A129263 Skylar (age 7) counts change by stacking all coins of the same type then arranging the stacks in a row. a(n) is the number of distinct Skylar stackings of n cents using any combination of pennies, nickels, dimes or quarters.
1, 1, 1, 1, 1, 2, 3, 3, 3, 3, 5, 7, 7, 7, 7, 10, 15, 15, 15, 15, 19, 25, 25, 25, 25, 31, 41, 41, 41, 41, 49, 63, 63, 63, 63, 74, 95, 95, 95, 95, 111, 147, 147, 147, 147, 166, 209, 209, 209, 209, 234, 293, 293, 293, 293, 322, 391, 391, 391, 391, 427, 515, 515, 515, 515
Offset: 0
Keywords
Examples
a(16) = 15 = 1+2*4+6*1 since the distinct Skylar stackings of 16 cents are: 16p, 11p1n, 1n11p, 6p2n, 2n6p, 1p3n, 3n1p, 1p1d, 1d1p, 1p1n1d, 1p1d1n, 1n1p1d, 1n1d1p, 1d1p1n, 1d1n1p
References
- Skylar Sutherland, student presentation at "The Undiscovered Country", a course for young mathematicians. Part of MIT's Educational Studies Program.
Crossrefs
Cf. A001299.
Formula
Let A_v(x,y) = 1-y+y/(1-x)^v and A(x,y) = A_1(x,y)A_5(x,y)A_10(x,y)A_25(x,y). Let A^(k)(x,y) denote the k-th partial derivative of A(x,y) w.r.t. y. The generating function of a(n) is A(x) = Sum A^(k)(x,0) for k from 0 to 4.
Comments