A136494 - OEIS

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A136494

Number of permutation symmetries in the binary expansion of n.

1, 1, 1, 2, 2, 2, 2, 6, 6, 4, 4, 6, 4, 6, 6, 24, 24, 12, 12, 12, 12, 12, 12, 24, 12, 12, 12, 24, 12, 24, 24, 120 (list; graph; refs; listen; history; internal format)

	OFFSET	`0,4`
	COMMENTS	`One can find the number of 0s in 1s in a string recursively by finding the number of 0s and 1s in disjoint substrings. Then just follow the formula.`
	FORMULA	`a(n) = A093659! * A139329!`
	EXAMPLE	`a(14) = 6 because there are 3! permutation symmetries of 1s * the 0! permutation symmetries of 0s.`
	CROSSREFS	`Cf. A139329, A093659.` `Sequence in context: A151704 A110023 A116863 * A048764 A038714 A139554` `Adjacent sequences: A136491 A136492 A136493 * A136495 A136496 A136497`
	KEYWORD	`base,nonn`
	AUTHOR	`Max Sills (maxwell.sills(AT)case.edu), Apr 13 2008`

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Last modified February 17 13:11 EST 2012. Contains 206031 sequences.