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A144115
Total number of Fibonacci parts in all partitions of n.
9
1, 3, 6, 11, 19, 32, 49, 77, 114, 169, 241, 345, 480, 667, 910, 1237, 1656, 2213, 2918, 3840, 5003, 6497, 8368, 10751, 13711, 17441, 22052, 27806, 34879, 43645, 54355, 67535, 83571, 103171, 126907, 155766, 190554, 232629, 283158, 343969, 416716, 503900, 607807
OFFSET
1,2
COMMENTS
a(n) is also the sum of the differences between the sum of f-th largest and the sum of (f+1)-st largest elements in all partitions of n for all Fibonacci parts f. - Omar E. Pol, Oct 27 2012
LINKS
FORMULA
G.f.: Sum_{i>=2} x^Fibonacci(i)/(1 - x^Fibonacci(i)) / Product_{j>=1} (1 - x^j). - Ilya Gutkovskiy, Jan 24 2017
EXAMPLE
From Omar E. Pol, Nov 20 2011 (Start):
For n = 6 we have:
--------------------------------------
. Number of
Partitions Fibonacci parts
--------------------------------------
6 .......................... 0
3 + 3 ...................... 2
4 + 2 ...................... 1
2 + 2 + 2 .................. 3
5 + 1 ...................... 2
3 + 2 + 1 .................. 3
4 + 1 + 1 .................. 2
2 + 2 + 1 + 1 .............. 4
3 + 1 + 1 + 1 .............. 4
2 + 1 + 1 + 1 + 1 .......... 5
1 + 1 + 1 + 1 + 1 + 1 ...... 6
------------------------------------
Total ..................... 32
So a(6) = 32. (End)
MAPLE
b:= proc(n, i) option remember; `if`(n=0, [1, 0], `if`(i<1, 0,
b(n, i-1)+ (p-> p+`if`((t-> issqr(t+4) or issqr(t-4)
)(5*i^2), [0, p[1]], 0))(b(n-i, min(n-i, i)))))
end:
a:= n-> b(n$2)[2]:
seq(a(n), n=1..60); # Alois P. Heinz, Jun 24 2009, revised Aug 06 2019
MATHEMATICA
Clear[b]; b[_] = False; l = {0, 1}; For[k=1, k <= 100, k++, b[l[[1]]] = True; l = {l[[2]], l[[1]] + l[[2]]}]; aa[n_, i_] := aa[n, i] = Module[{g, h}, If[n==0, {1, 0}, If[i==0 || n<0, {0, 0}, g = aa[n, i-1]; h = aa[n-i, i]; {g[[1]] + h[[1]], g[[2]] + h[[2]] + If[b[i], h[[1]], 0]}]]]; a[n_] := aa[n, n][[2]]; Table[a[n], {n, 1, 60}] (* Jean-François Alcover, Jul 30 2015, after Alois P. Heinz *)
KEYWORD
nonn
AUTHOR
Omar E. Pol, Sep 11 2008
EXTENSIONS
More terms from Alois P. Heinz, Jun 24 2009
STATUS
approved