A120297 -id:A120297 - cp's OEIS frontend

A119997 Sum of all matrix elements of n X n matrix M[i,j] = (-1)^(i+j)*Fibonacci[i+j-1].

1, 1, 4, 5, 17, 32, 97, 225, 628, 1573, 4225, 10880, 28769, 74849, 196708, 513765, 1347025, 3523360, 9229441, 24154625, 63251156, 165571781, 433507969, 1134881280, 2971250497, 7778684737, 20365103812, 53316141125, 139584105233, 365434903328, 956722661665

Offset: 1

Alexander Adamchuk, Aug 03 2006

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.

			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

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).

Cf. A120297, A000045, A038872, A001924, A062381, A038872, A045468, A061446, A001578, A072183.

Table[Sum[Sum[(-1)^(i+j)*Fibonacci[i+j-1],{i,1,n}],{j,1,n}],{n,1,50}]

a(n) = sum(i=1, n, sum(j=1, n, (-1)^(i+j)*fibonacci(i+j-1))) \\ Colin Barker, Mar 26 2015

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

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

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].

Original entry on oeis.org

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

Alexander Adamchuk, Aug 07 2006

nonn

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.

			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...
...

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

Cf. A120297, A000032, A000045, A045468.

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}]

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]

A129762 Sum of all elements of n X n X n cubic array M[i,j,k] = Fibonacci[i+j+k-2].

Original entry on oeis.org

1, 13, 104, 615, 3149, 14912, 67537, 297945, 1293832, 5564911, 23795465, 101383680, 431003105, 1829784725, 7761645928, 32906509335, 139466630773, 590979780544, 2503927125041, 10608105770625, 44940061502216

Offset: 1

Alexander Adamchuk, May 15 2007, Oct 11 2007

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.

Index entries for linear recurrences with constant coefficients, signature (9,-26,24,6,-14,1,1).

Cf. A120297 = Sum of all matrix elements of n X n matrix M[i, j] = Fibonacci[i+j-1]. Cf. A000045, A045468, A001924, A062381.

[Fibonacci(3*n+4) - 3*Fibonacci(2*n+4) + 3*Fibonacci(n+4) - 3: n in [1..30]]; // Vincenzo Librandi, Apr 21 2011

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 *)

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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A119997 Sum of all matrix elements of n X n matrix M[i,j] = (-1)^(i+j)*Fibonacci[i+j-1].

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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].

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A129762 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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