%I #17 Sep 08 2022 08:45:30
%S 1,13,104,615,3149,14912,67537,297945,1293832,5564911,23795465,
%T 101383680,431003105,1829784725,7761645928,32906509335,139466630773,
%U 590979780544,2503927125041,10608105770625,44940061502216
%N Sum of all elements of n X n X n cubic array M[i,j,k] = Fibonacci[i+j+k-2].
%C 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,...}.
%C 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.
%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (9,-26,24,6,-14,1,1).
%F a(n) = Sum[ Sum[ Sum[ Fibonacci[i+j+k-2], {i,1,n} ], {j,1,n} ], {k,1,n} ].
%F a(n) = Fibonacci[3n+4] - 3*Fibonacci[2n+4] + 3*Fibonacci[n+4] - 3.
%F 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
%F 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
%t Table[ Sum[ Sum[ Sum[ Fibonacci[i+j+k-2], {i,1,n} ], {j,1,n} ], {k,1,n} ], {n,1,30} ]
%t Table[ Fibonacci[3n+4] - 3*Fibonacci[2n+4] + 3*Fibonacci[n+4] - 3, {n,1,50} ]
%t LinearRecurrence[{9,-26,24,6,-14,1,1},{1,13,104,615,3149,14912,67537},30] (* _Harvey P. Dale_, Aug 22 2021 *)
%o (Magma) [Fibonacci(3*n+4) - 3*Fibonacci(2*n+4) + 3*Fibonacci(n+4) - 3: n in [1..30]]; // _Vincenzo Librandi_, Apr 21 2011
%Y Cf. A120297 = Sum of all matrix elements of n X n matrix M[i, j] = Fibonacci[i+j-1]. Cf. A000045, A045468, A001924, A062381.
%K nonn,easy
%O 1,2
%A _Alexander Adamchuk_, May 15 2007, Oct 11 2007