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.

Showing 1-6 of 6 results.

A305309 Array read by rows: a(n, k) = A048996(n, k) * A118851(n, k), n >= 1, k = 1..A000041(n).

Original entry on oeis.org

1, 2, 1, 3, 4, 1, 4, 6, 4, 6, 1, 5, 8, 12, 9, 12, 8, 1, 6, 10, 16, 9, 12, 36, 8, 12, 24, 10, 1, 7, 12, 20, 24, 15, 48, 27, 36, 16, 72, 32, 15, 40, 12, 1, 8, 14, 24, 30, 16, 18, 60, 72, 48, 54, 20, 96, 54, 144, 16, 20, 120, 80, 18, 60, 14, 1, 9, 16, 28, 36, 40, 21, 72, 90, 48, 60, 144, 27, 24, 120, 144, 192, 216, 96, 25, 160, 90, 360, 80, 24, 180, 160, 21, 84, 16, 1, 10, 18, 32, 42, 48, 25, 24, 84, 108, 120, 72, 180, 96, 108, 28, 144, 180, 96, 240, 576, 108, 128, 216, 30, 200, 240, 480, 540, 480, 32, 30, 240, 135, 720, 240, 28, 252, 280, 24, 112, 18, 1
Offset: 1

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Author

Wolfdieter Lang, May 31 2018

Keywords

Comments

The Data section here is longer than usual. Do not shorten it! - N. J. A. Sloane, Jan 10 2019
The length of row n is A000041(n), the number of partitions of n.
Partitions follow the Abramowitz-Stegun (A-St) order (see the link).
The row sums give A001906(n) = Fibonacci(2*n).
The triangle T(n, m) obtained by summing in row n the entries of the columns k with identical part number m is A078812(n, m) = binomial(n+m-1, 2*m-1) (with offsets n >= 1, m = 1..n). The array of the number of parts m = m(n,k) = A036043(n, k) in A-St order.
This array is the elementwise product of the array A048996, the composition numbers, and A118851, the products of the parts of partitions, both arrays are in A-St order.
Therefore a(n, k) is the sum of the number of products of the block lengths of all the A048996(n, k) set partition of [n] := {1,2, ..., n} with m = m(n, k) blocks consisting of consecutive numbers corresponding to the k-th partition of n in A-St order. Because the block structure depends only on the exponents (signature) of the underlying partition this leads to the product of the two array entries. Equivalently, one can consider compositions. Then a(n, k) gives the sum of the products of the parts of all A048996(n, k) compositions originating from the k-th partition of n.
This array is the result of an attempt to understand the comment of Kevin Long, May 11 2018, on A001906.
This array is similar to A085643 but some pairs of numbers like (27, 36), (72,48), (54,144), ... are there swapped.

Examples

			For the rows n = 1..10, and comments on compositions and set partitions with blocks of consecutive numbers, see the link.
Example: n = 5, k = 4: the partition is (1^2, 3^1) = [1,1,3] with m = m(n,k) = 3. The A048996(5, 4) = 3 compositions are 1 + 1 + 3, 1 + 3 + 1 and 3 + 1 + 1. The corresponding three consecutive 3-block partitions of [5] := {1, 2, ..., 5} are {1}, {2}, {3,4,5} and {1}, {2,3,4}, {5} and {1,2,3}, {4}, {5}, Therefore, a(5, 4) = 1*1*3 + 1*3*1 + 3*1*1 = 3*3 = 9. For the compositions one has the same sum from the products of the parts.
		

Crossrefs

Formula

a(n, k) = A048996(n, k) * A118851(n, k), n >= 1, k = 1..A000041(n).

A182779 a(n) = A049019(n) * A118851(n). Irregular table read by rows.

Original entry on oeis.org

1, 1, 2, 2, 3, 12, 6, 4, 24, 24, 72, 24, 5, 40, 120, 180, 360, 480, 120, 6, 60, 240, 180, 360, 2160, 720, 1440, 4320, 3600, 720, 7, 84, 420, 840, 630, 5040, 3780, 7560, 3360, 30240, 20160, 12600, 50400, 30240, 5040
Offset: 0

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Author

Alford Arnold, Dec 01 2010

Keywords

Comments

The sequences have shape A000041 and their respective row sums are A000670, A006906 and A006153.

Examples

			For n = 3 the values are (3,12,6) = (1,6,6)*(3,2,1).
Table starts:
1;
1;
2, 2;
3, 12, 6;
4, (24, 24), 72, 24;
5, (40, 120), (180, 360), 480, 120;
6, (60, 240, 180), (360, 2160, 720), (1440, 4320), 3600, 720;
7, (84, 420, 840), (630, 5040, 3780, 7560), (3360, 30240, 20160), (12600, 50400), 30240, 5040;
		

Crossrefs

Cf. A006153 (related to function composition), A133314 (signed version of A049019).

Formula

a(n) = A049019(n) * A118851(n).

Extensions

a(0) = 1 prepended by Peter Luschny, May 31 2020

A006906 a(n) is the sum of products of terms in all partitions of n.

Original entry on oeis.org

1, 1, 3, 6, 14, 25, 56, 97, 198, 354, 672, 1170, 2207, 3762, 6786, 11675, 20524, 34636, 60258, 100580, 171894, 285820, 480497, 791316, 1321346, 2156830, 3557353, 5783660, 9452658, 15250216, 24771526, 39713788, 64011924, 102199026, 163583054, 259745051
Offset: 0

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Author

Keywords

Comments

a(0) = 1 since the only partition of 0 is the empty partition. The product of its terms is the empty product, namely 1.
Same parity as A000009. - Jon Perry, Feb 12 2004

Examples

			Partitions of 0 are {()} whose products are {1} whose sum is 1.
Partitions of 1 are {(1)} whose products are {1} whose sum is 1.
Partitions of 2 are {(2),(1,1)} whose products are {2,1} whose sum is 3.
Partitions of 3 are 3 => {(3),(2,1),(1,1,1)} whose products are {3,2,1} whose sum is 6.
Partitions of 4 are {(4),(3,1),(2,2),(2,1,1),(1,1,1,1)} whose products are {4,3,4,2,1} whose sum is 14.
		

References

  • G. Labelle, personal communication.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Haskell
    a006906 n = p 1 n 1 where
       p _ 0 s = s
       p k m s | mReinhard Zumkeller, Dec 07 2011
  • Maple
    A006906 := proc(n)
        option remember;
        if n = 0 then
            1;
        else
            add( A078308(k)*procname(n-k),k=1..n)/n ;
        end if;
    end proc: # R. J. Mathar, Dec 14 2011
    # second Maple program:
    b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0,
           b(n, i-1) +add(b(n-i*j, i-1)*(i^j), j=1..n/i)))
        end:
    a:= n-> b(n, n):
    seq(a(n), n=0..40);  # Alois P. Heinz, Feb 25 2013
  • Mathematica
    (* a[n,k]=sum of products of partitions of n into parts <= k *) a[0,0]=1; a[n_,0]:=0; a[n_,k_]:=If[k>n, a[n,n], a[n,k] = a[n,k-1] + k a[n-k,k] ]; a[n_]:=a[n,n] (* Dean Hickerson, Aug 19 2007 *)
    Table[Total[Times@@@IntegerPartitions[n]],{n,0,35}] (* Harvey P. Dale, Jan 14 2013 *)
    nmax = 40; CoefficientList[Series[Product[1/(1 - k*x^k), {k, 1, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Dec 15 2015 *)
    nmax = 40; CoefficientList[Series[Exp[Sum[PolyLog[-j, x^j]/j, {j, 1, nmax}]], {x, 0, nmax}], x] (* Vaclav Kotesovec, Dec 15 2015 *)

Formula

The limit of a(n+3)/a(n) is 3. However, the limit of a(n+1)/a(n) does not exist. In fact, the sequence {a(n+1)/a(n)} has three limit points, which are about 1.4422447, 1.4422491 and 1.4422549. (See the Links entry.) - Dean Hickerson, Aug 19 2007
a(n) ~ c(n mod 3) 3^(n/3), where c(0)=97923.26765718877..., c(1)=97922.93936857030... and c(2)=97922.90546334208... - Dean Hickerson, Aug 19 2007
G.f.: 1 / Product_{k>=1} (1-k*x^k).
G.f.: 1 + Sum_{n>=1} n*x^n / Product_{k=1..n} (1-k*x^k) = 1 + Sum_{n>=1} n*x^n / Product_{k>=n} (1-k*x^k). - Joerg Arndt, Mar 23 2011
a(n) = (1/n)*Sum_{k=1..n} A078308(k)*a(n-k). - Vladeta Jovovic, Nov 22 2002
O.g.f.: exp( Sum_{n>=1} Sum_{k>=1} k^n * x^(n*k) / n ). - Paul D. Hanna, Sep 18 2017
O.g.f.: exp( Sum_{n>=1} Sum_{k=1..n} A008292(n,k)*x^(n*k)/(n*(1-x^n)^(n+1)) ), where A008292 is the Eulerian numbers. - Paul D. Hanna, Sep 18 2017

Extensions

More terms from Vladeta Jovovic, Oct 04 2001
Edited by N. J. A. Sloane, May 19 2007

A124758 Product of the parts of the compositions in standard order.

Original entry on oeis.org

1, 1, 2, 1, 3, 2, 2, 1, 4, 3, 4, 2, 3, 2, 2, 1, 5, 4, 6, 3, 6, 4, 4, 2, 4, 3, 4, 2, 3, 2, 2, 1, 6, 5, 8, 4, 9, 6, 6, 3, 8, 6, 8, 4, 6, 4, 4, 2, 5, 4, 6, 3, 6, 4, 4, 2, 4, 3, 4, 2, 3, 2, 2, 1, 7, 6, 10, 5, 12, 8, 8, 4, 12, 9, 12, 6, 9, 6, 6, 3, 10, 8, 12, 6, 12, 8, 8, 4, 8, 6, 8, 4, 6, 4, 4, 2, 6, 5, 8, 4, 9, 6
Offset: 0

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Author

Keywords

Comments

The standard order of compositions is given by A066099.
A composition of n is a finite sequence of positive integers summing to n. The k-th composition in standard order (row k of 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. - Gus Wiseman, Apr 03 2020

Examples

			Composition number 11 is 2,1,1; 2*1*1 = 2, so a(11) = 2.
The table starts:
  1
  1
  2 1
  3 2 2 1
  4 3 4 2 3 2 2 1
  5 4 6 3 6 4 4 2 4 3 4 2 3 2 2 1
The 146-th composition in standard order is (3,3,2), with product 18, so a(146) = 18. - _Gus Wiseman_, Apr 03 2020
		

Crossrefs

Cf. A066099, A118851, A011782 (row lengths), A001906 (row sums).
The lengths of standard compositions are given by A000120.
The version for prime indices is A003963.
The version for binary indices is A096111.
Taking the sum instead of product gives A070939.
The sum of binary indices is A029931.
The sum of prime indices is A056239.
Taking GCD instead of product gives A326674.
Positions of first appearances are A331579.

Programs

  • Mathematica
    stc[n_]:=Differences[Prepend[Join@@Position[Reverse[IntegerDigits[n,2]],1],0]]//Reverse;
    Table[Times@@stc[n],{n,0,100}] (* Gus Wiseman, Apr 03 2020 *)

Formula

For a composition b(1),...,b(k), a(n) = Product_{i=1}^k b(i).
a(A164894(n)) = a(A246534(n)) = n!. - Gus Wiseman, Apr 03 2020
a(A233249(n)) = a(A333220(n)) = A003963(n). - Gus Wiseman, Apr 03 2020
From Mikhail Kurkov, Jul 11 2021: (Start)
Some conjectures:
a(2n+1) = a(n) for n >= 0.
a(2n) = (1 + 1/A001511(n))*a(n) = 2*a(n) + a(n - 2^f(n)) - a(2n - 2^f(n)) for n > 0 with a(0)=1 where f(n) = A007814(n).
From the 1st formula for a(2n) we get a(4n+2) = 2*a(n), a(4n) = 2*a(2n) - a(n).
Sum_{k=0..2^n - 1} a(k) = A001519(n+1) for n >= 0.
a((4^n - 1)/3) = A011782(n) for n >= 0.
a(2^m*(2^n - 1)) = m + 1 for n > 0, m >= 0. (End)

A339095 Triangle read by rows: T(n,k) is the number of partitions of n with product of parts equal to k, 1 <= k <= A000792(n).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 0, 2, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 0, 2, 1, 1, 1, 2, 1, 2, 1, 3, 1, 1, 0, 3, 0, 0, 1, 3, 0, 1, 1, 1, 1, 2, 1, 2, 1, 3, 2, 1, 0, 3, 0, 1, 1, 3, 0, 2, 0, 2, 0, 0, 0, 2, 0, 0, 1
Offset: 1

Views

Author

Abhimanyu Kumar, Nov 23 2020

Keywords

Comments

The product of the parts of a partition is called its norm.

Examples

			For n=6 the partitions and their counts for each norm are given in the table below.
  Relevant partition(s)  | Norm | Count
  1+1+1+1+1+1+1          | 1    | 1
  2+1+1+1+1              | 2    | 1
  3+1+1+1                | 3    | 1
  4+1+1, 2+2+1+1         | 4    | 2
  5+1                    | 5    | 1
  6, 3+2+1               | 6    | 2
  4+2, 2+2+2             | 8    | 2
  3+3                    | 9    | 1
The number of partitions of 6 with norm value 4 are 2, expressed as T(6,4)=2. Similarly, T(6,7)=0 because there is no partition of 6 with norm 7.
So the 6th row is 1, 1, 1, 2, 1, 2, 0, 2, 1.
First few rows of the array are:
  1;
  1, 1;
  1, 1, 1;
  1, 1, 1, 2;
  1, 1, 1, 2, 1, 1;
  1, 1, 1, 2, 1, 2, 0, 2, 1;
  1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 0, 2;
  1, 1, 1, 2, 1, 2, 1, 3, 1, 1, 0, 3, 0, 0, 1, 3, 0, 1;
  ...
		

References

  • Abhimanyu Kumar and Meenakshi Rana, On the treatment of partitions as factorization and further analysis, Journal of the Ramanujan Mathematical Society 35(3), 263-276 (2020).

Crossrefs

Cf. A000041 (row sums), A000792 (row lengths), A001055, A118851, A212721.

Programs

  • PARI
    row(n) = {my(list = List()); forpart(p=n, listput(list, vecprod(Vec(p)));); my(vlist = Vec(list)); my(v = vector(vecmax(vlist))); for (i=1, #vlist, v[vlist[i]]++); v;} \\ Michel Marcus, Nov 26 2020

Formula

Let the number of partitions of n having the norm value k refer to the norm counting function T(n,k). The following properties hold true:
Max_{n=1..oo} T(n,k) = A001055(k).
Sum_{k=1..A000792(n)} T(n,k) = A000041(n).
Sum_{n>=1} T(n+1,k) - T(n,k) = A001055(k) - 1.
G.f.: Sum_{n>=0} Sum_{k>=1} ((q^n)/(k^s))*T(n,k) = Product_{m>=1}(1-((q^m)/(m^s)))^(-1).
n*Sum_{k=1..A000792(n)} T(n,k) = Sum_{m=1..n} (Sum_{k=1..A000792(n-m)} T(n-m,i)*k^(-s))*(Sum_{d|m} (d/m)^(d*s-1))

Extensions

More terms from Michel Marcus, Nov 26 2020

A092991 Least product of the parts of the partitions of n where that product has the maximum number of divisors.

Original entry on oeis.org

1, 1, 2, 2, 4, 6, 6, 12, 12, 24, 36, 48, 60, 60, 120, 180, 240, 360, 360, 720, 1080, 1440, 2160, 2880, 2520, 6480, 5040, 7560, 10080, 15120, 20160, 30240, 45360, 60480, 75600, 120960, 151200, 226800, 302400, 453600, 604800, 907200, 1209600, 1814400
Offset: 0

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Author

Amarnath Murthy, Mar 28 2004

Keywords

Comments

Let P be the set of all products of partitions of n and t = max_{m in P} tau(m). Then a(n) = min_{m in P and tau(m) = t} m. Note that the sequence is not monotonic; the first decrease is a(26) = 5040 < 6480 = a(25) and the second is a(49) = 3326400 < 10886400 = a(48). - Franklin T. Adams-Watters, Jun 14 2006
All terms are in A025487. - David A. Corneth, Apr 30 2024

Examples

			a(9) = 24 corresponding to the partition (2,2,2,3).
a(8) = 12 corresponding to the partition (1,3,4). Another partition (3,3,2) gives a product 18 with same number of divisors 6 but 18>12 hence a(8) = 12.
		

Crossrefs

Programs

  • Mathematica
    a[n_] := Module[{t = Transpose[{t = Times @@@ IntegerPartitions[n], DivisorSigma[0, t]}]}, MaximalBy[SortBy[t, Last], Last, 1][[1, 1]]]; Array[a, 50, 0] (* Amiram Eldar, Apr 13 2024 *)

Extensions

Corrected and extended by Franklin T. Adams-Watters, Jun 14 2006
Showing 1-6 of 6 results.