A115729 -id:A115729 - cp's OEIS frontend

A115728 Number of subpartitions of partitions in Abramowitz and Stegun order.

1, 2, 3, 3, 4, 5, 4, 5, 7, 6, 7, 5, 6, 9, 9, 10, 9, 9, 6, 7, 11, 12, 13, 10, 14, 13, 10, 12, 11, 7, 8, 13, 15, 16, 14, 19, 17, 16, 16, 19, 16, 14, 15, 13, 8, 9, 15, 18, 19, 18, 24, 21, 15, 23, 22, 26, 21, 19, 22, 23, 24, 19, 15, 18, 18, 15, 9, 10, 17, 21, 22, 22, 29

Offset: 0

Franklin T. Adams-Watters, Mar 11 2006

nonn

subpart([n^k]) = C(n+k,k); subpart([1,2,3,...,n]) = C_n = A000108(n). The b(i,j) defined in the formula for sequences [1,2,3,...] form A009766.

Row sums are A297388. Row lengths are A000041. - Geoffrey Critzer, Jan 10 2021

			Partition 5 in A&S order is [2,1]; it has 5 subpartitions: [], [1], [2], [1^2] and [2,1] itself.
1
2
3, 3
4, 5, 4
5, 7, 6, 7,  5
6, 9, 9, 10, 9, 9, 6

M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].

Cf. A115729, A036036, A000108, A009766, A007318, A297388, A000041.

/* Expects input as vector in increasing order - e.g. [1,1,2,3] */
subpart(p)=local(i,j,v,n);n=matsize(p)[2];if(n==0,1,v=vector(p[n]+1);v[1] =1;for(i=1,n,for(j=1,p[i],v[j+1]+=v[j]));for(j=1,p[n],v[j+1]+=v[j]);v[p[n ]+1])

/* Given Partition p(), Find Subpartitions s(): */ {s(n)=polcoeff(x^n-sum(k=0, n-1, s(k)*x^k*(1-x+x*O(x^n))^p(k)),n)} \\ Paul D. Hanna, Jul 03 2006

For a partition P = [p_1,...,p_n] with the p_i in increasing order, define b(i,j) to be the number of subpartitions of [p_1,...,p_i] with the i-th part = j (b(i,0) is subpartitions with less than i parts). Then b(1,j)=1 for j<=p_1, b(i+1,j) = Sum_{k=0..j} b(i,k) for 0<=j<=p_{i+1}; and the total number of subpartitions is sum_{k=1..p_n} b(n,k).

For a partition P = {p(n)}, the number of subpartitions {s(n)} of P can be determined by the g.f.: 1/(1-x) = Sum_{n>=0} s(n)*x^n*(1-x)^p(n). - Paul D. Hanna, Jul 03 2006

A107876 Triangular matrix T, read by rows, that satisfies: [T^k](n,k) = T(n,k-1) for n>=k>0, or, equivalently, (column k of T^k) = SHIFT_LEFT(column k-1 of T) when zeros above the diagonal are ignored.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 7, 7, 3, 1, 1, 37, 37, 15, 4, 1, 1, 268, 268, 106, 26, 5, 1, 1, 2496, 2496, 975, 230, 40, 6, 1, 1, 28612, 28612, 11100, 2565, 425, 57, 7, 1, 1, 391189, 391189, 151148, 34516, 5570, 707, 77, 8, 1, 1, 6230646, 6230646, 2401365, 544423

Offset: 0

Paul D. Hanna, Jun 04 2005

Remarkably, T equals the product of these triangular matrices: T = A107862^-1*A107867 = A107867^-1*A107870 = A107870^-1*A107873; reversing the order of these products yields triangle A101479.
Column m of T^k is the number of subpartitions of the initial terms of the sequence (k-1)+n(m-1)+n(n-1)/2 (ignoring 0's above the diagonal). E.g., column 4 of T^3 is 1,3,15,106,975,.... The sequence above is 2,5,9,14,20,.... subp([]) = 1, subp([2]) = 3, subp([2,5]) = 15, subp([2,5,9]) = 106, etc. The matrix product of T^(k-1) * T computes the number of such subpartitions by looking at the first part index where the subpartition is maxed - for [2,5,9,14,20] the third term (9 maxed) has subp([1,4]) for the first two values (not maxed), times subp([5,11]) for the last two values (possibly maxed). - Franklin T. Adams-Watters, Jun 26 2006
T(n,k) is the number of Dyck paths whose sequence of ascent lengths is exactly k,k+1,...,n, for example the T(4,3) = 3 paths are UUUdUUUUd^6, UUUddUUUUd^5 and UUUdddUUUUd^4. - David Scambler, May 30 2012

			G.f. for column 1:
1 = T(1,1)*(1-x)^1 + T(2,1)*x*(1-x)^2 + T(3,1)*x^2*(1-x)^4 + T(4,1)*x^3*(1-x)^7 + T(5,1)*x^4*(1-x)^11 + T(6,1)*x^5*(1-x)^16 +...
= 1*(1-x)^1 + 1*x*(1-x)^2 + 2*x^2*(1-x)^4 + 7*x^3*(1-x)^7 + 37*x^4*(1-x)^11 + 268*x^5*(1-x)^16 +...
G.f. for column 2:
1 = T(2,2)*(1-x)^1 + T(3,2)*x*(1-x)^3 + T(4,2)*x^2*(1-x)^6 + T(5,2)*x^3*(1-x)^10 + T(6,2)*x^4*(1-x)^15 + T(7,2)*x^5*(1-x)^21 +...
= 1*(1-x)^1 + 1*x*(1-x)^3 + 3*x^2*(1-x)^6 + 15*x^3*(1-x)^10 + 106*x^4*(1-x)^15 + 975*x^5*(1-x)^21 +...
Triangle T begins:
       1;
       1,      1;
       1,      1,      1;
       2,      2,      1,     1;
       7,      7,      3,     1,    1;
      37,     37,     15,     4,    1,   1;
     268,    268,    106,    26,    5,   1,  1;
    2496,   2496,    975,   230,   40,   6,  1, 1;
   28612,  28612,  11100,  2565,  425,  57,  7, 1, 1;
  391189, 391189, 151148, 34516, 5570, 707, 77, 8, 1, 1; ...
where column 1 of T = SHIFT_LEFT(column 0 of T).
Matrix square T^2 begins:
     1;
     2,   1;
     3,   2,   1;
     7,   5,   2,  1;
    26,  19,   7,  2,  1;
   141, 104,  37,  9,  2, 1;
  1034, 766, 268, 61, 11, 2, 1; ...
Compare column 2 of T^2 with column 1 of T.
Matrix inverse begins:
   1;
  -1,    1;
   0,   -1,   1;
   0,   -1,  -1,   1;
   0,   -3,  -2,  -1,  1;
   0,  -15,  -9,  -3, -1,  1;
   0, -106, -61, -18, -4, -1, 1; ...
Compare column 1 of T^-1 with column 2 of T and
compare column 2 of T^-1 with column 3 of T^2.

Alois P. Heinz, Rows n = 0..50, flattened

Cf. A107862, A107865, A107867, A107870, A107877 (column 1), A107878 (column 2), A107879 (column 3), A107880 (matrix square), A107884 (matrix cube), A107889 (matrix inverse).
Cf. A115728, A115729, A101479 (dual triangle).
T(2n,n) gives A300954.

max = 10;
A107862 = Table[Binomial[If[nA107867 = Table[Binomial[If[nA107862].A107867;
Table[t[[n, k]], {n, 1, max+1}, {k, 1, n}] // Flatten (* Jean-François Alcover, Dec 12 2012, after first comment, fixed by Vaclav Kotesovec, Jun 13 2018 *)

{T(n,k)=polcoeff(1-sum(j=0,n-k-1, T(j+k,k)*x^j*(1-x+x*O(x^n))^(1+(k+j)*(k+j-1)/2-k*(k-1)/2)),n-k)}
for(n=0,10,for(k=0,n,print1(T(n,k),", ")); print(""))

/* Print the Triangular Matrix to the Power p: */
{T(n,k,p)=polcoeff(1- sum(j=0,n-k-1,T(j+k,k,p)*x^j*(1-x+x*O(x^n))^(j*(j-1)/2+j*k+p)),n-k)}
for(n=0,10,for(k=0,n,print1(T(n,k,1),", ")); print(""))

G.f. for column k of T^m, the m-th matrix power of this triangle T:
(1) 1 = Sum_{j>=0} T(k+j, k) * x^j * (1-x)^(1+(k+j)*(k+j-1)/2-k*(k-1)/2) for m=1.
(2) 1 = Sum_{j>=0} [T^m](k+j, k)*x^j*(1-x)^(m+(k+j)*(k+j-1)/2-k*(k-1)/2) for all m and k>=0.
(3) 1 = Sum_{j>=0} [T^m](k+j, k)*x^j / C(x)^(m-j+(k+j)*(k+j-1)/2-k*(k-1)/2) where C(x)=2/(1+sqrt(1-4*x)) is g.f. for A000108 (Catalan numbers).
Matrix inverse of this triangle T satisfies:
(4) [T^-1](n,k) = -[T^k](n,k+1) for n>k>=0.

A238690 Let each integer m (1 <= m <= n) be factorized as m = prime_m(1)prime_m(2)...*prime_m(bigomega(m)), with the primes sorted in nonincreasing order. Then a(n) is the number of values of m such that each prime_m(i) <= prime_n(i).

Original entry on oeis.org

1, 2, 3, 3, 4, 5, 5, 4, 6, 7, 6, 7, 7, 9, 9, 5, 8, 9, 9, 10, 12, 11, 10, 9, 10, 13, 10, 13, 11, 14, 12, 6, 15, 15, 14, 12, 13, 17, 18, 13, 14, 19, 15, 16, 16, 19, 16, 11, 15, 16, 21, 19, 17, 14, 18, 17, 24, 21, 18, 19, 19, 23, 22, 7, 22, 24, 20, 22, 27, 23, 21

Offset: 1

Matthew Vandermast, Apr 28 2014

nonn

Equivalently, a(n) equals the number of values of m such that each value of A238689 T(m,k) <= A238689 T(n,k). (Since the prime factorization of 1 is the empty factorization, we consider each prime_1(i) not to be greater than prime_n(i) for all positive integers n.)
Suppose we say that n "covers" m iff both m and n are factorized as described in the sequence definition and each prime_m(i) <= prime_n(i). At least three sequences (A037019, A108951 and A181821) have the property that a(m) divides a(n) iff n "covers" m.  These sequences are also divisibility sequences (i.e., sequences with the property that a(m) divides a(n) if m divides n), since any positive integer "covers" each of its divisors.
For any positive integers m and k, the following integer sequences (with n >= 0) are arithmetic progressions:
1. The sequence b(n) = a(m*(2^n)).
2. The sequence b(n) = a(m*(prime(n+k))) if prime(k) >= A006530(m).
Also, a(n) = the number of distinct prime signatures that occur among the divisors of any integer m such that A181819(m) = n and/or A238745(m) = n.
Number of skew partitions whose numerator has Heinz number n, where a skew partition is a pair y/v of integer partitions such that the diagram of v fits inside the diagram of y. The Heinz number of an integer partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k). - Gus Wiseman, Feb 24 2018

			The prime factorizations of integers 1 through 9, with prime factors sorted from largest to smallest:
1 - the empty factorization (no prime factors)
2 = 2
3 = 3
4 = 2*2
5 = 5
6 = 3*2
7 = 7
8 = 2*2*2
9 = 3*3
To find a(9), we consider 9 = 3*3. There are 6 positive integers (1, 2, 3, 4, 6 and 9) which satisfy the following criteria:
1) The largest prime factor, if one exists, is not greater than 3;
2) The second-largest prime factor, if one exists, is not greater than 3;
3) The total number of prime factors (counting repeated factors) does not exceed 2.
Therefore, a(9) = 6.
From _Gus Wiseman_, Feb 24 2018: (Start)
Heinz numbers of the a(15) = 9 partitions contained within the partition (32) are 1, 2, 3, 4, 5, 6, 9, 10, 15. The a(15) = 9 skew partitions are (32)/(), (32)/(1), (32)/(11), (32)/(2), (32)/(21), (32)/(22), (32)/(3), (32)/(31), (32)/(32).
Corresponding diagrams are:
  o o o   . o o   . o o   . . o   . . o   . . o   . . .   . . .   . . .
  o o     o o     . o     o o     . o     . .     o o     . o     . .    (End)

Amiram Eldar, Table of n, a(n) for n = 1..10000

Rearrangement of A115728, A115729 and A238746. A116473(n) is the number of times n appears in the sequence.

Cf. A000041, A000085, A000720, A056239, A063834, A112798, A122111, A153452, A215366, A238689, A259478, A259480, A296150, A296188, A296561, A297388, A299925, A299926, A299966, A299967.

undptns[y_]:=Select[Tuples[Range[0,#]&/@y],OrderedQ[#,GreaterEqual]&];
primeMS[n_]:=If[n===1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
Table[Length[undptns[Reverse[primeMS[n]]]],{n,100}] (* Gus Wiseman, Feb 24 2018 *)

a(n) = A085082(A108951(n)) = A085082(A181821(n)).
a(n) = a(A122111(n)).
a(prime(n)) = a(2^n) = n+1.
a((prime(n))^m) = a((prime(m))^n) = binomial(n+m, n).
a(A002110(n)) = A000108(n+1).
A000005(n) <= a(n) <= n.

A107354 To compute a(n) we first write down 2^n 1's in a row. Each row takes the right half of the previous row and each element in it equals sum of the elements in the previous row starting at the middle. The single element in the last row is a(n).

Original entry on oeis.org

1, 1, 2, 7, 44, 516, 11622, 512022, 44588536, 7718806044, 2664170119608, 1836214076324153, 2529135272371085496, 6964321029630556852944, 38346813253279804426846032, 422247020982575523983378003936, 9298487213328788062025571134762096

Offset: 0

Max Alekseyev, May 24 2005

Number of subpartitions of partition [1,3,7,...,2^n-1]. - Franklin T. Adams-Watters, Mar 11 2006
Can also be computed summing forwards:
  1
  1,1
  1,2,2, 2
  1,3,5, 7, 7, 7, 7, 7
  1,4,9,16,23,30,37,44,44,44,44,44,44,44,44,44

			For n=4, the array looks like this:
  1..1..1..1..1..1..1..1..1..1..1..1..1..1..1..1
  ........................1..2..3..4..5..6..7..8
  ....................................5.11.18.26
  .........................................18.44
  ............................................44
  Therefore a(4)=44.
For n=5, we can illustrate the recurrence by:
a(5) = 516 = C(19, 4) - ( 1*C(17, 4) + 2*C(14, 3) + 7*C(9, 2) ) = C(16+4-1, 4) - ( 1*C(16-2+4-1, 4) + 2*C(16-4+3-1, 3) + 7*C(16-8+2-1, 2) ).

Alois P. Heinz, Table of n, a(n) for n = 0..82 (first 26 terms from Reinhard Zumkeller)

Cf. A105996; variants: A109055 - A109061; subpartitions defined: A115728, A115729.

Column k=2 of A355576.

a107354 n = head $ snd $ until ((== 1) . fst)
                               f (2^n, replicate (2^n) 1) where
   f (len, xs) = (len', scanl1 (+) $ drop len' xs) where
      len' = len `div` 2
-- Feasible only for small n.
-- Reinhard Zumkeller, Nov 20 2011

a:= proc(n) option remember; `if`(n=0, 1, -add(
      a(j)*(-1)^(n-j)*binomial(2^j, n-j), j=0..n-1))
    end:
seq(a(n), n=0..16);  # Alois P. Heinz, Jul 08 2022

f[n_] := If[n == 0, 1, Binomial[2^(n - 1) + n - 2, n - 1] - Sum[ f[k]*Binomial[2^(n - 1) - 2^k + n - k - 1, n - k], {k, n - 2}]]; Table[ f[n], {n, 0, 15}] (* Robert G. Wilson v, May 25 2005 *)
Table[NestWhile[Accumulate[Drop[#,Ceiling[Length[#]/2]]]&,PadRight[{},2^n+1,1], Length[ #]> 1&],{n,0,16}]//Flatten (* Harvey P. Dale, Jun 24 2018 *)

{a(n)=if(n==0,1,binomial(2^(n-1)+n-2,n-1)- sum(k=1,n-2,a(k)*binomial(2^(n-1)-2^k+n-k-1,n-k)))} \\ Paul D. Hanna, May 24 2005

{a(n)=polcoeff(x^n-sum(k=0, n-1, a(k)*x^k*(1-x+x*O(x^n))^(2^k-1) ), n)} \\ Paul D. Hanna, May 24 2005

a(n) = C(2^(n-1)+n-2,n-1) - Sum_{k=1..n-2} a(k)*C(2^(n-1)-2^k+n-k-1,n-k) for n>=2, with a(0)=1, a(1)=1, where C = binomial. - Paul D. Hanna, May 24 2005
The first number in row 3 is 2^(n-2)+1. - Ralf Stephan, May 24 2005
G.f.: 1/(1-x) = Sum_{n>=0} a(n)*x^n*(1-x)^(2^n-1) (g.f. of subpartitions). - Paul D. Hanna, Jul 03 2006
G.f.: 1 = Sum_{n>=0} a(n)*x^n/(1+x)^(2^n+n). - Paul D. Hanna, Jul 03 2006

Edited by Paul D. Hanna, Jul 03 2006

A107877 Column 1 of triangle A107876.

Original entry on oeis.org

1, 1, 2, 7, 37, 268, 2496, 28612, 391189, 6230646, 113521387, 2332049710, 53384167192, 1348601249480, 37291381915789, 1120914133433121, 36406578669907180, 1271084987848923282, 47487293697623885913, 1890771531272515677250, 79947079338974990793060

Offset: 0

Paul D. Hanna, Jun 04 2005, Apr 10 2007

nonn

Also number of subpartitions of partition consisting of first n-1 triangular numbers; e.g., a(4) = subp([1,3,6]) = 37. - Franklin T. Adams-Watters, Jun 26 2006
Number of length-n restricted growth strings (RGS) [s(0),s(1),...,s(n-1)] where s(0)=0 and s(k) <= s(k-1)+k, see Fxtbook link and example. - Joerg Arndt, Apr 30 2011
Number of Dyck paths whose ascent lengths are exactly {1,2,...,n+1}; for example, the a(2) = 2 paths are uduuduuudddd and uduudduuuddd. - David Scambler, May 30 2012
Number of types of cells of a fine mixed subdivision of the Tesler flow polytope. - Alejandro H. Morales, Oct 11 2017

			1 = 1*(1-x)^1 + 1*x*(1-x)^2 + 2*x^2*(1-x)^4 + 7*x^3*(1-x)^7 + 37*x^4*(1-x)^11 + 268*x^5*(1-x)^16 + 2496*x^6*(1-x)^22 + ...
Also equals the final term in rows of the triangle where row n+1 equals the partial sums of row n with the final term repeated n+1 times, starting with a '1' in row 0, as illustrated by:
1;
1, 1;
1, 2,  2,  2;
1, 3,  5,  7,  7,  7,   7;
1, 4,  9, 16, 23, 30,  37,  37,  37,  37,  37;
1, 5, 14, 30, 53, 83, 120, 157, 194, 231, 268, 268, 268, 268, 268, 268; ...
Restricted growth strings: a(0)=1 corresponds to the empty string; a(1)=1 to [0];
a(2) = 2 to [00] and [01]; a(3)=7 to
  1:  [ 0 0 0 ],
  2:  [ 0 0 1 ],
  3:  [ 0 0 2 ],
  4:  [ 0 1 0 ],
  5:  [ 0 1 1 ],
  6:  [ 0 1 2 ],
  7:  [ 0 1 3 ].
[_Joerg Arndt_, Apr 30 2011]

R. P. Stanley, Enumerative Combinatorics volume 1, 2nd edition, Cambridge University Press, 2011, Ch. 3

Alois P. Heinz, Table of n, a(n) for n = 0..390
Joerg Arndt, Matters Computational (The Fxtbook), section 17.3.6, pp. 368-369
K. Mészáros, A. H. Morales, Volumes and Ehrhart polynomials of flow polytopes, arXiv:1710.00701 [math.CO], 2017, sections 6.1 and 7.

Cf. A107876, A107878, A107879, A115728, A115729.

Cf. A305605, A305601.

b:= proc(n, y) option remember; `if`(n=0, 1, add(
      b(n-1, y+i-n), i=max(1, n-y)..n*(n-1)/2+1-y))
    end:
a:= n-> b(n+1, 0):
seq(a(n), n=0..25);  # Alois P. Heinz, Nov 26 2016
# second Maple program:
a:= n-> LinearAlgebra:-Determinant(Matrix(n,(i,j)->
        binomial(binomial(n+1-i,2)+1,i-j+1))):
seq(a(n), n=0..25); # Alejandro H. Morales, Aug 31 2017

a[ n_, k_: 1, j_: 1] := If[ n < 2, Boole[n >= 0], a[n, k, j] = Sum[a[n - 1, i, j + 1], {i, k + j}]]; (* Michael Somos, Nov 26 2016 *)

{a(n)=polcoeff(1-sum(k=0,n-1,a(k)*x^k*(1-x+x*O(x^n))^(1+k*(k+1)/2)),n)}

G.f.: 1 = Sum_{k>=0} a(k)*x^k*(1-x)^(1 + k*(k+1)/2).
G.f.: 1 = Sum_{k>=0} a(k)*x^k/(1+x)^((k+1)*(k+2)/2).
From Benedict W. J. Irwin, Nov 26 2016: (Start)
Conjecture: a(n) can be expressed with a series of nested sums,
a(3) = Sum_{i=1..2} i+2,
a(4) = Sum_{i=1..2} Sum_{j=1..i+2} j+3,
a(5) = Sum_{i=1..2} Sum_{j=1..i+2} Sum_{k=1..j+3} k+4,
a(6) = Sum_{i=1..2} Sum_{j=1..i+2} Sum_{k=1..j+3} Sum_{l=1..k+4} l+5. (End)
Determinantal formula: a(n) = Det(A) where A is the n X n matrix with entries A(i,j) = binomial(binomial(n+1-i,2)+1,i-j+1). This follows by the formula by MacMahon (see EC1 Ex 3.63) for the number of such subpartitions. -  Alejandro H. Morales, Aug 31 2017

A109055 To compute a(n) we first write down 3^n 1's in a row. Each row takes the rightmost 3rd part of the previous row and each element in it equals sum of the elements of the previous row starting with the first of the rightmost 3rd part. The single element in the last row is a(n).

Original entry on oeis.org

1, 1, 3, 24, 541, 35649, 6979689, 4085743032, 7166723910237, 37698139930450365, 594816080266215640710, 28154472624850002001979592, 3997853576535778666975681355079, 1703042427700923785323670557504832751, 2176429411666209822350337722381643148477248

Offset: 0

Augustine O. Munagi, Jun 17 2005

nonn

Comment from Franklin T. Adams-Watters, Jul 13 2006: This is the number of subpartitions of the sequence 3^n-1. As such it can also be computed adding forward, with 3^n terms in the n-th line:
..........................................................................
1 1.......................................................................
2.3.3..3..3..3..3..3......................................................
3.6.9.12.15.18.21.24.24.24.24.24.24.24.24.24.24.24.24.24.24.24.24.24.24.24

			For example, for n=3 the array looks like this:
1..1..1..1..1........1..1..1..1..1..1..1..1..1..1
........................1..2..3..4..5..6..7..8..9
..........................................7.15.24
...............................................24
Therefore a(3)=24.

Alois P. Heinz, Table of n, a(n) for n = 0..65

Cf. A107354, A109056, A109057, A109058, A109059, A109060, A109061, A109062.
Cf. A115728, A115729.
Column k=3 of A355576.

proc(n::nonnegint) local f,a; if n=0 or n=1 then return 1; end if; f:=L->[seq(add(L[i],i=2*nops(L)/3+1..j),j=2*nops(L)/3+1..nops(L))]; a:=f([seq(1,j=1..3^n)]); while nops(a)>3 do a:=f(a) end do; a[3]; end proc;

A[n_, k_] := A[n, k] = If[n == 0, 1, -Sum[A[j, k]*(-1)^(n - j)*Binomial[If[j == 0, 1, k^j], n - j], {j, 0, n - 1}]];
a[n_] := A[n, 3];
Table[a[n], {n, 0, 14}] (* Jean-François Alcover, Apr 01 2024, after Alois P. Heinz in A355576 *)

More terms from Paul D. Hanna

A115725 Number of partitions with maximum rectangle <= n.

Original entry on oeis.org

1, 2, 5, 10, 26, 42, 118, 171, 389, 692, 1442, 1854, 5534, 6895, 11910, 21116, 44278, 52568, 118734, 138670, 300326, 492507, 728514, 829244, 2167430, 2987124, 4167602, 6092588, 11308432, 12554900, 29925267, 33023589, 57950313, 81424281, 106214784, 148101088

Offset: 0

Franklin T. Adams-Watters, Mar 11 2006

nonn

A partition has maximum rectangle <= n iff it is a subpartition of row n of A010766.

			The 10 partitions with maximum rectangle <= 3: 0: []; 1: [1]; 2: [2], [1^2], [2,1]; 3: [3], [1^3], [3,1], [2,1^2], [3,1^2].

Alois P. Heinz, Table of n, a(n) for n = 0..100
Eric Weisstein's World of Mathematics, Ferrers Diagram.

Cf. A115728, A115729, A115724, A010766.

a(n) = subpart([A115728 (or A115729), [] is row n of A010766.

a(n) = Sum_{k>=0} A182114(k,n). - Alois P. Heinz, Nov 02 2012

A028378 Concatenate rows of triangle in A028364 (removing duplicates).

Original entry on oeis.org

0, 1, 2, 3, 5, 7, 9, 14, 19, 23, 28, 42, 56, 66, 76, 90, 132, 174, 202, 227, 255, 297, 429, 561, 645, 715, 785, 869, 1001, 1430, 1859, 2123, 2333, 2529, 2739, 3003, 3432, 4862, 6292, 7150, 7810, 8398, 8986, 9646, 10504, 11934, 16796, 21658

Offset: 0

Wouter Meeussen

a(n+1) is the number of subpartitions of the smallest partition of n into distinct parts (minimizing the size of the largest part). The sequence of partitions is [], [1], [2], [2,1], [3,1], [3,2], [3,2,1], [4,2,1], .... - Franklin T. Adams-Watters, Apr 11 2006

Cf. A028364, A028376.

Cf. A115728, A115729.

A116473 Number of partitions with exactly n subpartitions.

Original entry on oeis.org

1, 1, 2, 2, 3, 3, 4, 2, 6, 5, 4, 4, 6, 5, 8, 6, 5, 6, 9, 5, 10, 8, 6, 6, 12, 6, 8, 10, 8, 12, 10, 4, 12, 14, 12, 10, 13, 4, 10, 12, 10, 12, 12, 8, 16, 15, 10, 8, 16, 13, 16, 14, 9, 14, 22, 10, 12, 12, 8, 12, 20, 16, 14, 16, 19, 16, 14, 8, 17, 23, 10, 12, 16, 10, 25, 18, 14, 16, 22, 16, 22

Offset: 1

Franklin T. Adams-Watters, Mar 17 2006

nonn

Conjecture that lim_{n->infinity} a(n) = infinity. Definitely a(n) > 2 for n > 8.

Niccolò Castronuovo, On the number of fixed points of the map gamma, arXiv:2102.02739 [math.NT], 2021. Mentions this sequence.

Cf. A115728, A115729.

A238746 Number of distinct prime signatures that occur among the divisors of the n-th prime signature number (A025487(n)).

Original entry on oeis.org

1, 2, 3, 3, 4, 5, 5, 7, 4, 6, 6, 9, 7, 7, 9, 11, 10, 8, 12, 9, 13, 5, 10, 13, 9, 15, 14, 15, 9, 14, 16, 10, 18, 19, 17, 13, 18, 10, 19, 11, 16, 21, 12, 15, 24, 19, 17, 22, 16, 22, 12, 23, 24, 6, 19, 20, 29, 21, 21, 26, 22, 25, 13, 30, 27, 11, 26, 25, 19, 34

Offset: 1

Matthew Vandermast, Apr 28 2014

nonn

Also the number of members of A025487 that divide A025487(n).

			5 members of A025487 divide A025487(6) = 12 (namely, 1, 2, 4, 6 and 12); therefore, a(6) = 5.

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A085082, A025487, A181822, A322584.
Rearrangement of A115728, A115729 and A238690.
A116473(n) is the number of times n appears in the sequence.

lpsQ[n_] := n == 1 || (Max@ Differences[(f = FactorInteger[n])[[;;,2]]] < 1 && f[[-1, 1]] == Prime[Length[f]]); lps = Select[Range[6000], lpsQ]; c[n_] := Count[Divisors[n], ?(MemberQ[lps, #] &)]; c /@ lps  (* _Amiram Eldar, Jan 21 2024 *)

a(n) = A085082(A025487(n)) = A085082(A181822(n)).

a(n) = A322584(A025487(n)). - Amiram Eldar, Jan 21 2024

cp's OEIS Frontend

A115728 Number of subpartitions of partitions in Abramowitz and Stegun order.

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A107876 Triangular matrix T, read by rows, that satisfies: [T^k](n,k) = T(n,k-1) for n>=k>0, or, equivalently, (column k of T^k) = SHIFT_LEFT(column k-1 of T) when zeros above the diagonal are ignored.

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A238690 Let each integer m (1 <= m <= n) be factorized as m = prime_m(1)*prime_m(2)*...*prime_m(bigomega(m)), with the primes sorted in nonincreasing order. Then a(n) is the number of values of m such that each prime_m(i) <= prime_n(i).

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A107354 To compute a(n) we first write down 2^n 1's in a row. Each row takes the right half of the previous row and each element in it equals sum of the elements in the previous row starting at the middle. The single element in the last row is a(n).

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A107877 Column 1 of triangle A107876.

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A109055 To compute a(n) we first write down 3^n 1's in a row. Each row takes the rightmost 3rd part of the previous row and each element in it equals sum of the elements of the previous row starting with the first of the rightmost 3rd part. The single element in the last row is a(n).

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A115725 Number of partitions with maximum rectangle <= n.

A238690 Let each integer m (1 <= m <= n) be factorized as m = prime_m(1)prime_m(2)...*prime_m(bigomega(m)), with the primes sorted in nonincreasing order. Then a(n) is the number of values of m such that each prime_m(i) <= prime_n(i).