A106344 Triangle read by rows: T(n,k) = binomial(k,n-k) mod 2.
1, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1
Offset: 0
Examples
Triangle begins 1; 0, 1; 0, 1, 1; 0, 0, 0, 1; 0, 0, 1, 1, 1; 0, 0, 0, 1, 0, 1;
Links
- G. C. Greubel, Rows n = 0..50, flattened
- Thomas Baruchel, Flattening Karatsuba's Recursion Tree into a Single Summation, SN Computer Science (2020) Vol. 1, Article No. 48.
- George Beck and Karl Dilcher, A Matrix Related to Stern Polynomials and the Prouhet-Thue-Morse Sequence, arXiv:2106.10400 [math.CO], 2021. See (1.6) p. 2.
Crossrefs
Programs
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GAP
Flat(List([0..15], n-> List([0..n], k-> (Binomial(k,n-k) mod 2) ))); # G. C. Greubel, Feb 07 2020
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Magma
[ Binomial(k,n-k) mod 2: k in [0..n], n in [0..15]]; // G. C. Greubel, Feb 07 2020
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Maple
seq(seq(`mod`(binomial(k, n-k), 2), k = 0..n), n = 0..15); # G. C. Greubel, Feb 07 2020
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Mathematica
Table[Mod[Binomial[k, n-k], 2], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Apr 18 2017 *)
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PARI
T(n,k) = binomial(k,n-k)%2; for(n=0,15, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Feb 07 2020
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Sage
[[ mod(binomial(k,n-k), 2) for k in (0..n)] for n in (0..15)] # G. C. Greubel, Feb 07 2020
Comments