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 A100310 Modulo 2 binomial transform of 7^n. 6
 1, 8, 50, 400, 2402, 19216, 120100, 960800, 5764802, 46118416, 288240100, 2305920800, 13847054404, 110776435232, 692352720200, 5538821761600, 33232930569602, 265863444556816, 1661646528480100, 13293172227840800 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS 7^n may be retrieved through 7^n=sum{k=0..n, (-1)^A010060(n-k)*mod(binomial(n,k),2)A100310(k)}. LINKS V. Shevelev, On Stephan's conjectures concerning Pascal triangle modulo 2 and their polynomial generalization, J. of Algebra Number Theory: Advances and Appl., 7 (2012), no.1, 11-29. FORMULA a(n)=sum{k=0..n, mod(binomial(n, k), 2)7^k}. From Vladimir Shevelev, Dec 26-27 2013: (Start) sum{n>=0}1/a(n)^r = prod{k>=0}(1 + 1/(7^(2^k)+1)^r), sum{n>=0}(-1)^A000120(n)/a(n)^r = prod{k>=0}(1 - 1/(7^(2^k)+1)^r), where r>0 is a real number. In particular, sum{n>=0}1/a(n) = prod{k>=0}(1 + 1/(7^(2^k)+1)) = 1.1479779...; sum{n>=0}(-1)^A000120(n)/a(n) = 6/7. a(2^n) = 7^(2^n)+1, n>=0. Note that analogs of Stephan's limit formulas (see Shevelev link) reduce to the relations: a(2^t*n+2^(t-1)) = 48*(7^(2^(t-1)+1))/(7^(2^(t-1))-1) * a(2^t*n+2^(t-1)-2), t>=2. In particular, for t=2,3,4, we have the following formulas: a(4*n+2) = 50 * a(4*n); a(8*n+4) = 1201/25 * a(8*n+2); a(16*n+8)= 2882401/60050 * a(16*n+6), etc. (End) CROSSREFS Cf. A001316, A001317, A038183, A100307, A100308, A100309, A100311. Sequence in context: A133129 A103458 A238841 * A124963 A295167 A195231 Adjacent sequences:  A100307 A100308 A100309 * A100311 A100312 A100313 KEYWORD easy,nonn AUTHOR Paul Barry, Dec 06 2004 STATUS approved

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Last modified August 13 00:27 EDT 2020. Contains 336441 sequences. (Running on oeis4.)