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
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.
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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;
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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 *)
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
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.
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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;
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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 *)
A109057
To compute a(n) we first write down 5^n 1's in a row. Each row takes the rightmost 5th 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 5th part. The single element in the last row is a(n).
Original entry on oeis.org
1, 1, 5, 115, 12885, 7173370, 19940684251, 277078842941900, 19249144351745111125, 6686277384080730564862875, 11612516024884420913314995604000, 100841213012622614260440382077516990500, 4378443591626306255827149380635713364079323075
Offset: 0
For example, for n=3 the array, from 2nd row, follows:
1..2..3.....14..15..16..17..18..19..20..21..22..23..24..25
........................................21..43..66..90.115
.......................................................115
Therefore a(3)=115.
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proc(n::nonnegint) local f,a; if n=0 or n=1 then return 1; end if; f:=L->[seq(add(L[i],i=4*nops(L)/5+1..j),j=4*nops(L)/5+1..nops(L))]; a:=f([seq(1,j=1..5^n)]); while nops(a)>5 do a:=f(a) end do; a[5]; end proc;
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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, 5];
Table[a[n], {n, 0, 12}] (* Jean-François Alcover, Apr 02 2024, after Alois P. Heinz in A355576 *)
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
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.
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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;
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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 *)
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
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.
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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;
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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 *)
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
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.
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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;
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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 *)
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