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A285551
Volume of each square prism building the next 3-dimensional box in A100538 where side lengths form the Padovan spiral number sequence (A134816), starting with 1 X 1 X 1, 1 X 1 X 2, 2 X 2 X 2, 2 X 2 X 3, 4 X 4 X 5, ...
2
1, 2, 8, 12, 36, 80, 175, 441, 972, 2304, 5376, 12348, 29008, 67081, 156065, 363350, 843144, 1962396, 4560200, 10600000, 24648975, 57288465, 133194600, 309636096, 719790336, 1673379352, 3890033728, 9043304417, 21023197601, 48872682810, 113615800200, 264124052396
OFFSET
1,2
LINKS
Robert Dickau, Padovan's Spiral Numbers, Wolfram Demonstrations Project Published: August 19, 2010.
FORMULA
a(n) = A000931(n+5)^2*A000931(n+6).
a(n) = A100538(n+1) - A100538(n).
MATHEMATICA
A[n_]:=Sum[Binomial[k, n - 2k], {k, 0, Floor[n/2]}]; a000931[n_]:=If[n==0, 1, If[n<3, 0, A[n - 3]]]; a[n_]:=a000931[n + 5]^2*a000931[n + 6]; Table[a[n], {n, 0, 50}] (* Indranil Ghosh, Apr 26 2017 *)
LinearRecurrence[{1, 2, 3, -2, 4, -4, -1, -1, 0, -1}, {1, 2, 8, 12, 36, 80, 175, 441, 972, 2304}, 40] (* Vincenzo Librandi, Jul 19 2017 *)
PROG
(PARI) A(n) = sum(k=0, n\2, binomial(k, n - 2*k));
a000931(n) = if(n==0, 1, if(n<3, 0, A(n - 3)));
a(n) = a000931(n + 5)^2*a000931(n + 6); \\ Indranil Ghosh, Apr 26 2017
(Python)
from sympy import binomial
def A(n): return sum([binomial(k, n - 2*k) for k in range(int(n/2) + 1)])
def a000931(n): return 1 if n==0 else 0 if n<3 else A(n - 3)
def a(n): return a000931(n + 5)**2*a000931(n + 6) # Indranil Ghosh, Apr 26 2017
CROSSREFS
KEYWORD
nonn
AUTHOR
Peter M. Chema, Apr 25 2017
STATUS
approved