A010551 -id:A010551 - cp's OEIS frontend

A179079 Permanent of the n X n matrix whose (i,j)-element is (i+j-1) modulo 3.

1, 4, 9, 34, 192, 1032, 6936, 62496, 530712, 5005152, 61710336, 707802624, 8714718720, 133983590400, 1938416832000, 29588291712000, 544216485888000, 9509523337728000, 173318541516288000, 3711395156281344000, 76000261811572224000, 1610876530967703552000, 39351073330327191552000

Offset: 1

John W. Layman, Jan 04 2011

nonn

Eric Weisstein's World of Mathematics, Permanent

Cf. A010551.

A179079 := proc(n) M := Matrix(n,n,(i,j) -> (i+j-1) mod 3) ; LinearAlgebra[Permanent](M) ; end proc: # R. J. Mathar, Jan 04 2011

Permanent[m_List] := With[{v = Array[x, Length[m]]}, Coefficient[Times @@ (m.v), Times @@ v]]; Table[Permanent[Table[Mod[i+j-1,3], {i,1,n}, {j,1,n}]], {n,1,25}]

A293783 Triangle of numbers of squares {i^2}, i = 0,1..ceiling(n/2), in permutations of {1..n} in A293857.

Original entry on oeis.org

0, 1, 0, 1, 2, 0, 2, 8, 0, 4, 0, 24, 0, 12, 0, 108, 0, 36, 576, 0, 720, 0, 144, 4608, 0, 4032, 0, 576, 0, 31680, 0, 31680, 0, 2880, 0, 288000, 0, 201600, 0, 14400, 2505600, 0, 2764800, 0, 1987200, 0, 86400, 30067200, 0, 28512000, 0, 14515200, 0, 518400

Offset: 1

Peter J. C. Moses and Vladimir Shevelev, Oct 18 2017

From Shevelev's comment in A008794 it follows that the last entries of rows, corresponding to the maximal possible i^2, form sequence A010551, n >= 1. Also note that the last entry of each row is the gcd of all its entries (for a proof see comment in A293857). - Vladimir Shevelev, Oct 26 2017

			Triangle begins
         0,      1;
         0,      1;
         2,      0,        2;
         8,      0,        4;
         0,     24,        0,     12;
         0,    108,        0,     36;
       576,      0,      720,      0,      144;
      4608,      0,     4032,      0,      576;
         0,  31680,        0,  31680,        0,  2880;
         0, 288000,        0, 201600,        0, 14400;
   2505600,      0,  2764800,      0,  1987200,     0,  86400;
  30067200,      0, 28512000,      0, 14515200,     0, 518400;
The compressed triangle resulting from the division of each entry by the last entry of its row begins as follows. If i is the index of the row, starting with i = 1 then this last entry is floor(i/2)! * (i - floor(i/2))!.
    0,   1;
    0,   1;
    1,   0,   1;
    2,   0,   1;
    0,   2,   0,   1;
    0,   3,   0,   1;
    4,   0,   5,   0,   1;
    8,   0,   7,   0,   1;
    0,  11,   0,  11,   0,   1;
    0,  20,   0,  14,   0,   1;
   29,   0,  32,   0,  23,   0,   1;
   58,   0,  55,   0,  28,   0,   1;
    0,  88,   0,  94,   0,  46,   0,   1;
    0, 169,   0, 146,   0,  53,   0,   1;
  263,   0, 282,   0, 283,   0,  86,   0,   1;
  526,   0, 515,   0, 383,   0,  97,   0,   1;
For the sense of the entries of this triangle see the [Shevelev] link (with a continuation there). Let B(n,i) be the set of permutations C of 1..n for which c_1 - c_2 + ... + (-1)^(n-1)*c_n = i^2, i >= 0. Then |B(n,i)| is the entry in the n-th row and i-th column of the first triangle. Let us call two permutations C_1 and C_2 equivalent if one of them is obtained from another by a permutation of its elements with odd indices and/or separately with even indices. Let b(n,i) be the entry in the n-th row and i-th column of the second triangle. Then b(n,i) is the maximal possible number of pairwise non-equivalent permutations which could be chosen in B(n,i). On the other hand, it is the smallest number of non-equivalent permutations in B(n,i) such that every other permutation in B(n,i) is equivalent to one of them. So in some sense b(n,i) is the dimension of B(n,i). In particular, b(n,i) = 0 corresponds to empty B(n,i). - _Vladimir Shevelev_, Nov 13 2017

Peter J. C. Moses, Table of the first 100 rows
Vladimir Shevelev, Basis in subsets of permutations described in A293783, SeqFan post Oct 27 2017.

Cf. A008794, A010551, A293857, A293984.

a293783=Flatten[Table[PadLeft[Riffle[#,Table[0,{Floor[(n-1)/4]}]/.{}->0],1+Floor[(1+n)/2]](Floor[n/2]!*(n-Floor[n/2])!)&[Reverse[Map[SeriesCoefficient[QBinomial[n,Floor[(n+1)/2],q],{q,0,#}]&,Map[2#(Floor[(n+1)/2] - #)&,Range[0,Floor[(n+1)/4]]]]]],{n,20}]] (* Peter J. C. Moses, Nov 01 2017 *)

Row sums of triangle give A293857.

If C = {c_1..c_n} is a permutation of {1..n}, then c_1 - c_2 + ... has the same parity as 1 + 2 + ... + n = n*(n+1)/2. So adjacent rows in the triangle for odd and even n have the same positions of 0's. These positions follow through one, beginning from the first position for n == 1,2 (mod 4) and from the second position for n == 3,0 (mod 4). - David A. Corneth and Vladimir Shevelev, Oct 19 2017

A171645 Partial products of Product_{n=1..inf.} (p(n)/p(n-1)p(n)/p(n-1)), = 222(3/2)(3/2)(5/3)(5/3)(7/5)(7/5)(11/7)(11/7)...; p = primes, A000040, a(1) = 2.

Original entry on oeis.org

2, 4, 8, 12, 18, 30, 50, 70, 98, 154, 242, 286, 338, 442, 578, 646, 722, 874, 1058, 1334, 1682, 1798, 1922, 2294, 2738

Offset: 1

Gary W. Adamson, Dec 13 2009

nonn

Analogous formulas using A000041 terms = A171646; Fibonacci numbers, A006498; factorials, A010551.

			a(10) = 154 = 2*2*2*(3/2)*(3/2)*(5/3)*(5/3)*(7/5)*(7/5)*(11/7).

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A171646, A006498, A010551.

FoldList[Times,Join[{2,2,2},Flatten[{#[[2]]/#[[1]],#[[2]]/#[[1]]}&/@Partition[Prime[Range[20]],2,1]]]] (* Harvey P. Dale, Oct 02 2024 *)

Partial products of Product_{n=1..inf.} (p(n)/p(n-1)*p(n)/p(n-1)), =
2*2*2*(3/2)*(3/2)*(5/3)*(5/3)*(7/5)*(7/5)*(11/7)*(11/7)*...; p = primes,
A000040, a(1) = 2.
a(n)=2*A057602(n-1). [From R. J. Mathar, Dec 15 2009]

A171646 a(1) = 1, then partial products of Product_{n>=1} (p(n)/p(n-1)p(n)/p(n-1)) = 11*1(2)(2)(3/2)(3/2)(5/3)(5/3)(7/5)(7/5)...; p = partition numbers, A000041 starting (1, 2, 3, 5, ...).

Original entry on oeis.org

1, 1, 1, 2, 4, 6, 9, 15, 25, 35, 49, 77, 121, 165, 225, 330, 484, 660, 900, 1260, 1764, 2352, 3136, 4312, 5929, 7777, 10201, 13635, 18225, 23760, 30976, 40656, 53361, 68607, 88209, 114345, 148225, 188650, 240100, 307230

Offset: 1

Gary W. Adamson, Dec 13 2009

nonn

A006498 = analogous sequence using the Fibonacci numbers.
A171645 = .............................Primes, analogous formula.
A010551 = .............................Factorial numbers, analogous formula.

			a(12) = 77 = 1*1*1*2*2*(3/2)*(3/2)*(5/3)*(5/3)*(7/5)*(7/5)*(11/7).

Cf. A000041, A171645, A006498, A010551.

A171646t := proc(n)
    local nh;
    nh := floor(n/2) ;
    combinat[numbpart](nh)/combinat[numbpart](nh-1) ;
end proc:
A171646 := proc(n)
    mul(A171646t(i),i=2..n) ;
end proc:
1,seq(A171646(n),n=2..40) ; # R. J. Mathar, Jul 21 2015

a(1) = 1, then partial products of Product_{n>=1} (p(n)/p(n-1)*p(n)/p(n-1)) = 1*1*1*(2)*(2)*(3/2)*(3/2)*(5/3)*(5/3)*(7/5)*(7/5)*...; p = partition numbers, A000041 starting (1, 2, 3, 5, ...).

Corrected by R. J. Mathar, Jul 21 2015

A203151 (n-1)-st elementary symmetric function of {1,1,2,2,3,3,4,4,5,5,...,Floor[(n+1)/2]}.

Original entry on oeis.org

1, 2, 5, 12, 40, 132, 564, 2400, 12576, 65760, 408960, 2540160, 18299520, 131725440, 1079205120, 8836853760, 81157386240, 745047797760, 7582159872000, 77138417664000, 861690783744000, 9623448705024000, 117074735382528000

Offset: 1

Clark Kimberling, Dec 29 2011

nonn

Column 3 of A246117. - Peter Bala, Aug 15 2014
From R. J. Mathar, Oct 01 2016 (Start):
The k-th elementary symmetric functions of the repeated integers 1,1,2,2,..[(n+1)/2], form a triangle T(n,k), 0<=k<=n, n>=0:
1
1 1
1 2 1
1 4 5 2
1 6 13 12 4
1 9 31 51 40 12
which is a row-reversed version of A246117.  This here is the first subdiagonal. The diagonal is A010551. The 2nd column is A002620, the 3rd A203246. (End)

			Let esf abbreviate "elementary symmetric function".  Then
0th esf of {2}:  1;
1st esf of {1,1}:  1+1=2;
2nd esf of {1,1,2} is 1*1+1*2+1*2=5.

Cf. A203152, A246117.

f[k_] := Floor[(k + 1)/2]; t[n_] := Table[f[k], {k, 1, n}]
a[n_] := SymmetricPolynomial[n - 1, t[n]]
Table[a[n], {n, 1, 22}] (* A203151 *)

A354208 Number of parity-alternating permutations of [n] avoiding the pattern 321.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 6, 11, 22, 44, 89, 185, 382, 808, 1702, 3635, 7779, 16736, 36229, 78466, 171238, 373203, 819186, 1795611, 3958662, 8721086, 19294525, 42691298, 94733886, 210379132, 468084856, 1042703207, 2325575076, 5193931583, 11609749877, 25986720374, 58203955771

Offset: 0

Per W. Alexandersson, Jun 06 2022

nonn

A permutation is parity-alternating if it sends odd integers to odd integers, and even integers to even integers. It avoids 321 if there is no subsequence a..b..c with a > b > c. The values are computed by Michael Albert, see MathOverflow link.

			For n=4, the two permutations are 1234, 3412.
For n=5, we have 12345, 34125, 14523.
For n=6, we have 123456, 341256, 145236, 125634, 561234, 345612.

Peter J. Taylor, Table of n, a(n) for n = 0..50
Per Alexandersson, Samuel Asefa Fufa, Frether Getachew and Dun Qiu, Pattern-avoidance and Fuss-Catalan numbers, arXiv:2201.08168 [math.CO], 2022. See also J. Int. Seq. (2023) Vol. 26, Art. 23.4.2.
MathOverflow, 321-avoiding and parity-alternating permutations, Jun 06 2022.
Nathan Williams, Oberwolfach Problem Session: Enumerative Combinatorics 2022, Univ. Texas Dallas (2023). See example 3.4, where this sequence is misidentified by typographical error.

Cf. A000108 (321-avoiding permutations), A010551 (parity-alternating permutations).

Offset corrected and terms a(30) and beyond from Peter J. Taylor, Jun 10 2022

A080392 Numbers k such that A000984(k) mod k = 0 and A080383(k) != 7.

Original entry on oeis.org

2, 420, 920, 1122, 1218, 1892, 1978, 2444, 2914, 3198, 3782, 4028, 4136, 4292, 4664, 4958, 4960, 5330, 5762, 5986, 6020, 6032, 6710, 6834, 6864, 6882, 6954, 6956, 6968, 7106, 7130, 7140, 7238, 7254, 7448, 7616, 8178, 8190, 8400, 8692, 9462, 9506, 10712, 11060, 11288

Offset: 1

Labos Elemer, Mar 17 2003

nonn

Numbers arising in A067348 and not present in A080385.

Even numbers n such that n divides binomial(n, [n/2]) and A010551(n) does not divide j!*(n-j)! exactly 7 times for j = 0..n. - Peter Luschny, Aug 04 2017

			A080383(2) = 3;
A080383(420) = 11;
A080383(920) = 11;
A080383(1122) = 9;
A080383(1218) = 9.

David A. Corneth, Table of n, a(n) for n = 1..274 (Terms <= 60000)

Cf. A000984, A001405, A067348, A080383, A080385.

isa := proc(n)  local bn, bm;
if n mod 2 = 0 then bn := binomial(n, iquo(n,2)):
if modp(bn, n) = 0 then
   bm := (n, j) -> `if`(modp(bn, binomial(n, j)) = 0, 1, 0):
   return 1 <> add(bm(n, j), j=2..iquo(n,2)-1)
fi fi; false end:
select(isa, [$1..5000]); # Peter Luschny, Aug 04 2017

Do[s=Count[Table[IntegerQ[Binomial[n, Floor[n/2]]/ Binomial[n, j]], {j, 0, n}], True]; s1=IntegerQ[Binomial[n, n/2]/n]; If[ !Equal[s, 7] && Equal[s1, True], Print[n]], {n, 1, 10000}]
(* Second program: *)
Select[Range@ 5000, Function[n, And[Divisible[Binomial[n, n/2], n], Count[Table[Divisible[Binomial[n, Floor[n/2]], Binomial[n, j]], {j, 0, n}], True] != 7]]] (* Michael De Vlieger, Jul 30 2017 *)

More terms from Michael De Vlieger, Jul 30 2017

A139769 T(n,k) = [x^k] Product_{m=1..n} d/dx Sum_{i=1..m} x^i; triangle read by rows, n >= 0, 0 <= k <= A161680(n).

Original entry on oeis.org

1, 1, 1, 2, 1, 4, 7, 6, 1, 6, 18, 36, 49, 46, 24, 1, 8, 33, 94, 204, 354, 497, 562, 501, 326, 120, 1, 10, 52, 188, 528, 1222, 2406, 4102, 6116, 7996, 9132, 9014, 7541, 5116, 2556, 720, 1, 12, 75, 326, 1105, 3106, 7513, 16014, 30558, 52752, 82938, 119230, 156983

Offset: 0

Roger L. Bagula, Jun 13 2008

Row sums are A006472(n+1).

T(n, binomial(n,2)-k) is the number of rank-k intervals in the middle order on permutations. (See Bouvel et al. reference.) - Bridget Tenner, May 24 2024

			Triangle T(n,k) begins:
  1;
  1;
  1, 2;
  1, 4,  7,  6;
  1, 6, 18, 36,  49,  46,  24;
  1, 8, 33, 94, 204, 354, 497, 562, 501, 326, 120;
  ...

Alois P. Heinz, Rows n = 0..50, flattened
Mathilde Bouvel, Luca Ferrari, and Bridget Eileen Tenner, Between weak and Bruhat: the middle order on permutations, arXiv:2405.08943 [math.CO], 2024.

Cf. A000142, A008302 (Mahonian numbers), A006472, A010551, A161680, A259459.

a := Table[CoefficientList[Product[Sum[D[x^i, x], {i, 1, m}], {m, 1, n}], x], {n, 0, 7}]; Flatten[a]

From Alois P. Heinz, May 24 2024: (Start)
|Sum_{k=0..binomial(n,2)} (-1)^k T(n,k)| = A010551(n).
Sum_{k=0..binomial(n,2)} (binomial(n,2)-k)*T(n,k) = A259459(n-2) for n>=2. (End)

Edited by Alois P. Heinz, May 24 2024

A356639 Number of integer sequences b with b(1) = 1, b(m) > 0 and b(m+1) - b(m) > 0, of length n which transform under the map S into a nonnegative integer sequence. The transform c = S(b) is defined by c(m) = Product_{k=1..m} b(k) / Product_{k=2..m} (b(k) - b(k-1)).

Original entry on oeis.org

1, 1, 3, 17, 155, 2677, 73327, 3578339, 329652351

Offset: 1

Thomas Scheuerle, Aug 19 2022

This sequence can be calculated by a recursive algorithm:
Let B1 be an array of finite length, the "1" denotes that it is the first generation. Let B1' be the reversed version of B1. Let C be the element-wise product C = B1 * B1'. Then B2 is a concatenation of taking each element of B1 and add all divisors of the corresponding element in C. If we start with B1 = {1} then we get this sequence of arrays: B2 = {2}, B3 = {3, 4, 6}, ... . a(n) is the length of the array Bn. In short the length of Bn+1 and so a(n+1) is the sum over A000005(Bn * Bn').
The transform used in the definition of this sequence is its own inverse, so if c = S(b) then b = S(c). The eigensequence is 2^n = S(2^n).
There exist some transformation pairs of infinite sequences in the database:
  A000027 <--> A000142; A000079 <--> A000079; A000045 <--> A001654;
  A026549 <--> A038754; A100071 <--> A001405; A058295 <--> A------;
  A111286 <--> A098011; A093968 <--> A205825; A166447 <--> A------;
  A098558 <--> A004277; A010551 <--> A087811; A100538 <--> A329227;
  A079352 <--> A------; A082458 <--> A------; A008233 <--> A264635;
  A138278 <--> A------; A006501 <--> A264557; A336496 <--> A------;
  A019464 <--> A------; A062112 <--> A------; A171647 <--> A359039;
  A279312 <--> A------; A031923 <--> A------.
These transformation pairs are conjectured:
  A137326 <--> A------; A066332 <--> A300902; A208147 <--> A308546;
  A057895 <--> A------; A349080 <--> A------; A019442 <--> A------;
  A349079 <--> A------.
("A------" means not yet in the database.)
Some sequences in the lists above may need offset adjustment to force a beginning with 1,2,... in the transformation.
If we allowed signed rational numbers, further interesting transformation pairs could be observed. For example, 1/n will transform into factorials with alternating sign. 2^(-n) transforms into ones with alternating sign and 1/A000045(n) into A000045 with alternating sign.

			a(4) = 17. The 17 transformation pairs of length 4 are:
  {1, 2, 3, 4}  = S({1, 2, 6, 24}).
  {1, 2, 3, 5}  = S({1, 2, 6, 15}).
  {1, 2, 3, 6}  = S({1, 2, 6, 12}).
  {1, 2, 3, 9}  = S({1, 2, 6, 9}).
  {1, 2, 3, 12} = S({1, 2, 6, 8}).
  {1, 2, 3, 21} = S({1, 2, 6, 7}).
  {1, 2, 4, 5}  = S({1, 2, 4, 20}).
  {1, 2, 4, 6}  = S({1, 2, 4, 12}).
  {1, 2, 4, 8}  = S({1, 2, 4, 8}).
  {1, 2, 4, 12} = S({1, 2, 4, 6}).
  {1, 2, 4, 20} = S({1, 2, 4, 5}).
  {1, 2, 6, 7}  = S({1, 2, 3, 21}).
  {1, 2, 6, 8}  = S({1, 2, 3, 12}).
  {1, 2, 6, 9}  = S({1, 2, 3, 9}).
  {1, 2, 6, 12} = S({1, 2, 3, 6}).
  {1, 2, 6, 15} = S({1, 2, 3, 5}).
  {1, 2, 6, 24} = S({1, 2, 3, 4}).
b(1) = 1 by definition, b(2) = 1+1 as 1 has only 1 as divisor.
a(3) = A000005(b(2)*b(2)) = 3.
The divisors of b(2) are 1,2,4. So b(3) can be b(2)+1, b(2)+2 and b(2)+4.
a(4) = A000005((b(2)+1)*(b(2)+4)) + A000005((b(2)+2)*(b(2)+2)) + A000005((b(2)+4)*(b(2)+1)) = 17.

Cf. A000005.
Cf. A000027, A000045, A000079, A000142, A001405.
Cf. A001654, A004277, A006501, A008233, A019442.
Cf. A010551, A019464, A026549, A031923, A038754.
Cf. A057895, A058295, A062112, A066332, A100071.
Cf. A079352, A082458, A087811, A093968, A098011.
Cf. A098558, A100538, A111286, A166447, A137326.
Cf. A138278, A171647, A205825, A208147, A264557.
Cf. A264635, A308546, A329227, A336496, A349079.
Cf. A349080, A359039.

A335109 Triangle read by rows: T(n,k) is the number of permutations of length n with each cycle of the permutation containing only elements that are identical (mod k), where 1 <= k <= n.

Original entry on oeis.org

1, 2, 1, 6, 2, 1, 24, 4, 2, 1, 120, 12, 4, 2, 1, 720, 36, 8, 4, 2, 1, 5040, 144, 24, 8, 4, 2, 1, 40320, 576, 72, 16, 8, 4, 2, 1, 362880, 2880, 216, 48, 16, 8, 4, 2, 1, 3628800, 14400, 864, 144, 32, 16, 8, 4, 2, 1

Offset: 1

Dennis P. Walsh, May 23 2020

Let [n] denote {1,2,...,n} and let [n](j,k) denote the subset of [n] consisting of all elements of [n] that equal j mod k. The cardinality of [n](j,k) equals ceiling(n/k) for j = 1..(n mod k) and equals floor(n/k) for j > (n mod k). Therefore, upon permuting the elements of each [n](j,k) subset, we obtain T(n,k) = (ceiling(n/k)!)^(n mod k)*(floor(n/k)!)^(k-(n mod k)).

			Triangle begins:
    1;
    2  1;
    6  2 1;
   24  4 2 1;
  120 12 4 2 1;
  ...
T(6,3) counts the 8 permutations of [6] where all cycle-mates are identical mod 3, namely, (1 4)(2 5)(3 6), (1 4)(2 5)(3)(6), (1 4)(2)(5)(3 6), (1)(4)(2 5)(3 6), (1 4)(2)(5)(3)(6), (1)(4)(2 5)(3)(6), (1)(4)(2)(5)(3 6) and (1)(2)(3)(4)(5)(6).

Per Alexandersson, Frether Getachew Kebede, Samuel Asefa Fufa, and Dun Qiu, Pattern-Avoidance and Fuss-Catalan Numbers, J. Int. Seq. (2023) Vol. 26, Art. 23.4.2.

Cf. A000142, A010551, A264557, A264635, A264656.

Cf. A275062.

seq(seq((ceil(n/k)!)^(n mod k)*(floor(n/k)!)^(k-(n mod k)), k=1..n), n=1..10);

Table[(Ceiling[n/k]!)^Mod[n, k]*(Floor[n/k]!)^(k - Mod[n, k]), {n, 10}, {k, n}] // Flatten (* Michael De Vlieger, Jun 28 2020 *)

T(n,k) = (ceiling(n/k)!)^(n mod k)*(floor(n/k)!)^(k-(n mod k)) for 1 <= k <= n.
T(n,1) = A000142(n).
T(n,2) = A010551(n) for n > 1.
T(n,3) = A264557(n) for n > 2.
T(n,4) = A264635(n) for n > 3.
T(n,5) = A264656(n) for n > 4.
T(n,k) = Product_{i=0..k-1} floor((n+i)/k)!. - Alois P. Heinz, May 23 2020

cp's OEIS Frontend

A179079 Permanent of the n X n matrix whose (i,j)-element is (i+j-1) modulo 3.

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A293783 Triangle of numbers of squares {i^2}, i = 0,1..ceiling(n/2), in permutations of {1..n} in A293857.

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A171645 Partial products of Product_{n=1..inf.} (p(n)/p(n-1)*p(n)/p(n-1)), = 2*2*2*(3/2)*(3/2)*(5/3)*(5/3)*(7/5)*(7/5)*(11/7)*(11/7)*...; p = primes, A000040, a(1) = 2.

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A171646 a(1) = 1, then partial products of Product_{n>=1} (p(n)/p(n-1)*p(n)/p(n-1)) = 1*1*1*(2)*(2)*(3/2)*(3/2)*(5/3)*(5/3)*(7/5)*(7/5)*...*; p = partition numbers, A000041 starting (1, 2, 3, 5, ...).

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A203151 (n-1)-st elementary symmetric function of {1,1,2,2,3,3,4,4,5,5,...,Floor[(n+1)/2]}.

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Mathematica

A354208 Number of parity-alternating permutations of [n] avoiding the pattern 321.

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A080392 Numbers k such that A000984(k) mod k = 0 and A080383(k) != 7.

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A139769 T(n,k) = [x^k] Product_{m=1..n} d/dx Sum_{i=1..m} x^i; triangle read by rows, n >= 0, 0 <= k <= A161680(n).

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A171645 Partial products of Product_{n=1..inf.} (p(n)/p(n-1)p(n)/p(n-1)), = 222(3/2)(3/2)(5/3)(5/3)(7/5)(7/5)(11/7)(11/7)...; p = primes, A000040, a(1) = 2.

A171646 a(1) = 1, then partial products of Product_{n>=1} (p(n)/p(n-1)p(n)/p(n-1)) = 11*1(2)(2)(3/2)(3/2)(5/3)(5/3)(7/5)(7/5)...; p = partition numbers, A000041 starting (1, 2, 3, 5, ...).