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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A003586 3-smooth numbers: numbers of the form 2^i*3^j with i, j >= 0.

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 27, 32, 36, 48, 54, 64, 72, 81, 96, 108, 128, 144, 162, 192, 216, 243, 256, 288, 324, 384, 432, 486, 512, 576, 648, 729, 768, 864, 972, 1024, 1152, 1296, 1458, 1536, 1728, 1944, 2048, 2187, 2304, 2592, 2916, 3072, 3456, 3888
Offset: 1

Views

Author

Paul Zimmermann, Dec 11 1996

Keywords

Comments

This sequence is easily confused with A033845, which gives numbers of the form 2^i*3^j with i, j >= 1. Don't simply say "numbers of the form 2^i*3^j", but specify which sequence you mean. - N. J. A. Sloane, May 26 2024
These numbers were once called "harmonic numbers", see Lenstra links. - N. J. A. Sloane, Jul 03 2015
Successive numbers k such that phi(6k) = 2k. - Artur Jasinski, Nov 05 2008
Where record values greater than 1 occur in A088468: A160519(n) = A088468(a(n)). - Reinhard Zumkeller, May 16 2009
Also numbers that are divisible by neither 6k - 1 nor 6k + 1, for all k > 0. - Robert G. Wilson v, Oct 26 2010
Also numbers m such that the rooted tree with Matula-Goebel number m has m antichains. The Matula-Goebel number of a rooted tree can be defined in the following recursive manner: to the one-vertex tree there corresponds the number 1; to a tree T with root degree 1 there corresponds the t-th prime number, where t is the Matula-Goebel number of the tree obtained from T by deleting the edge emanating from the root; to a tree T with root degree m>=2 there corresponds the product of the Matula-Goebel numbers of the m branches of T. The vertices of a rooted tree can be regarded as a partially ordered set, where u<=v holds for two vertices u and v if and only if u lies on the unique path between v and the root. An antichain is a nonempty set of mutually incomparable vertices. Example: m=4 is in the sequence because the corresponding rooted tree is \/=ARB (R is the root) having 4 antichains (A, R, B, AB). - Emeric Deutsch, Jan 30 2012
A204455(3*a(n)) = 3, and only for these numbers. - Wolfdieter Lang, Feb 04 2012
The number of terms less than or equal to n is Sum_{i=0..floor(log_2(n))} floor(log_3(n/2^i) + 1), or Sum_{i=0..floor(log_3(n))} floor(log_2(n/3^i) + 1), which requires fewer terms to compute. - Robert G. Wilson v, Aug 17 2012
Named 3-friables in French. - Michel Marcus, Jul 17 2013
In the 14th century Levi Ben Gerson proved that the only pairs of terms which differ by 1 are (1,2), (2,3), (3,4), and (8,9); see A235365, A235366, A236210. - Jonathan Sondow, Jan 20 2014
Range of values of A000005(n) (and also A181819(n)) for cubefree numbers n. - Matthew Vandermast, May 14 2014
A036561 is a permutation of this sequence. - L. Edson Jeffery, Sep 22 2014
Also the sorted union of A000244 and A007694. - Lei Zhou, Apr 19 2017
The sum of the reciprocals of the 3-smooth numbers is equal to 3. Brief proof: 1 + 1/2 + 1/3 + 1/4 + 1/6 + 1/8 + 1/9 + ... = (Sum_{k>=0} 1/2^k) * (Sum_{m>=0} 1/3^m) = (1/(1-1/2)) * (1/(1-1/3)) = (2/(2-1)) * (3/(3-1)) = 3. - Bernard Schott, Feb 19 2019
Also those integers k for which, for every prime p > 3, p^(2k) - 1 == 0 (mod 24k). - Federico Provvedi, May 23 2022
For n>1, the exponents’ parity {parity(i), parity(j)} of one out of four consecutive terms is {odd, odd}. Therefore, for n>1, at least one out of every four consecutive terms is a Zumkeller number (A083207). If for the term whose parity is {even, odd}, even also means nonzero, then this term is also a Zumkeller number (as is the case with the last of the four consecutive terms 1296, 1458, 1536, 1728). - Ivan N. Ianakiev, Jul 10 2022
Except the initial terms 2, 3, 4, 8, 9 and 16, these are numbers k such that k^6 divides 6^k. Except the initial terms 2, 3, 4, 6, 8, 9, 16, 18 and 27, these are numbers k such that k^12 divides 12^k. - Mohammed Yaseen, Jul 21 2022
In music theory, a comma is a ratio, close to 1 (typically less than 1.04), between two natural numbers divisible by only small primes (typically single digit). In this sequence, a(131) / a(130) = 531441 / 524288 ~ 1.013643 is the Pythagorean comma (A221363), the difference between 12 perfect fifths and 7 octaves. - Hal M. Switkay, Mar 23 2025

References

  • J.-M. De Koninck & A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 654 pp. 85, 287-8, Ellipses Paris 2004.
  • S. Ramanujan, Collected Papers, Ed. G. H. Hardy et al., Cambridge 1927; Chelsea, NY, 1962, p. xxiv.
  • R. Tijdeman, Some applications of Diophantine approximation, pp. 261-284 of Surveys in Number Theory (Urbana, May 21, 2000), ed. M. A. Bennett et al., Peters, 2003.

Links

Crossrefs

Cf. A051037, A002473, A051038, A080197, A080681, A080682, A117221, A105420, A062051, A117222, A117220, A090184, A131096, A131097, A186711, A186712, A186771, A088468, A061987, A080683 (p-smooth numbers with other values of p), A025613 (a subsequence).
Cf. also A235365, A235366, A236210, A036561.
Cf. also A000244, A007694. - Lei Zhou, Apr 19 2017
Cf. A191475 (successive values of i), A191476 (successive values of j), A022330 (indices of the pure terms 2^i), A022331 (indices of the pure terms 3^j). - N. J. A. Sloane, May 26 2024
Cf. A221363.

Programs

  • Haskell
    import Data.Set (Set, singleton, insert, deleteFindMin)
smooth :: Set Integer -> [Integer]
smooth s = x : smooth (insert (3*x) $ insert (2*x) s')
  where (x, s') = deleteFindMin s
a003586_list = smooth (singleton 1)
a003586 n = a003586_list !! (n-1)
-- Reinhard Zumkeller, Dec 16 2010
  • Magma
    [n: n in [1..4000] | PrimeDivisors(n) subset [2,3]]; // Bruno Berselli, Sep 24 2012
  • Maple
    A003586 := proc(n) option remember; if n = 1 then 1; else for a from procname(n-1)+1 do numtheory[factorset](a) minus {2,3} ; if % = {} then return a; end if; end do: end if; end proc: # R. J. Mathar, Feb 28 2011
with(numtheory): for i from 1 to 23328 do if(i/phi(i)=3)then print(i/6) fi od; # Gary Detlefs, Jun 28 2011
  • Mathematica
    a[1] = 1; j = 1; k = 1; n = 100; For[k = 2, k <= n, k++, If[2*a[k - j] < 3^j, a[k] = 2*a[k - j], {a[k] = 3^j, j++}]]; Table[a[i], {i, 1, n}] (* Hai He (hai(AT)mathteach.net) and Gilbert Traub, Dec 28 2004 *)
aa = {}; Do[If[EulerPhi[6 n] == 2 n, AppendTo[aa, n]], {n, 1, 1000}]; aa (* Artur Jasinski, Nov 05 2008 *)
fQ[n_] := Union[ MemberQ[{1, 5}, # ] & /@ Union@ Mod[ Rest@ Divisors@ n, 6]] == {False}; fQ[1] = True; Select[ Range@ 4000, fQ] (* Robert G. Wilson v, Oct 26 2010 *)
powerOfTwo = 12; Select[Nest[Union@Join[#, 2*#, 3*#] &, {1}, powerOfTwo-1], # < 2^powerOfTwo &] (* Robert G. Wilson v and T. D. Noe, Mar 03 2011 *)
fQ[n_] := n == 3 EulerPhi@ n; Select[6 Range@ 4000, fQ]/6 (* Robert G. Wilson v, Jul 08 2011 *)
mx = 4000; Sort@ Flatten@ Table[2^i*3^j, {i, 0, Log[2, mx]}, {j, 0, Log[3, mx/2^i]}] (* Robert G. Wilson v, Aug 17 2012 *)
f[n_] := Block[{p2, p3 = 3^Range[0, Floor@ Log[3, n] + 1]}, p2 = 2^Floor[Log[2, n/p3] + 1]; Min[ Select[ p2*p3, IntegerQ]]]; NestList[f, 1, 54] (* Robert G. Wilson v, Aug 22 2012 *)
Select[Range@4000, Last@Map[First, FactorInteger@#] <= 3 &] (* Vincenzo Librandi, Aug 25 2016 *)
Select[Range[4000],Max[FactorInteger[#][[All,1]]]<4&] (* Harvey P. Dale, Jan 11 2017 *)
  • PARI
    test(n)=for(p=2,3, while(n%p==0, n/=p)); n==1;
for(n=1,4000,if(test(n),print1(n",")))
  • PARI
    list(lim)=my(v=List(),N);for(n=0,log(lim\1+.5)\log(3),N=3^n;while(N<=lim,listput(v,N);N<<=1));vecsort(Vec(v)) \\ Charles R Greathouse IV, Jun 28 2011
  • PARI
    is_A003586(n)=n<5||vecmax(factor(n,5)[, 1])<5 \\ M. F. Hasler, Jan 16 2015
  • PARI
    list(lim)=my(v=List(), N); for(n=0, logint(lim\=1,3), N=3^n; while(N<=lim, listput(v, N); N<<=1)); Set(v) \\ Charles R Greathouse IV, Jan 10 2018
  • Python
    from itertools import count, takewhile
def aupto(lim):
    pows2 = list(takewhile(lambda x: xMichael S. Branicky, Jul 08 2022
  • Python
    from sympy import integer_log
def A003586(n):
    def bisection(f,kmin=0,kmax=1):
        while f(kmax) > kmax: kmax <<= 1
        while kmax-kmin > 1:
            kmid = kmax+kmin>>1
            if f(kmid) <= kmid:
                kmax = kmid
            else:
                kmin = kmid
        return kmax
    def f(x): return n+x-sum((x//3**i).bit_length() for i in range(integer_log(x,3)[0]+1))
    return bisection(f,n,n) # Chai Wah Wu, Sep 15 2024
  • Python
    # faster for initial segment of sequence
import heapq
from itertools import islice
def A003586gen(): # generator of terms
    v, oldv, h, psmooth_primes, = 1, 0, [1], [2, 3]
    while True:
        v = heapq.heappop(h)
        if v != oldv:
            yield v
            oldv = v
            for p in psmooth_primes:
                heapq.heappush(h, v*p)
print(list(islice(A003586gen(), 65))) # Michael S. Branicky, Sep 17 2024
(C++) // Returns A003586 <= threshold without approximations nor sorting
#include 
std::forward_list A003586(const int threshold) {
    std::forward_list sequence;
    auto start_it = sequence.before_begin();
    for (int i = 1; i <= threshold; i *= 2) {
        for (int inc = 1; std::next(start_it) != sequence.end() && inc <= i; inc *= 3)
            ++start_it;
        auto it = start_it;
        for (int j = 1; i * j <= threshold; j *= 3) {
            sequence.emplace_after(it, i * j);
            for (int inc = 1; std::next(it) != sequence.end() && inc <= i; inc *= 2)
                ++it;
        }
    }
    return sequence;
} // Eben Gino Lester, Apr 17 2025
  • Sage
    def isA003586(n) :
    return not any(d != 2 and d != 3 for d in prime_divisors(n))
@CachedFunction
def A003586(n) :
    if n == 1 : return 1
    k = A003586(n-1) + 1
    while not isA003586(k) : k += 1
    return k
[A003586(n) for n in (1..55)] # Peter Luschny, Jul 20 2012

Formula

An asymptotic formula for a(n) is roughly a(n) ~ 1/sqrt(6)*exp(sqrt(2*log(2)*log(3)*n)). - Benoit Cloitre, Nov 20 2001
A061987(n) = a(n + 1) - a(n), a(A084791(n)) = A084789(n), a(A084791(n) + 1) = A084790(n). - Reinhard Zumkeller, Jun 03 2003
Union of powers of 2 and 3 with n such that psi(n) = 2*n, where psi(n) = n*Product_(1 + 1/p) over all prime factors p of n = A001615(n). - Lekraj Beedassy, Sep 07 2004; corrected by Franklin T. Adams-Watters, Mar 19 2009
a(n) = 2^A022328(n)*3^A022329(n). - N. J. A. Sloane, Mar 19 2009
The characteristic function of this sequence is given by Sum_{n >= 1} x^a(n) = Sum_{n >= 1} moebius(6*n)*x^n/(1 - x^n). - Paul D. Hanna, Sep 18 2011
a(n) = A007694(n+1)/2. - Lei Zhou, Apr 19 2017

Extensions

Deleted claim that this sequence is union of 2^n (A000079) and 3^n (A000244) sequences -- this does not include the terms which are not pure powers. - Walter Roscello (wroscello(AT)comcast.net), Nov 16 2008

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

David W. Wilson, N. J. A. Sloane and J. H. Conway

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.

Links

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.
Cf. A023893, A062051, A105420, A131995, A040039, A018819, A088567, A089054, A115361, A168261, A171238, A179051, A008619.
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.
Binary indices: A000120, A001511, A029931, A048793, A070939, A272020.
Cf. A005940, A019565, A056239, A158704, A158705, A231204, A277319, A372688.

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

A023893 Number of partitions of n into prime power parts (1 included); number of nonisomorphic Abelian subgroups of symmetric group S_n.

Original entry on oeis.org

1, 1, 2, 3, 5, 7, 10, 14, 20, 27, 36, 48, 63, 82, 105, 134, 171, 215, 269, 335, 415, 511, 626, 764, 929, 1125, 1356, 1631, 1953, 2333, 2776, 3296, 3903, 4608, 5427, 6377, 7476, 8744, 10205, 11886, 13818, 16032, 18565, 21463, 24768, 28536
Offset: 0

Views

Author

Olivier Gérard

Keywords

Examples

			From _Gus Wiseman_, Jul 28 2022: (Start)
The a(0) = 1 through a(6) = 10 partitions:
  ()  (1)  (2)   (3)    (4)     (5)      (33)
           (11)  (21)   (22)    (32)     (42)
                 (111)  (31)    (41)     (51)
                        (211)   (221)    (222)
                        (1111)  (311)    (321)
                                (2111)   (411)
                                (11111)  (2211)
                                         (3111)
                                         (21111)
                                         (111111)
(End)

Links

Crossrefs

Cf. A009490, A023894 (first differences), A062297 (number of Abelian subgroups).
Cf. A018819, A062051, A131995.
The multiplicative version (factorizations) is A000688.
Not allowing 1's gives A023894, strict A054685, ranked by A355743.
The version for just primes (not prime-powers) is A034891, strict A036497.
The strict version is A106244.
These partitions are ranked by A302492.
A000041 counts partitions, strict A000009.
A001222 counts prime-power divisors.
A072233 counts partitions by sum and length.
A246655 lists the prime-powers (A000961 includes 1), towers A164336.
Cf. A001970, A055887, A063834, A085970, A279784, A295935, A356065.

Programs

  • Mathematica
    Table[Length[Select[IntegerPartitions[n],Count[Map[Length,FactorInteger[#]], 1] == Length[#] &]], {n, 0, 35}] (* Geoffrey Critzer, Oct 25 2015 *)
nmax = 50; Clear[P]; P[m_] := P[m] = Product[Product[1/(1-x^(p^k)), {k, 1, m}], {p, Prime[Range[PrimePi[nmax]]]}]/(1-x)+O[x]^nmax // CoefficientList[ #, x]&; P[1]; P[m=2]; While[P[m] != P[m-1], m++]; P[m] (* Jean-François Alcover, Aug 31 2016 *)
  • PARI
    lista(m) = {x = t + t*O(t^m); gf = prod(k=1, m, if (isprimepower(k), 1/(1-x^k), 1))/(1-x); for (n=0, m, print1(polcoeff(gf, n, t), ", "));} \\ Michel Marcus, Mar 09 2013
  • Python
    from functools import lru_cache
from sympy import factorint
@lru_cache(maxsize=None)
def A023893(n):
    @lru_cache(maxsize=None)
    def c(n): return sum((p**(e+1)-p)//(p-1) for p,e in factorint(n).items())+1
    return (c(n)+sum(c(k)*A023893(n-k) for k in range(1,n)))//n if n else 1 # Chai Wah Wu, Jul 15 2024

Formula

G.f.: (Product_{p prime} Product_{k>=1} 1/(1-x^(p^k))) / (1-x).

A039966 a(0) = 1; thereafter a(3n+2) = 0, a(3n) = a(3n+1) = a(n).

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
Offset: 0

Views

Author

N. J. A. Sloane, Dec 11 1999

Keywords

Comments

Number of partitions of n into distinct powers of 3.
Trajectory of 1 under the morphism: 1 -> 110, 0 -> 000. Thus 1 -> 110 ->110110000 -> 110110000110110000000000000 -> ... - Philippe Deléham, Jul 09 2005
Also, an example of a d-perfect sequence.
This is a composite of two earlier sequences contributed at different times by N. J. A. Sloane and by Reinhard Zumkeller, Mar 05 2005. Christian G. Bower extended them and found that they agreed for at least 512 terms. The proof that they were identical was found by Ralf Stephan, Jun 13 2005, based on the fact that they were both 3-regular sequences.

Examples

			The triples of elements (a(3k), a(3k+1), a(3k+2)) are (1,1,0) if a(k) = 1 and (0,0,0) if a(k) = 0.  So since a(2) = 0, a(6) = a(7) = a(8) = 0, and since a(3) = 1, a(9) = a(10) = 1 and a(11) = 0. - _Michael B. Porter_, Jul 11 2016

Links

Crossrefs

For generating functions Product_{k>=0} (1+a*x^(b^k)) for the following values of (a,b) see: (1,2) A000012 and A000027, (1,3) A039966 and A005836, (1,4) A151666 and A000695, (1,5) A151667 and A033042, (2,2) A001316, (2,3) A151668, (2,4) A151669, (2,5) A151670, (3,2) A048883, (3,3) A117940, (3,4) A151665, (3,5) A151671, (4,2) A102376, (4,3) A151672, (4,4) A151673, (4,5) A151674.
Cf. A062051, A000009, A000244, A004642.
Characteristic function of A005836 (and apart from offset of A003278).

Programs

  • Haskell
    a039966 n = fromEnum (n < 2 || m < 2 && a039966 n' == 1)
   where (n',m) = divMod n 3
-- Reinhard Zumkeller, Sep 29 2011
  • Maple
    a := proc(n) option remember; if n <= 1 then RETURN(1) end if; if n = 2 then RETURN(0) end if; if n mod 3 = 2 then RETURN(0) end if; if n mod 3 = 0 then RETURN(a(1/3*n)) end if; if n mod 3 = 1 then RETURN(a(1/3*n - 1/3)) end if end proc; # Ralf Stephan, Jun 13 2005
  • Mathematica
    (* first do *) Needs["DiscreteMath`Combinatorica`"] (* then *) s = Rest[ Sort[ Plus @@@ Table[UnrankSubset[n, Table[3^i, {i, 0, 4}]], {n, 32}]]]; Table[ If[ Position[s, n] == {}, 0, 1], {n, 105}] (* Robert G. Wilson v, Jun 14 2005 *)
CoefficientList[Series[Product[(1 + x^(3^k)), {k, 0, 5}], {x, 0, 111}], x] (* or *)
Nest[ Flatten[ # /. {0 -> {0, 0, 0}, 1 -> {1, 1, 0}}] &, {1}, 5] (* Robert G. Wilson v, Mar 29 2006 *)
Nest[ Join[#, #, 0 #] &, {1}, 5] (* Robert G. Wilson v, Jul 27 2014 *)
  • PARI
    {a(n)=local(A,m); if(n<0, 0, m=1; A=1+O(x); while(m<=n, m*=3; A=(1+x)*subst(A,x,x^3)); polcoeff(A,n))} /* Michael Somos, Jul 15 2005 */
  • PARI
    A039966(n)=vecmax(digits(n+!n,3))<2;
apply(A039966, [0..99]) \\ M. F. Hasler, Feb 15 2023
  • Python
    def A039966(n):
    while n > 2:
        n,r = divmod(n,3)
        if r==2: return 0
    return int(n!=2) # M. F. Hasler, Feb 15 2023

Formula

a(0) = 1, a(1) = 0, a(n) = b(n-2), where b is the sequence defined by b(0) = 1, b(3n+2) = 0, b(3n) = b(3n+1) = b(n). - Ralf Stephan
a(n) = A005043(n-1) mod 3. - Christian G. Bower, Jun 12 2005
a(n) = A002426(n) mod 3. - John M. Campbell, Aug 24 2011
a(n) = A000275(n) mod 3. - John M. Campbell, Jul 08 2016
Properties: 0 <= a(n) <= 1, a(A074940(n)) = 0, a(A005836(n)) = 1; A104406(n) = Sum(a(k), 1 <= k <= n). - Reinhard Zumkeller, Mar 05 2005
Euler transform of sequence b(n) where b(3^k) = 1, b(2*3^k) = -1 and zero otherwise. - Michael Somos, Jul 15 2005
G.f. A(x) satisfies A(x) = (1+x)*A(x^3). - Michael Somos, Jul 15 2005
G.f.: Product{k>=0} 1+x^(3^k). Exponents give A005836.

Extensions

Entry revised Jun 30 2005
Offset corrected by John M. Campbell, Aug 24 2011

A106244 Number of partitions into distinct prime powers.

Original entry on oeis.org

1, 1, 1, 2, 2, 3, 3, 4, 5, 6, 7, 8, 10, 11, 13, 14, 17, 19, 21, 24, 27, 30, 33, 37, 41, 46, 50, 56, 62, 68, 75, 82, 91, 99, 108, 118, 129, 141, 152, 166, 180, 196, 211, 229, 248, 267, 288, 310, 335, 360, 387, 415, 447, 479, 513, 549, 589, 630, 672, 719, 768, 820, 873, 930
Offset: 0

Views

Author

Reinhard Zumkeller, Apr 26 2005

Keywords

Comments

A054685(n) < a(n) < A023893(n) for n>2.

Examples

			a(10) = #{3^2+1,2^3+2,7+3,7+2+1,5+2^2+1,5+3+2,2^2+3+2+1} = 7.

Links

Crossrefs

Cf. A000586, A000607, A000961.
Cf. A062051, A105420, A131996.
Cf. A023893, A051613, A054685.

Programs

  • Haskell
    import Data.MemoCombinators (memo2, integral)
a106244 n = a106244_list !! n
a106244_list = map (p' 1) [0..] where
   p' = memo2 integral integral p
   p _ 0 = 1
   p k m = if m < pp then 0 else p' (k + 1) (m - pp) + p' (k + 1) m
           where pp = a000961 k
-- Reinhard Zumkeller, Nov 24 2015
  • Maple
    g:=(1+x)*(product(product(1+x^(ithprime(k)^j),j=1..5),k=1..20)): gser:=series(g,x=0,68): seq(coeff(gser,x,n),n=1..63); # Emeric Deutsch, Aug 27 2007
  • Mathematica
    m = 64; gf = (1+x)*Product[1+x^(Prime[k]^j), {j, 1, 5}, {k, 1, 18}] + O[x]^m; CoefficientList[gf, x] (* Jean-François Alcover, Mar 02 2019, from Maple *)
  • PARI
    lista(m) = {x = t + t*O(t^m); gf = (1+x)*prod(k=1, m, if (isprimepower(k),(1+x^k), 1)); for (n=0, m, print1(polcoeff(gf, n, t), ", "));} \\ Michel Marcus, Mar 02 2019

Formula

a(n) = A054685(n-1)+A054685(n). - Vladeta Jovovic, Apr 28 2005
G.f.: (1+x)*Product(Product(1+x^(p(k)^j), j=1..infinity),k=1..infinity), where p(k) is the k-th prime (offset 0). - Emeric Deutsch, Aug 27 2007

Extensions

Offset corrected and a(0)=1 added by Reinhard Zumkeller, Nov 24 2015

A254296 The number of partitions of n having the minimum number of summands such that all integers from 1 to n can be represented as the sum of the summands times one of {-1, 0, 1}.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 3, 2, 2, 2, 1, 1, 1, 10, 11, 12, 11, 12, 12, 11, 11, 12, 9, 9, 9, 7, 7, 7, 5, 5, 5, 3, 3, 3, 2, 2, 2, 1, 1, 1, 131, 136, 140, 133, 137, 140, 133, 136, 138, 129, 131, 134, 125, 126, 128, 117, 119, 120, 109, 110, 111, 101, 102, 102, 92, 92, 93, 81, 81, 81, 72, 72, 72, 63, 63, 63, 54, 54, 54, 47, 47, 47, 40, 40, 40, 33, 33, 33
Offset: 1

Views

Author

Md. Towhidul Islam, Jan 27 2015

Keywords

Comments

Define a feasible partition of an n-kilogram stone as an ordered partition of minimum possible m parts W_1 <= W_2 <= ... <= W_m broken from the stone such that all integral weights from 1 to n can be weighed in one weighing using the parts/weights on a two pan balance. The minimum m for any n is m=ceiling(log_3(2n)). This sequence gives the number of feasible partitions of n.
From Robert G. Wilson v, Feb 04 2015: (Start)
Records: 1, 2, 3, 10, 11, 12, 131, 136, 140, 3887, 3921, 3950, 262555, 263112, 263707, 42240104, 42262878, 42285095, 16821037273, 16823225535, 16825391023, ..., .
Possible values: 1, 2, 3, 5, 7, 9, 10, 11, 12, 15, 18, 23, 28, 33, 40, 47, 54, 63, 72, 81, 92, 93, 101, 102, 105, ..., .
First occurrence on k, or 0 if not present: 1, 5, 7, 0 29, 0, 26, 0, 23, 14, 15, 16, 0, 0, 98, 0, 0, 95, 0, 0, 0, 0, 92, ..., .
1 occurs at: 1, 2, 3, 4, 11, 12, 13, 38, 39, 40, 119, 120, 121, 362, 363, 364, 1091, 1092, 1093, 3278, 3279, 3280, 9839, 9840, 9841, ..., .
2 occurs at: 5, 6, 8, 9, 10, 35, 36, 37, 116, 117, 118, 359, 360, 361, 1088, 1089, 1090, 3275, 3276, 3277, 9836, 9837, 9838, ..., .
3 occurs at: 7, 32, 33, 34, 113, 114, 115, 356, 357, 358, 1085, 1086, 1087, 3272, 3273, 3274, 9833, 9834, 9835, ..., .
5 occurs at: 29, 30, 31, 110, 111, 112, 353, 354, 355, 1082, 1083, 1084, 3269, 3270, 3271, 9830, 9831, 9832, ..., . (End)

Examples

			For n=3, minimum number of weights m is 2. The only "feasible" set of weights is [1,2]. So, a(3)=1.
For n=7, m is 3. The "feasible" sets of weights are [1,1,5], [1,2,4], [1,3,3]. So, a(7)=3.
For n=19, m is 4. The "feasible" sets of weights are [1,1,4,13], [1,1,5,12], [1,2,3,13], [1,2,4,12], [1,2,5,11], [1,2,6,10], [1,2,7,9], [1,3,3,12], [1,3,4,11], [1,3,5,10], [1,3,6,9], [1,3,7,8]. There are no other "feasible" sets. So, a(19)=12.

Links

Crossrefs

Cf. A254296, A254430, A254431, A254432, A254433, A254435, A254436, A254437, A254438, A254439, A254440, A254442.
When we calculate a(n) for (3^(m-1)+1)/2+3^(m-2)+1 <= n <= (3^m-1)/2 starting from n=(3^m-1)/2 backwards, we get the sequence A062051 which is also the triplication of the terms of sequence A005704.

Programs

  • Mathematica
    okQ[v_] := Module[{s=0}, For[i=1, i <= Length[v], i++, If[v[[i]] > 2*s+1, Return[ False], s += v[[i]] ] ]; Return[True]]; a[n_] := With[{k = Ceiling[Log[3, 2n]]}, Select[Reverse /@ IntegerPartitions[n, {k}], okQ] // Length]; Table[a[n], {n, 1, 88}] (* Jean-François Alcover, Feb 03 2015, after Charles R Greathouse IV *)
  • PARI
    ok(v)=my(s);for(i=1,#v,if(v[i]>2*s+1,return(0),s+=v[i]));1
a(n)=my(k=ceil(log(2*n)/log(3))); #select(ok, partitions(n,,k)) \\ Charles R Greathouse IV, Feb 02 2015

Formula

Let us suppose, a(0)=1 and for (3^(m-1)+1)/2<=n<=(3^m-1)/2, m=ceiling(log_3(2n)).
Then for (3^(m-1)+1)/2<=n<=(3^(m-1)+1)/2+(3^(m-2)),a(n)=Sum{s=ceiling((n-1)/3..floor((2n+3^(m-2)-1)/4)}a(s)-Sum{d=ceiling((3n+2)/5)..(3^(m-1)-1)/2}Sum{p=ceiling((d-1)/3..2d-n-1}a(p)
and for (3^(m-1)+1)/2+3^(m-2)+1<=n<=(3^m-1)/2, a(n)=Sum_{s=ceiling((n-1)/3)..(3^(m-1)-1)/2}a(s).

A102430 Triangle read by rows where T(n,k) is the number of integer partitions of n > 1 into powers of k > 1.

Original entry on oeis.org

2, 2, 2, 4, 2, 2, 4, 2, 2, 2, 6, 3, 2, 2, 2, 6, 3, 2, 2, 2, 2, 10, 3, 3, 2, 2, 2, 2, 10, 5, 3, 2, 2, 2, 2, 2, 14, 5, 3, 3, 2, 2, 2, 2, 2, 14, 5, 3, 3, 2, 2, 2, 2, 2, 2, 20, 7, 4, 3, 3, 2, 2, 2, 2, 2, 2, 20, 7, 4, 3, 3, 2, 2, 2, 2, 2, 2, 2, 26, 7, 4, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2
Offset: 2

Views

Author

Marc LeBrun, Jan 08 2005

Keywords

Comments

All entries above main diagonal are = 1.

Examples

			The T(9,3)=5 partitions of 9 into powers of 3: 111111111, 1111113, 11133, 333, 9.
From _Gus Wiseman_, Jun 07 2019: (Start)
Triangle begins:
   2
   2  2
   4  2  2
   4  2  2  2
   6  3  2  2  2
   6  3  2  2  2  2
  10  3  3  2  2  2  2
  10  5  3  2  2  2  2  2
  14  5  3  3  2  2  2  2  2
  14  5  3  3  2  2  2  2  2  2
  20  7  4  3  3  2  2  2  2  2  2
  20  7  4  3  3  2  2  2  2  2  2  2
  26  7  4  3  3  3  2  2  2  2  2  2  2
  26  9  4  4  3  3  2  2  2  2  2  2  2  2
  36  9  6  4  3  3  3  2  2  2  2  2  2  2  2
  36  9  6  4  3  3  3  2  2  2  2  2  2  2  2  2
  46 12  6  4  4  3  3  3  2  2  2  2  2  2  2  2  2
Row n = 8 counts the following partitions:
  8          3311       44         5111       611        71         8
  44         311111     41111      11111111   11111111   11111111   11111111
  422        11111111   11111111
  2222
  4211
  22211
  41111
  221111
  2111111
  11111111
(End)

Links

Crossrefs

Cf. A102431, A102432, A102433, A102434, A001700, A018819, A062051, A008645, A008650, A008652, A008648.
Same as A308558 except for the k = 1 column.
Row sums are A102431.
First column (k = 2) is A018819.
Second column (k = 3) is A062051.
Cf. A000961, A001597, A007916, A023894, A052410, A112344.

Programs

  • Maple
    b:= proc(n, i, k) option remember; `if`(n=0, 1, `if`(i<0, 0,
      b(n, i-1, k)+(p-> `if`(p>n, 0, b(n-p, i, k)))(k^i)))
    end:
T:= (n, k)-> b(n, ilog[k](n), k):
seq(seq(T(n, k), k=2..n), n=2..20);  # Alois P. Heinz, Oct 12 2019
  • Mathematica
    Table[Length[Select[IntegerPartitions[n],And@@(IntegerQ[Log[k,#]]&/@#)&]],{n,2,10},{k,2,n}] (* Gus Wiseman, Jun 07 2019 *)

Formula

T(1, k) = 1, T(n, 1) = choose(2n-1, n), T(n>1, k>1) = T(n-1, k) + (T(n/k, k) if k divides n, else 0)

Extensions

Corrected and rewritten by Gus Wiseman, Jun 07 2019

A112344 Number of partitions of n into perfect powers with each part > 1.

Original entry on oeis.org

0, 0, 0, 1, 0, 0, 0, 2, 1, 0, 0, 2, 1, 0, 0, 4, 2, 1, 0, 4, 2, 1, 0, 6, 5, 2, 2, 6, 5, 2, 2, 10, 8, 5, 4, 13, 8, 5, 4, 17, 14, 8, 9, 20, 17, 8, 9, 26, 24, 15, 14, 34, 27, 19, 14, 40, 38, 27, 25, 48, 47, 31, 30, 58, 59, 44, 42, 75, 68, 55, 47, 91, 86, 70, 67, 110, 106, 81, 81, 130, 134, 104
Offset: 1

Views

Author

Reinhard Zumkeller, Sep 05 2005

Keywords

Examples

			a(20) = #{16+4, 8+8+4, 8+4+4+4, 4+4+4+4+4} = 4.

Links

Crossrefs

Cf. A001597, A000041, A001156, A078134, A062051, A112345.
Cf. A078635 (allowing 1).

Programs

  • Maple
    N:= 200: # to get a(1) to a(N)
Pows:= {seq(seq(k^p, p=2..floor(log[k](N))),k=2..floor(sqrt(N)))}:
g:= proc(n,q) option remember; if n = 0 then 1 else `+`(seq(procname(n-r,r), r=select(`<=`,Pows,min(q,n)))) fi end proc:
seq(g(n,n), n=1..N); # Robert Israel, Nov 04 2015
  • Mathematica
    M = 200; (* to get a(1) to a(M) *)
Pows = Table[k^p, {k, 2, Floor[Sqrt[M]]}, {p, 2, Floor[Log[k, M]]}] // Flatten // Union;
g[n_, q_] := g[n, q] = If[n == 0, 1, Plus @@ Table[g[n - r, r], {r, Select[Pows, # <= Min[q, n]&]}]];
Table[g[n, n], {n, 1, M}] (* Jean-François Alcover, Feb 03 2018, translated from Robert Israel's Maple code *)
  • PARI
    leastp(n) = {while(!ispower(n), n--; if (n==0, return (0))); n;}
a(n) = {pmax = leastp(n); if (! pmax, return (0)); nb = 0; forpart(p=n, nb += (#select(x->ispower(x), Vec(p)) == #p), [4, pmax]); nb;} \\ Michel Marcus, Nov 04 2015

Extensions

Name clarified by Sean A. Irvine, Jan 12 2025

A179051 Number of partitions of n into powers of 10 (cf. A011557).

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

Reinhard Zumkeller, Jun 27 2010

Keywords

Comments

A179052 and A008592 give record values and where they occur.

Examples

			a(19) = #{10 + 9x1, 19x1} = 2;
a(20) = #{10 + 10, 10 + 10x1, 20x1} = 3;
a(21) = #{10 + 10 + 1, 10 + 11x1, 21x1} = 3.

Links

Crossrefs

Number of partitions of n into powers of b: A018819 (b=2), A062051 (b=3).
Cf. A206245, A000041, A179051.
Cf. A133880, A132272.
Cf. A008592, A179052.

Programs

  • Haskell
    a179051 = p 1 where
   p _ 0 = 1
   p k m = if m < k then 0 else p k (m - k) + p (k * 10) m
-- Reinhard Zumkeller, Feb 05 2012
  • Mathematica
    terms = 10001;
CoefficientList[Product[1/(1 - x^(10^k)) + O[x]^terms,
     {k, 0, Log[10, terms] // Ceiling}], x]
(* Jean-François Alcover, Dec 12 2021, after Ilya Gutkovskiy *)

Formula

a(n) = A133880(n) for n < 90; a(n) = A132272(n) for n < 100.
a(10^n) = A145513(n).
a(10*n) = A179052(n).
A179052(n) = a(A008592(n));
a(n) = p(n,1) where p(n,k) = if k<=n then p(10*[(n-k)/10],k)+p(n,10*k) else 0^n.
G.f.: Product_{k>=0} 1/(1 - x^(10^k)). - Ilya Gutkovskiy, Jul 26 2017

A105420 Number of partitions of n into 3-smooth parts.

Original entry on oeis.org

1, 1, 2, 3, 5, 6, 10, 12, 18, 23, 31, 38, 53, 63, 82, 100, 128, 152, 194, 228, 284, 336, 410, 478, 586, 678, 814, 947, 1127, 1296, 1539, 1761, 2070, 2372, 2764, 3146, 3667, 4153, 4796, 5437, 6249, 7044, 8080, 9080, 10358, 11636, 13208, 14778, 16762, 18698
Offset: 0

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Author

Reinhard Zumkeller, Apr 07 2005

Keywords

Comments

See A062051 for partitions into distinct 3-smooth numbers.

Examples

			n=10: there are 11 partitions of 10 with at least one part not of the form 2^i*3^j: 10, 7+3, 7+2+1, 7+1+1+1, 5+5, 5+4+1, 5+3+2, 5+3+1+1, 5+2+2+1, 5+2+1+1+1 and 5+1+1+1+1+1, therefore a(10) = A000041(10) - 11 = 42 - 11 = 31.

Links

Crossrefs

Cf. A000041, A003586.
Cf. A117220, A117221.
Cf. A062051, A023893, A131995.

Programs

  • Mathematica
    nmax = 120;
S = Select[Range[nmax], Max[FactorInteger[#][[All, 1]]] <= 3 &];
P[n_] := IntegerPartitions[n, All, TakeWhile[S, # <= n &] ];
a[n_] := a[n] = P[n] // Length;
Table[Print[n, " ", a[n]]; a[n], {n, 0, nmax}] (* Jean-François Alcover, Oct 13 2021 *)

Formula

A117222(n) = a(A003586(n)). - Reinhard Zumkeller, Mar 04 2006
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