A117903
Diagonal sums of number triangle A117901.
Original entry on oeis.org
1, -1, 1, -2, 4, -2, -5, 14, -5, -26, 64, -26, -101, 254, -101, -410, 1024, -410, -1637, 4094, -1637, -6554, 16384, -6554, -26213, 65534, -26213, -104858, 262144, -104858, -419429
Offset: 0
- Harvey P. Dale, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (-1,-1,3,3,3,4,4,4).
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R:=PowerSeriesRing(Integers(), 50); Coefficients(R!( (1+x^2-5*x^3+3*x^4-3*x^5-x^6-2*x^7)/((1-4*x^3)*(1+x+x^2+x^3+x^4+x^5)) )); // G. C. Greubel, Oct 09 2021
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LinearRecurrence[{-1,-1,3,3,3,4,4,4},{1,-1,1,-2,4,-2,-5,14},40] (* Harvey P. Dale, Oct 04 2021 *)
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def A133851(n): return 4^(n/3) if (n%3==0) else 0
def A057079(n): return chebyshev_U(n, 1/2) + chebyshev_U(n-1, 1/2)
def A117903(n): return (1/30)*(28*(-1)^n + (15*(-1)^n - 1)* A057079(n) - 6*(2*A133851(n) - 5*A133851(n-1) + 2*A133851(n-2)))
[A117903(n) for n in (0..50)] # G. C. Greubel, Oct 09 2021
A117898
Number triangle 2^abs(L(C(n,2)/3) - L(C(k,2)/3))*[k<=n] where L(j/p) is the Legendre symbol of j and p.
Original entry on oeis.org
1, 1, 1, 2, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1
Offset: 0
Triangle begins
1;
1, 1;
2, 2, 1;
1, 1, 2, 1;
1, 1, 2, 1, 1;
2, 2, 1, 2, 2, 1;
1, 1, 2, 1, 1, 2, 1;
1, 1, 2, 1, 1, 2, 1, 1;
2, 2, 1, 2, 2, 1, 2, 2, 1;
1, 1, 2, 1, 1, 2, 1, 1, 2, 1;
1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1;
2, 2, 1, 2, 2, 1, 2, 2, 1, 2, 2, 1;
1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1;
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Flatten[CoefficientList[CoefficientList[Series[(1 +x(1+y) +x^2(2+2y+y^2) +x^3*y(1 +2y) +2x^4*y^2)/((1-x^3)(1-x^3*y^3)), {x,0,15}, {y,0,15}], x], y]] (* G. C. Greubel, May 03 2017 *)
T[n_, k_]:= 2^Abs[JacobiSymbol[Binomial[n, 2], 3] - JacobiSymbol[Binomial[k, 2], 3]]; Table[T[n, k], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Sep 27 2021 *)
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def T(n, k): return 2^abs(kronecker(binomial(n,2), 3) - kronecker(binomial(k,2), 3))
flatten([[T(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Sep 27 2021
A117902
Expansion of (1-x^2-2x^3)/(1-4x^3).
Original entry on oeis.org
1, 0, -1, 2, 0, -4, 8, 0, -16, 32, 0, -64, 128, 0, -256, 512, 0, -1024, 2048, 0, -4096, 8192, 0, -16384, 32768, 0, -65536, 131072, 0, -262144, 524288, 0, -1048576, 2097152, 0, -4194304, 8388608, 0, -16777216, 33554432, 0, -67108864, 134217728, 0, -268435456, 536870912, 0, -1073741824
Offset: 0
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[1] cat [n le 3 select (-1)^(n-1)*(n-1) else 4*Self(n-3): n in [1..50]]; // G. C. Greubel, Oct 09 2021
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LinearRecurrence[{0,0,4}, {1,0,-1,2}, 50] (* G. C. Greubel, Oct 09 2021 *)
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def A133851(n): return 4^(n/3) if (n%3==0) else 0
def A117902(n): return bool(n==0)/2 + 2*A133851(n-3) - A133851(n-2)
[A117902(n) for n in (0..50)] # G. C. Greubel, Oct 09 2021
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