Each row takes the rightmost 8th part of the previous row and each element in it equals sum of the elements of the previous row starting with the first of the rightmost 8th part. The single element in the last row is a(n).

Original entry on oeis.org

1, 1, 8, 484, 231736, 886208954, 27106585594040, 6632714300472863716, 12983632019302863224103688, 203325054125533158416534341556735, 25472733809776289439071490656049076425792, 25529963965104465687252347321830255523307055463168
Offset: 0

Views

Author

Augustine O. Munagi, Jun 17 2005

Keywords

Examples

			For example, for n=3 the array, from 2nd row, follows:
1..2..3.....53..54..55..56..57..58..59..60..61..62..63..64
............................57.115.174.234.295.357.420.484
.......................................................484
Therefore a(3)=484.
		

Crossrefs

Programs

  • Maple
    proc(n::nonnegint) local f,a; if n=0 or n=1 then return 1; end if; f:=L->[seq(add(L[i],i=7*nops(L)/8+1..j),j=7*nops(L)/8+1..nops(L))]; a:=f([seq(1,j=1..8^n)]); while nops(a)>8 do a:=f(a) end do; a[8]; end proc;
  • Mathematica
    A[n_, k_] := A[n, k] = If[n == 0, 1, -Sum[A[j, k]*(-1)^(n - j)* Binomial[If[j == 0, 1, k^j], n - j], {j, 0, n - 1}]];
    a[n_] := A[n, 8];
    Table[a[n], {n, 0, 12}] (* Jean-François Alcover, Apr 01 2024, after Alois P. Heinz in A355576 *)

Extensions

More terms from Alois P. Heinz, Jul 06 2022
Showing 1-7 of 7 results.