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 10 results.

A179052 Range and record values of number of partitions of n into powers of 10 (cf. A179051).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100, 105, 110, 115, 120, 125, 130, 135, 140, 145, 150, 156, 162, 168, 174, 180, 186, 192, 198, 204, 210, 217, 224, 231, 238
Offset: 0

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Author

Reinhard Zumkeller, Jun 27 2010

Keywords

Comments

a(n) = A008728(n) for n < 100;
a(n) = A179051(10*n);
a(10^n) = A145513(n+1);

A018819 Binary partition function: number of partitions of n into powers of 2.

Original entry on oeis.org

1, 1, 2, 2, 4, 4, 6, 6, 10, 10, 14, 14, 20, 20, 26, 26, 36, 36, 46, 46, 60, 60, 74, 74, 94, 94, 114, 114, 140, 140, 166, 166, 202, 202, 238, 238, 284, 284, 330, 330, 390, 390, 450, 450, 524, 524, 598, 598, 692, 692, 786, 786, 900, 900, 1014, 1014, 1154, 1154, 1294, 1294
Offset: 0

Views

Author

Keywords

Comments

First differences of A000123; also A000123 with terms repeated. See the relevant proof that follows the first formula below.
Among these partitions there is exactly one partition with all distinct terms, as every number can be expressed as the sum of the distinct powers of 2.
Euler transform of A036987 with offset 1.
a(n) is the number of "non-squashing" partitions of n, that is, partitions n = p_1 + p_2 + ... + p_k with 1 <= p_1 <= p_2 <= ... <= p_k and p_1 + p_2 + ... + p_i <= p_{i+1} for all 1 <= i < k. - N. J. A. Sloane, Nov 30 2003
Normally the OEIS does not include sequences like this where every term is repeated, but an exception was made for this one because of its importance. The unrepeated sequence A000123 is the main entry.
Number of different partial sums from 1 + [1, *2] + [1, *2] + ..., where [1, *2] means we can either add 1 or multiply by 2. E.g., a(6) = 6 because we have 6 = 1 + 1 + 1 + 1 + 1 + 1 = (1+1) * 2 + 1 + 1 = 1 * 2 * 2 + 1 + 1 = (1+1+1) * 2 = 1 * 2 + 1 + 1 + 1 + 1 = (1*2+1) * 2 where the connection is defined via expanding each bracket; e.g., this is 6 = 1 + 1 + 1 + 1 + 1 + 1 = 2 + 2 + 1 + 1 = 4 + 1 + 1 = 2 + 2 + 2 = 2 + 1 + 1 + 1 + 1 = 4 + 2. - Jon Perry, Jan 01 2004
Number of partitions p of n such that the number of compositions generated by p is odd. For proof see the Alekseyev and Adams-Watters link. - Vladeta Jovovic, Aug 06 2007
Differs from A008645 first at a(64). - R. J. Mathar, May 28 2008
Appears to be row sums of A155077. - Mats Granvik, Jan 19 2009
Number of partitions (p_1, p_2, ..., p_k) of n, with p_1 >= p_2 >= ... >= p_k, such that for each i, p_i >= p_{i+1} + ... + p_k. - John MCKAY (mckay(AT)encs.concordia.ca), Mar 06 2009 (these are the "non-squashing" partitions as nonincreasing lists).
Equals rightmost diagonal of triangle of A168261. Starting with offset 1 = eigensequence of triangle A115361 and row sums of triangle A168261. - Gary W. Adamson, Nov 21 2009
Equals convolution square root of A171238: (1, 2, 5, 8, 16, 24, 40, 56, 88, ...). - Gary W. Adamson, Dec 05 2009
Let B = the n-th convolution power of the sequence and C = the aerated variant of B. It appears that B/C = the binomial sequence beginning (1, n, ...). Example: Third convolution power of the sequence is (1, 3, 9, 19, 42, 78, 146, ...), with C = (1, 0, 3, 0, 9, 0, 19, ...). Then B/C = (1, 3, 6, 10, 15, 21, ...). - Gary W. Adamson, Aug 15 2016
From Gary W. Adamson, Sep 08 2016: (Start)
The limit of the matrix power M^k as n-->inf results in a single column vector equal to the sequence, where M is the following production matrix:
1, 0, 0, 0, 0, ...
1, 0, 0, 0, 0, ...
1, 1, 0, 0, 0, ...
1, 1, 0, 0, 0, ...
1, 1, 1, 0, 0, ...
1, 1, 1, 0, 0, ...
1, 1, 1, 1, 0, ...
1, 1, 1, 1, 0, ...
1, 1, 1, 1, 1, ...
... (End)
a(n) is the number of "non-borrowing" partitions of n, meaning binary subtraction of a smaller part from a larger part will never require place-value borrowing. - David V. Feldman, Jan 29 2020
From Gus Wiseman, May 25 2024: (Start)
Also the number of multisets of positive integers whose binary rank is n, where the binary rank of a multiset m is given by Sum_i 2^(m_i-1). For example, the a(1) = 1 through a(8) = 10 multisets are:
{1} {2} {12} {3} {13} {23} {123} {4}
{11} {111} {22} {122} {113} {1113} {33}
{112} {1112} {222} {1222} {223}
{1111} {11111} {1122} {11122} {1123}
{11112} {111112} {2222}
{111111} {1111111} {11113}
{11222}
{111122}
{1111112}
{11111111}
(End)

Examples

			G.f. = 1 + x + 2*x^2 + 2*x^3 + 4*x^4 + 4*x^5 + 6*x^6 + 6*x^7 + 10*x^8 + ...
a(4) = 4: the partitions are 4, 2 + 2, 2 + 1 + 1, 1 + 1 + 1 + 1.
a(7) = 6: the partitions are 4 + 2 + 1, 4 + 1 + 1 + 1, 2 + 2 + 2 + 1, 2 + 2 + 1 + 1 + 1, 2 + 1 + 1 + 1 + 1 + 1, 1 + 1 + 1 + 1 + 1 + 1 + 1.
From _Joerg Arndt_, Dec 17 2012: (Start)
The a(10) = 14 binary partitions of 10 are (in lexicographic order)
[ 1]  [ 1 1 1 1 1 1 1 1 1 1 ]
[ 2]  [ 2 1 1 1 1 1 1 1 1 ]
[ 3]  [ 2 2 1 1 1 1 1 1 ]
[ 4]  [ 2 2 2 1 1 1 1 ]
[ 5]  [ 2 2 2 2 1 1 ]
[ 6]  [ 2 2 2 2 2 ]
[ 7]  [ 4 1 1 1 1 1 1 ]
[ 8]  [ 4 2 1 1 1 1 ]
[ 9]  [ 4 2 2 1 1 ]
[10]  [ 4 2 2 2 ]
[11]  [ 4 4 1 1 ]
[12]  [ 4 4 2 ]
[13]  [ 8 1 1 ]
[14]  [ 8 2 ]
The a(11) = 14 binary partitions of 11 are obtained by appending 1 to each partition in the list.
The a(10) = 14 non-squashing partitions of 10 are (in lexicographic order)
[ 1]  [ 6 3 1 1 ]
[ 2]  [ 6 3 2 ]
[ 3]  [ 6 4 1 ]
[ 4]  [ 6 5 ]
[ 5]  [ 7 2 1 1 ]
[ 6]  [ 7 2 2 ]
[ 7]  [ 7 3 1 ]
[ 8]  [ 7 4 ]
[ 9]  [ 8 2 1 ]
[10]  [ 8 3 ]
[11]  [ 9 1 1 ]
[12]  [ 9 2 ]
[13]  [ 10 1 ]
[14]  [ 11 ]
The a(11) = 14 non-squashing partitions of 11 are obtained by adding 1 to the first part in each partition in the list.
(End)
From _David V. Feldman_, Jan 29 2020: (Start)
The a(10) = 14 non-borrowing partitions of 10 are (in lexicographic order)
[ 1] [1 1 1 1 1 1 1 1 1 1]
[ 2] [2 2 2 2 2]
[ 3] [3 1 1 1 1 1 1 1]
[ 4] [3 3 1 1 1 1]
[ 5] [3 3 2 2]
[ 6] [3 3 3 1]
[ 7] [5 1 1 1 1 1]
[ 8] [5 5]
[ 9] [6 2 2]
[10] [6 4]
[11] [7 1 1 1]
[12] [7 3]
[13] [9 1]
[14] [10]
The a(11) = 14 non-borrowing partitions of 11 are obtained either by adding 1 to the first even part in each partition (if any) or else appending a 1 after the last part.
(End)
For example, the five partitions of 4, written in nonincreasing order, are [1, 1, 1, 1], [2, 1, 1], [2, 2], [3, 1], [4]. The last four satisfy the condition, and a(4) = 4. The Maple program below verifies this for small values of n.
		

Crossrefs

A000123 is the main entry for the binary partition function and gives many more properties and references.
Cf. A115625 (labeled binary partitions), A115626 (labeled non-squashing partitions).
Convolution inverse of A106400.
Multiplicity of n in A048675, for distinct prime indices A087207.
Row lengths of A277905.
A118462 lists binary ranks of strict integer partitions, row sums A372888.
A372890 adds up binary ranks of integer partitions.

Programs

  • Haskell
    a018819 n = a018819_list !! n
    a018819_list = 1 : f (tail a008619_list) where
       f (x:xs) = (sum $ take x a018819_list) : f xs
    -- Reinhard Zumkeller, Jan 28 2012
    
  • Haskell
    import Data.List (intersperse)
    a018819 = (a018819_list !!)
    a018819_list = 1 : 1 : (<*>) (zipWith (+)) (intersperse 0) (tail a018819_list)
    -- Johan Wiltink, Nov 08 2018
    
  • Maple
    with(combinat); N:=8; a:=array(1..N); c:=array(1..N);
    for n from 1 to N do p:=partition(n); np:=nops(p); t:=0;
    for s to np do r:=p[s]; r:=sort(r,`>`); nr:=nops(r); j:=1;
    # while jsum(r[k],k=j+1..nr) do j:=j+1;od; # gives A040039
    while j= sum(r[k],k=j+1..nr) do j:=j+1;od; # gives A018819
    if j=nr then t:=t+1;fi od; a[n]:=t; od; # John McKay
  • Mathematica
    max = 59; a[0] = a[1] = 1; a[n_?OddQ] := a[n] = a[n-1]; a[n_?EvenQ] := a[n] = a[n-1] + a[n/2]; Table[a[n], {n, 0, max}]
    (* or *) CoefficientList[Series[1/Product[(1-x^(2^j)), {j, 0, Log[2, max] // Ceiling}], {x, 0, max}], x] (* Jean-François Alcover, May 17 2011, updated Feb 17 2014 *)
    a[ n_] := If[n<1, Boole[n==0], a[n] = a[n-1] + If[EvenQ@n, a[Quotient[n,2]], 0]]; (* Michael Somos, May 04 2022 *)
    Table[Count[IntegerPartitions[n],?(AllTrue[Log2[#],IntegerQ]&)],{n,0,60}] (* _Harvey P. Dale, Jun 20 2024 *)
  • PARI
    { n=15; v=vector(n); for (i=1,n,v[i]=vector(2^(i-1))); v[1][1]=1; for (i=2,n, k=length(v[i-1]); for (j=1,k, v[i][j]=v[i-1][j]+1; v[i][j+k]=v[i-1][j]*2)); c=vector(n); for (i=1,n, for (j=1,2^(i-1), if (v[i][j]<=n, c[v[i][j]]++))); c } /* Jon Perry */
    
  • PARI
    {a(n) = my(A, m); if( n<1, n==0, m=1; A = 1 + O(x); while(m<=n, m*=2; A = subst(A, x, x^2) / (1 - x)); polcoeff(A, n))}; /* Michael Somos, Aug 25 2003 */
    
  • PARI
    {a(n) = if( n<1, n==0, if( n%2, a(n-1), a(n/2)+a(n-1)))}; /* Michael Somos, Aug 25 2003 */
    
  • Python
    from functools import lru_cache
    @lru_cache(maxsize=None)
    def A018819(n): return 1 if n == 0 else A018819(n-1) + (0 if n % 2 else A018819(n//2)) # Chai Wah Wu, Jan 18 2022

Formula

a(2m+1) = a(2m), a(2m) = a(2m-1) + a(m). Proof: If n is odd there is a part of size 1; removing it gives a partition of n - 1. If n is even either there is a part of size 1, whose removal gives a partition of n - 1, or else all parts have even sizes and dividing each part by 2 gives a partition of n/2.
G.f.: 1 / Product_{j>=0} (1-x^(2^j)).
a(n) = (1/n)*Sum_{k = 1..n} A038712(k)*a(n-k), n > 1, a(0) = 1. - Vladeta Jovovic, Aug 22 2002
a(2*n) = a(2*n + 1) = A000123(n). - Michael Somos, Aug 25 2003
a(n) = 1 if n = 0, Sum_{j = 0..floor(n/2)} a(j) if n > 0. - David W. Wilson, Aug 16 2007
G.f. A(x) satisfies A(x^2) = (1-x) * A(x). - Michael Somos, Aug 25 2003
G.f. A(x) satisfies 0 = f(A(x), A(x^2), A(x^4)) where f(u, v, w) = u^2*w - 2*u*v^2 + v^3. - Michael Somos, Apr 10 2005
G.f. A(x) satisfies 0 = f(A(x), A(x^2), A(x^3), A(x^6)) where f(u1, u2, u3, u6) = u6 * u1^3 - 3*u3*u2*u1^2 + 3*u3*u2^2*u1 - u3*u2^3. - Michael Somos, Oct 15 2006
G.f.: 1/( Sum_{n >= 0} x^evil(n) - x^odious(n) ), where evil(n) = A001969(n) and odious(n) = A000069(n). - Paul D. Hanna, Jan 23 2012
Let A(x) by the g.f. and B(x) = A(x^k), then 0 = B*((1-A)^k - (-A)^k) + (-A)^k, see fxtbook link. - Joerg Arndt, Dec 17 2012
G.f.: Product_{n>=0} (1+x^(2^n))^(n+1), see the fxtbook link. - Joerg Arndt, Feb 28 2014
G.f.: 1 + Sum_{i>=0} x^(2^i) / Product_{j=0..i} (1 - x^(2^j)). - Ilya Gutkovskiy, May 07 2017

A008592 Multiples of 10: a(n) = 10*n.

Original entry on oeis.org

0, 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120, 130, 140, 150, 160, 170, 180, 190, 200, 210, 220, 230, 240, 250, 260, 270, 280, 290, 300, 310, 320, 330, 340, 350, 360, 370, 380, 390, 400, 410, 420, 430, 440, 450, 460, 470, 480, 490, 500, 510, 520, 530
Offset: 0

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Author

Keywords

Comments

Number of 3 X n binary matrices avoiding simultaneously the right angled numbered polyomino patterns (ranpp) (00;1), (01,1) and (11;0). An occurrence of a ranpp (xy;z) in a matrix A=(a(i,j)) is a triple (a(i1,j1), a(i1,j2), a(i2,j1)) where i11 and n>1. - Sergey Kitaev, Nov 12 2004
If Y is a 5-subset of an n-set X then, for n>=5, a(n-4) is the number of 3-subsets of X having at least two elements in common with Y. - Milan Janjic, Dec 08 2007
Complement of A067251; A168184(a(n)) = 0. - Reinhard Zumkeller, Nov 30 2009
Where record values occur for the number of partitions of n into powers of 10: A179052(n) = A179051(a(n)). - Reinhard Zumkeller, Jun 27 2010
Numbers ending in 0. - Wesley Ivan Hurt, Apr 10 2016

Crossrefs

Programs

Formula

From Vincenzo Librandi, Dec 24 2010: (Start)
G.f.: 10*x/(x-1)^2.
a(n) = 2*a(n-1) - a(n-2) for n > 1. (End)
a(n) = Sum_{i=2n-2..2n+2} i. - Wesley Ivan Hurt, Apr 11 2016
E.g.f.: 10*x*exp(x). - Stefano Spezia, May 31 2021

A133880 n modulo p repeated p times (where p=10).

Original entry on oeis.org

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

Views

Author

Hieronymus Fischer, Oct 10 2007

Keywords

Comments

Periodic with length p^2=100.
a(n) = A179051(n) for n < 90. - Reinhard Zumkeller, Jun 27 2010

Crossrefs

Programs

Formula

The following formulas are given for a general parameter p (p=10 for this sequence).
a(n)=(1+floor(n/p)) mod p.
a(n)=1+floor(n/p)-p*floor((n+p)/p^2).
a(n)=(((n+p) mod p^2)-(n mod p))/p.
a(n)=((n+p-(n mod p))/p) mod p.
G.f. g(x)=((p-1)x^(p^2)-px^(p(p-1))+1)/((1-x)(1-x^p)(1-x^(p^2))).
G.f. g(x)=(1-x^p)*sum{0<=k<(p-1), (k+1)*x^(k*p)}/((1-x)(1-x^(p^2))).

A062051 Number of partitions of n into powers of 3.

Original entry on oeis.org

1, 1, 1, 2, 2, 2, 3, 3, 3, 5, 5, 5, 7, 7, 7, 9, 9, 9, 12, 12, 12, 15, 15, 15, 18, 18, 18, 23, 23, 23, 28, 28, 28, 33, 33, 33, 40, 40, 40, 47, 47, 47, 54, 54, 54, 63, 63, 63, 72, 72, 72, 81, 81, 81, 93, 93, 93, 105, 105, 105, 117, 117, 117, 132, 132, 132, 147, 147, 147, 162
Offset: 0

Views

Author

Amarnath Murthy, Jun 06 2001

Keywords

Comments

Number of different partial sums of 1+[1,*3]+[1,*3]+..., where [1,*3] means we can either add 1 or multiply by 3. E.g., a(6)=3 because we have 6=1+1+1+1+1+1=(1+1)*3=1*3+1+1+1. - Jon Perry, Jan 01 2004
Also number of partitions of n into distinct 3-smooth parts. E.g., a(10) = #{9+1, 8+2, 6+4, 6+3+1, 4+3+2+1} = #{9+1, 3+3+3+1, 3+3+1+1+1+1, 3+1+1+1+1+1+1+1, 1+1+1+1+1+1+1+1+1+1} = 5. - Reinhard Zumkeller, Apr 07 2005
Starts to differ from A008650 at a(81). - R. J. Mathar, Jul 31 2010
If m=ceiling(log_3(2k)) and n=(3^m+1)/2-k for k in the range (3^(m-1)+1)/2+(3^(m-2))<=k<=(3^m-1)/2, this sequence gives the number of "feasible" partitions described in the sequence A254296. For instance, the terms starting at 121st term of A254296 backwards to 68th term of A254296 provide the first 54 terms of this sequence. - Md. Towhidul Islam, Mar 01 2015
From Gary W. Adamson, Sep 03 2016: (Start)
Let M =
1, 0, 0, 0, 0, ...
1, 0, 0, 0, 0, ...
1, 0, 0, 0, 0, ...
1, 1, 0, 0, 0, ...
1, 1, 0, 0, 0, ...
1, 1, 0, 0, 0, ...
1, 1, 1, 0, 0, ...
1, 1, 1, 0, 0, ...
..., where the leftmost column is all 1's, and all other columns are 1's shifted down thrice. Lim_{k=1..inf} M^k has a single nonzero column, which gives the sequence. (End)

Examples

			a(4) = 2 and the partitions are 3+1, 1+1+1+1;
a(9) = 5 and the partitions are 9; 3+3+3; 3+3+1+1+1; 3+1+1+1+1+1+1; 1+1+1+1+1+1+1+1+1.
		

Crossrefs

Programs

  • Mathematica
    nn=70;a=Product[1/(1-x^(3^i)),{i,0,4}];CoefficientList[Series[a,{x,0,nn}],x] (* Geoffrey Critzer, Oct 30 2012 *)
  • PARI
    { n=15; v=vector(n); for (i=1,n,v[i]=vector(2^(i-1))); v[1][1]=1; for (i=2,n, k=length(v[i-1]); for (j=1,k, v[i][j]=v[i-1][j]+1; v[i][j+k]=v[i-1][j]*3)); c=vector(n); for (i=1,n, for (j=1,2^(i-1), if (v[i][j]<=n, c[v[i][j]]++))); c } \\ Jon Perry
    
  • Python
    from functools import lru_cache
    @lru_cache(maxsize=None)
    def A062051(n): return A062051(n-1)+(0 if n%3 else A062051(n//3)) if n>2 else 1 # Chai Wah Wu, Sep 21 2022

Formula

a(n) = A005704([n/3]).
G.f.: Product_{k>=0} 1/(1-x^(3^k)). - R. J. Mathar, Jul 31 2010
If m = ceiling(log_3(2k)), define n = (3^m + 1)/2 - k for k in the range (3^(m-1)+1)/2 + (3^(m-2)) <= k <= (3^m-1)/2. Then, a(n) = Sum_{s=ceiling((k-1)/3)..(3^(m-1)-1)/2} a(s). This gives the first 2(3^(m-1))/3 terms. - Md. Towhidul Islam, Mar 01 2015
G.f.: 1 + Sum_{i>=0} x^(3^i) / Product_{j=0..i} (1 - x^(3^j)). - Ilya Gutkovskiy, May 07 2017

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), Jun 11 2001

A132272 a(n) = Product_{k>0} (1 + floor(n/10^k)).

Original entry on oeis.org

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

Views

Author

Hieronymus Fischer, Aug 20 2007

Keywords

Comments

If n is written in base-10 as n=d(m)d(m-1)d(m-2)...d(2)d(1)d(0) (where d(k) is the digit at position k) then a(n) is also the product (1+d(m)d(m-1)d(m-2)...d(2)d(1))*(1+d(m)d(m-1)d(m-2)...d(2))*...*(1+d(m)d(m-1)d(m-2))*(1+d(m)d(m-1))*(1+d(m)).
a(n) = A179051(n) for n < 100. [From Reinhard Zumkeller, Jun 27 2010]

Examples

			a(121)=(1+floor(121/10^1))*(1+floor(121/10^2))=13*2=26; a(132)=28 since 132=132(base-10) and so
a(132)=(1+13)*(1+1)(base-10)=14*2=28.
		

Crossrefs

For the product of terms floor(n/p^k) see A098844, A067080, A132027-A132033, A132263, A132264.

Programs

  • Maple
    A132272 := proc(n)
        a := 1;
        for k from 1 do
            f := floor(n/10^k) ;
            if f <=0 then
                return a;
            else
                a := a*(1+f) ;
            end if;
        end do:
    end proc:
    seq(A132272(n),n=1..120) ; # R. J. Mathar, Jun 13 2025
  • Mathematica
    Table[Product[1+Floor[n/10^k],{k,n}],{n,0,100}] (* Harvey P. Dale, Jul 20 2024 *)

Formula

The following formulas are given for a general parameter p considering the product of terms 1+floor(n/p^k) for 0
Recurrence: a(n)=(1+floor(n/p))*a(floor(n/p)); a(pn)=(1+n)*a(n); a(n*p^m)=product{0<=k
a(k*p^m-j)=k^m*p^(m(m-1)/2), for 0=1. a(p^m)=p^(m(m-1)/2)*product{0<=k
a(n)=A132271(floor(n/p))=A132271(n)/(1+n).
Asymptotic behavior: a(n)=O(n^((log_p(n)-1)/p)); this follows from the inequalities below.
a(n)<=A067080(n)/(n+1)*product{0<=k<=floor(log_p(n)), 1+1/p^k}.
a(n)>=A067080(n)/((n+1)*product{0
a(n)A000217(log_p(n))/(n+1), where c=product{k>0, 1+1/p^k}=2.2244691382741012... (for p=10 see constant A132325).
a(n)>n^((1+log_p(n))/2)/(n+1)=p^A000217(log_p(n))/(n+1).
lim sup n*a(n)/A067080(n)=2*product{k>0, 1+1/p^k}=2.2244691382741012..., for n-->oo (for p=10 see constant A132325).
lim inf n*a(n)/A067080(n)=1/product{k>0, 1-1/p^k}=1/0.8900100999989990000001000..., for n-->oo (for p=10 s. constant A132038).
lim inf a(n)/n^((1+log_p(n))/2)=1, for n-->oo.
lim sup a(n)/n^((1+log_p(n))/2)=2*product{k>0, 1+1/p^k}=2.2244691382741012..., for n-->oo (for p=10 see constant A132325).
lim inf a(n+1)/a(n)=2*product{k>0, 1+1/p^k}=2.2244691382741012... for n-->oo (for p=10 see constant A132325).

A329624 Number of iterations of A329623 for starting value n before a repeated value appears, or -1 if this never happens.

Original entry on oeis.org

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

Author

Scott R. Shannon, Nov 19 2019

Keywords

Comments

This sequence gives the number of iterations of A329623 for start value n before a repeated value first appears. Unlike the "ghost iteration" of A329200 the only fixed points are the single digits 0 to 9. See A328865 for the first repeating value.
Due to A329623(n) being significantly larger than n for some values of n, the iterative sequence can grow to infinity for some n. The first value to do so is n = 1373. This appears due to the occurrence of the digit string '62637' at the end of the term of the third iteration. These five digits reappear every second iteration at the end of the term, but with more and more digits preceding it. A329917 lists other divergent n values.
The smallest value, for n >= 10, which acts as an end point before repeating is 9, which is the final value for many starting values.
The digit string '8091' forms the basis of a very long convergent series for some values of n. The digit string consisting of an arbitrary number of copies of '80' followed by '91' will eventually converge to 8091, then 891, then 91, which finally converges in ten more iterations. We can thus form a number of arbitrary length using this rule which is guaranteed to converge. This sequence appears naturally with the starting value n = 139100 which converges to 9 after 136 iterations after reaching a term with 72 digits after 20 iterations. See the linked file below.
From M. F. Hasler, Dec 03 2019: (Start)
It seems the a(n) = -1 are conjectural, i.e., we have no proof that the terms for which the trajectory seems to "explode" do not eventually end up in a cycle. For example, the 8th iterate of 1373 is 5218725017016262626273. If the 2nd digit is changed from 2 to 0, then the further iterates appear to explode up to a length of 157 bits, but finally end up in a 2-cycle of 41-digit numbers (26...26273, 62...62637).
The "repeating values" are members of cycles, listed in A328142. Only fixed points 1, ..., 9 and 4*(10^k-1)/9 + 11, k >=3, and 6 infinite families of 2-cycles are known.
(End)
The first escape value is a(1373) = -1 (without proof). - Georg Fischer, Jul 16 2020

Examples

			a(1) = 1 as A329623(1) = 1, so a repeating value occurs after 1 iteration.
a(10) = 2 as A329623(10) = 9 and A329623(9) = 9, so a repeating value occurs after 2 iterations.
a(128) = 3, as A329623(128) = 182, A329623(182) = 728, A329623(728) = 182, so a repeating value occurs after 3 iterations.
		

Crossrefs

Sequences A324160, A226233, A179051, A140438, A132272 are unrelated; they begin with the same numbers as this sequence but differ after a(110) = 10, which ends the pattern of incrementing numbers, 2 through 11, repeated ten times.

Programs

  • PARI
    A329624(n,L=n^10,U=[n])=-!for(i=1,oo,setsearch(U,n=A329623(n))&&return(i); nM. F. Hasler, Dec 02 2019

A145513 Number of partitions of 10^n into powers of 10.

Original entry on oeis.org

1, 2, 12, 562, 195812, 515009562, 10837901390812, 1899421190329234562, 2851206628197445401265812, 37421114946843687272702534859562, 4362395890943439751990308572939648140812, 4573514084633441973328831327010967245403925484562, 43557001521047571730475817291330175020887917015964570015812
Offset: 0

Author

Alois P. Heinz, Oct 11 2008

Keywords

Comments

a(n) = A179051(10^n); for n>0: a(n) = A179052(10^(n-1)). - Reinhard Zumkeller, Jun 27 2010

Examples

			a(1) = 2, because there are 2 partitions of 10^1 into powers of 10: [1,1,1,1,1,1,1,1,1,1], [10].
		

Crossrefs

Cf. 10th column of A145515, A007318.

Programs

  • Haskell
    import Data.MemoCombinators (memo2, list, integral)
    a145513 n = a145513_list !! n
    a145513_list = f [1] where
       f xs = (p' xs $ last xs) : f (1 : map (* 10) xs)
       p' = memo2 (list integral) integral p
       p  0 = 1; p []  = 0
       p ks'@(k:ks) m = if m < k then 0 else p' ks' (m - k) + p' ks m
    -- Reinhard Zumkeller, Nov 27 2015
  • Maple
    g:= proc(b,n,k) option remember; local t; if b<0 then 0 elif b=0 or n=0 or k<=1 then 1 elif b>=n then add(g(b-t, n, k) *binomial(n+1, t) *(-1)^(t+1), t=1..n+1); else g(b-1, n, k) +g(b*k, n-1, k) fi end: a:= n-> g(1,n,10): seq(a(n), n=0..13);
  • Mathematica
    g[b_, n_, k_] := g[b, n, k] = Module[{t}, Which[b < 0, 0, b == 0 || n == 0 || k <= 1, 1, b >= n, Sum[g[b - t, n, k]*Binomial[n + 1, t] *(-1)^(t + 1), {t, 1, n + 1}], True, g[b - 1, n, k] + g[b*k, n - 1, k]]]; a[n_] := g[1, n, 10]; Table[a[n], {n, 0, 13}] (* Jean-François Alcover, Feb 01 2017, after Alois P. Heinz *)

Formula

a(n) = [x^(10^n)] 1/Product_{j>=0} (1-x^(10^j)).

A206245 Number of partitions of n into repunit powers, cf. A083278.

Original entry on oeis.org

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

Author

Reinhard Zumkeller, Feb 05 2012

Keywords

Comments

a(n) = A206244(n) for n <= 120, a(n) > A206244(n) for n > 120.

Crossrefs

Programs

  • Haskell
    a206245 = p a083278_list where
       p _      0 = 1
       p rps'@(rp:rps) n = if n < rp then 0 else p rps' (n - rp) + p rps n

A206244 Number of partitions of n into repunits (A002275).

Original entry on oeis.org

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

Author

Reinhard Zumkeller, Feb 05 2012

Keywords

Comments

a(n) = A206245(n) for n <= 120, a(n) < A206245(n) for n > 120.

Examples

			a(12)=2 is the first nontrivial term, from the partitions 12 = 1+1+...+1 = 11+1. - _N. J. A. Sloane_, Jul 26 2017
		

Crossrefs

Programs

  • Haskell
    a206244 = p $ tail a002275_list where
       p _             0 = 1
       p rus'@(ru:rus) n = if n < ru then 0 else p rus' (n - ru) + p rus n
  • Mathematica
    With[{nn = 50}, Table[Count[IntegerPartitions@ n, k_ /; ContainsAll[Array[Floor[10^#/9] &, IntegerLength[nn + 1]], Union@ k]], {n, 0, nn}]] (* Michael De Vlieger, Jul 26 2017 *)

Formula

G.f.: Product_{k>=1} 1/(1 - x^((10^k-1)/9)). - Ilya Gutkovskiy, Jul 26 2017
Showing 1-10 of 10 results.