A109055 -id:A109055 - cp's OEIS frontend

A107354 To compute a(n) we first write down 2^n 1's in a row. Each row takes the right half of the previous row and each element in it equals sum of the elements in the previous row starting at the middle. The single element in the last row is a(n).

1, 1, 2, 7, 44, 516, 11622, 512022, 44588536, 7718806044, 2664170119608, 1836214076324153, 2529135272371085496, 6964321029630556852944, 38346813253279804426846032, 422247020982575523983378003936, 9298487213328788062025571134762096

Offset: 0

Max Alekseyev, May 24 2005

Number of subpartitions of partition [1,3,7,...,2^n-1]. - Franklin T. Adams-Watters, Mar 11 2006
Can also be computed summing forwards:
  1
  1,1
  1,2,2, 2
  1,3,5, 7, 7, 7, 7, 7
  1,4,9,16,23,30,37,44,44,44,44,44,44,44,44,44

			For n=4, the array looks like this:
  1..1..1..1..1..1..1..1..1..1..1..1..1..1..1..1
  ........................1..2..3..4..5..6..7..8
  ....................................5.11.18.26
  .........................................18.44
  ............................................44
  Therefore a(4)=44.
For n=5, we can illustrate the recurrence by:
a(5) = 516 = C(19, 4) - ( 1*C(17, 4) + 2*C(14, 3) + 7*C(9, 2) ) = C(16+4-1, 4) - ( 1*C(16-2+4-1, 4) + 2*C(16-4+3-1, 3) + 7*C(16-8+2-1, 2) ).

Alois P. Heinz, Table of n, a(n) for n = 0..82 (first 26 terms from Reinhard Zumkeller)

Cf. A105996; variants: A109055 - A109061; subpartitions defined: A115728, A115729.

Column k=2 of A355576.

a107354 n = head $ snd $ until ((== 1) . fst)
                               f (2^n, replicate (2^n) 1) where
   f (len, xs) = (len', scanl1 (+) $ drop len' xs) where
      len' = len `div` 2
-- Feasible only for small n.
-- Reinhard Zumkeller, Nov 20 2011

a:= proc(n) option remember; `if`(n=0, 1, -add(
      a(j)*(-1)^(n-j)*binomial(2^j, n-j), j=0..n-1))
    end:
seq(a(n), n=0..16);  # Alois P. Heinz, Jul 08 2022

f[n_] := If[n == 0, 1, Binomial[2^(n - 1) + n - 2, n - 1] - Sum[ f[k]*Binomial[2^(n - 1) - 2^k + n - k - 1, n - k], {k, n - 2}]]; Table[ f[n], {n, 0, 15}] (* Robert G. Wilson v, May 25 2005 *)
Table[NestWhile[Accumulate[Drop[#,Ceiling[Length[#]/2]]]&,PadRight[{},2^n+1,1], Length[ #]> 1&],{n,0,16}]//Flatten (* Harvey P. Dale, Jun 24 2018 *)

{a(n)=if(n==0,1,binomial(2^(n-1)+n-2,n-1)- sum(k=1,n-2,a(k)*binomial(2^(n-1)-2^k+n-k-1,n-k)))} \\ Paul D. Hanna, May 24 2005

{a(n)=polcoeff(x^n-sum(k=0, n-1, a(k)*x^k*(1-x+x*O(x^n))^(2^k-1) ), n)} \\ Paul D. Hanna, May 24 2005

a(n) = C(2^(n-1)+n-2,n-1) - Sum_{k=1..n-2} a(k)*C(2^(n-1)-2^k+n-k-1,n-k) for n>=2, with a(0)=1, a(1)=1, where C = binomial. - Paul D. Hanna, May 24 2005
The first number in row 3 is 2^(n-2)+1. - Ralf Stephan, May 24 2005
G.f.: 1/(1-x) = Sum_{n>=0} a(n)*x^n*(1-x)^(2^n-1) (g.f. of subpartitions). - Paul D. Hanna, Jul 03 2006
G.f.: 1 = Sum_{n>=0} a(n)*x^n/(1+x)^(2^n+n). - Paul D. Hanna, Jul 03 2006

Edited by Paul D. Hanna, Jul 03 2006

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

Alois P. Heinz, Jul 07 2022

			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, ...

Alois P. Heinz, Antidiagonals n = 0..43, flattened

Columns k=1-9 give: A000012, A107354, A109055, A109056, A109057, A109058, A109059, A109060, A109061.
Rows n=1-4 give: A000012, A001477, A081436(k-1) for k>0, A354608.
Main diagonal gives A355561.

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);

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 *)

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

Augustine O. Munagi, Jun 17 2005

nonn

			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.

Alois P. Heinz, Table of n, a(n) for n = 0..46

Cf. A107354, A109055, A109056, A109057, A109058, A109059, A109060, A109062.

Column k=9 of A355576.

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;

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 *)

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

Augustine O. Munagi, Jun 17 2005

nonn

			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.

Alois P. Heinz, Table of n, a(n) for n = 0..58

Cf. A107354, A109055, A109057, A109058, A109059, A109060, A109061.

Column k=4 of A355576.

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;

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 *)

More terms from Alois P. Heinz, Jul 06 2022

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

Augustine O. Munagi, Jun 17 2005

nonn

			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.

Alois P. Heinz, Table of n, a(n) for n = 0..54

Cf. A107354, A109055, A109056, A109058, A109059, A109060, A109061, A109062.

Column k=5 of A355576.

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;

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 *)

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

Augustine O. Munagi, Jun 17 2005

nonn

			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.

Alois P. Heinz, Table of n, a(n) for n = 0..51

Cf. A107354, A109055, A109056, A109057, A109059, A109060, A109061, A109062.

Column k=6 of A355576.

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;

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 *)

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

Augustine O. Munagi, Jun 17 2005

nonn

			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.

Alois P. Heinz, Table of n, a(n) for n = 0..49

Cf. A107354, A109055, A109056, A109057, A109058, A109060, A109061, A109062.

Column k=7 of A355576.

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;

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 *)

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

Augustine O. Munagi, Jun 17 2005

nonn

			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.

Alois P. Heinz, Table of n, a(n) for n = 0..47

Cf. A107354, A109055, A109056, A109057, A109058, A109059, A109061, A109062.

Column k=8 of A355576.

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;

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 *)

More terms from Alois P. Heinz, Jul 06 2022

cp's OEIS Frontend

A107354 To compute a(n) we first write down 2^n 1's in a row. Each row takes the right half of the previous row and each element in it equals sum of the elements in the previous row starting at the middle. The single element in the last row is a(n).

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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.

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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).

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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).

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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).

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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).

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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).

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