A135911 -id:A135911 - cp's OEIS frontend

A045634 Number of ways in which n can be partitioned as a sum of a square and cube.

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

Offset: 0

Felice Russo

nonn

a(A022550(n))=0; a(A179509(n))=1; a(A022549(n))>0; a(A060861(n))=n. [From Reinhard Zumkeller, Jul 17 2010]

			a(9)=2 because 9=2^3+1^2 and 9=3^2+0^3.

T. D. Noe, Table of n, a(n) for n = 0..10000

Cf. A022549, A060861, A135910, A135911, A135912.

M:=100; M2:=M^2; t0:=array(0..M2);
for i from 0 to M2 do t0[i]:=0; od:
for a from 0 to M do for b from 0 to M do
i:=a^2+b^3; if i <= M2 then t0[i]:=t0[i]+1; fi; od: od:
[seq(t0[i],i=0..M2)];

max = 100; Clear[a]; a[_] = 0;
Do[n = i^2 + j^3; a[n] += 1, {i, 0, Sqrt[max]}, {j, 0, max^(1/3)}];
Table[a[n], {n, 0, max}] (* Jean-François Alcover, Aug 02 2018 *)

More terms from Erich Friedman

A111151 Numbers not of the form a^2 + b^3 + c^4 + d^5 for a,b,c,d >= 0.

Original entry on oeis.org

15, 23, 55, 62, 71, 471, 478, 510, 646, 806, 839, 879, 939, 1023, 1063, 1287, 2127, 5135, 6811, 7499, 9191, 26471

Offset: 1

David W. Wilson, Oct 19 2005

It is conjectured that this list is complete.
Comments from Richard C. Schroeppel: (Start)
"I can't put my hand on the book, but I think Vaughn has shown that every sufficiently large number is of the form a^2+b^3+c^5.
The key is that the sum 1/2 + 1/3 + 1/5 = 31/30 > 1, so the expected number of representations for a number N is, on average, K * N^1/30.
K is some mess of gamma(fractions). [It's also necessary to check that there's no modular exclusion.] I once tried to estimate how large an N was "sufficiently large" and if I remember correctly got around 10^60.
The results for a^2+b^3+c^4+d^5 and a^2+b^3+c^4+d^5+e^6 would follow immediately, although proving an upper bound is a big question. I'm not even sure if Vaughn's proof is constructive.
Christopher Landauer worked on the 2...6 problem long long ago; I think his program got up to a million or so.
There was a paper in Math. Comp. about five years ago, about the problem of 4 cubes. Beeler & I tried to find the empirical last-unrepresentable number, but couldn't reach it. The Math. Comp. paper found a likely candidate. It was around 10^13 or 14." (End)
No other terms < 10^8. - T. D. Noe, Mar 10 2008

Cf. A135911 (number of 4-tuples (x, y, z, t) of nonnegative integers such that x^2+y^3+z^4+t^5 = n).

Complement[Range[10^6], Flatten[Table[a^2 + b^3 + c^4 + d^5, {a, 0, 1000}, {b, 0, 100}, {c, 0, 31}, {d, 0, 15}]]] (* Robert G. Wilson v, Oct 19 2005 *)

A135910 Number of triples (x,y,z) of nonnegative integers such that x^2+y^3+z^4 = n.

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane, Mar 07 2008

nonn

T. D. Noe, Table of n, a(n) for n = 0..10000
K. B. Ford, The representation of numbers as sums of unlike powers II, J. Amer. Math. Soc., 9 (1996), 919-940.

Cf. A045634, A135911, A135912.

M:=10; M2:=M^2; t0:=array(0..M2);
for i from 0 to M2 do t0[i]:=0; od:
for a from 0 to M do for b from 0 to M do for c from 0 to M do
i:=a^2+b^3+c^4; if i <= M2 then t0[i]:=t0[i]+1; fi;
od: od: od: [seq(t0[i],i=0..M2)];

A135912 Number of 5-tuples (x,y,z,t,u) of nonnegative integers such that x^2+y^3+z^4+t^5+u^6 = n.

Original entry on oeis.org

1, 5, 10, 10, 6, 5, 6, 4, 2, 5, 10, 10, 6, 4, 3, 1, 2, 9, 15, 11, 4, 3, 3, 1, 2, 8, 13, 12, 10, 9, 5, 2, 5, 12, 15, 9, 5, 10, 12, 6, 3, 7, 10, 9, 10, 11, 6, 2, 4, 10, 14, 10, 8, 11, 8, 2, 2, 7, 10, 9, 9, 7, 2, 2, 9, 21, 26, 16, 9, 13, 11, 3, 3, 11, 16, 12, 9, 9, 5, 3, 8, 21, 29, 21, 14, 12, 7, 3, 4

Offset: 0

N. J. A. Sloane, Mar 07 2008

nonn

a(n) > 0 for n <= 10000. Is there any n for which a(n) = 0?
Note that there are many famous hard problems connected with sequences A045634, A135910, A135911 and the present entry (see the Ford reference).
The graph of this sequence suggests that a(n) is never zero. Checked to 10^5. - T. D. Noe, Mar 07 2008

T. D. Noe, Table of n, a(n) for n = 0..10000
K. B. Ford, The representation of numbers as sums of unlike powers II, J. Amer. Math. Soc., 9 (1996), 919-940.

Cf. A045634, A135910, A135911.

M:=100; M2:=M^2; t0:=array(0..M2); for i from 0 to M2 do t0[i]:=0; od:
for a from 0 to M do na:=a^2; for b from 0 to M do nb:=na+b^3;
if nb <= M2 then for c from 0 to M do nc:=nb+c^4; if nc <= M2 then for d from 0 to M2 do nd:=nc+d^5; if nd <= M2 then for e from 0 to M2 do i:=nd+e^6; if i <= M2 then t0[i]:=t0[i]+1; fi; od: fi; od; fi; od: fi; od: od:
[seq(t0[i],i=0..M2)];
for i from 0 to M2 do if t0[i]=0 then lprint(i); fi; od:

cp's OEIS Frontend

A045634 Number of ways in which n can be partitioned as a sum of a square and cube.

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A111151 Numbers not of the form a^2 + b^3 + c^4 + d^5 for a,b,c,d >= 0.

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A135910 Number of triples (x,y,z) of nonnegative integers such that x^2+y^3+z^4 = n.

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A135912 Number of 5-tuples (x,y,z,t,u) of nonnegative integers such that x^2+y^3+z^4+t^5+u^6 = n.

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