A131262 - OEIS

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A131262

a(n) = least index k such that A130654(k) = n.

1, 3, 14, 60, 248, 1008 (list; graph; refs; listen; history; internal format)

	OFFSET	`0,2`
	COMMENTS	`Also a(n) = least index k such that A092505(k) = A002430(k) / A046990(k) = 2^n.` `Note that` `a(0) = 1 = 1 - 0 = 2^0 - 0;` `a(1) = 3 = 4 - 1 = 2^2 - 1;` `a(2) = 14 = 16 - 2 = 2^4 - 2;` `a(3) = 60 = 64 - 4 = 2^6 - 4;` `a(4) = 248 = 256 - 8 = 2^8 - 8.` `Conjecture: a(n) = Sigma(2^n)*EulerPhi(2^n) = 2^(2n) - Floor(2^n/2) = A062354(2^n).` `If this conjecture is true the next term would be a(5) = 1008 = 1024 - 16 = 2^10 - 16.`
	FORMULA	`Conjecture: a(n) = Sigma(2^n)*EulerPhi(2^n) = 2^(2n) - Floor(2^n/2) = A062354(2^n).`
	EXAMPLE	`A130654(n) begins` `{0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 2, 1, 0, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 0, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 3, 1, 2, 1, 0, ...}.` `Thus a(0) = 1, a(1) = 3, a(2) = 14, a(3) = 60.`
	CROSSREFS	`Cf. A130654 = Exponent m such that 2^m = A092505(n) = A002430(n) / A046990(n). Cf. A092505 = A002430(n) / A046990(n), n>0. Cf. A002430 = Numerators in Taylor series for tan(x). Cf. A046990 = Numerators of Taylor series for log(1/cos(x)). Cf. A062354 = Sigma(n)EulerPhi(n).` `Sequence in context: A151235 A151236 A006224 A171499 A006502 A024037` `Adjacent sequences: A131259 A131260 A131261 * A131263 A131264 A131265`
	KEYWORD	`hard,more,nonn`
	AUTHOR	`Alexander Adamchuk (alex(AT)kolmogorov.com), Jun 24 2007`
	EXTENSIONS	`a(5) = 1008 from Alexander Adamchuk (alex(AT)kolmogorov.com), May 02 2010`

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Last modified February 17 10:03 EST 2012. Contains 206009 sequences.