OFFSET
1,3
COMMENTS
Prime p divides a(p-1) for p={5,11,19,29,31,41,59,61,71,...} = A038872[n] Primes congruent to {0, 1, 4} mod 5. Also odd primes where 5 is a square mod p. p^2 divides a(p-1) for prime p={11,19,29,31,41,59,61,71,...} = A045468[n] Primes congruent to {1, 4} mod 5. Square prime divisors of a(n) up to n=50 are{2,3,5,7,11,13,19,23,29,31,41,47,89,101,139,151,199,211,461,521,3571,9349}, It appears that square prime divisors of a(n) belong to A061446[n] Primitive part of Fibonacci(n), A001578[n] Smallest primitive prime factor of Fibonacci number F(n) and A072183[n] Sequence arising from factorization of the Fibonacci numbers. Sum[Sum[Fibonacci[i+j-1],{i,1,n}],{j,1,n}] = A120297[n]. Sum[Sum[i+j-1,{i,1,n}],{j,1,n}] = n^3. Sum[Sum[(-1)^(i+j)*(i+j-1),{i,1,n}],{j,1,n}] = n for odd n and = 0 for even n.
LINKS
Colin Barker, Table of n, a(n) for n = 1..1000 [Terms up to n=200 from Vincenzo Librandi]
Index entries for linear recurrences with constant coefficients, signature (3,1,-7,5,-1).
FORMULA
a(n) = Sum[Sum[(-1)^(i+j)*Fibonacci[i+j-1],{i,1,n}],{j,1,n}].
a(n) = 3*a(n-1)+a(n-2)-7*a(n-3)+5*a(n-4)-a(n-5) for n>5. - Colin Barker, Mar 26 2015
G.f.: -x*(x^3+2*x-1) / ((x-1)*(x^2-3*x+1)*(x^2-x-1)). - Colin Barker, Mar 26 2015
EXAMPLE
Matrix begins:
1 -1 2 -3 5
-1 2 -3 5 -8
2 -3 5 -8 13
-3 5 -8 13 -21
5 -8 13 -21 34
MATHEMATICA
Table[Sum[Sum[(-1)^(i+j)*Fibonacci[i+j-1], {i, 1, n}], {j, 1, n}], {n, 1, 50}]
PROG
(PARI) a(n) = sum(i=1, n, sum(j=1, n, (-1)^(i+j)*fibonacci(i+j-1))) \\ Colin Barker, Mar 26 2015
(PARI) Vec(-x*(x^3+2*x-1)/((x-1)*(x^2-3*x+1)*(x^2-x-1)) + O(x^100)) \\ Colin Barker, Mar 26 2015
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Alexander Adamchuk, Aug 03 2006
STATUS
approved