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A129762
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Sum of all elements of n X n X n cubic array M[i,j,k] = Fibonacci[i+j+k-2].
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0
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1, 13, 104, 615, 3149, 14912, 67537, 297945, 1293832, 5564911, 23795465, 101383680, 431003105, 1829784725, 7761645928, 32906509335, 139466630773, 590979780544, 2503927125041, 10608105770625, 44940061502216
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OFFSET
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1,2
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COMMENTS
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p^3 divides a(p-1) for prime p = {11,19,29,31,41,59,61,71,79,89,...} = A045468 Primes congruent to {1, 4} mod 5; also primes p that divide Fibonacci(p-1). a(n) is prime for n = {2,7,19,...}.
a(n) is prime for n = {2, 7, 19, 47, 175, 179, ...}. The formula a(n) = F(3n+4) - 3F(2n+4) + 3F(n+4) - 3 and its generalization for k-dimensional hypercubes with elements M(i,j,...) = F(i+j+...-k+1) was stated and proved by the user 1istik_figi in private communication at LiveJournal on Oct 10 2007. The k-dimensional formula is a(n) = Sum[(-1)^i*Binomial[k,i]*Fibonacci[(k-i)*n+k+1],{i,0,k}]. Conjecture: if prime p divides F(p-1) then p^k divides a(n) in k-dimensional case.
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LINKS
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FORMULA
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a(n) = Sum[ Sum[ Sum[ Fibonacci[i+j+k-2], {i,1,n} ], {j,1,n} ], {k,1,n} ].
a(n) = Fibonacci[3n+4] - 3*Fibonacci[2n+4] + 3*Fibonacci[n+4] - 3.
a(n) = 9*a(n-1) - 26*a(n-2) + 24*a(n-3) + 6*a(n-4) - 14*a(n-5) + a(n-6) + a(n-7). - Joerg Arndt, Apr 21 2011
G.f.: -x*(x^5 - 7*x^3 + 13*x^2 + 4*x + 1)/((x-1)*(x^2 - 3*x + 1)*(x^2 + x - 1)*(x^2 + 4*x - 1)). - Colin Barker, Aug 10 2012
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MATHEMATICA
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Table[ Sum[ Sum[ Sum[ Fibonacci[i+j+k-2], {i, 1, n} ], {j, 1, n} ], {k, 1, n} ], {n, 1, 30} ]
Table[ Fibonacci[3n+4] - 3*Fibonacci[2n+4] + 3*Fibonacci[n+4] - 3, {n, 1, 50} ]
LinearRecurrence[{9, -26, 24, 6, -14, 1, 1}, {1, 13, 104, 615, 3149, 14912, 67537}, 30] (* Harvey P. Dale, Aug 22 2021 *)
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PROG
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(Magma) [Fibonacci(3*n+4) - 3*Fibonacci(2*n+4) + 3*Fibonacci(n+4) - 3: n in [1..30]]; // Vincenzo Librandi, Apr 21 2011
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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