A277872 -id:A277872 - cp's OEIS frontend

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

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

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

A303827 Number of ways of writing n as a sum of powers of 4, each power being used at most 5 times.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 3, 3, 2, 2, 4, 4, 2, 2, 3, 3, 1, 1, 2, 2, 1, 1, 3, 3, 2, 2, 4, 4, 2, 2, 3, 3, 1, 1, 2, 2, 1, 1, 3, 3, 2, 2, 4, 4, 2, 2, 3, 3, 1, 1, 2, 2, 1, 1, 4, 4, 3, 3, 6, 6, 3, 3, 5, 5, 2, 2, 4, 4, 2, 2, 6, 6, 4, 4, 8, 8, 4, 4, 6, 6

Offset: 0

Seiichi Manyama, May 01 2018

			a(17) = 3 because 17=16+1=4+4+4+4+1=4+4+4+1+1+1+1+1.

Alois P. Heinz, Table of n, a(n) for n = 0..16383

Number of ways of writing n as a sum of powers of b, each power being used at most b+1 times: A117535 (b=3), this sequence (b=4), A303828 (b=5).

Cf. A277872.

b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<0, 0,
      add(b(n-j*4^i, i-1), j=0..min(5, n/4^i))))
    end:
a:= n-> b(n, ilog[4](n)):
seq(a(n), n=0..120);  # Alois P. Heinz, May 01 2018

m = 100; A[_] = 1;
Do[A[x_] = (1+x+x^2+x^3+x^4+x^5) * A[x^4] + O[x]^m // Normal, {m}];
CoefficientList[A[x], x] (* Jean-François Alcover, Oct 06 2019, after Ilya Gutkovskiy *)

G.f.: Product_{k>=0} (1-x^(6*4^k))/(1-x^(4^k)).

G.f. A(x) satisfies: A(x) = (1 + x + x^2 + x^3 + x^4 + x^5) * A(x^4). - Ilya Gutkovskiy, Jul 09 2019

A345008 a(0) = 1; a(4n) = a(n) - a(n-1), a(4n+1) = a(4n+2) = a(4n+3) = a(n).

Original entry on oeis.org

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

Offset: 0

Ilya Gutkovskiy, Jun 05 2021

sign

Cf. A005590, A277872, A309048, A345009.

a[0] = 1; a[n_] := Switch[Mod[n, 4], 0, a[n/4] - a[(n - 4)/4], 1, a[(n - 1)/4], 2, a[(n - 2)/4], 3, a[(n - 3)/4]]; Table[a[n], {n, 0, 105}]
nmax = 105; A[] = 1; Do[A[x] = (1 + x + x^2 + x^3 - x^4) A[x^4] + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]
nmax = 105; CoefficientList[Series[Product[(1 + x^(4^k) + x^(2 4^k) + x^(3 4^k) - x^(4^(k + 1))), {k, 0, Floor[Log[4, nmax]] + 1}], {x, 0, nmax}], x]

G.f. A(x) satisfies: A(x) = (1 + x + x^2 + x^3 - x^4) * A(x^4).

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

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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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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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A303827 Number of ways of writing n as a sum of powers of 4, each power being used at most 5 times.

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A345008 a(0) = 1; a(4*n) = a(n) - a(n-1), a(4*n+1) = a(4*n+2) = a(4*n+3) = a(n).

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A345008 a(0) = 1; a(4n) = a(n) - a(n-1), a(4n+1) = a(4n+2) = a(4n+3) = a(n).