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.

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A351008 Alternately strict partitions. Number of even-length integer partitions y of n such that y_i > y_{i+1} for all odd i.

Original entry on oeis.org

1, 0, 0, 1, 1, 2, 2, 3, 4, 5, 6, 8, 10, 12, 15, 19, 23, 28, 34, 41, 50, 60, 71, 85, 102, 120, 142, 168, 197, 231, 271, 316, 369, 429, 497, 577, 668, 770, 888, 1023, 1175, 1348, 1545, 1767, 2020, 2306, 2626, 2990, 3401, 3860, 4379, 4963, 5616, 6350, 7173, 8093
Offset: 0

Views

Author

Gus Wiseman, Jan 31 2022

Keywords

Comments

Write the series in the g.f. given below as Sum_{k >= 0} q^(1 + 3 + 5 + ... + 2*k-1 + 2*k)/Product_{i = 1..2*k} 1 - q^i. Since 1/Product_{i = 1..2*k} 1 - q^i is the g.f. for partitions with parts <= 2*k, we see that the k-th summand of the series is the g.f. for partitions with largest part 2*k in which every odd number less than 2*k appears at least once as a part. The partitions of this type are conjugate to (and hence equinumerous with) the partitions (y_1, y_2, ..., y_{2*k}) of even length 2*k having strict decrease y_i > y_(i+1) for all odd i < 2*k. - Peter Bala, Jan 02 2024

Examples

			The a(3) = 1 through a(13) = 12 partitions (A..C = 10..12):
  21   31   32   42   43   53     54     64     65     75     76
            41   51   52   62     63     73     74     84     85
                      61   71     72     82     83     93     94
                           3221   81     91     92     A2     A3
                                  4221   4321   A1     B1     B2
                                         5221   4331   4332   C1
                                                5321   5331   5332
                                                6221   5421   5431
                                                       6321   6331
                                                       7221   6421
                                                              7321
                                                              8221
a(10) = 6: the six partitions 64, 73, 82, 91, 4321 and 5221 listed above have conjugate partitions 222211, 2221111, 22111111, 211111111, 4321 and 43111, These are the partitions of 10 with largest part L even and such that every odd number less than L appears at least once as a part. - _Peter Bala_, Jan 02 2024
		

Crossrefs

The version for equal instead of unequal is A035363.
The alternately equal and unequal version is A035457, any length A351005.
This is the even-length case of A122129, opposite A122135.
The odd-length version appears to be A122130.
The alternately unequal and equal version is A351007, any length A351006.

Programs

  • Maple
    series(add(q^(n*(n+2))/mul(1 - q^k, k = 1..2*n), n = 0..10), q, 141):
    seq(coeftayl(%, q = 0, n), n = 0..140); # Peter Bala, Jan 03 2025
  • Mathematica
    Table[Length[Select[IntegerPartitions[n],EvenQ[Length[#]]&&And@@Table[#[[i]]!=#[[i+1]],{i,1,Length[#]-1,2}]&]],{n,0,30}]

Formula

Conjecture: a(n+1) = A122129(n+1) - A122130(n). - Gus Wiseman, Feb 21 2022
G.f.: Sum_{n >= 0} q^(n*(n+2))/Product_{k = 1..2*n} 1 - q^k = 1 + q^3 + q^4 + 2*q^5 + 2*q^6 + 3*q^7 + 4*q^8 + 5*q^9 + 6*q^10 + .... - Peter Bala, Jan 02 2024

A351012 Number of even-length integer partitions y of n such that y_i = y_{i+1} for all even i.

Original entry on oeis.org

1, 0, 1, 1, 3, 3, 5, 6, 9, 10, 13, 16, 21, 24, 29, 35, 43, 50, 60, 70, 83, 97, 113, 132, 156, 178, 206, 239, 275, 316, 365, 416, 477, 545, 620, 706, 806, 912, 1034, 1173, 1326, 1496, 1691, 1902, 2141, 2410, 2704, 3034, 3406, 3808, 4261, 4765, 5317, 5932, 6617
Offset: 0

Views

Author

Gus Wiseman, Feb 03 2022

Keywords

Examples

			The a(2) = 1 through a(8) = 9 partitions:
  (11)  (21)  (22)    (32)    (33)      (43)      (44)
              (31)    (41)    (42)      (52)      (53)
              (1111)  (2111)  (51)      (61)      (62)
                              (3111)    (2221)    (71)
                              (111111)  (4111)    (2222)
                                        (211111)  (3221)
                                                  (5111)
                                                  (311111)
                                                  (11111111)
		

Crossrefs

The ordered version (compositions) is A027383(n-2).
For odd instead of even indices we have A035363, any length A351004.
The version for unequal parts appears to be A122134, any length A122135.
This is the even-length case of A351003.
Requiring inequalities at odd positions gives A351007, any length A351006.

Programs

  • Mathematica
    Table[Length[Select[IntegerPartitions[n],EvenQ[Length[#]]&&And@@Table[#[[i]]==#[[i+1]],{i,2,Length[#]-1,2}]&]],{n,0,30}]

A357635 Numbers k such that the half-alternating sum of the partition having Heinz number k is 1.

Original entry on oeis.org

2, 8, 24, 32, 54, 128, 135, 162, 375, 384, 512, 648, 864, 875, 1250, 1715, 1944, 2048, 2160, 2592, 3773, 4374, 4802, 5000, 6000, 6144, 8192, 9317, 10368, 10935, 13122, 13824, 14000, 15000, 17303, 19208, 20000, 24167, 27440, 29282, 30375, 31104, 32768, 33750
Offset: 1

Views

Author

Gus Wiseman, Oct 28 2022

Keywords

Comments

We define the half-alternating sum of a sequence (A, B, C, D, E, F, G, ...) to be A + B - C - D + E + F - G - ...
The Heinz number of a partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k). This gives a bijective correspondence between positive integers and integer partitions.

Examples

			The terms together with their prime indices begin:
    2: {1}
    8: {1,1,1}
   24: {1,1,1,2}
   32: {1,1,1,1,1}
   54: {1,2,2,2}
  128: {1,1,1,1,1,1,1}
  135: {2,2,2,3}
  162: {1,2,2,2,2}
  375: {2,3,3,3}
  384: {1,1,1,1,1,1,1,2}
  512: {1,1,1,1,1,1,1,1,1}
  648: {1,1,1,2,2,2,2}
  864: {1,1,1,1,1,2,2,2}
  875: {3,3,3,4}
		

Crossrefs

The version for k = 0 is A000583, standard compositions A357625-A357626.
The version for original alternating sum is A345958.
Positions of ones in A357633, non-reverse A357629.
The skew version for k = 0 is A357636, non-reverse A357632.
These partitions are counted by A035444, skew A035544.
The non-reverse version is A357851, k = 0 version A357631.
A056239 adds up prime indices, row sums of A112798.
A316524 gives alternating sum of prime indices, reverse A344616.
A351005 = alternately equal and unequal partitions, compositions A357643.
A351006 = alternately unequal and equal partitions, compositions A357644.
A357641 counts comps w/ half-alt sum 0, even-length A357642.

Programs

  • Mathematica
    primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
    halfats[f_]:=Sum[f[[i]]*(-1)^(1+Ceiling[i/2]),{i,Length[f]}];
    Select[Range[1000],halfats[Reverse[primeMS[#]]]==1&]

A122130 Expansion of f(-x^4, -x^16) / psi(-x) in powers of x where psi() is a Ramanujan theta function and f(, ) is Ramanujan's general theta function.

Original entry on oeis.org

1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 11, 14, 18, 22, 27, 34, 41, 50, 61, 73, 88, 106, 126, 150, 179, 211, 249, 294, 345, 404, 473, 551, 642, 747, 865, 1002, 1159, 1336, 1539, 1771, 2033, 2331, 2670, 3052, 3485, 3976, 4527, 5150, 5854, 6642, 7530, 8529, 9647, 10902
Offset: 0

Views

Author

Michael Somos, Aug 21 2006, corrected Aug 21 2006

Keywords

Comments

Generating function arises naturally in Rodney Baxter's solution of the Hard Hexagon Model according to George Andrews.
Ramanujan theta functions: f(q) (see A121373), phi(q) (A000122), psi(q) (A010054), chi(q) (A000700).
From Gus Wiseman, Feb 19 2022: (Start)
This appears to be the number of odd-length alternately strict integer partitions of n + 1, i.e., partitions y such that y_i != y_{i+1} for all odd i. For example, the a(1) = 1 through a(9) = 7 partitions are:
(1) (2) (3) (4) (5) (6) (7) (8) (9)
(211) (311) (321) (322) (422) (432)
(411) (421) (431) (522)
(511) (521) (531)
(611) (621)
(711)
(32211)
The even-length version is A351008. Including even-length partitions appears to give A122129. Swapping strictly and weakly decreasing relations gives A351595. The constant instead of strict version is A351594. (End)
Wiseman's first conjecture above was proved by Connor, Proposition 2. - Peter Bala, Dec 22 2024

Examples

			G.f. = 1 + x + x^2 + 2*x^3 + 2*x^4 + 3*x^5 + 4*x^6 + 5*x^7 + 7*x^8 + 9*x^9 + ...
G.f. = q^31 + q^71 + q^111 + 2*q^151 + 2*q^191 + 3*q^231 + 4*q^271 + 5*q^311 + ...
		

References

  • G. E. Andrews, R. Askey and R. Roy, Special Functions, Cambridge University Press, 1999; Exercise 6(b), p. 591.
  • G. E. Andrews, q-series, CBMS Regional Conference Series in Mathematics, 66, Amer. Math. Soc. 1986, see p. 8, Eq. (1.8). MR0858826 (88b:11063)

Crossrefs

Programs

  • Mathematica
    nmax = 100; CoefficientList[Series[Product[1/((1-x^(2*k-1))*(1-x^(20*k-8))*(1-x^(20*k-12))), {k, 1, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Aug 30 2015 *)
    a[ n_] := SeriesCoefficient[ 1 / (QPochhammer[x, x^2] QPochhammer[x^8, x^20] QPochhammer[x^12, x^20]), {x, 0, n}]; (* Michael Somos, Nov 12 2016 *)
    a[ n_] := SeriesCoefficient[ Sqrt[2] x^(1/8) QPochhammer[ x^4, x^20] QPochhammer[ x^16, x^20] QPochhammer[x^20] / EllipticTheta[ 2, Pi/4, x^(1/2)], {x, 0, n}] // Simplify; (* Michael Somos, Nov 12 2016 *)
  • PARI
    {a(n) = if( n<1, n==0, polcoeff( sum(k=1, sqrtint(n+1), x^(k^2-1) / prod(i=1, 2*k-1, 1 - x^i, 1 + x * O(x^(n-k^2+1)))), n))};

Formula

Expansion of f(x, x^9) / f(-x^2, -x^3) in powers of x where f(, ) is Ramanujan's general theta function. - Michael Somos, Nov 12 2016
Expansion of f(-x^2) * f(-x^20) / (f(-x) * f(-x^8, -x^12)) in powers of x where f(-x) : = f(-x, -x^2) and f(, ) is Ramanujan's general theta function.
Euler transform of period 20 sequence [ 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, ...].
G.f.: Sum_{k>0} x^(k^2 - 1) / ((1 - x) * (1 - x^2) * ... * (1 - x^(2k-1))).
G.f.: 1/(Product_{k>0} (1-x^(2k-1))(1-x^(20k-8))(1-x^(20k-12))).
a(n) ~ (3-sqrt(5))^(1/4) * exp(Pi*sqrt(2*n/5)) / (4*sqrt(5)*n^(3/4)). - Vaclav Kotesovec, Aug 30 2015

A357645 Triangle read by rows where T(n,k) is the number of integer compositions of n with half-alternating sum k, where k ranges from -n to n in steps of 2.

Original entry on oeis.org

1, 0, 1, 0, 0, 2, 0, 0, 1, 3, 0, 0, 2, 2, 4, 0, 0, 3, 5, 3, 5, 0, 0, 4, 8, 10, 4, 6, 0, 0, 5, 11, 18, 18, 5, 7, 0, 0, 6, 14, 28, 36, 30, 6, 8, 0, 0, 7, 17, 41, 63, 65, 47, 7, 9, 0, 0, 8, 20, 58, 104, 126, 108, 70, 8, 10, 0, 0, 9, 23, 80, 164, 230, 230, 168, 100, 9, 11
Offset: 0

Views

Author

Gus Wiseman, Oct 12 2022

Keywords

Comments

We define the half-alternating sum of a sequence (A, B, C, D, E, F, G, ...) to be A + B - C - D + E + F - G - ...

Examples

			Triangle begins:
   1
   0   1
   0   0   2
   0   0   1   3
   0   0   2   2   4
   0   0   3   5   3   5
   0   0   4   8  10   4   6
   0   0   5  11  18  18   5   7
   0   0   6  14  28  36  30   6   8
   0   0   7  17  41  63  65  47   7   9
   0   0   8  20  58 104 126 108  70   8  10
Row n = 6 counts the following compositions:
  (114)   (123)    (132)     (141)  (6)
  (1113)  (213)    (222)     (231)  (15)
  (1122)  (1212)   (312)     (321)  (24)
  (1131)  (1221)   (1311)    (411)  (33)
          (2112)   (2211)           (42)
          (2121)   (3111)           (51)
          (11121)  (11112)
          (11211)  (12111)
                   (21111)
                   (111111)
		

Crossrefs

Row sums are A011782.
For original alternating sum we have A097805, unordered A344651.
Column k = n-4 appears to be A177787.
The case of partitions is A357637, skew A357638.
The central column k=0 is A357641 (aerated).
The skew-alternating version is A357646.
The reverse version for partitions is A357704, skew A357705.
A351005 = alternately equal and unequal partitions, compositions A357643.
A351006 = alternately unequal and equal partitions, compositions A357644.
A357621 gives half-alternating sum of standard compositions, skew A357623.
A357629 gives half-alternating sum of prime indices, skew A357630.
A357633 gives half-alternating sum of Heinz partition, skew A357634.

Programs

  • Mathematica
    halfats[f_]:=Sum[f[[i]]*(-1)^(1+Ceiling[i/2]),{i,Length[f]}];
    Table[Length[Select[Join@@Permutations/@IntegerPartitions[n],halfats[#]==k&]],{n,0,10},{k,-n,n,2}]

A357704 Triangle read by rows where T(n,k) is the number of reversed integer partitions of n with half-alternating sum k, where k ranges from -n to n in steps of 2.

Original entry on oeis.org

1, 0, 1, 0, 0, 2, 0, 0, 1, 2, 0, 0, 2, 0, 3, 0, 0, 2, 2, 0, 3, 0, 0, 3, 1, 3, 0, 4, 0, 0, 3, 2, 4, 2, 0, 4, 0, 0, 4, 2, 6, 2, 3, 0, 5, 0, 0, 4, 3, 5, 7, 3, 3, 0, 5, 0, 0, 5, 3, 8, 4, 10, 2, 4, 0, 6, 0, 0, 5, 4, 8, 6, 11, 9, 3, 4, 0, 6, 0, 0, 6, 4, 11, 5, 15, 8, 13, 3, 5, 0, 7
Offset: 0

Views

Author

Gus Wiseman, Oct 10 2022

Keywords

Comments

We define the half-alternating sum of a sequence (A, B, C, D, E, F, G, ...) to be A + B - C - D + E + F - G - ...

Examples

			Triangle begins:
  1
  0  1
  0  0  2
  0  0  1  2
  0  0  2  0  3
  0  0  2  2  0  3
  0  0  3  1  3  0  4
  0  0  3  2  4  2  0  4
  0  0  4  2  6  2  3  0  5
  0  0  4  3  5  7  3  3  0  5
  0  0  5  3  8  4 10  2  4  0  6
  0  0  5  4  8  6 11  9  3  4  0  6
  0  0  6  4 11  5 15  8 13  3  5  0  7
  0  0  6  5 11  8 13 19 10 13  4  5  0  7
  0  0  7  5 14  8 19 13 25  9 17  4  6  0  8
  0  0  7  6 14 11 19 17 29 23 13 18  5  6  0  8
Row n = 7 counts the following reversed partitions:
  .  .  (115)   (124)   (133)      (11113)   .  (7)
        (1114)  (1222)  (223)      (111112)     (16)
        (1123)          (11122)                 (25)
                        (1111111)               (34)
		

Crossrefs

Row sums are A000041.
Last entry of row n is A008619(n).
The central column in the non-reverse case is A035363, skew A035544.
For original reverse-alternating sum we have A344612.
For original alternating sum we have A344651, ordered A097805.
The non-reverse version is A357637, skew A357638.
The central column is A357639, skew A357640.
The non-reverse ordered version (compositions) is A357645, skew A357646.
The skew-alternating version is A357705.
A351005 = alternately equal and unequal partitions, compositions A357643.
A351006 = alternately unequal and equal partitions, compositions A357644.
A357621 gives half-alternating sum of standard compositions, skew A357623.
A357629 gives half-alternating sum of prime indices, skew A357630.
A357633 gives half-alternating sum of Heinz partition, skew A357634.

Programs

  • Mathematica
    halfats[f_]:=Sum[f[[i]]*(-1)^(1+Ceiling[i/2]),{i,Length[f]}];
    Table[Length[Select[Reverse/@IntegerPartitions[n],halfats[#]==k&]],{n,0,15},{k,-n,n,2}]

A357705 Triangle read by rows where T(n,k) is the number of reversed integer partitions of n with skew-alternating sum k, where k ranges from -n to n in steps of 2.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 2, 0, 1, 0, 2, 2, 0, 1, 0, 3, 1, 2, 0, 1, 0, 3, 2, 3, 2, 0, 1, 0, 4, 2, 4, 1, 3, 0, 1, 0, 4, 3, 3, 6, 2, 3, 0, 1, 0, 5, 3, 5, 3, 7, 2, 4, 0, 1, 0, 5, 4, 5, 4, 9, 7, 3, 4, 0, 1, 0, 6, 4, 7, 3, 12, 5, 10, 3, 5, 0, 1
Offset: 0

Views

Author

Gus Wiseman, Oct 10 2022

Keywords

Comments

We define the skew-alternating sum of a sequence (A, B, C, D, E, F, G, ...) to be A - B - C + D + E - F - G + ...

Examples

			Triangle begins:
  1
  0  1
  0  1  1
  0  2  0  1
  0  2  2  0  1
  0  3  1  2  0  1
  0  3  2  3  2  0  1
  0  4  2  4  1  3  0  1
  0  4  3  3  6  2  3  0  1
  0  5  3  5  3  7  2  4  0  1
  0  5  4  5  4  9  7  3  4  0  1
  0  6  4  7  3 12  5 10  3  5  0  1
  0  6  5  7  5 10 16  7 11  4  5  0  1
  0  7  5  9  5 14 11 18  7 14  4  6  0  1
Row n = 7 counts the following reversed partitions:
  .  (16)   (25)   (34)       (1123)  (1114)   .  (7)
     (115)  (223)  (1222)             (11113)
     (124)         (111112)           (11122)
     (133)         (1111111)
		

Crossrefs

Row sums are A000041.
First nonzero entry of each row is A004526.
The central column is A357640, half A357639.
For original alternating sum we have A344651, ordered A097805.
The half-alternating version is A357704.
The ordered non-reverse version (compositions) is A357646, half A357645.
The non-reverse version is A357638, half A357637.
A351005 = alternately equal and unequal partitions, compositions A357643.
A351006 = alternately unequal and equal partitions, compositions A357644.
A357621 gives half-alternating sum of standard compositions, skew A357623.
A357629 gives half-alternating sum of prime indices, skew A357630.
A357633 gives half-alternating sum of Heinz partition, skew A357634.

Programs

  • Mathematica
    skats[f_]:=Sum[f[[i]]*(-1)^(1+Ceiling[(i+1)/2]),{i,Length[f]}];
    Table[Length[Select[Reverse/@IntegerPartitions[n],skats[#]==k&]],{n,0,11},{k,-n,n,2}]

A351011 Numbers k such that the k-th composition in standard order has even length and alternately equal and unequal parts, i.e., all run-lengths equal to 2.

Original entry on oeis.org

0, 3, 10, 36, 43, 58, 136, 147, 228, 235, 528, 547, 586, 676, 698, 904, 915, 2080, 2115, 2186, 2347, 2362, 2696, 2707, 2788, 2795, 3600, 3619, 3658, 3748, 3770, 8256, 8323, 8458, 8740, 8747, 8762, 9352, 9444, 9451, 10768, 10787, 10826, 11144, 11155, 14368
Offset: 1

Views

Author

Gus Wiseman, Feb 03 2022

Keywords

Comments

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.

Examples

			The terms together with their binary expansions and standard compositions begin:
    0:           0  ()
    3:          11  (1,1)
   10:        1010  (2,2)
   36:      100100  (3,3)
   43:      101011  (2,2,1,1)
   58:      111010  (1,1,2,2)
  136:    10001000  (4,4)
  147:    10010011  (3,3,1,1)
  228:    11100100  (1,1,3,3)
  235:    11101011  (1,1,2,2,1,1)
  528:  1000010000  (5,5)
  547:  1000100011  (4,4,1,1)
  586:  1001001010  (3,3,2,2)
  676:  1010100100  (2,2,3,3)
  698:  1010111010  (2,2,1,1,2,2)
  904:  1110001000  (1,1,4,4)
  915:  1110010011  (1,1,3,3,1,1)
		

Crossrefs

The case of twins (binary weight 2) is A000120.
All terms are evil numbers A001969.
These compositions are counted by A003242 interspersed with 0's.
Partitions of this type are counted by A035457, any length A351005.
The Heinz numbers of these compositions are A062503.
Taking singles instead of twins gives A333489, complement A348612.
This is the anti-run case of A351010.
The strict case (distinct twins) is A351009, counted by A077957(n-2).
A011782 counts compositions.
A085207/A085208 represent concatenation of standard compositions.
A345167 ranks alternating compositions, counted by A025047.
A350355 ranks up/down compositions, counted by A025048.
A350356 ranks down/up compositions, counted by A025049.
A351014 counts distinct runs in standard compositions.
Selected statistics of standard compositions:
- Length is A000120.
- Sum is A070939.
- Heinz number is A333219.
- Number of distinct parts is A334028.
Selected classes of standard compositions:
- Partitions are A114994, strict A333256.
- Multisets are A225620, strict A333255.
- Strict compositions are A233564.
- Constant compositions are A272919.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join@@ Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Select[Range[0,1000],And@@(#==2&)/@Length/@Split[stc[#]]&]

A351595 Number of odd-length integer partitions y of n such that y_i > y_{i+1} for all even i.

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 2, 3, 4, 5, 6, 9, 10, 13, 16, 20, 24, 30, 35, 44, 52, 63, 74, 90, 105, 126, 148, 175, 204, 242, 280, 330, 382, 446, 515, 600, 690, 800, 919, 1060, 1214, 1398, 1595, 1830, 2086, 2384, 2711, 3092, 3506, 3988, 4516, 5122, 5788, 6552, 7388, 8345
Offset: 0

Views

Author

Gus Wiseman, Feb 25 2022

Keywords

Examples

			The a(1) = 1 through a(12) = 10 partitions (A..C = 10..12):
  1   2   3   4   5     6     7     8     9     A     B       C
                  221   321   331   332   432   442   443     543
                              421   431   441   532   542     552
                                    521   531   541   551     642
                                          621   631   632     651
                                                721   641     732
                                                      731     741
                                                      821     831
                                                      33221   921
                                                              43221
		

Crossrefs

The ordered version (compositions) is A000213 shifted right once.
All odd-length partitions are counted by A027193.
The opposite appears to be A122130, even-length A351008, any length A122129.
This appears to be the odd-length case of A122135, even-length A122134.
The case that is constant at odd indices:
- any length: A351005
- odd length: A351593
- even length: A035457
- opposite any length: A351006
- opposite odd length: A053251
- opposite even length: A351007
For equality instead of inequality:
- any length: A351003
- odd-length: A000009 (except at 0)
- even-length: A351012
- opposite any length: A351004
- opposite odd-length: A351594
- opposite even-length: A035363

Programs

  • Mathematica
    Table[Length[Select[IntegerPartitions[n],OddQ[Length[#]]&&And@@Table[#[[i]]>#[[i+1]],{i,2,Length[#]-1,2}]&]],{n,0,30}]

A351593 Number of odd-length integer partitions of n into parts that are alternately equal and strictly decreasing.

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 1, 2, 2, 2, 2, 4, 2, 4, 3, 5, 4, 6, 4, 8, 6, 9, 6, 12, 7, 14, 10, 16, 11, 20, 13, 24, 16, 28, 18, 34, 21, 40, 26, 46, 30, 56, 34, 64, 41, 75, 48, 88, 54, 102, 64, 118, 73, 138, 84, 159, 98, 182, 112, 210, 128, 242, 148, 276, 168, 318
Offset: 0

Views

Author

Gus Wiseman, Feb 23 2022

Keywords

Comments

Also odd-length partitions whose run-lengths are all 2's, except for the last, which is 1.

Examples

			The a(1) = 1 through a(15) = 6 partitions (A..F = 10..15):
  1  2  3  4  5    6  7    8    9    A    B      C    D      E    F
              221     331  332  441  442  443    552  553    554  663
                                          551         661    662  771
                                          33221       44221       44331
                                                                  55221
		

Crossrefs

The even-length ordered version is A003242, ranked by A351010.
The opposite version is A053251, even-length A351007, any length A351006.
This is the odd-length case of A351005, even-length A035457.
With only equalities we get:
- opposite any length: A351003
- opposite odd-length: A000009 (except at 0)
- opposite even-length: A351012
- any length: A351004
- odd-length: A351594
- even-length: A035363
Without equalities we get:
- opposite any length: A122129 (apparently)
- opposite odd-length: A122130 (apparently)
- opposite even-length: A351008
- any length: A122135 (apparently)
- odd-length: A351595
- even-length: A122134 (apparently)

Programs

  • Mathematica
    Table[Length[Select[IntegerPartitions[n],OddQ[Length[#]]&&And@@Table[If[EvenQ[i],#[[i]]!=#[[i+1]],#[[i]]==#[[i+1]]],{i,Length[#]-1}]&]],{n,0,30}]
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