%I A098925
%S A098925 1,1,1,1,2,1,1,3,1,3,4,1,1,6,5,1,4,10,6,1,1,10,15,7,1,5,20,21,8,1,1,15,
%T A098925 35,28,9,1,6,35,56,36,10,1,1,21,70,84,45,11,1,7,56,126,120,55,12,1,1,28,
%U A098925 126,210,165,66,13,1,8,84,252,330,220,78,14,1,1,36,210,462,495,286,91
%N A098925 Distribution of the number of ways for a child to climb a staircase having
r steps (one step or two steps at a time).
%C A098925 Note that the row sums in the example yield the terms of Fibonacci's
sequence(A000045). Were the child capable of taking three steps at
a time, the row sums of the resulting table would add to the tribonacci
sequence (A000073) etc.
%C A098925 Essentially the same as A030528 (without the 0's), where one can find
additional information. - Emeric Deutsch (deutsch(AT)duke.poly.edu),
Mar 29 2005
%D A098925 Mario Livio, The Golden Ratio, (2002) page 100.
%F A098925 a(n) = abs(A092865)
%e A098925 There are 13 ways for the child to climb a staircase with six steps since
the partitions of 6 into 1's and 2's are 222, 2211, 21111 and 111111;
and these can be permuted in 1 + 6 + 5 + 1 = 13 ways.
%e A098925 The general cases can be readily shown by displacing Pascal's Triangle
(A007318) as follows:
%e A098925 1
%e A098925 ..1
%e A098925 ..1..1
%e A098925 .....2..1
%e A098925 .....1..3..1
%e A098925 ........3..4..1
%e A098925 ........1..6..5..1
%p A098925 T:=(n,k)->sum((-1)^(n+i)*binomial(n,i)*binomial(i+k+1,2*k+1),i=0..n):
1,1,seq(seq(T(n,k),k=floor(n/2)..n),n=1..16); (Deutsch)
%Y A098925 Cf. A000045, A000073, A000078, A007318, A092865.
%Y A098925 Cf. A030528.
%Y A098925 Sequence in context: A035667 A102426 A092865 this_sequence A052920 A089141
A003687
%Y A098925 Adjacent sequences: A098922 A098923 A098924 this_sequence A098926 A098927
A098928
%K A098925 easy,nonn
%O A098925 0,5
%A A098925 Alford Arnold (Alford1940(AT)aol.com), Oct 19 2004
%E A098925 More terms from Emeric Deutsch (deutsch(AT)duke.poly.edu), Mar 29 2005
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