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Search: id:A099902
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| A099902 |
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Multiplies by 2 and shifts right under the XOR BINOMIAL transform (A099901). |
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+0
5
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| 1, 3, 7, 11, 23, 59, 103, 139, 279, 827, 1895, 2955, 5655, 14395, 24679, 32907, 65815, 197435, 460647, 723851, 1512983, 3881019, 6774887, 9142411, 18219287, 54002491, 123733863, 192940939, 369104407, 939538491, 1610637415, 2147516555
(list; graph; listen)
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OFFSET
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0,2
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COMMENT
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Equals the XOR BINOMIAL transform of A099901. Also, equals the main diagonal of the XOR difference triangle A099900, in which the central terms of the rows form the powers of 2.
Bisection of A101624. - Paul Barry (pbarry(AT)wit.ie), May 10 2005
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FORMULA
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a(n) = SumXOR_{k=0..n} (C(n-k+[k/2], [k/2])mod 2)*2^k for n>=0. a(n) = SumXOR_{i=0..n} (C(n, i)mod 2)*A099901(n-i), where SumXOR is the analogue of summation under the binary XOR operation, and C(i, j)mod 2 = A047999(i, j).
a(n) = Sum_{k=0..n} A047999(n-k+[k/2], [k/2]) * 2^k.
a(n)=sum{k=0..2n, (binomial(k, 2n-k) mod 2)*2^(2n-k)}; a(n)=sum{k=0..n, (binomial(2n-k, k) mod 2)*2^k}; - Paul Barry (pbarry(AT)wit.ie), May 10 2005
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PROGRAM
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(PARI) {a(n)=local(B); B=0; for(k=0, n, B=bitxor(B, binomial(n-k+k\2, k\2)%2*2^k)); B}
(PARI) a(n)=sum(k=0, n, binomial(n-k+k\2, k\2)%2*2^k)
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CROSSREFS
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Cf. A099884, A099900, A099901.
Sequence in context: A139253 A116606 A139814 this_sequence A092284 A024459 A001645
Adjacent sequences: A099899 A099900 A099901 this_sequence A099903 A099904 A099905
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KEYWORD
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eigen,nonn
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AUTHOR
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Paul D. Hanna (pauldhanna(AT)juno.com), Oct 30 2004
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