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 23 results. Next

A290247 Number of compositions (ordered partitions) of n^3 into cubes.

Original entry on oeis.org

1, 1, 2, 120, 290250, 107320441096, 21715974961480054078, 8487986089807555456140271121440, 22615863021403796876556069287242400147213424924, 1449638083412288206280215383952017948209203861522683138464747658192
Offset: 0

Views

Author

Ilya Gutkovskiy, Jul 24 2017

Keywords

Examples

			a(2) = 2 because 2^3 = 8 and we have [8], [1, 1, 1, 1, 1, 1, 1, 1].
		

Crossrefs

Programs

  • Maple
    b:= proc(n) option remember; local i; if n=0 then 1
          else 0; for i while i^3<=n do %+b(n-i^3) od fi
        end:
    a:= n-> b(n^3):
    seq(a(n), n=0..10);  # Alois P. Heinz, Aug 12 2017
  • Mathematica
    Table[SeriesCoefficient[1/(1 - Sum[x^k^3, {k, 1, n}]), {x, 0, n^3}], {n, 0, 9}]

Formula

a(n) = [x^(n^3)] 1/(1 - Sum_{k>=1} x^(k^3)).
a(n) = A023358(A000578(n)).

A323633 Expansion of 1/Sum_{k>=0} x^(k^3).

Original entry on oeis.org

1, -1, 1, -1, 1, -1, 1, -1, 0, 1, -2, 3, -4, 5, -6, 7, -7, 6, -4, 1, 3, -8, 14, -21, 28, -34, 38, -40, 38, -31, 18, 2, -29, 62, -99, 139, -178, 211, -232, 234, -210, 154, -62, -70, 242, -449, 680, -917, 1135, -1303, 1386, -1344, 1136, -725, 85, 794, -1898, 3183, -4571
Offset: 0

Views

Author

Ilya Gutkovskiy, Jan 21 2019

Keywords

Comments

Convolution inverse of A010057.

Crossrefs

Programs

  • Maple
    a:=series(1/add(x^(k^3),k=0..100),x=0,59): seq(coeff(a,x,n),n=0..58); # Paolo P. Lava, Apr 02 2019
  • Mathematica
    nmax = 58; CoefficientList[Series[1/Sum[x^k^3, {k, 0, nmax}], {x, 0, nmax}], x]
    a[0] = 1; a[n_] := a[n] = -Sum[Boole[IntegerQ[k^(1/3)]] a[n - k], {k, 1, n}]; Table[a[n], {n, 0, 58}]
  • PARI
    my(N=66, x='x+O('x^N)); Vec(1/sum(k=0, N^(1/3), x^k^3)) \\ Seiichi Manyama, Mar 19 2022
    
  • PARI
    a(n) = if(n==0, 1, -sum(k=1, n, ispower(k, 3)*a(n-k))); \\ Seiichi Manyama, Mar 19 2022

Formula

a(0) = 1; a(n) = -Sum_{k=1..n} A010057(k) * a(n-k). - Seiichi Manyama, Mar 19 2022

A331845 Number of compositions (ordered partitions) of n into distinct cubes.

Original entry on oeis.org

1, 1, 0, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 0, 2, 6, 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, 2, 0, 0, 0, 0, 0, 0, 2, 6, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 6, 0, 0, 0, 0, 0, 0, 6, 24
Offset: 0

Views

Author

Ilya Gutkovskiy, Jan 29 2020

Keywords

Examples

			a(36) = 6 because we have [27,8,1], [27,1,8], [8,27,1], [8,1,27], [1,27,8] and [1,8,27].
		

Crossrefs

Programs

  • Maple
    b:= proc(n, i, p) option remember;
          `if`((i*(i+1)/2)^2n, 0, b(n-i^3, i-1, p+1))+b(n, i-1, p)))
        end:
    a:= n-> b(n, iroot(n, 3), 0):
    seq(a(n), n=0..100);  # Alois P. Heinz, Jan 30 2020
  • Mathematica
    b[n_, i_, p_] := b[n, i, p] = If[(i(i+1)/2)^2 < n, 0, If[n == 0, p!, If[i^3 > n, 0, b[n-i^3, i-1, p+1]] + b[n, i-1, p]]];
    a[n_] := b[n, Floor[n^(1/3)], 0];
    a /@ Range[0, 100] (* Jean-François Alcover, Oct 31 2020, after Alois P. Heinz *)

A347714 Number of compositions (ordered partitions) of n into at most 2 cubes.

Original entry on oeis.org

1, 1, 1, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 2, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2
Offset: 0

Views

Author

Ilya Gutkovskiy, Sep 10 2021

Keywords

Crossrefs

A347715 Number of compositions (ordered partitions) of n into at most 3 cubes.

Original entry on oeis.org

1, 1, 1, 1, 0, 0, 0, 0, 1, 2, 3, 0, 0, 0, 0, 0, 1, 3, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 2, 3, 0, 0, 0, 0, 0, 2, 6, 0, 0, 0, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 3, 0, 0, 0, 0, 0, 0, 3, 0, 1, 2, 3, 0, 0, 0, 0, 0, 2, 6, 0, 0, 0, 0, 0, 0, 3, 1
Offset: 0

Views

Author

Ilya Gutkovskiy, Sep 10 2021

Keywords

Crossrefs

A347716 Number of compositions (ordered partitions) of n into at most 4 cubes.

Original entry on oeis.org

1, 1, 1, 1, 1, 0, 0, 0, 1, 2, 3, 4, 0, 0, 0, 0, 1, 3, 6, 0, 0, 0, 0, 0, 1, 4, 0, 1, 2, 3, 4, 0, 1, 0, 0, 2, 6, 12, 0, 0, 0, 0, 0, 3, 12, 0, 0, 0, 0, 0, 0, 4, 0, 0, 1, 3, 6, 0, 0, 0, 0, 0, 3, 12, 1, 2, 3, 4, 0, 0, 6, 0, 2, 6, 12, 0, 0, 0, 0, 0, 3, 13, 4, 0, 0
Offset: 0

Views

Author

Ilya Gutkovskiy, Sep 10 2021

Keywords

Crossrefs

A246885 Those n for which the coefficients of x^n in the reciprocal of 1+x+x^8+...+x^(i^3)+... are odd.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 9, 11, 13, 15, 16, 19, 20, 23, 29, 32, 34, 35, 37, 45, 47, 48, 49, 53, 54, 57, 58, 67, 69, 71, 73, 75, 85, 86, 99, 101, 107, 108, 109, 110, 115, 121, 123, 124, 127, 128, 129, 131, 132, 135, 137, 141, 143, 155, 157, 160, 161, 162, 163, 169, 177, 183, 189, 193, 195, 197, 199, 203
Offset: 1

Views

Author

David S. Newman, Sep 06 2014

Keywords

Comments

Numbers n such that the number of compositions of n into cubes (A023358) is odd. - Joerg Arndt, Sep 08 2014

Examples

			The reciprocal of 1+x+x^8+x^27+... begins 1 -x +x^2 -x^3 +x^4 -x^5 +x^6 -x^7 +x^9 -2*x^10 +...  So the first few values of a(n) are 0,1,2,3,4,5,6,7,9... .
		

Crossrefs

Cf. A023358.

Programs

  • Maple
    b:= proc(n) option remember; irem(`if`(n=0, 1,
          `if`(n<0, 0, add(b(n-i^3), i=1..iroot(n, 3)))), 2)
        end:
    a:= proc(n) option remember; local k; for k from 1+
          `if`(n=1, -1, a(n-1)) while b(k)=0 do od; k
        end:
    seq(a(n), n=1..80);  # Alois P. Heinz, Sep 08 2014
  • Mathematica
    iend=10;
    seq=Flatten[Position[Delete[Mod[CoefficientList[Series[1/Sum[x^(i^3),{i,0,iend}],{x,0,iend^3}],x],2],1],1]];
    Print[seq];

A278929 Expansion of 1/(1 - Sum_{k>=1} x^(prime(k)^3)).

Original entry on oeis.org

1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 2, 0, 0, 0, 0, 1, 0, 0, 3, 0, 0, 0, 0, 1, 0, 0, 4, 0, 0, 1, 0, 1, 0, 0, 5, 0, 0, 3, 0, 1, 0, 0, 6, 0, 0, 6, 0, 1, 0, 0, 7, 0, 0, 10, 0, 1, 1, 0, 8, 0, 0, 15, 0, 1, 4, 0, 9, 0, 0, 21
Offset: 0

Views

Author

Ilya Gutkovskiy, Dec 24 2016

Keywords

Comments

Number of compositions (ordered partitions) of n into cubes of primes (A030078).

Examples

			a(35) = 2 because we have [8, 27] and [27, 8].
		

Crossrefs

Programs

  • Maple
    N:= 200:
    Primes:= select(isprime, [2,seq(i,i=3..floor(N^(1/3)),2)]):
    G:= 1/(1- add(x^(Primes[i]^3),i=1..nops(Primes))):
    S:= series(G,x,N+1):
    seq(coeff(S,x,j),j=0..N); # Robert Israel, Jan 23 2019
  • Mathematica
    nmax = 120; CoefficientList[Series[1/(1 - Sum[x^Prime[k]^3, {k, 1, nmax}]), {x, 0, nmax}], x]

Formula

G.f.: 1/(1 - Sum_{k>=1} x^(prime(k)^3)).

A298936 Number of ordered ways of writing n^2 as a sum of n nonnegative cubes.

Original entry on oeis.org

1, 1, 0, 6, 6, 20, 120, 7, 1689, 6636, 36540, 64020, 963996, 2894892, 19555965, 176079995, 955611188, 6684303780, 42462792168, 292378003753, 1886275214112, 13384059605364, 87399249887334, 624073002367892, 5080120229014734, 37587589611771480
Offset: 0

Views

Author

Ilya Gutkovskiy, Jan 29 2018

Keywords

Examples

			a(3) = 6 because we have [8, 1, 0], [8, 0, 1], [1, 8, 0], [1, 0, 8], [0, 8, 1] and [0, 1, 8].
		

Crossrefs

Programs

  • Maple
    f:= n -> coeff(add(x^(k^3),k=0..floor(n^(2/3)))^n,x,n^2):
    map(f, [$0..30]); # Robert Israel, Jan 29 2018
  • Mathematica
    Table[SeriesCoefficient[Sum[x^k^3, {k, 0, Floor[n^(2/3) + 1]}]^n, {x, 0, n^2}], {n, 0, 25}]

Formula

a(n) = [x^(n^2)] (Sum_{k>=0} x^(k^3))^n.

A298937 Number of ordered ways of writing n^2 as a sum of n positive cubes.

Original entry on oeis.org

1, 1, 0, 0, 0, 0, 0, 7, 1, 0, 0, 9240, 34650, 1716, 48477, 551915, 6726720, 89973520, 102639744, 1824625081, 9915389400, 30143458884, 278196062760, 1995766236541, 6611689457736, 64547920386450, 236756174748626, 2315743488707806
Offset: 0

Views

Author

Ilya Gutkovskiy, Jan 29 2018

Keywords

Examples

			a(7) = 7 because we have [8, 8, 8, 8, 8, 8, 1], [8, 8, 8, 8, 8, 1, 8], [8, 8, 8, 8, 1, 8, 8], [8, 8, 8, 1, 8, 8, 8], [8, 8, 1, 8, 8, 8, 8], [8, 1, 8, 8, 8, 8, 8] and [1, 8, 8, 8, 8, 8, 8].
		

Crossrefs

Programs

  • Mathematica
    Join[{1}, Table[SeriesCoefficient[Sum[x^k^3, {k, 1, Floor[n^(2/3) + 1]}]^n, {x, 0, n^2}], {n, 1, 27}]]

Formula

a(n) = [x^(n^2)] (Sum_{k>=1} x^(k^3))^n.
Showing 1-10 of 23 results. Next