A214458 Let S_3(n) denote difference between multiples of 3 in interval [0,n) with even and odd binary digit sums. Then a(n)=(-1)^A000120(n)*(S_3(n)-3*S_3(floor(n/4))).
0, -1, -1, 1, 1, -1, -1, 0, 0, 0, 1, -1, 1, -2, -2, 2, 0, 0, 0, -1, 1, -1, 0, 0, 0, -1, -1, 1, 1, -1, -1, 0, 0, 0, 1, -1, 1, -2, -2, 2, 0, 0, 0, -1, 1, -1, 0, 0, 0, -1, -1, 1, 1, -1, -1, 0, 0, 0, 1, -1, 1, -2, -2, 2, 0, 0, 0, -1, 1, -1, 0, 0, 0, -1, -1, 1, 1
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Links
- J. Coquet, A summation formula related to the binary digits, Inventiones Mathematicae 73 (1983), pp. 107-115.
- D. J. Newman, On the number of binary digits in a multiple of three, Proc. Amer. Math. Soc. 21 (1969) 719-721.
- Vladimir Shevelev, Two algorithms for evaluation of the Newman digit sum, and a new proof of Coquet's theorem, arXiv:0709.0885 [math.NT], 2007-2012.
Formula
Recursion for evaluation S_3(n): S_3(n)=3*S_3(floor(n/4))+(-1)^A000120(n)*a(n). As a corollary, we have |S_3(n)-3*S_3(n/4)|<=2.
Comments