A054390 -id:A054390 - cp's OEIS frontend

A117535 Number of ways of writing n as a sum of powers of 3, each power being used at most 4 times.

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

Offset: 0

John W. Layman, Mar 27 2006

It seems that this sequence can be calculated by constructing an insertion tree in which the insertion rules depend on the "age" of a term at a particular stage of the calculation. See the link for a discussion of this concept.

			a(12) = 4 because 12=9+3=9+1+1+1=3+3+3+3=3+3+3+1+1+1.

Alois P. Heinz, Table of n, a(n) for n = 0..10000
P. de Castro et al., Counting binomial coefficients divisible by a prime power, Amer. Math. Monthly, 125 (2018), 531-540. See page 535.
John W. Layman, Ratio-Determined Insertion Sequences and the Tree of their Recurrence Types, June 2003 [Broken link]
John W. Layman, Ratio-Determined Insertion Sequences and the Tree of their Recurrence Types, June 2003 [local copy, corrected]
John W. Layman, Sequences Generated by Age-Determined Insertion Trees, Jan 2006
John W. Layman, Sequences Generated by Age-Determined Insertion Trees, Jan 2006 [Local copy]

Cf. A054390.

g:= product((1+x^(3^j)+x^(2*(3^j))+x^(3*(3^j))+x^(4*(3^j))), j=0..10): gser:= series(g,x=0,106): seq(coeff(gser,x,n), n=0..103); # Emeric Deutsch, Apr 02 2006
# second Maple program:
b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<0, 0,
      add(`if`(n-j*3^i<0, 0, b(n-j*3^i, i-1)), j=0..4)))
    end:
a:= n-> b(n, ilog[3](n)):
seq(a(n), n=0..100);  # Alois P. Heinz, Jun 21 2012

b[n_, i_] := b[n, i] = If[n == 0, 1, If[i < 0, 0, Sum[If[n - j*3^i < 0, 0, b[n - j*3^i, i - 1]], {j, 0, 4}]]]; a[n_] := b[n, Floor[Log[3, n]]]; Table[a[n], {n, 0, 100}] (* Jean-François Alcover, Dec 22 2016, after Alois P. Heinz *)

G.f.: product((1+x^(3^j)+x^(2*(3^j))+x^(3*(3^j))+x^(4*(3^j))), j=0..infinity). - Emeric Deutsch, Apr 02 2006
For n>=1, a(3*n+2) = a(n); a(3*n+1) = a(n) + a(n-1); a(3*n) = a(n) + a(n-1). - Tom Edgar, Jun 21 2017
G.f. A(x) satisfies: A(x) = (1 + x + x^2 + x^3 + x^4) * A(x^3). - Ilya Gutkovskiy, Jul 09 2019

A277872 Number of ways of writing n as a sum of powers of 4, each power being used at most four times.

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 3, 2, 2, 2, 3, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 3, 2, 2, 2, 3, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 3, 2, 2, 2, 3, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 4, 3, 3, 3, 5, 2, 2, 2, 4, 2, 2, 2, 4, 2, 2, 2, 5, 3, 3, 3, 4, 1

Offset: 0

Timothy B. Flowers, Nov 03 2016

nonn

Also known as the hyper 4-ary partition sequence, often denoted h_4(n).

Contains A002487 as a subsequence.

			a(72) = 4 because 72 = 64+4+4 = 64+4+1+1+1+1 = 16+16+16+16+4+4 = 16+16+16+16+4+1+1+1+1.

Timothy B. Flowers, Table of n, a(n) for n = 0..10000
K. Courtright and J. Sellers, Arithmetic properties for hyper m-ary partition functions, Integers, 4 (2004), A6.
Timothy B. Flowers, Extending a Recent Result on Hyper m-ary Partition Sequences, Journal of Integer Sequences, Vol. 20 (2017), #17.6.7.
T. B. Flowers and S. R. Lockard, Identifying an m-ary partition identity through an m-ary tree, Integers, 16 (2016), A10.

Cf. A002487, A054390, A277873.

n:=250;
r:=3;
(* To get up to n-th term, need r such that 4^r < n < 4^(r+1)  *)
h4 :=  CoefficientList[ Series[ Product[ (1 - q^(5*4^i))/(1 - q^(4^i)) , {i, 0, r}], {q, 0, n} ], q]

G.f.: Product_{j>=0} (1-x^(5*4^j))/(1-x^(4^j)).
G.f.: Product_{j>=0} (1+x^(4^j)+x^(2*4^j)+x^(3*4^j)+x^(4*4^j)).
a(0)=1 and for n>0, a(4n)=a(n)+a(n-1), a(4n+r)=a(n) for r=1,2,3.
G.f. A(x) satisfies: A(x) = (1 + x + x^2 + x^3 + x^4) * A(x^4). - Ilya Gutkovskiy, Jul 09 2019

A277873 Number of ways of writing n as a sum of powers of 5, each power being used at most five times.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 3, 2, 2, 2, 2, 3, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 3, 2, 2, 2, 2, 3, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 3, 2, 2, 2, 2, 3, 1, 1, 1, 1, 2

Offset: 0

Timothy B. Flowers, Nov 07 2016

nonn

Also known as the hyper 5-ary partition sequence, often denoted h_5(n).

Contains A002487 as a subsequence.

			a(140) = 4 because 140 = 125+5+5+5 = 125+5+5+1+1+1+1+1 = 25+25+25+25+25+5+5+5 = 25+25+25+25+25+5+5+1+1+1+1+1.

Timothy B. Flowers, Table of n, a(n) for n = 0..10000
K. Courtright and J. Sellers, Arithmetic properties for hyper m-ary partition functions, Integers, 4 (2004), A6.
Timothy B. Flowers, Extending a Recent Result on Hyper m-ary Partition Sequences, Journal of Integer Sequences, Vol. 20 (2017), #17.6.7.
T. B. Flowers and S. R. Lockard, Identifying an m-ary partition identity through an m-ary tree, Integers, 16 (2016), A10.

Cf. A002487, A054390, A277872.

b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<0, 0,
      add(b(n-j*5^i, i-1), j=0..min(5, n/5^i))))
    end:
a:= n-> b(n, ilog[5](n)):
seq(a(n), n=0..120);  # Alois P. Heinz, May 01 2018

n:=250; r:=3; (* To get up to n-th term, need r such that 5^r < n < 5^(r+1) *) h5 :=  CoefficientList[ Series[ Product[ (1 - q^(6*5^i))/(1 - q^(5^i)) , {i, 0, r}], {q, 0, n} ], q]

G.f.: Product_{j >= 0} (1-x^(6*5^j))/(1-x^(5^j)).
G.f.: Product_{j >= 0} Sum_{k=0..5} x^(k*5^j).
a(0)=1; for k>0, a(5*k) = a(k)+a(k-1) and a(5*k+r) = a(k) with r=1,2,3,4.
G.f. A(x) satisfies: A(x) = (1 + x + x^2 + x^3 + x^4 + x^5) * A(x^5). - Ilya Gutkovskiy, Jul 09 2019

A309048 Expansion of Product_{k>=0} (1 + x^(3^k) + x^(2*3^k) - x^(3^(k+1))).

Original entry on oeis.org

1, 1, 1, 0, 1, 1, 0, 1, 1, -1, 0, 0, 1, 1, 1, 0, 1, 1, -1, 0, 0, 1, 1, 1, 0, 1, 1, -2, -1, -1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, -1, 0, 0, 1, 1, 1, 0, 1, 1, -2, -1, -1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, -1, 0, 0, 1, 1, 1, 0, 1, 1, -3, -2, -2, 1, -1, -1, 0, -1, -1, 2, 1, 1, -1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1

Offset: 0

Ilya Gutkovskiy, Jul 09 2019

sign

Cf. A005590, A054390, A309047.

nmax = 109; CoefficientList[Series[Product[(1 + x^(3^k) + x^(2 3^k) - x^(3^(k + 1))), {k, 0, Floor[Log[3, nmax]] + 1}], {x, 0, nmax}], x]
nmax = 109; A[] = 1; Do[A[x] = (1 + x + x^2 - x^3) A[x^3] + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
a[0] = 1; a[n_] := Switch[Mod[n, 3], 0, a[n/3] - a[(n - 3)/3], 1, a[(n - 1)/3], 2, a[(n - 2)/3]]; Table[a[n], {n, 0, 109}]

G.f. A(x) satisfies: A(x) = (1 + x + x^2 - x^3) * A(x^3).

a(0) = 1; a(3*n) = a(n) - a(n-1), a(3*n+1) = a(n), a(3*n+2) = a(n).

A303824 Number of ways of writing n as a sum of powers of 6, each power being used at most six times.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 3, 2, 2, 2, 2, 2, 3, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 3, 2, 2, 2, 2, 2, 3, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2

Offset: 0

Seiichi Manyama, May 01 2018

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..10000

Number of ways of writing n as a sum of powers of b, each power being used at most b times: A054390 (b=3), A277872 (b=4), A277873 (b=5), this sequence (b=6), A303825 (b=7).

b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<0, 0,
      add(b(n-j*6^i, i-1), j=0..min(6, n/6^i))))
    end:
a:= n-> b(n, ilog[6](n)):
seq(a(n), n=0..120);  # Alois P. Heinz, May 01 2018

m = 100; A[_] = 1;
Do[A[x_] = Total[x^Range[0, 6]] A[x^6] + O[x]^m // Normal, {m}];
CoefficientList[A[x], x] (* Jean-François Alcover, Oct 19 2019 *)

def A(k, n)
  ary = [1]
  (1..n).each{|i|
    s = ary[i / k]
    s += ary[i / k - 1] if i % k == 0
    ary << s
  }
  ary
end
p A(6, 100)

G.f.: Product_{k>=0} (1-x^(7*6^k))/(1-x^(6^k)).
a(0)=1; for k>0, a(6*k) = a(k)+a(k-1) and a(6*k+r) = a(k) with r=1,2,3,4,5.
G.f. A(x) satisfies: A(x) = (1 + x + x^2 + x^3 + x^4 + x^5 + x^6) * A(x^6). - Ilya Gutkovskiy, Jul 09 2019

A303825 Number of ways of writing n as a sum of powers of 7, each power being used at most seven times.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 3, 2, 2, 2, 2, 2, 2, 3, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 3, 2, 2, 2, 2, 2, 2, 3

Offset: 0

Seiichi Manyama, May 01 2018

nonn

Seiichi Manyama, Table of n, a(n) for n = 0..10000

Number of ways of writing n as a sum of powers of b, each power being used at most b times: A054390 (b=3), A277872 (b=4), A277873 (b=5), A303824 (b=6), this sequence (b=7).

b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<0, 0,
      add(b(n-j*7^i, i-1), j=0..min(7, n/7^i))))
    end:
a:= n-> b(n, ilog[7](n)):
seq(a(n), n=0..120);  # Alois P. Heinz, May 01 2018

m = 120; A[_] = 1;
Do[A[x_] = Total[x^Range[0, 7]] A[x^7] + O[x]^m // Normal, {m}];
CoefficientList[A[x], x] (* Jean-François Alcover, Oct 19 2019 *)

def A(k, n)
  ary = [1]
  (1..n).each{|i|
    s = ary[i / k]
    s += ary[i / k - 1] if i % k == 0
    ary << s
  }
  ary
end
p A(7, 100)

G.f.: Product_{k>=0} (1-x^(8*7^k))/(1-x^(7^k)).
a(0)=1; for k>0, a(7*k) = a(k)+a(k-1) and a(7*k+r) = a(k) with r=1,2,3,4,5,6.
G.f. A(x) satisfies: A(x) = (1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + x^7) * A(x^7). - Ilya Gutkovskiy, Jul 09 2019

A309045 Expansion of Product_{k>=0} (1 + x^(3^k) + x^(2*3^k) + x^(3^(k+1)))^(2^k).

Original entry on oeis.org

1, 1, 1, 3, 2, 2, 5, 3, 3, 11, 8, 8, 19, 11, 11, 25, 14, 14, 41, 27, 27, 59, 32, 32, 70, 38, 38, 110, 72, 72, 158, 86, 86, 190, 104, 104, 289, 185, 185, 395, 210, 210, 455, 245, 245, 645, 400, 400, 829, 429, 429, 915, 486, 486, 1269, 783, 783, 1623, 840, 840, 1800, 960, 960, 2472

Offset: 0

Ilya Gutkovskiy, Jul 09 2019

nonn

The trisection equals the self-convolution of this sequence.

Cf. A054390, A237651, A309046.

nmax = 63; CoefficientList[Series[Product[(1 + x^(3^k) + x^(2 3^k) + x^(3^(k + 1)))^(2^k), {k, 0, Floor[Log[3, nmax]] + 1}], {x, 0, nmax}], x]
nmax = 63; A[] = 1; Do[A[x] = (1 + x + x^2 + x^3) A[x^3]^2 + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]

G.f.: Product_{k>=0} ((1 - x^(4*3^k))/(1 - x^(3^k)))^(2^k).

G.f. A(x) satisfies: A(x) = (1 + x + x^2 + x^3) * A(x^3)^2.

A309046 Expansion of Product_{k>=0} (1 + x^(3^k) + x^(2*3^k) + x^(3^(k+1)))^(3^k).

Original entry on oeis.org

1, 1, 1, 4, 3, 3, 9, 6, 6, 25, 19, 19, 58, 39, 39, 105, 66, 66, 211, 145, 145, 394, 249, 249, 630, 381, 381, 1114, 733, 733, 1903, 1170, 1170, 2889, 1719, 1719, 4827, 3108, 3108, 7869, 4761, 4761, 11574, 6813, 6813, 18489, 11676, 11676, 28839, 17163, 17163, 41013, 23850

Offset: 0

Ilya Gutkovskiy, Jul 09 2019

nonn

The trisection equals the three-fold convolution of this sequence with themselves.

Cf. A054390, A237651, A309045, A321344.

nmax = 52; CoefficientList[Series[Product[(1 + x^(3^k) + x^(2 3^k) + x^(3^(k + 1)))^(3^k), {k, 0, Floor[Log[3, nmax]] + 1}], {x, 0, nmax}], x]
nmax = 52; A[] = 1; Do[A[x] = (1 + x + x^2 + x^3) A[x^3]^3 + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]

G.f.: Product_{k>=0} ((1 - x^(4*3^k))/(1 - x^(3^k)))^(3^k).

G.f. A(x) satisfies: A(x) = (1 + x + x^2 + x^3) * A(x^3)^3.

A345006 a(0) = 1; a(3n) = a(n) + a(n-1), a(3n+1) = a(3*n+2) = -a(n).

Original entry on oeis.org

1, -1, -1, 0, 1, 1, -2, 1, 1, -1, 0, 0, 1, -1, -1, 2, -1, -1, -1, 2, 2, -1, -1, -1, 2, -1, -1, 0, 1, 1, -1, 0, 0, 0, 0, 0, 1, -1, -1, 0, 1, 1, -2, 1, 1, 1, -2, -2, 1, 1, 1, -2, 1, 1, -2, 1, 1, 1, -2, -2, 4, -2, -2, 1, 1, 1, -2, 1, 1, -2, 1, 1, 1, -2, -2, 1, 1, 1, -2, 1, 1, -1, 0, 0, 1, -1, -1, 2, -1, -1, 0, 1, 1, -1

Offset: 0

Ilya Gutkovskiy, Jun 05 2021

sign

Cf. A049347, A054390, A309048, A345007.

a[0] = 1; a[n_] := Switch[Mod[n, 3], 0, a[n/3] + a[(n - 3)/3], 1, -a[(n - 1)/3], 2, -a[(n - 2)/3]]; Table[a[n], {n, 0, 93}]
nmax = 93; A[] = 1; Do[A[x] = (1 - x - x^2 + x^3) A[x^3] + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
nmax = 93; CoefficientList[Series[Product[(1 - x^(3^k) - x^(2 3^k) + x^(3^(k + 1))), {k, 0, Floor[Log[3, nmax]] + 1}], {x, 0, nmax}], x]

G.f. A(x) satisfies: A(x) = (1 - x - x^2 + x^3) * A(x^3).

G.f.: Product_{k>=0} (1 - x^(3^k) - x^(2*3^k) + x^(3^(k+1))).

A345007 a(0) = 1; a(3n) = a(n) - a(n-1), a(3n+1) = a(3*n+2) = -a(n).

Original entry on oeis.org

1, -1, -1, -2, 1, 1, 0, 1, 1, -1, 2, 2, 3, -1, -1, 0, -1, -1, -1, 0, 0, 1, -1, -1, 0, -1, -1, -2, 1, 1, 3, -2, -2, 0, -2, -2, 1, -3, -3, -4, 1, 1, 0, 1, 1, 1, 0, 0, -1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, -1, -1, -2, 1, 1, 0, 1, 1, 1, 0, 0, -1, 1, 1, 0, 1, 1, -1, 2, 2, 3, -1, -1, 0, -1, -1, 2, -3, -3, -5, 2, 2

Offset: 0

Ilya Gutkovskiy, Jun 05 2021

sign

Cf. A054390, A177219, A309048, A345006.

a[0] = 1; a[n_] := Switch[Mod[n, 3], 0, a[n/3] - a[(n - 3)/3], 1, -a[(n - 1)/3], 2, -a[(n - 2)/3]]; Table[a[n], {n, 0, 95}]
nmax = 95; A[] = 1; Do[A[x] = (1 - x - x^2 - x^3) A[x^3] + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
nmax = 95; CoefficientList[Series[Product[(1 - x^(3^k) - x^(2 3^k) - x^(3^(k + 1))), {k, 0, Floor[Log[3, nmax]] + 1}], {x, 0, nmax}], x]

G.f. A(x) satisfies: A(x) = (1 - x - x^2 - x^3) * A(x^3).

G.f.: Product_{k>=0} (1 - x^(3^k) - x^(2*3^k) - x^(3^(k+1))).

cp's OEIS Frontend

A117535 Number of ways of writing n as a sum of powers of 3, each power being used at most 4 times.

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A277872 Number of ways of writing n as a sum of powers of 4, each power being used at most four times.

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A277873 Number of ways of writing n as a sum of powers of 5, each power being used at most five times.

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A309048 Expansion of Product_{k>=0} (1 + x^(3^k) + x^(2*3^k) - x^(3^(k+1))).

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A303824 Number of ways of writing n as a sum of powers of 6, each power being used at most six times.

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A303825 Number of ways of writing n as a sum of powers of 7, each power being used at most seven times.

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A309045 Expansion of Product_{k>=0} (1 + x^(3^k) + x^(2*3^k) + x^(3^(k+1)))^(2^k).

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A309046 Expansion of Product_{k>=0} (1 + x^(3^k) + x^(2*3^k) + x^(3^(k+1)))^(3^k).

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A345006 a(0) = 1; a(3n) = a(n) + a(n-1), a(3n+1) = a(3*n+2) = -a(n).

A345007 a(0) = 1; a(3n) = a(n) - a(n-1), a(3n+1) = a(3*n+2) = -a(n).