Euler was born on this date in 1707, and I'd like to propose in his honor a problem to be interpreted and solved in his spirit: find the sum of all the entries in Pascal's triangle. To those of you who object that the sum diverges, I'll point out that Euler was the guy who calculated 0!-1!+2!-3!+.... So don't give me any of that "But-the-sum-diverges!" stuff. To those of you who say "Well, the sum of the nth row (counting the top row as the n=0 row) is 2^n, so the sum of all the entries is the sum of the geometric series 1+2+4+8+..., which in Euler's formalistic mathematics is 1/(1-2) = -1; end of story," I'll say "That's good, except for the last three words," and then point out that Euler was the guy who calculated 0!-1!+2!-3!+... not just one way but half a dozen ways (see http://www.maa.org/editorial/euler/How%20Euler%20Did%20It%2032%20divergent%2...) and got the same answer every time. So while the geometric series calculation is fine, we want to see other ways to arrive at the answer -1. Here's another way to get -1: Decompose Pascal's triangle into tilted rows as indicated below (where the elements of distinct tilted rows get assigned different letters): a c b e d c g f e d i h g f e etc. Then we get a = 1, b = 1, c = 2, d = 3, e = 5, f = 8, etc.: the familiar way to write Fibonacci numbers as sums of binomial coefficients. Now 1+1+2+3+5+... is the x=1 evaluation of the power series 1+x+2x^2+3x^3+5x^4+... = 1/(1-x-x^2), and plugging in x=1 gives 1/(1-1-1) = -1. If you use other tilted rows, you get other rational functions playing the role of 1/(1-x-x^2), such as 1/(1-x-x^3) and 1/(1-x^2-x^3), but for every such rational function that I've tried, plugging in x=1 gives -1. Does every decomposition into tilted rows give -1? (This is really a question about one-parameter specializations of the two-variable generating function associated with the rational function 1/(1-x-y), and is probably not that hard, but I haven't done it; maybe one of you will.) Here's a start of a different way to calculate the sum that I don't know how to push through (and this is my real reason for posting): Looking down the middle of the array, we get the sum 1+2+6+20+..., which is the x=1 evaluation of the power series 1/sqrt(1-4x), which "is" 1/sqrt(-3). On either side we get the sum 1+3+10+35+..., with combined sum 2+6+20+..., which is 1/sqrt(-3) - 1. On either side of that we get ... Can anyone finish this calculation? All the columns I've examined look "sqrt(1-4x)-ish", but I don't see the unifying trick for getting all of them. And then there are the diagonals 1+1+1+1+..., 1+2+3+4+..., 1+3+6+10+..., etc. Euler knew about zeta function regularization and used it to evaluate 1+2+3+4+... as -1/12, and perhaps the trick extends to 1+3+6+10+... etc. That still leaves 1+1+1+1+..., which even Euler steered clear of, but perhaps there's a principled way of absorbing those infinitely many 1's into other sums. Jim Propp