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 1-10 of 11 results. Next

A014153 Expansion of 1/((1-x)^2*Product_{k>=1} (1-x^k)).

Original entry on oeis.org

1, 3, 7, 14, 26, 45, 75, 120, 187, 284, 423, 618, 890, 1263, 1771, 2455, 3370, 4582, 6179, 8266, 10980, 14486, 18994, 24757, 32095, 41391, 53123, 67865, 86325, 109350, 137979, 173450, 217270, 271233, 337506, 418662, 517795, 638565, 785350, 963320, 1178628
Offset: 0

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Author

Keywords

Comments

Number of partitions of n with three kinds of 1. E.g., a(2)=7 because we have 2, 1+1, 1+1', 1+1", 1'+1', 1'+1", 1"+1". - Emeric Deutsch, Mar 22 2005
Partial sums of the partial sums of the partition numbers A000041. Partial sums of A000070. Euler transform of 3,1,1,1,...
Also sum of parts, counted without multiplicity, in all partitions of n, offset 1. Also Sum phi(p), where the sum is taken over all parts p of all partitions of n, offset 1. - Vladeta Jovovic, Mar 26 2005
Equals row sums of triangle A141157. - Gary W. Adamson, Jun 12 2008
A014153 convolved with A010815 = (1, 2, 3, ...). n-th partial sum sequence of A000041 convolved with A010815 = (n-1)-th column of Pascal's triangle, starting (1, n, ...). - Gary W. Adamson, Nov 09 2008
From Omar E. Pol, May 25 2012: (Start)
a(n) is also the sum of all parts of the (n+1)st column of a version of the section model of partitions in which every section has its parts aligned to the right margin (cf. A210953, A210970, A135010).
Rows of triangle A210952 converge to this sequence. (End)
Using the above result (see Jovovic's comment) of Jovovic and Mertens's theorem on the average order of the phi function, we can obtain the estimate a(n-1) = (6/Pi^2)*n*p(n) + O(log(n)*A006128(n)), where p(n) is the partition function A000041(n). It can be shown that A006128(n) = O(sqrt(n)*log(n)*p(n)), so we have the asymptotic result a(n) ~ (6/Pi^2)*n*p(n). - Peter Bala, Dec 23 2013
a(n-2) is the number of partitions of 2n or 2n-1 with palindromicity 2; that is, partitions that can be listed in palindromic order except for a central sequence of two distinct parts. - Gregory L. Simay, Nov 01 2015
Convolution of A000041 and A000027. - Omar E. Pol, Jun 17 2021
Convolution of A002865 and the positive terms of A000217. Partial sums give A014160. - Omar E. Pol, Mar 01 2023

Crossrefs

Cf. A010815. - Gary W. Adamson, Nov 09 2008
Column k=3 of A292508.

Programs

  • Magma
    m:=45; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( 1/((1-x)^2*(&*[1-x^k: k in [1..50]])) )); // G. C. Greubel, Oct 15 2018
    
  • Maple
    with(numtheory):
    a:= proc(n) option remember;
          `if`(n=0, 1, add((2+sigma(j)) *a(n-j), j=1..n)/n)
        end:
    seq(a(n), n=0..40);  # Alois P. Heinz, Feb 13 2012
  • Mathematica
    a[n_] := a[n] = If[n == 0, 1, Sum[(2+DivisorSigma[1, j])*a[n-j], {j, 1, n}]/n]; Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Mar 03 2014, after Alois P. Heinz *)
    Table[Sum[(n-k)*PartitionsP[k],{k,0,n}],{n,1,50}] (* Vaclav Kotesovec, Jun 23 2015 *)
    t[n_, k_] := Sum[StirlingS1[n, j]* Binomial[i + j - 1, i]* PartitionsP[k - n - i], {j, 0, n}, {i, 0, k - n}]; Print@ Table[t[n, k], {k, 10}, {n, 0, k - 1}]; Table[t[2, k], {k, 3, 43}] (* George Beck, May 25 2016 *)
  • PARI
    x='x+O('x^45); Vec(1/((1-x)^2*prod(k=1,50, 1-x^k))) \\ G. C. Greubel, Oct 15 2018

Formula

Let t(n_, k_) = Sum_{i = 0..k} Sum_{j = 0..n} s(n, j)*C(i, j)*p(k - n - i), where s(n, j) are Stirling numbers of the first kind, C(i, j) are the number of compositions of i distinct objects into j parts, and p is the integer partition function. Then a(k) = t(2, k+2) (conjectured). The formula for t(n, k) is the same as at A126442 except that there the Stirling numbers are of the second kind. - George Beck, May 21 2016
a(n) = (n+1)*A000070(n+1) - A182738(n+1). - Vaclav Kotesovec, Nov 04 2016
a(n) ~ exp(sqrt(2*n/3)*Pi)*sqrt(3)/(2*Pi^2) * (1 + 23*Pi/(24*sqrt(6*n))). - Vaclav Kotesovec, Nov 04 2016

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

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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

A063746 Triangle read by rows giving number of partitions of k (k=0 .. n^2) with Ferrers plot fitting in an n X n box.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 3, 3, 3, 3, 2, 1, 1, 1, 1, 2, 3, 5, 5, 7, 7, 8, 7, 7, 5, 5, 3, 2, 1, 1, 1, 1, 2, 3, 5, 7, 9, 11, 14, 16, 18, 19, 20, 20, 19, 18, 16, 14, 11, 9, 7, 5, 3, 2, 1, 1, 1, 1, 2, 3, 5, 7, 11, 13, 18, 22, 28, 32, 39, 42, 48, 51, 55, 55, 58, 55, 55, 51, 48, 42, 39, 32, 28
Offset: 0

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Author

Wouter Meeussen, Aug 14 2001

Keywords

Comments

Seems to approximate a Gaussian distribution, the sum of all 1+n^2 terms in a row equals the central binomial coefficients.
a(n,k) is the number of sequences of n 0's and n 1's having major index equal to k (the major index is the sum of the positions of the 1's that are immediately followed by 0's). Equivalently, a(n,k) is the number of Grand Dyck paths of length 2n for which the sum of the positions of the valleys is k. Example: a(3,7)=2 because the only sequences of three 0's and three 1's with major index 7 are 010110 and 110010. The corresponding Grand Dyck paths are obtained by replacing a 0 by a U=(1,1) step and a 1 by a D=(1,-1) step. - Emeric Deutsch, Oct 02 2007
Also, number of n-multisets in [0..n] whose elements sum up to n. - M. F. Hasler, Apr 12 2012
Let P be the poset [n] X [n] ordered by the product order. Let J(P) be the set of all order ideals of P, ordered by inclusion. Then J(P) is a finite sublattice of Young's lattice and T(n,k) is the number of elements in J(P) that have rank k. - Geoffrey Critzer, Mar 26 2020

Examples

			From _M. F. Hasler_, Apr 12 2012: (Start)
The table reads:
n=0: 1  _  (k=0)
n=1: 1 1  _  (k=0..1)
n=2: 1 1 2 1 1  _  (k=0..4)
n=3: 1 1 2 3 3 3 3  2  1  1  _  (k=0..9)
n=4: 1 1 2 3 5 5 7  7  8  7  7  5  5  3  2  1  1  _  (k=0..16)
n=5: 1 1 2 3 5 7 9 11 14 16 18 19 20 20 19 18 16 ...  _  (k=0..25)
etc. (End)
Cycle index of S(3) is (1/6)*(x(1)^3+3*x(1)*x(2)+2*x(3)), so g.f. for 3rd row is (1/6)*((1+x+x^2+x^3)^3+3*(1+x+x^2+x^3)*(1+x^2+x^4+x^6)+2*(1+x^3+x^6+x^9)) = x^9+x^8+2*x^7+3*x^6+3*x^5+3*x^4+3*x^3+2*x^2+x+1.
a(3,7)=2 because the only partitions of 7 with Ferrers plot fitting into a 3 X 3 box are [3,3,1] and [3,2,2].
		

References

  • G. E. Andrews and K. Eriksson, Integer partitions, Cambridge Univ. Press, 2004, pp. 67-69.
  • D. M. Bressoud, Proofs and Confirmations, Camb. Univ. Press, 1999; exercise 3.2.3.
  • A. V. Yurkin, New binomial and new view on light theory, (book), 2013, 78 pages, no publisher listed.

Crossrefs

Row lengths are given by A002522. - M. F. Hasler, Apr 14 2012
Antidiagonal sums are given by A260894.
Row sums give A000984.

Programs

  • Maple
    for n from 0 to 15 do QBR[n]:=sum(q^i,i=0..n-1) od: for n from 0 to 15 do QFAC[n]:=product(QBR[j],j=1..n) od: qbin:=(n,k)->QFAC[n]/QFAC[k]/QFAC[n-k]: for n from 0 to 7 do P[n]:=sort(expand(simplify(qbin(2*n,n)))) od: for n from 0 to 7 do seq(coeff(P[n],q,j),j=0..n^2) od; # yields sequence in triangular form - Emeric Deutsch, Apr 23 2007
    # second Maple program:
    b:= proc(n, i, k) option remember;
          `if`(n=0, 1, `if`(i<1 or k<1, 0, b(n, i-1, k)+
          `if`(i>n, 0, b(n-i, i, k-1))))
        end:
    T:= n-> seq(b(k, min(n, k), n), k=0..n^2):
    seq(T(n), n=0..8); # Alois P. Heinz, Apr 05 2012
  • Mathematica
    Table[nn=n^2;CoefficientList[Series[Product[(1-x^(n+i))/(1-x^i),{i,1,n}],{x,0,nn}],x],{n,0,6}]//Grid (* Geoffrey Critzer, Sep 27 2013 *)
    Table[CoefficientList[QBinomial[2n,n,q] // FunctionExpand, q], {n,0,6}] // Flatten (* Peter Luschny, Jul 22 2016 *)
    b[n_, i_, k_] := b[n, i, k] = If[n == 0, 1, If[i < 1 || k < 1, 0, b[n, i - 1, k] + If[i > n, 0, b[n - i, i, k - 1]]]];
    T[n_] := Table[b[k, Min[n, k], n], {k, 0, n^2}];
    Table[T[n], {n, 0, 8}] // Flatten (* Jean-François Alcover, Nov 27 2020, after Alois P. Heinz *)
  • PARI
    T(n,k)=polcoeff(prod(i=0,n,sum(j=0,n,x^(j*i*(n^2+n+1)+j),O(x^(k*(n^2+n+1)+n+1)))),k*(n^2+n+1)+n)  /* Based on a more general formula due to R. Gerbicz. M. F. Hasler, Apr 12 2012 */

Formula

Table[T[k, n, n], {n, 0, 9}, {k, 0, n^2}] with T[ ] defined as in A047993.
G.f.: Consider a function; f(n) = 1 + sum(i_1=1, n, sum(i_2=0, i_1, ..., sum(i_n=0, i_(n-1), x^(sum(j=1, n, i_j))*(1+...+x^i_n))...)) Then the GF is f(1)+x^3.f(2)+x^8.f(3)+..., where after x^3 the increase is n^2+1 from f(n). - Jon Perry, Jul 13 2004
G.f. for n-th row is obtained if we set x(i) = 1+x^i+x^(2*i)+...+x^(n*i), i=1, 2, ..., n, in the cycle index Z(S(n);x(1), x(2), ..., x(n)) of the symmetric group S(n) of degree n. - Vladeta Jovovic, Dec 17 2004
G.f. of row n: the q-binomial coefficient [2n,n]. - Emeric Deutsch, Apr 23 2007
T(n,k)=1 for k=0,1,n^2-1,n^2. For all m>n, T(m,n)=T(n,n)=A000041(n), i.e., below the diagonal the columns remain constant, because there cannot be more than n nonzero elements with sum <= n. - M. F. Hasler, Apr 12 2012
T(n,2n) = A128552(n-2). - Geoffrey Critzer, Sep 27 2013
From Alois P. Heinz, Jan 09 2025: (Start)
Sum_{k=0..n} T(n,k) = A000070(n).
Sum_{k=0..n} k * T(n,k) = A182738(n).
Sum_{k=0..n^2} k * T(n,k) = A002544(n-1) for n>=1.
Sum_{k=0..n^2} (-1)^k * T(n,k) = A126869(n). (End)

A082775 Convolution of natural numbers >= 2 and the partition numbers (A000041).

Original entry on oeis.org

2, 5, 11, 21, 38, 64, 105, 165, 254, 381, 562, 813, 1162, 1636, 2279, 3139, 4285, 5794, 7776, 10353, 13694, 17992, 23502, 30520, 39433, 50687, 64855, 82607, 104785, 132375, 166608, 208921, 261090, 325196, 403779, 499818, 616928, 759335, 932135
Offset: 2

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Author

Alford Arnold, May 22 2003

Keywords

Comments

Contribution from George Beck, Jan 08 2011: (Start)
The number of multiset partitions of the n-multiset M={0,0,...,0,1,2} (with n-2 zeros) is sum_{k=0..(n-2)}( (n-k) * p(k) ) where p(k) is the number of partitions of k.
Proof:
For each k = 0, 1, ..., n-2, partition k zeros and add the remaining n-k-2 zeros to the block {1, 2}, to give p(k) partitions.
For each k, partition k zeros and add the remaining n-k-2 zeros to the two blocks {1} and {2} in all possible 1 + n-k-2 ways, which gives (1 + n-k-2) * p(k) partitions.
Together, the number of partitions of M is sum_{k=0..n-2}( (n-k) * p(k) ). (End)
A082775 is the special case of A126442 with n-k = 2.

Examples

			a(7) = 64 because (7,5,3,2,1,1) dot (2,3,4,5,6,7) = 14+15+12+10+6+7= 64.
		

Crossrefs

Column k=2 of A346426.

Programs

  • Mathematica
    f[n_] := Sum[(n - k) PartitionsP[k], {k, 0, n - 2}]; Array[f, 39, 2]

Formula

a(n) = a(n-1) + A000041(n) + A000070(n) for n>1. - Alford Arnold, Dec 10 2007
a(n) = n*A000070(n-2) - A182738(n-2) for n>2. - Vaclav Kotesovec, Jun 23 2015
a(n) ~ sqrt(3) * exp(Pi*sqrt(2*n/3)) / (2*Pi^2). - Vaclav Kotesovec, Jun 23 2015

Extensions

More terms from Ray Chandler, Oct 11 2003

A284870 Expansion of Sum_{i>=1} i*x^i/(1 - x) * Product_{j=1..i} 1/(1 - x^j).

Original entry on oeis.org

0, 1, 4, 10, 22, 42, 77, 131, 217, 345, 537, 812, 1211, 1767, 2547, 3615, 5078, 7043, 9687, 13185, 17815, 23867, 31766, 41972, 55146, 71997, 93519, 120813, 155358, 198811, 253374, 321509, 406436, 511802, 642264, 803140, 1001154, 1243966, 1541167, 1903754, 2345300, 2881404, 3531195, 4316632, 5264444, 6405389
Offset: 0

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Author

Ilya Gutkovskiy, Apr 04 2017

Keywords

Comments

Total number of parts in all partitions of all positive integers <= n.
Sum of largest parts of all partitions of all positive integers <= n.
From Omar E. Pol, Feb 16 2021: (Start)
Apart from initial zero this is as follows:
Convolution of A341062 and A014153.
Convolution of A000005 and A000070.
Convolution of nonzero terms of A006218 and A000041.
a(n) is also the total number of divisors of all terms in the n-th row of triangle A340581. These divisors are also all parts of all partitions of all positive integers <= n. (End)

Examples

			a(4) = 22 because we have 1 = 1, 2 = 2, 1 + 1 = 2, 3 = 3, 2 + 1 = 3, 1 + 1 + 1 = 3, 4 = 4, 3 + 1 = 4, 2 + 2 = 4, 2 + 1 + 1 = 4 and 1 + 1 + 1 + 1 = 4 therefore 1 + 1 + 2 + 1 + 2 + 3 + 1 + 2 + 2 + 3 + 4 = 22 (total number of parts) or 1 + 2 + 1 + 3 + 2 + 1 + 4 + 3 + 2 + 2 + 1 = 22 (sum of largest parts).
		

Crossrefs

Programs

  • Maple
    b:= proc(n, i) option remember; `if`(n=0 or i=1, [1, n],
          b(n, i-1) +(p-> p+[0, p[1]])(b(n-i, min(n-i, i))))
        end:
    a:= proc(n) a(n):= `if`(n<1, 0, a(n-1)+b(n$2)[2]) end:
    seq(a(n), n=0..45);  # Alois P. Heinz, Feb 16 2021
  • Mathematica
    nmax = 45; CoefficientList[Series[Sum[i x^i /(1 - x) Product[1/(1 - x^j), {j, 1, i}], {i, 1, nmax}], {x, 0, nmax}], x]
    nmax = 45; CoefficientList[Series[1/(1 - x) Sum[x^i /(1 - x^i), {i, 1, nmax}] Product[1/(1 - x^j), {j, 1, nmax}], {x, 0, nmax}], x]
    Accumulate[Table[Sum[DivisorSigma[0, k] PartitionsP[n - k], {k, 1, n}], {n, 0, 45}]]

Formula

G.f.: Sum_{i>=1} i*x^i/(1 - x) * Product_{j=1..i} 1/(1 - x^j).
G.f.: (1/(1 - x)) * Sum_{i>=1} x^i/(1 - x^i) * Product_{j>=1} 1/(1 - x^j).
a(n) = Sum_{k=0..n} A006128(k).
a(n) = A124920(n+1) - 1.
a(n) = Sum_{k=1..n} k * A299779(n,k). - Alois P. Heinz, May 14 2018

A340527 Triangle read by rows: T(n,k) = A024916(n-k+1)*A000041(k-1), 1 <= k <= n.

Original entry on oeis.org

1, 4, 1, 8, 4, 2, 15, 8, 8, 3, 21, 15, 16, 12, 5, 33, 21, 30, 24, 20, 7, 41, 33, 42, 45, 40, 28, 11, 56, 41, 66, 63, 75, 56, 44, 15, 69, 56, 82, 99, 105, 105, 88, 60, 22, 87, 69, 112, 123, 165, 147, 165, 120, 88, 30, 99, 87, 138, 168, 205, 231, 231, 225, 176, 120, 42, 127, 99, 174
Offset: 1

Views

Author

Omar E. Pol, Jan 10 2021

Keywords

Comments

Conjecture 1: T(n,k) is the sum of divisors of the terms that are in the k-th blocks of the first n rows of triangle A176206.
Conjecture 2: the sum of row n equals A182738(n), the sum of all parts of all partitions of all positive integers <= n.
Conjecture 3: T(n,k) is also the volume (or number of cubes) of the k-th block of a symmetric tower in which the terraces are the symmetric representation of sigma (n..1) starting from the base respectively (cf. A237270, A237593), hence the total area of the terraces is 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 levels give the partition numbers A000041, that is A000041(0)..A000041(n-1).
This symmetric tower has the property that its volume (or total number of cubes) equals A182738(n), the sum of all parts of all partitions of all positive integers <= n.
For another symmetric tower of the same family and whose volume equals A066186(n) see A339106 and A221529.
The above three conjectures are connected due to the correspondence between divisors and partitions (cf. A336811).

Examples

			Triangle begins:
   1;
   4,   1;
   8,   4,   2;
  15,   8,   8,   3;
  21,  15,  16,  12,   5;
  33,  21,  30,  24,  20,   7;
  41,  33,  42,  45,  40,  28,  11;
  56,  41,  66,  63,  75,  56,  44,  15;
  69,  56,  82,  99, 105, 105,  88,  60,  22;
  87,  69, 112, 123, 165, 147, 165, 120,  88,  30;
  99,  87, 138, 168, 205, 231, 231, 225, 176, 120,  42;
...
For n = 6 the calculation of every term of row 6 is as follows:
--------------------------
k   A000041         T(6,k)
1      1  *  33   =   33
2      1  *  21   =   21
3      2  *  15   =   30
4      3  *   8   =   24
5      5  *   4   =   20
6      7  *   1   =    7
.          A024916
--------------------------
The sum of row 6 is 33 + 21 + 30 + 24 + 20 + 7 = 135, equaling A182738(6).
		

Crossrefs

Columns 1 and 2 give A024916.
Column 3 gives A327329.
Leading diagonal gives A000041.
Row sums give A182738.

A340531 Irregular triangle read by rows T(n,k), (n >= 1, k >= 1), in which row n has length is A000070(n-1) and every column k is A024916, the sum of all divisors of all numbers <= n.

Original entry on oeis.org

1, 4, 1, 8, 4, 1, 1, 15, 8, 4, 4, 1, 1, 1, 21, 15, 8, 8, 4, 4, 4, 1, 1, 1, 1, 1, 33, 21, 15, 15, 8, 8, 8, 4, 4, 4, 4, 4, 1, 1, 1, 1, 1, 1, 1, 41, 33, 21, 21, 15, 15, 15, 8, 8, 8, 8, 8, 4, 4, 4, 4, 4, 4, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 56, 41, 33, 33, 21, 21, 21, 15, 15, 15, 15, 15
Offset: 1

Views

Author

Omar E. Pol, Jan 10 2021

Keywords

Comments

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 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).
T(n,k) is the volume (the number of cells) in the k-th level starting from the base.
This polycube has the property that the volume (the total number of cells) equals A182738(n), the sum of all parts of all partitions of all positive integers <= n.
A dissection of the symmetric tower is a three-dimensional spiral whose top view is described in A239660.
Other triangles related to the volume of this polycube are A340527 and A340579.
The symmetric tower is a member of the family of the stepped pyramid described in A245092.
For another symmetric tower of the same family and whose volume equals A066186(n) see A340423.
The sum of row n of triangle equals A182738(n). That property is due to the correspondence between divisors and parts. For more information see A336811.

Examples

			Triangle begins:
   1;
   4,  1;
   8,  4,  1,  1;
  15,  8,  4,  4, 1, 1, 1;
  21, 15,  8,  8, 4, 4, 4, 1, 1, 1, 1, 1;
  33, 21, 15, 15, 8, 8, 8, 4, 4, 4, 4, 4, 1, 1, 1, 1, 1, 1, 1;
...
For n = 5 the length of row 5 is A000070(4) = 12.
The sum of row 5 is 21 + 15 + 8 + 8 + 4 + 4 + 4 + 1 + 1 + 1 + 1 + 1 = 69, equaling A182738(5).
		

Crossrefs

Row sums give A182738.
Cf. A340527 (a regular version).
Members of the same family are: A176206, A337209, A339258, A340530.

Formula

a(m) = A024916(A176206(m)), assuming A176206 has offset 1.
T(n,k) = A024916(A176206(n,k)), assuming A176206 has offset 1.

A086716 Convolution of triangular numbers with partition numbers.

Original entry on oeis.org

1, 5, 15, 36, 75, 143, 255, 433, 707, 1119, 1725, 2602, 3851, 5607, 8046, 11399, 15963, 22123, 30369, 41328, 55792, 74763, 99496, 131566, 172931, 226027, 293864, 380160, 489480, 627428
Offset: 1

Views

Author

Jon Perry, Jul 29 2003

Keywords

Comments

Partial sum operator applied to partition numbers 4 times.

Crossrefs

Programs

Formula

a(n) = ((n+1)*(n+2)*(A000070(n)-1) - (2*n+3)*A182738(n) + A259279(n))/2. - Vaclav Kotesovec, Jun 23 2015
a(n) ~ 3*sqrt(n) * exp(Pi*sqrt(2*n/3)) / (sqrt(2)*Pi^3). - Vaclav Kotesovec, Jun 23 2015

A340579 Triangle read by rows: T(n,k) = A000203(n-k+1)*A000070(k-1), 1 <= k <= n.

Original entry on oeis.org

1, 3, 2, 4, 6, 4, 7, 8, 12, 7, 6, 14, 16, 21, 12, 12, 12, 28, 28, 36, 19, 8, 24, 24, 49, 48, 57, 30, 15, 16, 48, 42, 84, 76, 90, 45, 13, 30, 32, 84, 72, 133, 120, 135, 67, 18, 26, 60, 56, 144, 114, 210, 180, 201, 97, 12, 36, 52, 105, 96, 228, 180, 315, 268, 291, 139, 28, 24, 72, 91
Offset: 1

Views

Author

Omar E. Pol, Jan 12 2021

Keywords

Comments

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).
T(n,k) is the total volume (or total number of cubes) exactly below the symmetric representation of sigma(n-k+1). In other words: T(n,k) is the total volume (the total number of cubes) exactly below the terraces that are in the k-th level that contains terraces starting from the base.
This symmetric tower has the property that its volume (the total number of cubes) equals A182738(n), the sum of all parts of all partitions of all positive integers <= n. That is due to the correspondence between divisors and partitions (cf. A336811).
The growth of the volume represents the convolution of A000203 and A000070.
The symmetric tower is a member of the family of the pyramid described in A245092.
For another symmetric tower of the same family and whose volume equals A066186(n) see A221529 and A339106.

Examples

			Triangle begins:
   1;
   3,   2;
   4,   6,   4;
   7,   8,  12,   7;
   6,  14,  16,  21,  12;
  12,  12,  28,  28,  36,  19;
   8,  24,  24,  49,  48,  57,  30;
  15,  16,  48,  42,  84,  76,  90,  45;
  13,  30,  32,  84,  72, 133, 120, 135,  67;
  18,  26,  60,  56, 144, 114, 210, 180, 201,  97;
  12,  36,  52, 105,  96, 228, 180, 315, 268, 291, 139;
...
For n = 6 the calculation of every term of row 6 is as follows:
-------------------------
k   A000070        T(6,k)
1      1  *  12  =   12
2      2  *  6   =   12
3      4  *  7   =   28
4      7  *  4   =   28
5     12  *  3   =   36
6     19  *  1   =   19
.         A000203
-------------------------
The sum of row 6 is 12 + 12 + 28 + 28 + 36 + 19 = 135, equaling A182738(6).
		

Crossrefs

Programs

  • PARI
    row(n) = vector(n, k, sigma(n-k+1)*sum(i=0, k-1, numbpart(i))); \\ Michel Marcus, Jul 23 2021

A259279 a(n) = Sum_{k=0..n} k^2 * A000041(k).

Original entry on oeis.org

0, 1, 9, 36, 116, 291, 687, 1422, 2830, 5260, 9460, 16236, 27324, 44393, 70853, 110453, 169589, 255422, 380162, 557052, 807852, 1157124, 1642092, 2305987, 3213187, 4436937, 6083673, 8277963, 11192875, 15032040, 20075640, 26650802, 35200178, 46245905, 60476265
Offset: 0

Views

Author

Vaclav Kotesovec, Jun 23 2015

Keywords

Crossrefs

Programs

  • Mathematica
    Table[Sum[k^2*PartitionsP[k],{k,0,n}],{n,0,50}]

Formula

a(n) ~ n^(3/2) * exp(Pi*sqrt(2*n/3)) / (Pi*2^(3/2)).
Showing 1-10 of 11 results. Next