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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.

%I A129263 #2 Sep 24 2013 09:23:05
%S A129263 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,
%T A129263 41,41,49,63,63,63,63,74,95,95,95,95,111,147,147,147,147,166,209,209,
%U A129263 209,209,234,293,293,293,293,322,391,391,391,391,427,515,515,515,515
%N 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.
%C A129263 Sequence definition and Scratch program to compute the 100 terms due to Skylar Sutherland. Generating function contributed by Andrew V. Sutherland. Related to A001299, but distinguishes permutations of coin types.
%D A129263 Skylar Sutherland, student presentation at "The Undiscovered Country", a course for young mathematicians. Part of MIT's Educational Studies Program.
%F A129263 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.
%e A129263 a(16) = 15 = 1+2*4+6*1 since the distinct Skylar stackings of 16 cents are:
%e A129263 16p, 11p1n, 1n11p, 6p2n, 2n6p, 1p3n, 3n1p, 1p1d, 1d1p, 1p1n1d, 1p1d1n, 1n1p1d, 1n1d1p, 1d1p1n, 1d1n1p
%Y A129263 Cf. A001299.
%K A129263 nonn
%O A129263 0,6
%A A129263 _Andrew V. Sutherland_, Aug 20 2007