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 A101122 XOR BINOMIAL transform of A101119. 4
 7, 17, 0, 34, 0, 0, 0, 68, 0, 0, 0, 0, 0, 0, 0, 159, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 257, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 514, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Nonzero terms form A101121 and occur at positions 2^k for k >= 0. A101119 equals the nonzero differences of A006519 and A003484. See A099884 for the definition of the XOR BINOMIAL transform. LINKS Table of n, a(n) for n=1..101. FORMULA a(n) = SumXOR_{k=0..n} (C(n, k) mod 2)*A101119(k), where SumXOR is summation under XOR. A101119(n) = SumXOR_{k=0..n} (C(n, k) mod 2)*a(k). a(2^(n-1)) = A101121(n) for n >= 1 and a(k)=0 when k is not a power of 2. PROG (PARI) {a(n)=local(B); B=0; for(i=0, n-1, B=bitxor(B, binomial(n-1, i)%2* (16*2^valuation(n-i, 2)-2^(valuation(n-i, 2)%4)-8*(valuation(n-i, 2)\4)-8))); B} (Python) from operator import xor from functools import reduce def A101122(n): return reduce(xor, (((1<<(m:=(~(k+1)&k).bit_length()+4))-((m&-4)<<1)-(1<<(m&3)))&-int(not k&~(n-1)) for k in range(n))) # Chai Wah Wu, Jul 10 2022 CROSSREFS Cf. A003484, A006519, A101119, A101120, A101121. Sequence in context: A332573 A129422 A184062 * A090535 A107778 A122735 Adjacent sequences: A101119 A101120 A101121 * A101123 A101124 A101125 KEYWORD nonn AUTHOR Simon Plouffe and Paul D. Hanna, Dec 02 2004 STATUS approved

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Last modified July 12 19:59 EDT 2024. Contains 374252 sequences. (Running on oeis4.)