OFFSET
2,1
COMMENTS
In the case of n = 1, there are solutions for all k. In particular, 1^k is always a k-th power and -(1^k) is a k-th power for odd k. As a formula: T(1,k) = 1 + (k mod 2). This row is not included in the sequence.
EXAMPLE
Triangle begins:
k = 1 2 3 4 5
n= 2: 4;
n= 3: 8, 1;
n= 4: 16, 1;
n= 5: 32, 0, 2;
n= 6: 64, 6;
n= 7: 128, 8;
n= 8: 256, 16, 4;
n= 9: 512, 26;
n=10: 1024, 17, 10;
n=11: 2048, 67, 4, 3;
n=12: 4096, 100, 10;
n=13: 8192, 137, 34, 6;
n=14: 16384, 426, 28, 1;
n=15: 32768, 661, 96, 6;
n=16: 65536, 1351, 146, 16, 8;
n=17: 131072, 2637, 230, 15;
n=18: 262144, 3831, 258, 40;
n=19: 524288, 8095, 1130, 50;
n=20: 1048576, 15241, 854, 77, 6;
...
The T(6,2) = 6 solutions are:
- 1^2 - 2^2 + 3^2 - 4^2 + 5^2 + 6^2 = 49 = 7^2,
- 1^2 - 2^2 + 3^2 + 4^2 + 5^2 - 6^2 = 9 = 3^2,
- 1^2 - 2^2 + 3^2 + 4^2 + 5^2 + 6^2 = 81 = 9^2,
+ 1^2 - 2^2 + 3^2 - 4^2 - 5^2 + 6^2 = 1 = 1^2,
+ 1^2 + 2^2 - 3^2 + 4^2 + 5^2 - 6^2 = 1 = 1^2,
+ 1^2 + 2^2 + 3^2 - 4^2 - 5^2 + 6^2 = 9 = 3^2.
PROG
(PARI)f(k, u)=my(x=0, v=vector(#u)); for(i=1, #u, u[i]=if(u[i]==0, -1, 1); v[i]=i^k); u*v~
is(k, u)=my(x=f(k, u)); ispower(x, k)
T(n, k)=my(u=vector(n, i, [0, 1]), nbsol=0); if(k%2==1, u[1]=[1, 1]); forvec(X=u, if(is(k, X), nbsol++)); if(k%2==1, nbsol*=2); nbsol
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Jean-Marc Rebert, Jan 26 2024
STATUS
approved