A000097 -id:A000097 - cp's OEIS frontend

A344619 The a(n)-th composition in standard order (A066099) has alternating sum 0.

0, 3, 10, 13, 15, 36, 41, 43, 46, 50, 53, 55, 58, 61, 63, 136, 145, 147, 150, 156, 162, 165, 167, 170, 173, 175, 180, 185, 187, 190, 196, 201, 203, 206, 210, 213, 215, 218, 221, 223, 228, 233, 235, 238, 242, 245, 247, 250, 253, 255, 528, 545, 547, 550, 556, 568

Offset: 1

Gus Wiseman, Jun 03 2021

nonn

The alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i.

The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

			The sequence of terms together with the corresponding compositions begins:
    0: ()
    3: (1,1)
   10: (2,2)
   13: (1,2,1)
   15: (1,1,1,1)
   36: (3,3)
   41: (2,3,1)
   43: (2,2,1,1)
   46: (2,1,1,2)
   50: (1,3,2)
   53: (1,2,2,1)
   55: (1,2,1,1,1)
   58: (1,1,2,2)
   61: (1,1,1,2,1)
   63: (1,1,1,1,1,1)
  136: (4,4)
  145: (3,4,1)
  147: (3,3,1,1)
  150: (3,2,1,2)
  156: (3,1,1,3)

The version for Heinz numbers of partitions is A000290, counted by A000041.
These are the positions of zeros in A344618.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A116406 counts compositions with alternating sum >= 0.
A124754 gives the alternating sum of standard compositions.
A316524 is the alternating sum of the prime indices of n.
A344604 counts wiggly compositions with twins.
A344610 counts partitions by sum and positive reverse-alternating sum.
A344611 counts partitions of 2n with reverse-alternating sum >= 0.
A344616 gives the alternating sum of reversed prime indices.
All of the following pertain to compositions in standard order:
- The length is A000120.
- Converting to reversed ranking gives A059893.
- The rows are A066099.
- The sum is A070939.
- The runs are counted by A124767.
- The reversed version is A228351.
- Strict compositions are ranked by A233564.
- Constant compositions are ranked by A272919.
- The Heinz number is A333219.
- Anti-run compositions are ranked by A333489.
Cf. A000070, A000097, A003242, A006330, A028260, A119899, A239830, A344605, A344607, A344650, A344739.

ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
stc[n_]:=Reverse[Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]]
Select[Range[0,100],ats[stc[#]]==0&]

A345197 Concatenation of square matrices A(n), each read by rows, where A(n)(k,i) is the number of compositions of n of length k with alternating sum i, where 1 <= k <= n, and i ranges from -n + 2 to n in steps of 2.

Original entry on oeis.org

1, 0, 1, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 2, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 2, 3, 0, 0, 2, 2, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 2, 3, 4, 0, 0, 3, 4, 3, 0, 0, 0, 0, 2, 3, 0, 0, 0, 0, 1, 0, 0, 0

Offset: 0

Gus Wiseman, Jul 03 2021

The alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i.

			The matrices for n = 1..7:
  1   0 1   0 0 1   0 0 0 1   0 0 0 0 1   0 0 0 0 0 1   0 0 0 0 0 0 1
      1 0   1 1 0   1 1 1 0   1 1 1 1 0   1 1 1 1 1 0   1 1 1 1 1 1 0
            0 1 0   0 1 2 0   0 1 2 3 0   0 1 2 3 4 0   0 1 2 3 4 5 0
                    0 1 0 0   0 2 2 0 0   0 3 4 3 0 0   0 4 6 6 4 0 0
                              0 0 1 0 0   0 0 2 3 0 0   0 0 3 6 6 0 0
                                          0 0 1 0 0 0   0 0 3 3 0 0 0
                                                        0 0 0 1 0 0 0
Matrix n = 5 counts the following compositions:
           i=-3:        i=-1:          i=1:            i=3:        i=5:
        -----------------------------------------------------------------
   k=1: |    0            0             0               0          (5)
   k=2: |   (14)         (23)          (32)            (41)         0
   k=3: |    0          (131)       (221)(122)   (311)(113)(212)    0
   k=4: |    0       (1211)(1112)  (2111)(1121)         0           0
   k=5: |    0            0          (11111)            0           0

Gus Wiseman, A raster plot of the zeros in A(16).

The number of nonzero terms in each matrix appears to be A000096.
The number of zeros in each matrix appears to be A000124.
Row sums and column sums both appear to be A007318 (Pascal's triangle).
The matrix sums are A131577.
Antidiagonal sums appear to be A163493.
The reverse-alternating version is also A345197 (this sequence).
Antidiagonals are A345907.
Traces are A345908.
A000041 counts partitions of 2n with alternating sum 0, ranked by A000290.
A011782 counts compositions.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A316524 gives the alternating sum of prime indices (reverse: A344616).
A344610 counts partitions by sum and positive reverse-alternating sum.
A344611 counts partitions of 2n with reverse-alternating sum >= 0.
Other tetrangles: A318393, A318816, A320808, A321912.
Compositions of n, 2n, or 2n+1 with alternating/reverse-alternating sum k:
- k = 0:  counted by A088218, ranked by A344619/A344619.
- k = 1:  counted by A000984, ranked by A345909/A345911.
- k = -1: counted by A001791, ranked by A345910/A345912.
- k = 2:  counted by A088218, ranked by A345925/A345922.
- k = -2: counted by A002054, ranked by A345924/A345923.
- k >= 0: counted by A116406, ranked by A345913/A345914.
- k <= 0: counted by A058622(n-1), ranked by A345915/A345916.
- k > 0:  counted by A027306, ranked by A345917/A345918.
- k < 0:  counted by A294175, ranked by A345919/A345920.
- k != 0: counted by A058622, ranked by A345921/A345921.
- k even: counted by A081294, ranked by A053754/A053754.
- k odd:  counted by A000302, ranked by A053738/A053738.
Cf. A000070, A000097, A000346, A007318, A008549, A025047, A032443, A034871, A114121, A120452, A238279, A239830, A344604.

ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
Table[Length[Select[Join@@Permutations/@IntegerPartitions[n],Length[#]==k&&ats[#]==i&]],{n,0,6},{k,1,n},{i,-n+2,n,2}]

A344611 Number of integer partitions of 2n with reverse-alternating sum >= 0.

Original entry on oeis.org

1, 2, 4, 8, 15, 27, 48, 81, 135, 220, 352, 553, 859, 1313, 1986, 2969, 4394, 6439, 9357, 13479, 19273, 27353, 38558, 53998, 75168, 104022, 143172, 196021, 267051, 362086, 488733, 656802, 879026, 1171747, 1555997, 2058663, 2714133, 3566122, 4670256, 6096924, 7935184

Offset: 0

Gus Wiseman, May 30 2021

nonn

The reverse-alternating sum of a partition (y_1,...,y_k) is Sum_i (-1)^(k-i) y_i.
Also the number of reversed integer partitions of 2n with alternating sum >= 0.
The reverse-alternating sum of a partition is equal to (-1)^(k-1) times the number of odd parts in the conjugate partition, where k is the number of parts. So a(n) is the number of partitions of 2n whose conjugate parts are all even or whose length is odd. By conjugation, this is also the number of partitions of 2n whose parts are all even or whose greatest part is odd.

			The a(0) = 1 through a(4) = 15 partitions:
  ()  (2)   (4)     (6)       (8)
      (11)  (22)    (33)      (44)
            (211)   (222)     (332)
            (1111)  (321)     (422)
                    (411)     (431)
                    (2211)    (521)
                    (21111)   (611)
                    (111111)  (2222)
                              (3311)
                              (22211)
                              (32111)
                              (41111)
                              (221111)
                              (2111111)
                              (11111111)

The non-reversed version is A058696 (partitions of 2n).
The ordered version appears to be A114121.
Odd bisection of A344607.
Row sums of A344610.
The strict case is A344650.
A000041 counts partitions of 2n with alternating sum 0, ranked by A000290.
A000070 counts partitions with alternating sum 1.
A000097 counts partitions with alternating sum 2.
A103919 counts partitions by sum and alternating sum.
A120452 counts partitions of 2n with reverse-alternating sum 2.
A316524 is the alternating sum of the prime indices of n (reverse: A344616).
A325534/A325535 count separable/inseparable partitions.
A344612 counts partitions by sum and rev-alt sum (strict: A344739).
A344618 gives reverse-alternating sums of standard compositions.
A344741 counts partitions of 2n with reverse-alternating sum -2.
Cf. A001250, A027187, A028260, A116406, A119899, A124754, A152146, A239829, A344608, A344609, A344649, A344651, A344654.

sats[y_]:=Sum[(-1)^(i-Length[y])*y[[i]],{i,Length[y]}];
Table[Length[Select[IntegerPartitions[n],sats[#]>=0&]],{n,0,30,2}]

Conjecture: a(n) <= A160786(n). The difference is 0, 0, 0, 0, 1, 2, 4, 9, 16, 28, 48, 79, ...

More terms from Bert Dobbelaere, Jun 12 2021

A120452 Number of partitions of n-1 boys and one girl with no couple.

Original entry on oeis.org

1, 1, 3, 5, 9, 14, 23, 34, 52, 75, 109, 153, 216, 296, 407, 549, 739, 981, 1300, 1702, 2224, 2879, 3716, 4761, 6083, 7721, 9774, 12306, 15450, 19307, 24064, 29867, 36978, 45614, 56130, 68846, 84250, 102793, 125148, 151955, 184123, 222553, 268482

Offset: 1

Yasutoshi Kohmoto, Jul 20 2006

From Gus Wiseman, Jun 08 2021: (Start)
Also the number of:
- integer partitions of 2n with reverse-alternating sum 2;
- reversed integer partitions of 2n with alternating sum 2;
- integer partitions of 2n with exactly two odd parts, one of which is the greatest;
- odd-length integer partitions of 2n whose conjugate partition has exactly two odd parts.
Note that integer partitions of 2n with alternating or reverse-alternating sum 0 are counted by A000041, ranked by A000290.
(End)

			n=5:
If partitions have no pair "o*", then a(5)=9 ("o" means a boy, "*" means a girl): {o, o, o, o, *}, {o, o, *, oo}, {*, oo, oo}, {o, *, ooo}, {o, o, oo*}, {oo, oo*}, {*, oooo}, {o, ooo*}, {oooo*}.
From _Gus Wiseman_, Jun 08 2021: (Start)
The a(1) = 1 through a(6) = 14 partitions of 2n with reverse-alternating sum 2:
  (2)  (211)  (222)    (332)      (442)        (552)
              (321)    (431)      (541)        (651)
              (21111)  (22211)    (22222)      (33222)
                       (32111)    (32221)      (33321)
                       (2111111)  (33211)      (43221)
                                  (43111)      (44211)
                                  (2221111)    (54111)
                                  (3211111)    (2222211)
                                  (211111111)  (3222111)
                                               (3321111)
                                               (4311111)
                                               (222111111)
                                               (321111111)
                                               (21111111111)
For example, the partition (43221) has reverse-alternating sum 1 - 2 + 2 - 3 + 4 = 2, so is counted under a(6).
The a(1) = 1 through a(6) = 14 partitions of 2n with exactly two odd parts, one of which is the greatest:
  (11)  (31)  (33)   (53)    (55)     (75)
              (51)   (71)    (73)     (93)
              (321)  (332)   (91)     (111)
                     (521)   (532)    (543)
                     (3221)  (541)    (552)
                             (721)    (732)
                             (3322)   (741)
                             (5221)   (921)
                             (32221)  (5322)
                                      (5421)
                                      (7221)
                                      (33222)
                                      (52221)
                                      (322221)
(End)

Vaclav Kotesovec, Table of n, a(n) for n = 1..10000

Cf. A000041, A000070, A002865.
A diagonal of A103919.
A diagonal of A344612.
A000097 counts partitions of 2n with alternating sum 2.
A001700/A088218 appear to count compositions with reverse-alternating sum 2.
A058696 counts partitions of 2n, ranked by A300061.
A344610 counts partitions of 2n by sum and positive reverse-alternating sum.
A344611 counts partitions of 2n with reverse-alternating sum >= 0.
A344741 counts partitions of 2n with reverse-alternating sum -2.
Cf. A001250, A027187, A119899, A124754, A239830, A316524, A325535, A344618, A344651, A344739.

a[n_] := Total[PartitionsP[Range[0, n-3]]] + PartitionsP[n-1];
Array[a, 50] (* Jean-François Alcover, Jun 05 2021 *)

a(n) = A000070(n-2) + A002865(n-1). - Fung Cheok Yin (cheokyin_restart(AT)yahoo.com.hk), Aug 15 2006
a(n) = A000070(n-1) - A000041(n-2) = A000070(n-3) + A000041(n-1). - Max Alekseyev, Aug 23 2006
a(n) ~ exp(Pi*sqrt(2*n/3)) / (2^(3/2)*Pi*sqrt(n)) * (1 - 37*Pi/(24*sqrt(6*n))). - Vaclav Kotesovec, Oct 25 2016

More terms from Fung Cheok Yin (cheokyin_restart(AT)yahoo.com.hk), Aug 15 2006

More terms from Max Alekseyev, Aug 23 2006

A024786 Number of 2's in all partitions of n.

Original entry on oeis.org

0, 1, 1, 3, 4, 8, 11, 19, 26, 41, 56, 83, 112, 160, 213, 295, 389, 526, 686, 911, 1176, 1538, 1968, 2540, 3223, 4115, 5181, 6551, 8191, 10269, 12756, 15873, 19598, 24222, 29741, 36532, 44624, 54509, 66261, 80524, 97446, 117862, 142029, 171036, 205290, 246211

Offset: 1

Clark Kimberling

nonn

Also number of partitions of n-1 with a distinguished part different from all the others. [Comment corrected by Emeric Deutsch, Aug 13 2008]
In general the number of times that j appears in the partitions of n equals Sum_{kA024787, ..., A024794, for j = 2,...,10; it generalizes the formula given for A000070 for j=1. - Jose Luis Arregui (arregui(AT)posta.unizar.es), Apr 05 2002
Equals row sums of triangle A173238. - Gary W. Adamson, Feb 13 2010
The sums of two successive terms give A000070. - Omar E. Pol, Jul 12 2012
a(n) is also the difference between the sum of second largest and the sum of third largest elements in all partitions of n. More generally, the number of occurrences of k in all partitions of n equals the difference between the sum of k-th largest and the sum of (k+1)st largest elements in all partitions of n. And more generally, the sum of the number of occurrences of k, k+1, k+2..k+m in all partitions of n equals the difference between the sum of k-th largest and the sum of (k+m+1)st largest elements in all partitions of n. - Omar E. Pol, Oct 25 2012
Number of singletons in all partitions of n-1. A singleton in a partition is a part that occurs exactly once. Example: a(5) = 4 because in the partitions of 4, namely [1,1,1,1], [1,1,2'], [2,2], [1',3'], [4'] we have 4 singletons (marked by '). - Emeric Deutsch, Sep 12 2016
a(n) is also the number of non-isomorphic vertex-transitive cover graphs of lattice quotients of essential lattice congruences of the weak order on the symmetric group S_{n-1}. See Table 1 in the Hoang/Mütze reference in the Links section. - Torsten Muetze, Nov 28 2019
Assuming a partition is in weakly decreasing order, a(n) is also the number of times -1 occurs in the differences of the partitions of n+1. - George Beck, Mar 28 2023

			From _Omar E. Pol_, Oct 25 2012: (Start)
For n = 7 we have:
--------------------------------------
.                             Number
Partitions of 7               of 2's
--------------------------------------
7 .............................. 0
4 + 3 .......................... 0
5 + 2 .......................... 1
3 + 2 + 2 ...................... 2
6 + 1 .......................... 0
3 + 3 + 1 ...................... 0
4 + 2 + 1 ...................... 1
2 + 2 + 2 + 1 .................. 3
5 + 1 + 1 ...................... 0
3 + 2 + 1 + 1 .................. 1
4 + 1 + 1 + 1 .................. 0
2 + 2 + 1 + 1 + 1 .............. 2
3 + 1 + 1 + 1 + 1 .............. 0
2 + 1 + 1 + 1 + 1 + 1 .......... 1
1 + 1 + 1 + 1 + 1 + 1 + 1 ...... 0
------------------------------------
.  24 - 13 =                    11
.
The difference between the sum of the second column and the sum of the third column of the set of partitions of 7 is 24 - 13 = 11 and equals the number of 2's in all partitions of 7, so a(7) = 11.
(End)

J. Riordan, Combinatorial Identities, Wiley, 1968, p. 184.

Alois P. Heinz and Vaclav Kotesovec, Table of n, a(n) for n = 1..10000 (terms 1..1000 from Alois P. Heinz)
David Benson, Radha Kessar, and Markus Linckelmann, Hochschild cohomology of symmetric groups in low degrees, arXiv:2204.09970 [math.GR], 2022.
Philip Cuthbertson, Fixed hooks in arbitrary columns of partitions, Integers (2025) Vol. 25, Art. No. A28. See p. 3.
Manosij Ghosh Dastidar and Sourav Sen Gupta, Generalization of a few results in Integer Partitions, arXiv preprint arXiv:1111.0094 [cs.DM], 2011.
Emeric Deutsch et al., Problem 11237, Amer. Math. Monthly, 115 (No. 7, 2008), 666-667. [From _Emeric Deutsch_, Aug 13 2008]
Hung Phuc Hoang and Torsten Mütze, Combinatorial generation via permutation languages. II. Lattice congruences, arXiv:1911.12078 [math.CO], 2019.
Joseph Vandehey, Digital problems in the theory of partitions, Integers (2024) Vol. 24A, Art. No. A18. See p. 3.

Cf. A066633, A024787, A024788, A024789, A024790, A024791, A024792, A024793, A024794, A173238.
Column 2 of A060244.
First differences of A000097.

b:= proc(n, i) option remember; local f, g;
      if n=0 or i=1 then [1, 0]
    else f:= b(n, i-1); g:= `if`(i>n, [0$2], b(n-i, i));
         [f[1]+g[1], f[2]+g[2]+`if`(i=2, g[1], 0)]
      fi
    end:
a:= n-> b(n, n)[2]:
seq(a(n), n=1..50);  # Alois P. Heinz, May 18 2012

Table[ Count[ Flatten[ IntegerPartitions[n]], 2], {n, 1, 50} ]
(* Second program: *)
b[n_, i_] := b[n, i] = Module[{f, g}, If[n==0 || i==1, {1, 0}, f = b[n, i - 1]; g = If[i>n, {0, 0}, b[n-i, i]]; {f[[1]] + g[[1]], f[[2]] + g[[2]] + If[i == 2, g[[1]], 0]}]]; a[n_] := b[n, n][[2]]; Table[a[n], {n, 1, 50}] (* Jean-François Alcover, Sep 22 2015, after Alois P. Heinz *)
Join[{0}, (1/((1 - x^2) QPochhammer[x]) + O[x]^50)[[3]]] (* Vladimir Reshetnikov, Nov 22 2016 *)
Table[Sum[(1 + (-1)^k)/2 * PartitionsP[n-k], {k, 2, n}], {n, 1, 50}] (* Vaclav Kotesovec, Aug 27 2017 *)

from sympy import npartitions
def A024786(n): return sum(npartitions(n-(k<<1)) for k in range(1,(n>>1)+1)) # Chai Wah Wu, Oct 25 2023

a(n) = Sum_{k=1..floor(n/2)} A000041(n-2k). - Christian G. Bower, Jun 22 2000
a(n) = Sum_{kA000041, P(0) = 1. - Jose Luis Arregui (arregui(AT)posta.unizar.es), Apr 05 2002
G.f.: (x^2/((1-x)*(1-x^2)^2))*Product_{j>=3} 1/(1-x^j) from Riordan reference second term, last eq.
a(n) = A006128(n-1) - A194452(n-1). - Omar E. Pol, Nov 20 2011
a(n) = A181187(n,2) - A181187(n,3). - Omar E. Pol, Oct 25 2012
a(n) ~ exp(Pi*sqrt(2*n/3)) / (2^(5/2) * Pi * sqrt(n)) * (1 - 25*Pi/(24*sqrt(6*n)) + (25/48 + 433*Pi^2/6912)/n). - Vaclav Kotesovec, Mar 07 2016, extended Nov 05 2016
a(n) = Sum_{k} k * A116595(n-1,k). - Emeric Deutsch, Sep 12 2016
G.f.: x^2/((1 - x)*(1 - x^2)) * Sum_{n >= 0} x^(2*n)/( Product_{k = 1..n} 1 - x^k ); that is, convolution of A004526 (partitions into 2 parts, or, modulo offset differences, partitions into parts <= 2) and A002865 (partitions into parts >= 2). - Peter Bala, Jan 17 2021

A026810 Number of partitions of n in which the greatest part is 4.

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 2, 3, 5, 6, 9, 11, 15, 18, 23, 27, 34, 39, 47, 54, 64, 72, 84, 94, 108, 120, 136, 150, 169, 185, 206, 225, 249, 270, 297, 321, 351, 378, 411, 441, 478, 511, 551, 588, 632, 672, 720, 764, 816, 864, 920, 972, 1033, 1089, 1154, 1215, 1285, 1350

Offset: 0

Clark Kimberling

Also number of partitions of n into exactly 4 parts.
Also the number of weighted cubic graphs on 4 nodes (=the tetrahedron) with weight n. - R. J. Mathar, Nov 03 2018
From Gus Wiseman, Jun 27 2021: (Start)
Also the number of strict integer partitions of 2n with alternating sum 4, or (by conjugation) partitions of 2n covering an initial interval of positive integers with exactly 4 odd parts. The strict partitions with alternating sum 4 are:
  (4)  (5,1)  (6,2)    (7,3)    (8,4)      (9,5)      (10,6)
              (5,2,1)  (5,3,2)  (5,4,3)    (6,5,3)    (7,6,3)
                       (6,3,1)  (6,4,2)    (7,5,2)    (8,6,2)
                                (7,4,1)    (8,5,1)    (9,6,1)
                                (6,3,2,1)  (6,4,3,1)  (6,5,4,1)
                                           (7,4,2,1)  (7,4,3,2)
                                                      (7,5,3,1)
                                                      (8,5,2,1)
                                                      (6,4,3,2,1)
(End)

			From _Gus Wiseman_, Jun 27 2021: (Start)
The a(4) = 1 through a(10) = 9 partitions of length 4:
  (1111)  (2111)  (2211)  (2221)  (2222)  (3222)  (3322)
                  (3111)  (3211)  (3221)  (3321)  (3331)
                          (4111)  (3311)  (4221)  (4222)
                                  (4211)  (4311)  (4321)
                                  (5111)  (5211)  (4411)
                                          (6111)  (5221)
                                                  (5311)
                                                  (6211)
                                                  (7111)
(End)

G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, p. 275.
D. E. Knuth, The Art of Computer Programming, vol. 4,fascicle 3, Generating All Combinations and Partitions, Section 7.2.1.4., p. 56, exercise 31.

Washington Bomfim, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (1,1,0,0,-2,0,0,1,1,-1).

Cf. A001400, A026811, A026812, A026813, A026814, A026815, A026816, A069905 (3 positive parts), A002621 (partial sums), A005044 (first differences).
A non-strict version is A000710 or A088218.
This is column k = 2 of A152146.
A reverse version is A343941.
Cf. A000041, A000070, A000097, A067659, A103919, A120452, A236559, A239830, A306145, A343942, A344616, A344649, A344651.

[Round((n^3+3*n^2-9*n*(n mod 2))/144): n in [0..60]]; // Vincenzo Librandi, Oct 14 2015

A049347 := proc(n)
        op(1+(n mod 3),[1,-1,0]) ;
end proc:
A056594 := proc(n)
        op(1+(n mod 4),[1,0,-1,0]) ;
end proc:
A026810 := proc(n)
        1/288*(n+1)*(2*n^2+4*n-13+9*(-1)^n) ;
        %-A049347(n)/9 ;
        %+A056594(n)/8 ;
end proc: # R. J. Mathar, Jul 03 2012

Table[Count[IntegerPartitions[n], {4, _}], {n, 0, 60}]
LinearRecurrence[{1, 1, 0, 0, -2, 0, 0, 1, 1, -1}, {0, 0, 0, 0, 1, 1, 2, 3, 5, 6}, 60] (* Vincenzo Librandi, Oct 14 2015 *)
Table[Length[IntegerPartitions[n, {4}]], {n, 0, 60}] (* Eric Rowland, Mar 02 2017 *)
CoefficientList[Series[x^4/Product[1 - x^k, {k, 1, 4}], {x, 0, 60}], x] (* Robert A. Russell, May 13 2018 *)

for(n=0, 60, print(n, " ", round((n^3 + 3*n^2 -9*n*(n % 2))/144))); \\ Washington Bomfim, Jul 03 2012

x='x+O('x^60); concat([0, 0, 0, 0], Vec(x^4/((1-x)*(1-x^2)*(1-x^3)*(1-x^4)))) \\ Altug Alkan, Oct 14 2015

vector(60, n, n--; (n+1)*(2*n^2+4*n-13+9*(-1)^n)/288 + real(I^n)/8 - ((n+2)%3-1)/9) \\ Altug Alkan, Oct 26 2015

print1(0,", "); for(n=1,60,j=0;forpart(v=n,j++,,[4,4]); print1(j,", ")) \\ Hugo Pfoertner, Oct 01 2018

G.f.: x^4/((1-x)*(1-x^2)*(1-x^3)*(1-x^4)) = x^4/((1-x)^4*(1+x)^2*(1+x+x^2)*(1+x^2)).
a(n+4) = A001400(n). - Michael Somos, Apr 07 2012
a(n) = round( (n^3 + 3*n^2 -9*n*(n mod 2))/144 ). - Washington Bomfim, Jan 06 2021 and Jul 03 2012
a(n) = (n+1)*(2*n^2+4*n-13+9*(-1)^n)/288 -A049347(n)/9 +A056594(n)/8. - R. J. Mathar, Jul 03 2012
From Gregory L. Simay, Oct 13 2015: (Start)
a(n) = (n^3 + 3*n^2 - 9*n)/144 + a(m) - (m^3 + 3*m^2 - 9*m)/144 if n = 12k + m and m is odd. For example, a(23) = a(12*1 + 11) = (23^3 + 3*23^2 - 9*23)/144 + a(11) - (11^3 + 3*11^2 - 9*11)/144 = 94.
a(n) = (n^3 + 3*n^2)/144 + a(m) - (m^3 + 3*m^2)/144 if n = 12k + m and m is even. For example, a(22) = a(12*1 + 10) = (22^3 + 3*22^2)/144 + a(10) - (10^3 + 3*10^2)/144 = 84. (End)
a(n) = A008284(n,4). - Robert A. Russell, May 13 2018
From Gregory L. Simay, Jul 28 2019: (Start)
a(2n+1) = a(2n) + a(n+1) - a(n-3) and
a(2n) = a(2n-1) + a(n+2) - a(n-2). (End)

A344608 Number of integer partitions of n with reverse-alternating sum < 0.

Original entry on oeis.org

0, 0, 0, 1, 1, 3, 3, 7, 7, 14, 15, 27, 29, 49, 54, 86, 96, 146, 165, 242, 275, 392, 449, 623, 716, 973, 1123, 1498, 1732, 2274, 2635, 3411, 3955, 5059, 5871, 7427, 8620, 10801, 12536, 15572, 18065, 22267, 25821, 31602, 36617, 44533, 51560, 62338, 72105, 86716

Offset: 0

Gus Wiseman, May 30 2021

nonn

The reverse-alternating sum of a partition (y_1,...,y_k) is Sum_i (-1)^(k-i) y_i.
Also the number of reversed of integer partitions of n with alternating sum < 0.
No integer partitions have alternating sum < 0, so the non-reversed version is all zeros.
Is this sequence weakly increasing? Note: a(2n + 2) = A236914(n), a(2n) = A344743(n).
A formula for the reverse-alternating sum of a partition is: (-1)^(k-1) times the number of odd parts in the conjugate partition, where k is the number of parts. So a(n) is the number of integer partitions of n of even length whose conjugate parts are not all odd. Partitions of the latter type are counted by A086543. By conjugation, a(n) is also the number of integer partitions of n of even maximum whose parts are not all odd.

			The a(3) = 1 through a(9) = 14 partitions:
  (21)  (31)  (32)    (42)    (43)      (53)      (54)
              (41)    (51)    (52)      (62)      (63)
              (2111)  (3111)  (61)      (71)      (72)
                              (2221)    (3221)    (81)
                              (3211)    (4211)    (3222)
                              (4111)    (5111)    (3321)
                              (211111)  (311111)  (4221)
                                                  (4311)
                                                  (5211)
                                                  (6111)
                                                  (222111)
                                                  (321111)
                                                  (411111)
                                                  (21111111)

The opposite version (rev-alt sum > 0) is A027193, ranked by A026424.
The strict case (for n > 2) is A067659 (odd bisection: A344650).
The Heinz numbers of these partitions are A119899 (complement: A344609).
The bisections are A236914 (odd) and A344743 (even).
The ordered version appears to be A294175 (even bisection: A008549).
The complement is counted by A344607 (even bisection: A344611).
A000041 counts partitions of 2n with alternating sum 0, ranked by A000290.
A027187 counts partitions with alternating sum <= 0, ranked by A028260.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A120452 counts partitions with rev-alternating sum 2 (negative: A344741).
A316524 is the alternating sum of the prime indices of n (reverse: A344616).
A325534/A325535 count separable/inseparable partitions.
A344604 counts wiggly compositions with twins.
A344610 counts partitions by sum and positive reverse-alternating sum.
A344618 gives reverse-alternating sums of standard compositions.
Cf. A000070, A000097, A000346, A003242, A006330, A071322, A239829, A239830, A344649, A344651, A344654/A344740, A344739.

sats[y_]:=Sum[(-1)^(i-Length[y])*y[[i]],{i,Length[y]}];
Table[Length[Select[IntegerPartitions[n],sats[#]<0&]],{n,0,30}]

A163493 Number of binary strings of length n which have the same number of 00 and 01 substrings.

Original entry on oeis.org

1, 2, 2, 3, 6, 9, 15, 30, 54, 97, 189, 360, 675, 1304, 2522, 4835, 9358, 18193, 35269, 68568, 133737, 260802, 509132, 995801, 1948931, 3816904, 7483636, 14683721, 28827798, 56637969, 111347879, 219019294, 431043814, 848764585, 1672056525, 3295390800, 6497536449

Offset: 0

Christopher Carl Heckman, Jul 29 2009

nonn

A variation of problem 11424 in the American Mathematical Monthly. Terms were brute-force calculated using Maple 10.
Proposed Problem 11610 in the Dec 2011 A.M.M.
From Gus Wiseman, Jul 27 2021: (Start)
Also the antidiagonal sums of the matrices counting integer compositions by length and alternating sum (A345197). So a(n) is the number of integer compositions of n + 1 of length (n - s + 3)/2, where s is the alternating sum of the composition. For example, the a(0) = 1 through a(6) = 7 compositions are:
  (1)  (2)   (3)   (4)    (5)     (6)     (7)
       (11)  (21)  (31)   (41)    (51)    (61)
                   (121)  (122)   (123)   (124)
                          (221)   (222)   (223)
                          (1112)  (321)   (322)
                          (1211)  (1122)  (421)
                                  (1221)  (1132)
                                  (2112)  (1231)
                                  (2211)  (2122)
                                          (2221)
                                          (3112)
                                          (3211)
                                          (11131)
                                          (12121)
                                          (13111)
For a bijection with the main (binary string) interpretation, take the run-lengths of each binary string of length n + 1 that satisfies the condition and starts with 1.
(End)

			1 + 2*x + 2*x^2 + 3*x^3 + 6*x^4 + 9*x^5 + 15*x^6 + 30*x^7 + 54*x^8 + 97*x^9 + ...
From _Gus Wiseman_, Jul 27 2021: (Start)
The a(0) = 1 though a(6) = 15 binary strings:
  ()  (0)  (1,0)  (0,0,1)  (0,0,1,0)  (0,0,1,1,0)  (0,0,0,1,0,1)
      (1)  (1,1)  (1,1,0)  (0,0,1,1)  (0,0,1,1,1)  (0,0,1,0,0,1)
                  (1,1,1)  (0,1,0,0)  (0,1,1,0,0)  (0,0,1,1,1,0)
                           (1,0,0,1)  (1,0,0,1,0)  (0,0,1,1,1,1)
                           (1,1,1,0)  (1,0,0,1,1)  (0,1,0,0,0,1)
                           (1,1,1,1)  (1,0,1,0,0)  (0,1,1,1,0,0)
                                      (1,1,0,0,1)  (1,0,0,1,1,0)
                                      (1,1,1,1,0)  (1,0,0,1,1,1)
                                      (1,1,1,1,1)  (1,0,1,1,0,0)
                                                   (1,1,0,0,1,0)
                                                   (1,1,0,0,1,1)
                                                   (1,1,0,1,0,0)
                                                   (1,1,1,0,0,1)
                                                   (1,1,1,1,1,0)
                                                   (1,1,1,1,1,1)
(End)

Alois P. Heinz, Table of n, a(n) for n = 0..3328 (first 501 terms from R. H. Hardin)
Shalosh B. Ekhad and Doron Zeilberger, Automatic Solution of Richard Stanley's Amer. Math. Monthly Problem #11610 and ANY Problem of That Type, arXiv preprint arXiv:1112.6207 [math.CO], 2011. See subpages for rigorous derivations of g.f., recurrence, asymptotics for this sequence. [From _N. J. A. Sloane_, Apr 07 2012]
R. Stanley, Problem 11610, Amer. Math. Monthly, 118 (2011), 937; 120 (2013), 943-944.

Antidiagonal sums of the matrices A345197.
Row sums of A345907.
Taking diagonal instead of antidiagonal sums gives A345908.
A011782 counts compositions (or binary strings).
A097805 counts compositions by alternating (or reverse-alternating) sum.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A316524 gives the alternating sum of prime indices (reverse: A344616).
Compositions of n, 2n, or 2n+1 with alternating/reverse-alternating sum k:
- k = 0:  counted by A088218, ranked by A344619/A344619.
- k = 1:  counted by A000984, ranked by A345909/A345911.
- k = -1: counted by A001791, ranked by A345910/A345912.
- k = 2:  counted by A088218, ranked by A345925/A345922.
- k = -2: counted by A002054, ranked by A345924/A345923.
- k >= 0: counted by A116406, ranked by A345913/A345914.
- k <= 0: counted by A058622(n-1), ranked by A345915/A345916.
- k > 0:  counted by A027306, ranked by A345917/A345918.
- k < 0:  counted by A294175, ranked by A345919/A345920.
- k != 0: counted by A058622, ranked by A345921/A345921.
- k even: counted by A081294, ranked by A053754/A053754.
- k odd:  counted by A000302, ranked by A053738/A053738.
Cf. A000041, A000070, A000096, A000097, A000124, A000346, A007318, A008549, A025047, A131577, A238279.

with(combinat): count := proc(n) local S, matches, A, k, i; S := subsets(\{seq(i, i=1..n)\}): matches := 0: while not S[finished] do A := S[nextvalue](): k := 0: for i from 1 to n-1 do: if not (i in A) and not (i+1 in A) then k := k + 1: fi: if not (i in A) and (i+1 in A) then k := k - 1: fi: od: if (k = 0) then matches := matches + 1: fi: end do; return(matches); end proc:
# second Maple program:
b:= proc(n, l, t) option remember; `if`(n-abs(t)<0, 0, `if`(n=0, 1,
      add(b(n-1, i, t+`if`(l=0, (-1)^i, 0)), i=0..1)))
    end:
a:= n-> b(n, 1, 0):
seq(a(n), n=0..36);  # Alois P. Heinz, Mar 20 2024

a[0] = 1; a[n_] := Sum[Binomial[2*k - 1, k]*Binomial[n - 2*k, k] + Binomial[2*k, k]*Binomial[n - 2*k - 1, k], {k, 0, n/3}];
Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Nov 28 2017, after Joel B. Lewis *)
Table[Length[Select[Tuples[{0,1},n],Count[Partition[#,2,1],{0,0}]==Count[Partition[#,2,1],{0,1}]&]],{n,0,10}] (* Gus Wiseman, Jul 27 2021 *)
a[0]:=1; a[n_]:=(1 + 3*HypergeometricPFQ[{1/2, 1-3*n/8, (1-n)/3, (2-n)/3, -n/3},{1, (1-n)/2, 1-n/2, -3*n/8}, -27])/2; Array[a,37,0] (* Stefano Spezia, Apr 26 2024 *)

from math import comb
def A163493(n): return 2+sum((x:=comb((k:=m<<1)-1,m)*comb(n-k,m))+(x*(n-3*m)<<1)//(n-k) for m in range(1,n//3+1)) if n else 1 # Chai Wah Wu, May 01 2024

G.f.: 1/2/(1-x) + (1+2*x)/2/sqrt((1-x)*(1-2*x)*(1+x+2*x^2)). - Richard Stanley, corrected Apr 29 2011
G.f.: (1 + sqrt( 1 + 4*x / ((1 - x) * (1 - 2*x) * (1 + x + 2*x^2)))) / (2*(1 - x)). - Michael Somos, Jan 30 2012
a(n) = sum( binomial(2*k-1, k)*binomial(n-2*k,k) + binomial(2*k, k)*binomial(n-2*k-1, k), k=0..floor(n/3)). - Joel B. Lewis, May 21 2011
Conjecture: -n*a(n) +(2+n)*a(n-1) +(3n-12)*a(n-2) +(12-n)*a(n-3) +(2n-18)*a(n-4)+(56-12n)*a(n-5) +(8n-40)*a(n-6)=0. - R. J. Mathar, Nov 28 2011
G.f. y = A(x) satisfies x = (1 - x) * (1 - 2*x) * (1 + x + 2*x^2) * y * (y * (1 - x) - 1). - Michael Somos, Jan 30 2012
Sequence a(n) satisfies 0 = a(n) * (n^2-2*n) + a(n-1) * (-3*n^2+8*n-2) + a(n-2) * (3*n^2-10*n+2) + a(n-3) * (-5*n^2+18*n-6) + a(n-4) * (8*n^2-34*n+22) + a(n-5) * (-4*n^2+20*n-16) except if n=1 or n=2. - Michael Somos, Jan 30 2012
a(n) = (1 + 3*hypergeom([1/2, 1-3*n/8, (1-n)/3, (2-n)/3, -n/3],[1, (1-n)/2, 1-n/2, -3*n/8],-27))/2 for n > 0. - Stefano Spezia, Apr 26 2024
a(n) ~ 2^n / sqrt(Pi*n). - Vaclav Kotesovec, Apr 26 2024

A277579 Number of partitions of n for which the number of even parts is equal to the positive alternating sum of the parts.

Original entry on oeis.org

1, 0, 1, 1, 1, 2, 3, 3, 4, 6, 7, 9, 13, 15, 19, 25, 31, 38, 48, 59, 74, 90, 111, 136, 166, 201, 246, 297, 357, 431, 522, 621, 745, 892, 1063, 1263, 1503, 1780, 2109, 2491, 2941, 3463, 4077, 4783, 5616, 6576, 7689, 8981, 10486, 12207, 14209, 16516, 19178, 22231

Offset: 0

Emeric Deutsch and Alois P. Heinz, Oct 20 2016

nonn

In the first Maple program (improvable) AS gives the positive alternating sum of a finite sequence s, EP gives the number of even terms of a finite sequence of positive integers.
For the specified value of n, the second Maple program lists the partitions of n counted by a(n).
Also the number of integer partitions of n with as many even parts as odd parts in the conjugate partition. - Gus Wiseman, Jul 26 2021

			a(9) = 6: [2,1,1,1,1,1,1,1], [3,2,1,1,1,1], [3,3,2,1], [4,2,2,1], [4,3,1,1], [5,4].
a(10) = 7: [1,1,1,1,1,1,1,1,1,1], [3,2,2,1,1,1], [3,3,1,1,1,1], [4,2,1,1,1,1], [4,3,2,1], [5,5], [6,4].
a(11) = 9: [2,1,1,1,1,1,1,1,1,1], [3,2,1,1,1,1,1,1], [3,3,2,1,1,1], [3,3,3,2], [4,2,2,1,1,1], [4,3,1,1,1,1], [5,2,2,2], [5,4,1,1], [6,5].

Alois P. Heinz, Table of n, a(n) for n = 0..1000

The sign-sensitive version is A035457 (aerated version of A000009).
Comparing odd parts to odd conjugate parts gives A277103.
Comparing product of parts to product of conjugate parts gives A325039.
Comparing the rev-alt sum to that of the conjugate gives A345196.
A000041 counts partitions of 2n with alternating sum 0, ranked by A000290.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A120452 counts partitions of 2n with rev-alt sum 2 (negative: A344741).
A124754 gives alternating sums of standard compositions (reverse: A344618).
A316524 is the alternating sum of the prime indices of n (reverse: A344616).
A344610 counts partitions by sum and positive reverse-alternating sum.
A344611 counts partitions of 2n with reverse-alternating sum >= 0.
Cf. A000070, A000097, A000700, A006330, A027187, A027193, A236559, A257991, A325534, A325535, A344607, A344651.

with(combinat): AS := proc (s) options operator, arrow: abs(add((-1)^(i-1)*s[i], i = 1 .. nops(s))) end proc: EP := proc (s) local ct, j: ct := 0: for j to nops(s) do if `mod`(s[j], 2) = 0 then ct := ct+1 else  end if end do: ct end proc: a := proc (n) local P, c, k: P := partition(n): c := 0: for k to nops(P) do if AS(P[k]) = EP(P[k]) then c := c+1 else  end if end do: c end proc: seq(a(n), n = 0 .. 30);
n := 8: with(combinat): AS := proc (s) options operator, arrow: abs(add((-1)^(i-1)*s[i], i = 1 .. nops(s))) end proc: EP := proc (s) local ct, j: ct := 0: for j to nops(s) do if `mod`(s[j], 2) = 0 then ct := ct+1 else  end if end do: ct end proc: P := partition(n): C := {}: for k to nops(P) do if AS(P[k]) = EP(P[k]) then C := `union`(C, {P[k]}) else  end if end do: C;
# alternative Maple program:
b:= proc(n, i, s, t) option remember; `if`(n=0,
      `if`(s=0, 1, 0), `if`(i<1, 0, b(n, i-1, s, t)+
      `if`(i>n, 0, b(n-i, i, s+t*i-irem(i+1, 2), -t))))
    end:
a:= n-> b(n$2, 0, 1):
seq(a(n), n=0..60);

b[n_, i_, s_, t_] := b[n, i, s, t] = If[n == 0, If[s == 0, 1, 0], If[i<1, 0, b[n, i-1, s, t] + If[i>n, 0, b[n-i, i, s + t*i - Mod[i+1, 2], -t]]]]; a[n_] := b[n, n, 0, 1]; Table[a[n], {n, 0, 60}] (* Jean-François Alcover, Dec 21 2016, translated from Maple *)
conj[y_]:=If[Length[y]==0,y,Table[Length[Select[y,#>=k&]],{k,1,Max[y]}]]; Table[Length[Select[IntegerPartitions[n],Count[#,?EvenQ]==Count[conj[#],?OddQ]&]],{n,0,15}] (* Gus Wiseman, Jul 26 2021 *)

def a(n):
    AS = lambda s: abs(sum((-1)^i*t for i,t in enumerate(s)))
    EP = lambda s: sum((t+1)%2 for t in s)
    return sum(AS(p) == EP(p) for p in Partitions(n))
print([a(n) for n in (0..30)]) # Peter Luschny, Oct 21 2016

A277103 Number of partitions of n for which the number of odd parts is equal to the positive alternating sum of the parts.

Original entry on oeis.org

1, 1, 0, 1, 3, 3, 1, 3, 10, 10, 4, 10, 27, 27, 13, 28, 69, 69, 37, 72, 161, 162, 96, 171, 361, 364, 230, 388, 768, 777, 522, 836, 1581, 1605, 1128, 1739, 3145, 3203, 2345, 3495, 6094, 6225, 4712, 6831, 11511, 11794, 9198, 13010, 21293, 21875, 17496, 24239

Offset: 0

Emeric Deutsch, Oct 18 2016

nonn

It follows by conjugation that the partition statistics "alternating sum" and "number of odd parts" are equidistributed. Consequently, the self-conjugate partitions satisfy the required condition.
In the first Maple program (improvable) AS gives the positive alternating sum of a finite sequence s, OP gives the number of odd terms of a finite sequence of positive integers.
For the specified value of n, the second Maple program lists the partitions of n counted by a(n).
Number of integer partitions of n with the same number of odd parts as their conjugate. - Gus Wiseman, Jun 27 2021

			a(3) = 1 because we have [2,1]. The partitions [3] and [1,1,1] do not qualify.
a(4) = 3 because we have [3,1], [2,2], and [2,1,1]. The partitions [4] and [1,1,1,1] do not qualify.

Alois P. Heinz, Table of n, a(n) for n = 0..1000

Comparing even parts to odd conjugate parts gives A277579.
Comparing product of parts to product of conjugate parts gives A325039.
The reverse version is A345196.
A000041 counts partitions of 2n with alternating sum 0, ranked by A000290.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A120452 counts partitions of 2n with rev-alt sum 2 (negative: A344741).
A124754 gives alternating sums of standard compositions (reverse: A344618).
A316524 is the alternating sum of the prime indices of n (reverse: A344616).
A344610 counts partitions by sum and positive reverse-alternating sum.
A344611 counts partitions of 2n with reverse-alternating sum >= 0.
Cf. A000070, A000097, A000700, A006330, A027187, A027193, A236559, A257991, A325534, A325535, A344607, A344651.

with(combinat): AS := proc (s) options operator, arrow: abs(add((-1)^(i-1)*s[i], i = 1 .. nops(s))) end proc: OP := proc (s) local ct, j: ct := 0: for j to nops(s) do if `mod`(s[j], 2) = 1 then ct := ct+1 else  end if end do: ct end proc: a := proc (n) local P, c, k: P := partition(n): c := 0: for k to nops(P) do if AS(P[k]) = OP(P[k]) then c := c+1 else end if end do: c end proc: seq(a(n), n = 0 .. 50);
n := 8: with(combinat): AS := proc (s) options operator, arrow: abs(add((-1)^(i-1)*s[i], i = 1 .. nops(s))) end proc: OP := proc (s) local ct, j: ct := 0: for j to nops(s) do if `mod`(s[j], 2) = 1 then ct := ct+1 else  end if end do: ct end proc: P := partition(n): C := {}: for k to nops(P) do if AS(P[k]) = OP(P[k]) then C := `union`(C, {P[k]}) else  end if end do: C;
# alternative Maple program:
b:= proc(n, i, s, t) option remember; `if`(n=0,
      `if`(s=0, 1, 0), `if`(i<1, 0, b(n, i-1, s, t)+
      `if`(i>n, 0, b(n-i, i, s+t*i-irem(i, 2), -t))))
    end:
a:= n-> b(n$2, 0, 1):
seq(a(n), n=0..60);  # Alois P. Heinz, Oct 19 2016

b[n_, i_, s_, t_] := b[n, i, s, t] = If[n == 0, If[s == 0, 1, 0], If[i<1, 0, b[n, i-1, s, t] + If[i>n, 0, b[n-i, i, s + t*i - Mod[i, 2], -t]]]]; a[n_] := b[n, n, 0, 1]; Table[a[n], {n, 0, 60}] (* Jean-François Alcover, Dec 21 2016, after Alois P. Heinz *)
conj[y_]:=If[Length[y]==0,y,Table[Length[Select[y,#>=k&]],{k,1,Max[y]}]]; Table[Length[Select[IntegerPartitions[n],Count[#,?OddQ]==Count[conj[#],?OddQ]&]],{n,0,15}] (* Gus Wiseman, Jun 27 2021 *)

cp's OEIS Frontend

A344619 The a(n)-th composition in standard order (A066099) has alternating sum 0.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

A345197 Concatenation of square matrices A(n), each read by rows, where A(n)(k,i) is the number of compositions of n of length k with alternating sum i, where 1 <= k <= n, and i ranges from -n + 2 to n in steps of 2.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A344611 Number of integer partitions of 2n with reverse-alternating sum >= 0.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Formula

Extensions

A120452 Number of partitions of n-1 boys and one girl with no couple.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

A024786 Number of 2's in all partitions of n.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Maple

Mathematica

Python

Formula

A026810 Number of partitions of n in which the greatest part is 4.

Views

Author

Keywords

Comments

Examples

References

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

PARI

PARI

PARI

Formula

A344608 Number of integer partitions of n with reverse-alternating sum < 0.

Views

Author

Keywords

Comments