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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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1, 13, 104, 615, 3149, 14912, 67537, 297945, 1293832, 5564911, 23795465, 101383680, 431003105, 1829784725, 7761645928, 32906509335, 139466630773, 590979780544, 2503927125041, 10608105770625, 44940061502216
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,2
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COMMENTS
| 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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FORMULA
| 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) +1*a(n-6) +1*a(n-7) [Joerg Arndt, Apr 21 2011]
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MATHEMATICA
| 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} ]
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PROG
| (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
| Cf. A120297 = Sum of all matrix elements of n X n matrix M[i, j] = Fibonacci[i+j-1]. Cf. A000045, A045468, A001924, A062381.
Sequence in context: A087398 A080440 A159352 * A023011 A022641 A000590
Adjacent sequences: A129759 A129760 A129761 * A129763 A129764 A129765
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KEYWORD
| nonn
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AUTHOR
| Alexander Adamchuk (alex(AT)kolmogorov.com), May 15 2007, Oct 11 2007
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