A120537 Sum of all matrix elements of n X n matrix M[i,j] = Lucas[i+j-1], (i,j = 1..n), where Lucas[n] = A000032[n] = Fibonacci[n-1] + Fibonacci[n+1].

1, 11, 44, 145, 431, 1216, 3329, 8955, 23836, 63041, 166079, 436480, 1145441, 3003211, 7869644, 20614545, 53988271, 141373376, 370169249, 969194875, 2537513276, 6643503361, 17393253119, 45536670720, 119217430081

Offset: 1

Author

Alexander Adamchuk, Aug 07 2006

Keywords

• nonn

Comments

5 divides a(4k). a(n) is prime for n = {2,5,7,17,19,31,439,545,...}. p^2 divides a(p-1) for p = {11,19,29,31,41,59,61,71,...} = A045468[n] Primes congruent to {1, 4} mod 5, also odd primes where 5 is a square mod p except 5. Square prime divisors of a(n) up to n=70 are p = {2,3,7,11,13,19,23,29,31,41,47,59,61,71,89,101,139,151,199,233,281,461,521,911,1597,2207,3571,5779,9349,9901,19489,3010349,...} that appear to be the prime factors of Fibonacci numbers.

Examples

			Matrix begins:
1 3 4 7 11...
3 4 7 11 18...
4 7 11 18 29...
7 11 18 29 47...
11 18 29 47 76...
...
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..200

Crossrefs

Cf. A120297, A000032, A000045, A045468.

Programs

Mathematica

Table[Sum[Sum[Fibonacci[i+j-2]+Fibonacci[i+j],{i,1,n}],{j,1,n}],{n,1,70}] Table[(Fibonacci[2n+2]+Fibonacci[2n+4])-2(Fibonacci[n+2]+Fibonacci[n+4])+4,{n,1,70}]

Formula

a(n) = Sum[ Sum[ Fibonacci[i+j-2] + Fibonacci[i+j],{i,1,n}],{j,1,n}]. a(n) = Lucas[2n+3] - 2*Lucas[n+3] + 4, where Lucas[k] = Fibonacci[k-1] + Fibonacci[k+1].

G.f.:(1+x^3-4*x^2+6*x)/((x-1)*(x^2+x-1)*(x^2-3*x+1)) [From Maksym Voznyy (voznyy(AT)mail.ru), Aug 14 2009]