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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A340581 Irregular triangle read by rows in which row n has length A014153(n-1) and every column k lists the positive integers A000027, n >= 1, k >= 1.

Original entry on oeis.org

1, 2, 1, 1, 3, 2, 2, 1, 1, 1, 1, 4, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 5, 4, 4, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 6, 5, 5, 4, 4, 4, 4, 3, 3, 3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 1

Views

Author

Omar E. Pol, Jan 14 2021

Keywords

Comments

Row n lists in nonincreasing order the first A014153(n-1) terms of A176206.
In other words: row n lists in nonincreasing order the terms of the first n rows of triangle A176206.
Conjecture: all divisors of all terms in row n are also all parts of all partitions of all positive integers <= n.
The conjecture is in accordance with the conjectures in A336811 and in A176206.
A336811 contains the most elementary conjecture about the correspondence divisors/partitions.
The connection with A336811 (the main sequence) is as follows: A336811 --> A176206 --> this sequence.

Examples

			Triangle begins:
1;
2, 1, 1;
3, 2, 2, 1, 1, 1, 1;
4, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1;
5, 4, 4, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1;
...
For n = 4, by definition the length of row 6 is A014153(4-1) = A014153(3) = 14, so the row 4 of triangle has 14 terms. Since every column lists the positive integers A000027 so the row 4 is [4, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1].
Then we have that the divisors of the numbers of the 4th row are:
.
4th row of the triangle ----------> 4  3  3  2  2  2  2  1  1  1  1  1  1  1
                                    2  1  1  1  1  1  1
                                    1
.
There are fourteen 1's, five 2's, two 3's and one 4.
In total there are 14 + 5 + 2 + 1 = 22 divisors.
On the other hand all partitions of all positive integers <= 4 are as shown below:
.
.    Partition   Partitions    Partitions     Partitions
.       of 1        of 2          of 3           of 4
.
.                                             4
.                                             2  2
.                               3             3  1
.                   2           2  1          2  1  1
.        1          1  1        1  1  1       1  1  1  1
.
In these partitions there are fourteen 1's, five 2's, two 3's and one 4.
In total there are 14 + 5 + 2 + 1 = A284870(4) = 22 parts.
Finally in accordance with the conjecture we can see that all divisors of all numbers in the 4th row of the triangle are the same positive integers as all parts of all partitions of all positive integers <= 4.
		

Crossrefs

A000070 a(n) = Sum_{k=0..n} p(k) where p(k) = number of partitions of k (A000041).

Original entry on oeis.org

1, 2, 4, 7, 12, 19, 30, 45, 67, 97, 139, 195, 272, 373, 508, 684, 915, 1212, 1597, 2087, 2714, 3506, 4508, 5763, 7338, 9296, 11732, 14742, 18460, 23025, 28629, 35471, 43820, 53963, 66273, 81156, 99133, 120770, 146785, 177970, 215308, 259891, 313065, 376326, 451501
Offset: 0

Views

Author

Keywords

Comments

Also the total number of all different integers in all partitions of n + 1. E.g., a(3) = 7 because the partitions of 4 comprise the sets {1},{1, 2},{2},{1, 3},{4} of different integers and their total number is 7. - Thomas Wieder, Apr 10 2004
With offset 1, also the number of 1's in all partitions of n. For example, 3 = 2+1 = 1+1+1, a(3) = (zero 1's) + (one 1's) + (three 1's), so a(3) = 4. - Naohiro Nomoto, Jan 09 2002. See the Riordan reference p. 184, last formula, first term, for a proof based on Fine's identity given in Riordan, p. 182 (20).
Also, number of partitions of n into parts when there are two kinds of parts of size one.
Also number of graphical forest partitions of 2n+2.
a(n) = count 2 for each partition of n and 1 for each decrement. E.g., the partitions of 4 are 4 (2), 31 (3), 22 (2), 211 (3) and 1111 (2). 2 + 3 + 2 + 3 + 2 = 12. This is related to the Ferrers representation. We can see that taking the Ferrers diagram for each partition of n and adding a new * to all available columns, we generate each partition of n+1, but with repeats (A058884). - Jon Perry, Feb 06 2004
Also the number of 1-transitions among all integer partitions of n. A 1-transition is the removal of a digit "1" from a partition containing at least one "1" and subsequent addition of that "1" to another digit in that partition. This other digit may be a "1" also, but all digits of equal amount are considered as undistinquishable (unlabeled). E.g., for n=6 one has the partition [1113] for which the following two 1-transitions are possible: [1113] --> [123] and [1113] --> [114]. The 1-transitions of n form a partial order (poset). For n=6 one has 12 1-transitions: [111111] --> [11112], [11112] --> [1113], [11112] --> [1122], [1113] --> [114], [1113] --> [123], [1122] --> [123], [1122] --> [222], [123] --> [33], [123] --> [24], [114] --> [15], [114] --> [24], [15] --> [6]. - Thomas Wieder, Mar 08 2005
Also number of partitions of 2n+1 where one of the parts is greater than n (also where there are more than n parts) and of 2n+2 where one of the parts is greater than n+1 (or with more than n+1 parts). - Henry Bottomley, Aug 01 2005
Equals left border of triangle A137633 - Gary W. Adamson, Jan 31 2008
Equals row sums of triangle A027293. - Gary W. Adamson, Oct 26 2008
Convolved with A010815 = [1,1,1,...]. n-th partial sum of A000041 convolved with A010815 = the binomial sequence starting (1, n, ...). - Gary W. Adamson, Nov 09 2008
Equals A036469 convolved with A035363. - Gary W. Adamson, Jun 09 2009
a(A004526(n)) = A025065(n). - Reinhard Zumkeller, Jan 23 2010
a(n) = if n <= 1 then A054225(1,n) else A054225(n,1). - Reinhard Zumkeller, Nov 30 2011
Also the total number of 1's among all hook-lengths in all partitions of n. E.g., a(4)=7 because hooks of the partitions of n = 4 comprise the multisets {4,3,2,1}, {4,2,1,1}, {3,2,2,1}, {4,1,2,1}, {4,3,2,1} and their total number of 1's is 7. - T. Amdeberhan, Jun 03 2012
With offset 1, a(n) is also the difference between the sum of largest and the sum of second 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
a(0) = 1 and 2*a(n-1) >= a(n) for all n > 0. Hence a(n) is a complete sequence. - Frank M Jackson, Apr 08 2013
a(n) is the number of conjugacy classes in the order-preserving, order-decreasing and (order-preserving and order-decreasing) injective transformation semigroups. - Ugbene Ifeanyichukwu, Jun 03 2015
a(n) is also the number of unlabeled subgraphs of the n-cycle C_n. For example, for n = 3, there are 3 unlabeled subgraphs of the triangle C_3 with 0 edges, 2 with 1 edge, 1 with 2 edges, and 1 with 3 edges (C_3 itself), so a(3) = 3 + 2 + 1 + 1 = 7. - John P. McSorley, Nov 21 2016
a(n) is also the number of partitions of 2n with all parts either even or equal to 1. Proof: the number of such partitions of 2n with exactly 2k 1's is p(n-k), for k = 0,..,n. Summing over k gives the formula. - Leonard Chastkofsky, Jul 24 2018
a(n) is the total number of polygamma functions that appear in the expansion of the (n+1)st derivative of x! with respect to x. More specifically, a(n) is the number of times the string "PolyGamma" appears in the expansion of D[x!, {x, n + 1}] in Mathematica. For example, D[x!, {x, 3 + 1}] = Gamma[1 + x] PolyGamma[0, 1 + x]^4 + 6 Gamma[1 + x] PolyGamma[0, 1 + x]^2 PolyGamma[1, 1 + x] + 3 Gamma[1 + x] PolyGamma[1, 1 + x]^2 + 4 Gamma[1 + x] PolyGamma[0, 1 + x] PolyGamma[2, 1 + x] + Gamma[1 + x] PolyGamma[3, 1 + x], and we see that the string "PolyGamma" appears a total of a(3) = 7 times in this expansion. - John M. Campbell, Aug 11 2018
With offset 1, also the number of integer partitions of 2n that do not comprise the multiset of vertex-degrees of any multigraph (i.e., non-multigraphical partitions); see A209816 for multigraphical partitions. - Gus Wiseman, Oct 26 2018
Also a(n) is the number of partitions of 2n+1 with exactly one odd part.
Delete the odd part 2k+1, k=0, ..., n, to get a partition of 2n-2k into even parts. There are as many unrestricted partitions of n-k; now sum those numbers from 0 to n to get a(n). - George Beck, Jul 22 2019
In the Young's lattice, a(n) is the number of branches that connect the (n-1)-th layer to the n-th layer. - Shouvik Datta, Sep 19 2021
a(n) is the number of multiset partitions of the multiset {r^n, s^1}, equivalently, factorization patterns of any number m=p^n*q^1 where p and q are primes. - Joerg Arndt, Jan 01 2024
a(n) is the number of positive integers whose divisors are the parts of the partitions of n + 1. - Omar E. Pol, Nov 07 2024

Examples

			G.f. = 1 + 2*x + 4*x^2 + 7*x^3 + 12*x^4 + 19*x^5 + 30*x^6 + 45*x^7 + 67*x^8 + ...
From _Omar E. Pol_, Oct 25 2012: (Start)
For n = 5 consider the partitions of n+1:
--------------------------------------
.                         Number
Partitions of 6           of 1's
--------------------------------------
6 .......................... 0
3 + 3 ...................... 0
4 + 2 ...................... 0
2 + 2 + 2 .................. 0
5 + 1 ...................... 1
3 + 2 + 1 .................. 1
4 + 1 + 1 .................. 2
2 + 2 + 1 + 1 .............. 2
3 + 1 + 1 + 1 .............. 3
2 + 1 + 1 + 1 + 1 .......... 4
1 + 1 + 1 + 1 + 1 + 1 ...... 6
------------------------------------
35-16 =                     19
.
The difference between the sum of the first column and the sum of the second column of the set of partitions of 6 is 35 - 16 = 19 and equals the number of 1's in all partitions of 6, so the 6th term of this sequence is a(5) = 19.
(End)
From _Gus Wiseman_, Oct 26 2018: (Start)
With offset 1, the a(1) = 1 through a(6) = 19 partitions of 2*n whose greatest part is > n:
  (2)  (4)   (6)    (8)     (A)      (C)
       (31)  (42)   (53)    (64)     (75)
             (51)   (62)    (73)     (84)
             (411)  (71)    (82)     (93)
                    (521)   (91)     (A2)
                    (611)   (622)    (B1)
                    (5111)  (631)    (732)
                            (721)    (741)
                            (811)    (822)
                            (6211)   (831)
                            (7111)   (921)
                            (61111)  (A11)
                                     (7221)
                                     (7311)
                                     (8211)
                                     (9111)
                                     (72111)
                                     (81111)
                                     (711111)
With offset 1, the a(1) = 1 through a(6) = 19 partitions of 2*n whose number of parts is > n:
  (11)  (211)   (2211)    (22211)     (222211)      (2222211)
        (1111)  (3111)    (32111)     (322111)      (3222111)
                (21111)   (41111)     (331111)      (3321111)
                (111111)  (221111)    (421111)      (4221111)
                          (311111)    (511111)      (4311111)
                          (2111111)   (2221111)     (5211111)
                          (11111111)  (3211111)     (6111111)
                                      (4111111)     (22221111)
                                      (22111111)    (32211111)
                                      (31111111)    (33111111)
                                      (211111111)   (42111111)
                                      (1111111111)  (51111111)
                                                    (222111111)
                                                    (321111111)
                                                    (411111111)
                                                    (2211111111)
                                                    (3111111111)
                                                    (21111111111)
                                                    (111111111111)
(End)
From _Joerg Arndt_, Jan 01 2024: (Start)
The a(5) = 19 multiset partitions of the multiset {1^5, 2^1} are:
   1:  {{1, 1, 1, 1, 1, 2}}
   2:  {{1, 1, 1, 1, 1}, {2}}
   3:  {{1, 1, 1, 1, 2}, {1}}
   4:  {{1, 1, 1, 1}, {1, 2}}
   5:  {{1, 1, 1, 1}, {1}, {2}}
   6:  {{1, 1, 1, 2}, {1, 1}}
   7:  {{1, 1, 1, 2}, {1}, {1}}
   8:  {{1, 1, 1}, {1, 1, 2}}
   9:  {{1, 1, 1}, {1, 1}, {2}}
  10:  {{1, 1, 1}, {1, 2}, {1}}
  11:  {{1, 1, 1}, {1}, {1}, {2}}
  12:  {{1, 1, 2}, {1, 1}, {1}}
  13:  {{1, 1, 2}, {1}, {1}, {1}}
  14:  {{1, 1}, {1, 1}, {1, 2}}
  15:  {{1, 1}, {1, 1}, {1}, {2}}
  16:  {{1, 1}, {1, 2}, {1}, {1}}
  17:  {{1, 1}, {1}, {1}, {1}, {2}}
  18:  {{1, 2}, {1}, {1}, {1}, {1}}
  19:  {{1}, {1}, {1}, {1}, {1}, {2}}
(End)
		

References

  • H. Gupta, An asymptotic formula in partitions. J. Indian Math. Soc., (N. S.) 10 (1946), 73-76.
  • H. Gupta et al., Tables of Partitions. Royal Society Mathematical Tables, Vol. 4, Cambridge Univ. Press, 1958, p. 90.
  • R. Honsberger, Mathematical Gems III, M.A.A., 1985, p. 6.
  • D. E. Knuth, The Art of Computer Programming, Vol. 4A, Table A-1, page 778. - N. J. A. Sloane, Dec 30 2018
  • A. M. Odlyzko, Asymptotic Enumeration Methods, p. 19
  • J. Riordan, Combinatorial Identities, Wiley, 1968, p. 199.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • Stanley, R. P., Exercise 1.26 in Enumerative Combinatorics, Vol. 1. Cambridge, England: Cambridge University Press, p. 59, 1999.

Crossrefs

A diagonal of A066633.
Also second column of A126442. - George Beck, May 07 2011
Row sums of triangle A092905.
Also row sums of triangle A261555. - Omar E. Pol, Sep 14 2016
Also row sums of triangle A278427. - John P. McSorley, Nov 25 2016
Column k=2 of A292508.

Programs

  • GAP
    List([0..45],n->Sum([0..n],k->NrPartitions(k))); # Muniru A Asiru, Jul 25 2018
    
  • Haskell
    a000070 = p a028310_list where
       p _          0 = 1
       p ks'@(k:ks) m = if m < k then 0 else p ks' (m - k) + p ks m
    -- Reinhard Zumkeller, Nov 06 2012
    
  • Maple
    with(combinat): a:=n->add(numbpart(j),j=0..n): seq(a(n), n=0..44); # Zerinvary Lajos, Aug 26 2008
  • Mathematica
    CoefficientList[ Series[1/(1 - x)*Product[1/(1 - x^k), {k, 75}], {x, 0, 45}], x] (* Robert G. Wilson v, Jul 13 2004 *)
    Table[ Count[ Flatten@ IntegerPartitions@ n, 1], {n, 45}] (* Robert G. Wilson v, Aug 06 2008 *)
    Join[{1}, Accumulate[PartitionsP[Range[50]]]+1] (* _Harvey P. Dale, Mar 12 2013 *)
    a[ n_] := SeriesCoefficient[ 1 / (1 - x) / QPochhammer[ x], {x, 0, n}]; (* Michael Somos, Nov 09 2013 *)
    Accumulate[PartitionsP[Range[0,49]]] (* George Beck, Oct 23 2014; typo fixed by Virgile Andreani, Jul 10 2016 *)
  • PARI
    {a(n) = if( n<0, 0, polcoeff( 1 / prod(m=1, n, 1 - x^m, 1 + x * O(x^n)) / (1 - x), n))}; /* Michael Somos, Nov 08 2002 */
    
  • PARI
    x='x+O('x^66); Vec(1/((1-x)*eta(x))) /* Joerg Arndt, May 15 2011 */
    
  • PARI
    a(n) = sum(k=0, n, numbpart(k)); \\ Michel Marcus, Sep 16 2016
    
  • Python
    from itertools import accumulate
    def A000070iter(n):
        L = [0]*n; L[0] = 1
        def numpart(n):
            S = 0; J = n-1; k = 2
            while 0 <= J:
                T = L[J]
                S = S+T if (k//2)%2 else S-T
                J -= k  if (k)%2 else k//2
                k += 1
            return S
        for j in range(1, n): L[j] = numpart(j)
        return accumulate(L)
    print(list(A000070iter(100))) # Peter Luschny, Aug 30 2019
    
  • Python
    # Using function A365676Row. Compare also A365675.
    from itertools import accumulate
    def A000070List(size: int) -> list[int]:
        return [sum(accumulate(reversed(A365676Row(n)))) for n in range(size)]
    print(A000070List(45))  # Peter Luschny, Sep 16 2023
  • Sage
    def A000070_list(leng):
        p = [number_of_partitions(n) for n in range(leng)]
        return [add(p[:k+1]) for k in range(leng)]
    A000070_list(45) # Peter Luschny, Sep 15 2014
    

Formula

Euler transform of [ 2, 1, 1, 1, 1, 1, 1, ...].
log(a(n)) ~ -3.3959 + 2.44613*sqrt(n). - Robert G. Wilson v, Jan 11 2002
a(n) = (1/n)*Sum_{k=1..n} (sigma(k)+1)*a(n-k), n > 1, a(0) = 1. - Vladeta Jovovic, Aug 22 2002
G.f.: (1/(1 - x))*Product_{m >= 1} 1/(1 - x^m).
a(n) seems to have the same parity as A027349(n+1). Comment from James Sellers, Mar 08 2006: that is true.
a(n) = A000041(2n+1) - A110618(2n+1) = A000041(2n+2) - A110618(2n+2). - Henry Bottomley, Aug 01 2005
Row sums of triangle A133735. - Gary W. Adamson, Sep 22 2007
a(n) = A092269(n+1) - A195820(n+1). - Omar E. Pol, Oct 20 2011
a(n) = A181187(n+1,1) - A181187(n+1,2). - Omar E. Pol, Oct 25 2012
From Peter Bala, Dec 23 2013: (Start)
Gupta gives the asymptotic result a(n-1) ~ sqrt(6/Pi^2)* sqrt(n)*p(n), where p(n) is the partition function A000041(n).
Let P(2,n) denote the set of partitions of n into parts k >= 2.
a(n-2) = Sum_{parts k in all partitions in P(2,n)} phi(k), where phi(k) is the Euler totient function (see A000010). Using this result and Mertens's theorem on the average order of the phi function, leads to the asymptotic result
a(n-2) ~ (6/Pi^2)*n*(p(n) - p(n-1)) = (6/Pi^2)*A138880(n) as n -> infinity. (End)
a(n) ~ exp(Pi*sqrt(2*n/3)) / (2^(3/2)*Pi*sqrt(n)) * (1 + 11*Pi/(24*sqrt(6*n)) + (73*Pi^2 - 1584)/(6912*n)). - Vaclav Kotesovec, Oct 26 2016
a(n) = A024786(n+2) + A024786(n+1). - Vaclav Kotesovec, Nov 05 2016
G.f.: exp(Sum_{k>=1} (sigma_1(k) + 1)*x^k/k). - Ilya Gutkovskiy, Aug 21 2018
a(n) = A025065(2n). - Gus Wiseman, Oct 26 2018
a(n - 1) = A000041(2n) - A209816(n). - Gus Wiseman, Oct 26 2018

A116861 Triangle read by rows: T(n,k) is the number of partitions of n such that the sum of the parts, counted without multiplicities, is equal to k (n>=1, k>=1).

Original entry on oeis.org

1, 1, 1, 1, 0, 2, 1, 1, 1, 2, 1, 0, 2, 1, 3, 1, 1, 3, 1, 1, 4, 1, 0, 3, 2, 2, 2, 5, 1, 1, 3, 3, 2, 4, 2, 6, 1, 0, 5, 2, 3, 4, 4, 3, 8, 1, 1, 4, 3, 4, 7, 4, 5, 3, 10, 1, 0, 5, 3, 4, 7, 7, 6, 6, 5, 12, 1, 1, 6, 4, 3, 12, 6, 8, 7, 9, 5, 15, 1, 0, 6, 4, 5, 10, 10, 9, 10, 11, 10, 7, 18, 1, 1, 6, 4, 5, 15, 11, 13, 9, 16, 11, 13, 8, 22
Offset: 1

Views

Author

Emeric Deutsch, Feb 27 2006

Keywords

Comments

Conjecture: Reverse the rows of the table to get an infinite lower-triangular matrix b with 1's on the main diagonal. The third diagonal of the inverse of b is minus A137719. - George Beck, Oct 26 2019
Proof: The reversed rows yield the matrix I+N where N is strictly lower triangular, N[i,j] = 0 for j >= i, having its 2nd diagonal equal to the 2nd column (1, 0, 1, 0, 1, ...): N[i+1,i] = A000035(i), i >= 1, and 3rd diagonal equal to the 3rd column of this triangle, (2, 1, 2, 3, 3, 3, ...): N[i+2,i] = A137719(i), i >= 1. It is known that (I+N)^-1 = 1 - N + N^2 - N^3 +- .... Here N^2 has not only the second but also the 3rd diagonal zero, because N^2[i+2,i] = N[i+2,i+1]*N[i+1,i] = A000035(i+1)*A000035(i) = 0. Therefore the 3rd diagonal of (I+N)^-1 is equal to -A137719 without leading 0. - M. F. Hasler, Oct 27 2019
From Gus Wiseman, Aug 27 2023: (Start)
Also the number of ways to write n-k as a nonnegative linear combination of a strict integer partition of k. Also the number of ways to write n as a (strictly) positive linear combination of a strict integer partition of k. Row n=7 counts the following:
7*1 . 1*2+5*1 1*3+4*1 1*3+2*2 1*5+2*1 1*7
2*2+3*1 2*3+1*1 1*4+3*1 1*3+1*2+2*1 1*4+1*3
3*2+1*1 1*5+1*2
1*6+1*1
1*4+1*2+1*1
(End)

Examples

			T(10,7) = 4 because we have [6,1,1,1,1], [4,3,3], [4,2,2,1,1] and [4,2,1,1,1,1] (6+1=4+3=4+2+1=7).
Triangle starts:
  1;
  1, 1;
  1, 0, 2;
  1, 1, 1, 2;
  1, 0, 2, 1, 3;
  1, 1, 3, 1, 1,  4;
  1, 0, 3, 2, 2,  2, 5;
  1, 1, 3, 3, 2,  4, 2, 6;
  1, 0, 5, 2, 3,  4, 4, 3, 8;
  1, 1, 4, 3, 4,  7, 4, 5, 3, 10;
  1, 0, 5, 3, 4,  7, 7, 6, 6,  5, 12;
  1, 1, 6, 4, 3, 12, 6, 8, 7,  9,  5, 15;
  ...
		

Crossrefs

Cf. A000041 (row sums), A000009 (diagonal), A014153.
Cf. A114638 (count partitions with #parts = sum(distinct parts)).
Column 1: A000012, column 2: A000035(1..), column 3: A137719(1..).
For subsets instead of partitions we have A026820.
This statistic is ranked by A066328.
The central diagonal is T(2n,n) = A364910(n), non-strict A364907.
Partial sums of columns are columns of A364911.
Same as A364916 (offset 0) with rows reversed.
A008284 counts partitions by length, strict A008289.
A364912 counts linear combinations of partitions.
A364913 counts combination-full partitions, strict A364839.

Programs

  • Maple
    g:= -1+product(1+t^j*x^j/(1-x^j), j=1..40): gser:= simplify(series(g,x=0,18)): for n from 1 to 14 do P[n]:=sort(coeff(gser,x^n)) od: for n from 1 to 14 do seq(coeff(P[n],t^j),j=1..n) od; # yields sequence in triangular form
    # second Maple program:
    b:= proc(n, i) option remember; local f, g, j;
          if n=0 then [1] elif i<1 then [ ] else f:= b(n, i-1);
             for j to n/i do
               f:= zip((x, y)->x+y, f, [0$i, b(n-i*j, i-1)[]], 0)
             od; f
          fi
        end:
    T:= n-> subsop(1=NULL, b(n, n))[]:
    seq(T(n), n=1..20);  # Alois P. Heinz, Feb 27 2013
  • Mathematica
    max = 14; s = Series[-1+Product[1+t^j*x^j/(1-x^j), {j, 1, max}], {x, 0, max}, {t, 0, max}] // Normal; t[n_, k_] := SeriesCoefficient[s, {x, 0, n}, {t, 0, k}]; Table[t[n, k], {n, 1, max}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jan 17 2014 *)
    Table[Length[Select[IntegerPartitions[n],Total[Union[#]]==k&]],{n,0,10},{k,0,n}] (* Gus Wiseman, Aug 29 2023 *)
  • PARI
    A116861(n,k,s=0)={forpart(X=n,vecsum(Set(X))==k&&s++,k);s} \\ M. F. Hasler, Oct 27 2019

Formula

G.f.: -1 + Product_{j>=1} (1 + t^j*x^j/(1-x^j)).
Sum_{k=1..n} T(n,k) = A000041(n).
T(n,n) = A000009(n).
Sum_{k=1..n} k*T(n,k) = A014153(n-1).
T(n,1) = 1. T(n,2) = A000035(n+1). T(n,3) = A137719(n-2). - R. J. Mathar, Oct 27 2019
T(n,4) = A002264(n-1) + A121262(n). - R. J. Mathar, Oct 28 2019

A000097 Number of partitions of n if there are two kinds of 1's and two kinds of 2's.

Original entry on oeis.org

1, 2, 5, 9, 17, 28, 47, 73, 114, 170, 253, 365, 525, 738, 1033, 1422, 1948, 2634, 3545, 4721, 6259, 8227, 10767, 13990, 18105, 23286, 29837, 38028, 48297, 61053, 76926, 96524, 120746, 150487, 187019, 231643, 286152, 352413, 432937, 530383, 648245
Offset: 0

Views

Author

Keywords

Comments

Also number of partitions of 2*n with exactly 2 odd parts (offset 1). - Vladeta Jovovic, Jan 12 2005
Also number of transitions from one partition of n+2 to another, where a transition consists of replacing any two parts with their sum. Remove all 1' and 2' from the partition, replacing them with ((number of 2') + 1) and ((number of 1') + (number of 2') + 1); these are the two parts being summed. Number of partitions of n into parts of 2 kinds with at most 2 parts of the second kind, or of n+2 into parts of 2 kinds with exactly 2 parts of the second kind. - Franklin T. Adams-Watters, Mar 20 2006
From Christian Gutschwager (gutschwager(AT)math.uni-hannover.de), Feb 10 2010: (Start)
a(n) is also the number of pairs of partitions of n+2 which differ by only one box (for bijection see the first Gutschwager link).
a(n) is also the number of partitions of n with two parts marked.
a(n) is also the number of partitions of n+1 with two different parts marked. (End)
Convolution of A000041 and A008619. - Vaclav Kotesovec, Aug 18 2015
a(n) = P(/2,n), a particular case of P(/k,n) defined as follows: P(/0,n) = A000041(n) and P(/k,n) = P(/k-1, n) + P(/k-1,n-k) + P(/k-1, n-2k) + ... Also, P(/k,n) = the convolution of A000041 and the partitions of n with exactly k parts, and g.f. P(/k,n) = (g.f. for P(n)) * 1/(1-x)...(1-x^k). - Gregory L. Simay, Mar 22 2018
a(n) is also the sum of binomial(D(p),2) in partitions p of (n+3), where D(p)= number of different sizes of parts in p. - Emily Anible, Apr 03 2018
Also partitions of 2*(n+1) with alternating sum 2. Also partitions of 2*(n+1) with reverse-alternating sum -2 or 2. - Gus Wiseman, Jun 21 2021
Define the distance graph of the partitions of n using the distance function in A366156 as follows: two vertices (partitions) share an edge if and only if the distance between the vertices is 2. Then a(n) is the number of edges in the distance graph of the partitions of n. - Clark Kimberling, Oct 12 2023

Examples

			a(3) = 9 because we have 3, 2+1, 2+1', 2'+1, 2'+1', 1+1+1, 1+1+1', 1+1'+1' and 1'+1'+1'.
From _Gus Wiseman_, Jun 22 2021: (Start)
The a(0) = 1 through a(4) = 9 partitions of 2*(n+1) with exactly 2 odd parts:
  (1,1)  (3,1)    (3,3)      (5,3)
         (2,1,1)  (5,1)      (7,1)
                  (3,2,1)    (3,3,2)
                  (4,1,1)    (4,3,1)
                  (2,2,1,1)  (5,2,1)
                             (6,1,1)
                             (3,2,2,1)
                             (4,2,1,1)
                             (2,2,2,1,1)
The a(0) = 1 through a(4) = 9 partitions of 2*(n+1) with alternating sum 2:
  (2)  (3,1)    (4,2)        (5,3)
       (2,1,1)  (2,2,2)      (3,3,2)
                (3,2,1)      (4,3,1)
                (3,1,1,1)    (3,2,2,1)
                (2,1,1,1,1)  (4,2,1,1)
                             (2,2,2,1,1)
                             (3,2,1,1,1)
                             (3,1,1,1,1,1)
                             (2,1,1,1,1,1,1)
(End)
		

References

  • H. Gupta et al., Tables of Partitions. Royal Society Mathematical Tables, Vol. 4, Cambridge Univ. Press, 1958, p. 90.
  • J. Riordan, Combinatorial Identities, Wiley, 1968, p. 199.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

First differences are in A024786.
Third column of Riordan triangle A008951 and of triangle A103923.
The case of reverse-alternating sum 1 or alternating sum 0 is A000041.
The case of reverse-alternating sum -1 or alternating sum 1 is A000070.
The normal case appears to be A004526 or A065033.
The strict case is A096914.
The case of reverse-alternating sum 2 is A120452.
The case of reverse-alternating sum -2 is A344741.
A001700 counts compositions with alternating sum 2.
A035363 counts partitions into even parts.
A058696 counts partitions of 2n.
A103919 counts partitions by sum and alternating sum (reverse: A344612).
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.
Shift of A093695.

Programs

  • Maple
    with(numtheory): etr:= proc(p) local b; b:=proc(n) option remember; local d,j; if n=0 then 1 else add(add(d*p(d), d=divisors(j)) *b(n-j), j=1..n)/n fi end end: a:= etr(n->`if`(n<3,2,1)): seq(a(n), n=0..40); # Alois P. Heinz, Sep 08 2008
  • Mathematica
    CoefficientList[Series[1/((1 - x) (1 - x^2) Product[1 - x^k, {k, 1, 100}]), {x, 0, 100}], x] (* Ben Branman, Mar 07 2012 *)
    etr[p_] := Module[{b}, b[n_] := b[n] = If[n == 0, 1, Sum[Sum[d*p[d], {d, Divisors[j]}]*b[n - j], {j, 1, n}]/n]; b]; a = etr[If[# < 3, 2, 1]&]; Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Apr 09 2014, after Alois P. Heinz *)
    (1/((1 - x) (1 - x^2) QPochhammer[x]) + O[x]^50)[[3]] (* Vladimir Reshetnikov, Nov 22 2016 *)
    Table[Length@IntegerPartitions[n,All,Join[{1,2},Range[n]]],{n,0,15}] (* Robert Price, Jul 28 2020 and Jun 21 2021 *)
    T[n_, 0] := PartitionsP[n];
    T[n_, m_] /; (n >= m (m + 1)/2) := T[n, m] = T[n - m, m - 1] + T[n - m, m];
    T[, ] = 0;
    a[n_] := T[n + 3, 2];
    Table[a[n], {n, 0, 60}] (* Jean-François Alcover, May 30 2021 *)
    ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];Table[Length[Select[IntegerPartitions[n],ats[#]==2&]],{n,0,30,2}] (* Gus Wiseman, Jun 21 2021 *)
  • PARI
    my(x = 'x + O('x^66)); Vec( 1/((1-x)*(1-x^2)*eta(x)) ) \\ Joerg Arndt, Apr 29 2013

Formula

Euler transform of 2 2 1 1 1 1 1...
G.f.: 1/( (1-x) * (1-x^2) * Product_{k>=1} (1-x^k) ).
a(n) = Sum_{j=0..floor(n/2)} A000070(n-2*j), n>=0.
a(n) = A014153(n)/2 + A087787(n)/4 + A000070(n)/4. - Vaclav Kotesovec, Nov 05 2016
a(n) ~ sqrt(3) * exp(Pi*sqrt(2*n/3)) / (4*Pi^2) * (1 + 35*Pi/(24*sqrt(6*n))). - Vaclav Kotesovec, Aug 18 2015, extended Nov 05 2016
a(n) = A120452(n) + A344741(n). - Gus Wiseman, Jun 21 2021

Extensions

More terms from Pab Ter (pabrlos(AT)yahoo.com), May 04 2004
Edited by Emeric Deutsch, Mar 23 2005
More terms from Franklin T. Adams-Watters, Mar 20 2006
Edited by Charles R Greathouse IV, Apr 20 2010

A176206 Irregular triangle T(n,k) (n >= 1, k >= 1) read by rows: row n has length A000070(n-1) and every column k gives the positive integers.

Original entry on oeis.org

1, 2, 1, 3, 2, 1, 1, 4, 3, 2, 2, 1, 1, 1, 5, 4, 3, 3, 2, 2, 2, 1, 1, 1, 1, 1, 6, 5, 4, 4, 3, 3, 3, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 7, 6, 5, 5, 4, 4, 4, 3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 8, 7, 6, 6, 5, 5, 5, 4, 4, 4, 4, 4, 3, 3, 3, 3, 3, 3, 3
Offset: 1

Views

Author

Alford Arnold, Apr 11 2010

Keywords

Comments

The original definition was: An irregular table: Row n begins with n, counts down to 1 and repeats the intermediate numbers as often as given by the partition numbers.
Row n contains a decreasing sequence where n-k is repeated A000041(k) times, k = 0..n-1.
From Omar E. Pol, Nov 23 2020: (Start)
Row n lists in nonincreasing order the first A000070(n-1) terms of A336811.
In other words: row n lists in nonincreasing order the terms from the first n rows of triangle A336811.
Conjecture: all divisors of all terms in row n are also all parts of all partitions of n.
For more information see the example and A336811 which contains the most elementary conjecture about the correspondence divisors/partitions.
Row sums give A014153.
A338156 lists the divisors of every term of this sequence.
The n-th row of A340581 lists in nonincreasing order the terms of the first n rows of this triangle.
For a regular triangle with the same row sums see A141157. (End)
From Omar E. Pol, Jul 31 2021: (Start)
The number of k's in row n is equal to A000041(n-k), 1 <= k <= n.
The number of terms >= k in row n is equal to A000070(n-k), 1 <= k <= n.
The number of k's in the first n rows (or in the first A014153(n-1) terms of the sequence) is equal to A000070(n-k), 1 <= k <= n.
The number of terms >= k in the first n rows (or in the first A014153(n-1) terms of the sequence) is equal to A014153(n-k), 1 <= k <= n. (End)

Examples

			Triangle begins:
  1;
  2, 1;
  3, 2, 1, 1;
  4, 3, 2, 2, 1, 1, 1;
  5, 4, 3, 3, 2, 2, 2, 1, 1, 1, 1, 1;
  6, 5, 4, 4, 3, 3, 3, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1;
  7, 6, 5, 5, 4, 4, 4, 3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, ...
  ... Extended by _Omar E. Pol_, Nov 23 2020
From _Omar E. Pol_, Jan 25 2020: (Start)
For n = 5, by definition the length of row 5 is A000070(5-1) = A000070(4) = 12, so the row 5 of triangle has 12 terms. Since every column lists the positive integers A000027 so the row 5 is [5, 4, 3, 3, 2, 2, 2, 1, 1, 1, 1, 1].
Then we have that the divisors of the numbers of the 5th row are:
.
5th row of triangle -----> 5  4  3  3  2  2  2  1  1  1  1  1
                           1  2  1  1  1  1  1
                              1
.
There are twelve 1's, four 2's, two 3's, one 4 and one 5.
In total there are 12 + 4 + 2 + 1 + 1 = 20 divisors.
On the other hand the partitions of 5 are as shown below:
.
.      5
.      3  2
.      4  1
.      2  2  1
.      3  1  1
.      2  1  1  1
.      1  1  1  1  1
.
There are twelve 1's, four 2's, two 3's, one 4 and one 5, as shown also in the 5th row of triangle A066633.
In total there are 12 + 4 + 2 + 1 + 1 = A006128(5) = 20 parts.
Finally in accordance with the conjecture we can see that all divisors of all numbers in the 5th row of the triangle are the same positive integers as all parts of all partitions of 5. (End)
		

Crossrefs

Cf. A000027 (columns), A000070 (row lengths), A338156 (divisors), A340061 (mirror).

Programs

  • Mathematica
    Table[Flatten[Table[ConstantArray[n-k,PartitionsP[k]],{k,0,n-1}]],{n,10}] (* Paolo Xausa, May 30 2022 *)

Extensions

New name, changed offset, edited and more terms from Omar E. Pol, Nov 22 2020

A292508 Number A(n,k) of partitions of n with k kinds of 1; square array A(n,k), n>=0, k>=0, read by antidiagonals.

Original entry on oeis.org

1, 1, 0, 1, 1, 1, 1, 2, 2, 1, 1, 3, 4, 3, 2, 1, 4, 7, 7, 5, 2, 1, 5, 11, 14, 12, 7, 4, 1, 6, 16, 25, 26, 19, 11, 4, 1, 7, 22, 41, 51, 45, 30, 15, 7, 1, 8, 29, 63, 92, 96, 75, 45, 22, 8, 1, 9, 37, 92, 155, 188, 171, 120, 67, 30, 12, 1, 10, 46, 129, 247, 343, 359, 291, 187, 97, 42, 14
Offset: 0

Views

Author

Alois P. Heinz, Sep 17 2017

Keywords

Comments

Partial sum operator applied to column k gives column k+1.
A(n,k) is also defined for k < 0. All given formulas and programs can be applied also if k is negative.

Examples

			Square array A(n,k) begins:
  1,  1,  1,   1,   1,    1,    1,    1,     1, ...
  0,  1,  2,   3,   4,    5,    6,    7,     8, ...
  1,  2,  4,   7,  11,   16,   22,   29,    37, ...
  1,  3,  7,  14,  25,   41,   63,   92,   129, ...
  2,  5, 12,  26,  51,   92,  155,  247,   376, ...
  2,  7, 19,  45,  96,  188,  343,  590,   966, ...
  4, 11, 30,  75, 171,  359,  702, 1292,  2258, ...
  4, 15, 45, 120, 291,  650, 1352, 2644,  4902, ...
  7, 22, 67, 187, 478, 1128, 2480, 5124, 10026, ...
		

Crossrefs

Rows n=0-4 give: A000012, A001477, A000124, A004006(k+1), A027927(k+3).
Main diagonal gives A292463.
A(n,n+1) gives A292613.

Programs

  • Maple
    A:= proc(n, k) option remember; `if`(n=0, 1, add(
          (numtheory[sigma](j)+k-1)*A(n-j, k), j=1..n)/n)
        end:
    seq(seq(A(n, d-n), n=0..d), d=0..14);
    # second Maple program:
    A:= proc(n, k) option remember; `if`(n=0, 1, `if`(k<1,
          A(n, k+1)-A(n-1, k+1), `if`(k=1, combinat[numbpart](n),
          A(n-1, k)+A(n, k-1))))
        end:
    seq(seq(A(n, d-n), n=0..d), d=0..14);
    # third Maple program:
    b:= proc(n, i, k) option remember; `if`(n=0 or i<2,
          binomial(k+n-1, n), add(b(n-i*j, i-1, k), j=0..n/i))
        end:
    A:= (n, k)-> b(n$2, k):
    seq(seq(A(n, d-n), n=0..d), d=0..14);
  • Mathematica
    b[n_, i_, k_] := b[n, i, k] = If[n == 0 || i < 2, Binomial[k + n - 1, n], Sum[b[n - i*j, i - 1, k], {j, 0, n/i}]];
    A[n_, k_] := b[n, n, k];
    Table[A[n, d - n], {d, 0, 14}, {n, 0, d}] // Flatten (* Jean-François Alcover, May 17 2018, translated from 3rd Maple program *)

Formula

G.f. of column k: 1/(1-x)^k * 1/Product_{j>1} (1-x^j).
Column k is Euler transform of k,1,1,1,... .
For fixed k>=0, A(n,k) ~ 2^((k-5)/2) * 3^((k-2)/2) * n^((k-3)/2) * exp(Pi*sqrt(2*n/3)) / Pi^(k-1). - Vaclav Kotesovec, Oct 24 2018

A340061 Irregular triangle read by rows T(n,k) in which row n lists n blocks, where the m-th block consists of A000041(n-m) copies of m, with n >= 1 and m >= 1.

Original entry on oeis.org

1, 1, 2, 1, 1, 2, 3, 1, 1, 1, 2, 2, 3, 4, 1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 4, 5, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5, 6, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 5, 5, 6, 7, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2
Offset: 1

Views

Author

Omar E. Pol, Dec 28 2020

Keywords

Comments

Conjecture: all divisors of all terms of row n are also all parts of all partitions of n.
The conjecture gives a correspondence between divisors and partitions (see example).
It is conjectured that every section of the set of partitions of n has essentially the same correspondence. For more information see A336811.

Examples

			Triangle begins:
  1;
  1, 2;
  1, 1, 2, 3;
  1, 1, 1, 2, 2, 3, 4;
  1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 4, 5;
  1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5, 6;
  1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, ...
  ...
For n = 6 the 6th row of the triangle consists of:
  p(5) = 7 copies of 1, that is, [1, 1, 1, 1, 1, 1, 1],
  p(4) = 5 copies of 2, that is, [2, 2, 2, 2, 2],
  p(3) = 3 copies of 3, that is, [3, 3, 3],
  p(2) = 2 copies of 4, that is, [4, 4],
  p(1) = 1 copy   of 5, that is, [5],
  p(0) = 1 copy   of 6, that is, [6],
where p(j) is the j-th partition number A000041(j).
About the conjecture we have that the divisors of the terms of the 6th row are:
                                                                     1
                                                            1, 1,    2
                                    1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 3
  6th row -->  1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5, 6
There are nineteen 1's, eight 2's, four 3's, two 4's, one 5 and one 6.
In total there are 19 + 8 + 4 + 2 + 1 + 1 = 35 divisors.
On the other hand the partitions of 6 are:
      Diagram          Parts
    _ _ _ _ _ _
   |_ _ _      |       6
   |_ _ _|_    |       3 3
   |_ _    |   |       4 2
   |_ _|_ _|_  |       2 2 2
   |_ _ _    | |       5 1
   |_ _ _|_  | |       3 2 1
   |_ _    | | |       4 1 1
   |_ _|_  | | |       2 2 1 1
   |_ _  | | | |       3 1 1 1
   |_  | | | | |       2 1 1 1 1
   |_|_|_|_|_|_|       1 1 1 1 1 1
There are nineteen 1's, eight 2's, four 3's, two 4's, one 5 and one 6, as shown also the 6th row of A066633.
In total there are 19 + 8 + 4 + 2 + 1 + 1 = A006128(6) = 35 parts.
In accordance with the conjecture we can see that all divisors of all terms of the 6th row of triangle are the same positive integers as all parts of all partitions of 6.
		

Crossrefs

Mirror of A176206.
Row sums give A014153.
Row n has length A000070(n-1).
Right border gives A000027.

Programs

  • Mathematica
    A340061row[n_]:=Flatten[Table[ConstantArray[m,PartitionsP[n-m]],{m,n}]];Array[A340061row,10] (* Paolo Xausa, Sep 01 2023 *)

A340793 Sequence whose partial sums give A000203.

Original entry on oeis.org

1, 2, 1, 3, -1, 6, -4, 7, -2, 5, -6, 16, -14, 10, 0, 7, -13, 21, -19, 22, -10, 4, -12, 36, -29, 11, -2, 16, -26, 42, -40, 31, -15, 6, -6, 43, -53, 22, -4, 34, -48, 54, -52, 40, -6, -6, -24, 76, -67, 36, -21, 26, -44, 66, -48, 48, -40, 10, -30, 108, -106, 34, 8
Offset: 1

Views

Author

Omar E. Pol, Jan 21 2021

Keywords

Comments

Essentially a duplicate of A053222.
Convolved with the nonzero terms of A000217 gives A175254, the volume of the stepped pyramid described in A245092.
Convolved with the nonzero terms of A046092 gives A244050, the volume of the stepped pyramid described in A244050.
Convolved with A000027 gives A024916.
Convolved with A000041 gives A138879.
Convolved with A000070 gives the nonzero terms of A066186.
Convolved with the nonzero terms of A002088 gives A086733.
Convolved with A014153 gives A182738.
Convolved with A024916 gives A000385.
Convolved with A036469 gives the nonzero terms of A277029.
Convolved with A091360 gives A276432.
Convolved with A143128 gives the nonzero terms of A000441.
For the correspondence between divisors and partitions see A336811.

Crossrefs

Programs

  • Maple
    a:= n-> (s-> s(n)-s(n-1))(numtheory[sigma]):
    seq(a(n), n=1..77);  # Alois P. Heinz, Jan 21 2021
  • Mathematica
    Join[{1}, Differences @ Table[DivisorSigma[1, n], {n, 1, 100}]] (* Amiram Eldar, Jan 21 2021 *)
  • PARI
    a(n) = if (n==1, 1, sigma(n)-sigma(n-1)); \\ Michel Marcus, Jan 22 2021

Formula

a(n) = A053222(n-1) for n>1. - Michel Marcus, Jan 22 2021

A182738 Partial sums of A066186.

Original entry on oeis.org

1, 5, 14, 34, 69, 135, 240, 416, 686, 1106, 1722, 2646, 3959, 5849, 8489, 12185, 17234, 24164, 33474, 46014, 62646, 84690, 113555, 151355, 200305, 263641, 344911, 449015, 581400, 749520, 961622, 1228790, 1563509, 1982049
Offset: 1

Views

Author

Omar E. Pol, Jan 22 2011

Keywords

Comments

a(n) is also the volume of a three-dimensional version of the section model of partitions: the 3D illustrations in A135010 show boxes with face areas of 1 X 1, 2 X 2, 3 X 3, 4 X 5, 5 X 7 units along the m and p(m) axis, which is sequence A066186. Assuming that the boxes are 1 unit deep, the total volume of all boxes up to layer n is a(n). See the first two links.
From Omar E. Pol, Jan 20 2021: (Start)
a(n) is the sum of all parts of all partitions of all positive integers <= n.
Convolution of A000203 and A000070.
Convolution of A024916 and A000041.
Convolution of A175254 and A002865.
Convolution of A340793 and A014153.
Row sums of triangles A340527, A340531, A340579.
Consider a symmetric tower (a polycube) in which the terraces are the symmetric representation of sigma (n..1) respectively starting from the base (cf. A237270, A237593). The total area of the terraces equals A024916(n), the same as the area of the base.
The levels of the terraces starting from the base are the first n terms of A000070, that is A000070(0)..A000070(n-1), hence the differences between two successive levels give the partition numbers A000041, that is A000041(0)..A000041(n-1).
a(n) is the volume (or the total number of unit cubes) of the polycube.
That is due to the correspondence between divisors and partitions (cf. A336811).
The symmetric tower is a member of the family of the pyramid described in A245092.
The growth of the volume of the polycube represents every convolution mentioned above. (End)

Examples

			a(6) = 135 because the volume V(6) = p(1) + 2*p(2) + 3*p(3) + 4*p(4) + 5*p(5) + 6*p(6) = 1 + 2*2 + 3*3 + 4*5 + 5*7 + 6*11 = 1 + 4 + 9 + 20 + 35 + 66 = 135 where p(n) = A000041(n).
		

Crossrefs

Programs

  • Mathematica
    With[{no=35},Accumulate[PartitionsP[Range[no]]Range[no]]] (* Harvey P. Dale, Feb 02 2011 *)

Formula

a(n) = n*A000070(n) - A014153(n-1). - Vaclav Kotesovec, Jun 23 2015
a(n) ~ sqrt(n) * exp(Pi*sqrt(2*n/3)) / (Pi*2^(3/2)) * (1 + (11*Pi/(24*sqrt(6)) - sqrt(6)/Pi)/sqrt(n) + (73*Pi^2/6912 - 3/16)/n). - Vaclav Kotesovec, Jun 23 2015, extended Nov 04 2016
G.f.: x*f'(x)/(1 - x), where f(x) = Product_{k>=1} 1/(1 - x^k). - Ilya Gutkovskiy, Apr 10 2017

A222730 Total sum T(n,k) of parts <= n of multiplicity k in all partitions of n; triangle T(n,k), n>=0, 0<=k<=n, read by rows.

Original entry on oeis.org

0, 0, 1, 3, 2, 1, 11, 6, 0, 1, 36, 10, 3, 0, 1, 79, 21, 3, 1, 0, 1, 186, 33, 7, 3, 1, 0, 1, 345, 59, 9, 4, 1, 1, 0, 1, 672, 89, 20, 4, 4, 1, 1, 0, 1, 1163, 145, 22, 11, 4, 2, 1, 1, 0, 1, 2026, 212, 44, 13, 6, 4, 2, 1, 1, 0, 1, 3273, 325, 56, 21, 8, 6, 2, 2, 1, 1, 0, 1
Offset: 0

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Author

Alois P. Heinz, Mar 03 2013

Keywords

Comments

For k > 0, column k is asymptotic to sqrt(3) * (2*k+1) * exp(Pi*sqrt(2*n/3)) / (2 * k^2 * (k+1)^2 * Pi^2) ~ 6 * (2*k+1) * n * p(n) / (k^2 * (k+1)^2 * Pi^2), where p(n) is the partition function A000041(n). - Vaclav Kotesovec, May 29 2018

Examples

			The partitions of n=4 are [1,1,1,1], [2,1,1], [2,2], [3,1], [4].  Parts <= 4 with multiplicity m=0 sum up to (2+3+4)+(3+4)+(1+3+4)+(2+4)+(1+2+3) = 36, for m=1 the sum is 2+(3+1)+4 = 10, for m=2 the sum is 1+2 = 3, for m=3 the sum is 0, for m=4 the sum is 1 => row 4 = [36, 10, 3, 0, 1].
Triangle T(n,k) begins:
    0;
    0,  1;
    3,  2,  1;
   11,  6,  0, 1;
   36, 10,  3, 0, 1;
   79, 21,  3, 1, 0, 1;
  186, 33,  7, 3, 1, 0, 1;
  345, 59,  9, 4, 1, 1, 0, 1;
  672, 89, 20, 4, 4, 1, 1, 0, 1;
		

Crossrefs

Programs

  • Maple
    b:= proc(n, p) option remember; `if`(n=0 and p=0, [1, 0],
          `if`(p=0, [0$(n+2)], add((l-> subsop(m+2=p*l[1]+l[m+2], l))
              ([b(n-p*m, p-1)[], 0$(p*m)]), m=0..n/p)))
        end:
    T:= n-> subsop(1=NULL, b(n, n))[]:
    seq(T(n), n=0..14);
  • Mathematica
    b[n_, p_] := b[n, p] = If[n == 0 && p == 0, {1, 0}, If[p == 0, Array[0&, n+2], Sum[Function[l, ReplacePart[l, m+2 -> p*l[[1]] + l[[m+2]]]][Join[b[n - p*m, p-1] , Array[0&, p*m]]], {m, 0, n/p}]]]; Rest /@ Table[b[n, n], {n, 0, 14}] // Flatten (* Jean-François Alcover, Dec 16 2013, translated from Maple *)

Formula

Sum_{k=0..n} k*T(n,k) = A066186(n) = n*A000041(n).
Sum_{k=1..n} T(n,k) = A014153(n-1) for n>0.
Sum_{k=0..n} T(n,k) = n*(n+1)/2*A000041(n) = A000217(n)*A000041(n).
(2 * Sum_{k=0..n} T(n,k)) / (Sum_{k=0..n} k*T(n,k)) = n+1 for n>0.
T(2*n+1,n+1) = A002865(n).
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