US president George Bush will host a state dinner for the Queen and Prince Philip today as the royal tour of the US continues.
The British royals will attend a reception in the state dining room in the White House which will include 134 guests, said to be a "diverse representation of guests from across the country".
A member of the Bush administration will serve as the table host at each of the 13 tables although the royal couple will not sit together. In keeping with White House tradition, couples will be seated at different tables.
Vice president Dick Cheney and secretary of state Condoleezza Rice are expected to be among the guests.
The royal tour began in Virginia last week as the Queen and Prince Philip visited Richmond, Williamsburg and Jamestown to mark the 400th anniversary of the Jamestown settlement.
"The United States has no closer ally and friend than Great Britain," a statement from the White House said.
"Our nations share an exceptionally close relationship based on deep historical and cultural ties, a common language, shared values and interests, and a commitment to defend freedom around the world."
President Bush and his wife Laura visited the Queen in England in November 2003 – hospitality given by the Queen to seven of the president's predecessors.
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