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0, 1, 4, 21, 64, 321, 1092, 5461, 16384, 81921, 278532, 1392661, 4210752, 21053761, 71582788, 357913941, 1073741824, 5368709121, 18253611012, 91268055061, 275951648832, 1379758244161, 4691178030148, 23455890150741
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,3
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COMMENTS
| a(n) = (A038192(n) - 3)/12 = (A038183(n) - 1)/4, reflecting the fact that, if you look at each row of the Pascal triangle's parity as a binary number (cf. A001317), then the numbers in odd rows are thrice the numbers in even rows.
Conjecture: a(2^k) = 2^(2^(k+1)-2). [This conjecture is true - Vladimir Shevelev (shevelev(AT)bgu.ac.il), Nov 28 2010]
Conjectures: lim(n->inf, a(2n+1)/a(2n)) = 5, lim(n->inf, a(4n+2)/a(4n+1)) = 17/5, lim(n->inf, a(8n+4)/a(8n+3)) = 257/85 etc. [This follows from the formula, for n>=0, t>=1: ( 4*a(2^t*n+2^(t-1))+1 )/( 4*a(2^t*n+2^(t-1)-1)+1 ) = 3*F_t/(F_t-2), where F_t= A000215(t) - Vladimir Shevelev (shevelev(AT)bgu.ac.il), Nov 28 2010]
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LINKS
| V. Shevelev,On Stephan's conjectures concerning Pascal triangle modulo 2
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PROG
| (PARI) a(n)=(sum(k=0, 2*n+1, (binomial(2*n+1, k)%2)*2^k)-3)/12
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CROSSREFS
| Sequence in context: A201446 A135559 A131478 * A095668 A078800 A184706
Adjacent sequences: A089890 A089891 A089892 * A089894 A089895 A089896
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KEYWORD
| nonn
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AUTHOR
| Ralf Stephan (ralf(AT)ark.in-berlin.de), Jan 10 2004
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