Using “Pascal’s” triangle to sum kth powers of consecutive integers Al-Bahir fi'l Hisab (Shining Treatise on Calculation), alSamaw'al, Iraq, 1144 Siyuan Yujian (Jade Mirror of the Four Unknowns), Zhu Shijie, China, 1303 Maasei Hoshev (The Art of the Calculator), Levi ben Gerson, France, 1321 Ganita Kaumudi (Treatise on Calculation), Narayana Pandita, India, 1356 1 x 0 1 1 x 1 1 x 1 x 2 1 2x x 2 1 x 1 3x 3x x 1 x 1 4x 6x 2 4x 3 x 4 1 x 1 5x 10x 10 x 5x x 3 4 5 2 2 3 3 4 5 1 x 0 1 1 x 1 1 x 1 x 2 1 2x x 2 1 x 1 3x 3x x 1 x 1 4x 6x 2 4x 3 x 4 1 x 1 5x 10x 10 x 5x x 3 4 5 2 2 3 3 4 5 1 x 0 1 1 x 1 1 x 1 x 2 1 2x x 2 1 x 1 3x 3x x 1 x 1 4x 6x 2 4x 3 x 4 1 x 1 5x 10x 10 x 5x x 3 4 5 2 2 k k 1 k 2 k k k 3 3 4 5 n n 1 k k 1 1 2 k k 0 0 k 1 k k k 0 n n 1 k k 1 1 2 k k 0 0 k 1 k k k n n 1 k k 1 0 Note that the binomial coefficient j choose k is a polynomial in j of degree k. 1 j Pk j j j 1 j 2 k k! j k 1 All the coefficients are positive integers. Can we find a simple way of generating them? Can we discover what they count? HP(k,i ) is the House-Painting number 1 2 3 5 6 7 4 8 It is the number of ways of painting k houses using exactly i colors. j j j HP(k,k) HP(k,k 1) k k 1 k j HP(k,k 2) k 2 j HP(k,1) 1 j k is the number of ways of painting k houses when we have j colors to choose from at each house, and we don' t care whether or not all the colors are used. 1 2 1 6 6 HP(k,k) k! 1 24 36 14 1 HP(k,1) 1 HP(k,i ) is the House-Painting number 1 2 3 5 6 7 4 8 HP(k,i) i HPk 1,i HPk 1,i 1 1 HP(k,k) k! 2 1 6 6 1 24 36 14 + X4 HP(k,1) 1 1 120 240 150 30 1 1 HP(k,k) k! 2 1 6 6 1 24 36 14 + HP(k,1) 1 X3 1 120 240 150 30 1 1 HP(k,k) k! 2 1 6 6 1 24 36 14 HP(k,1) 1 + 1 X2 120 240 150 30 1 1 1 2 1 6 6 1 24 36 14 1 120 240 150 30 1 1 1 1 3 1 1 6 7 1 10 25 1 15 1 HP(k,i) is always divisible by i! (number of ways of permuting the colors) HP(k,i) / i! = S(k,i) = Stirling number of the second kind 1 2 k k 0 0 k 1 k k k 0 n n 1 k k 1 1 n 1nn 1n 2 n 3 5 3 n 1nn 1n 2 2 7 n 1n n 1 3 1 n 1n 2