A001924 -id:A001924 - cp's OEIS frontend

A327990 The Fibonacci Codes. Irregular triangle T(n, k) with n >= 0 and 0 <= k < A000045(n+1).

0, 1, 3, 1, 7, 3, 1, 9, 15, 7, 3, 1, 19, 17, 31, 9, 15, 7, 3, 1, 21, 39, 35, 33, 63, 19, 17, 31, 9, 15, 7, 3, 1, 43, 41, 79, 37, 71, 67, 65, 127, 21, 39, 35, 33, 63, 19, 17, 31, 9, 15, 7, 3, 1

Offset: 0

Peter Luschny, Oct 08 2019

The Fibonacci codes are binary strings enumerated in an irregular triangle FC(n, k). The first few are shown below in the Example section.
The Fibonacci codes are for n > 1 defined recursively FC(n) = C(n) concatenated with FC(n-1), where C(n) are the conjugates of the compositions of n which do not have '1' as a part and the parts of which were reduced by 1. The recurrence is based in FC(0) = '' (empty string) and FC(1) = '0'.
The Fibonacci numbers are defined F(n) = A309896(2,n) = A000045(n+1) for n >= 0. Row FC(n) contains F(n) codes. A nonzero code is a code that does not consist entirely of zeros. The number of nonzero codes in row n is A001924(n-3) for n>=3.
Fibonacci codes are represented here through
    T(n, k) = Sum_{j=0..m} (c[j] + 1)*2^j,
where c = FC(n, k) and m = length(FC(n, k)).

			The Fibonacci codes start:
[0] [[]]
[1] [[0]]
[2] [[00][0]]
[3] [[000][00][0]]
[4] [[010][0000][000][00][0]]
[5] [[0010][0100][00000][010][0000][000][00][0]]
[6] [[0110][00010][00100][01000][000000][0010][0100][00000][010][0000][000][00][0]]
[7] [[00110][01010][000010][01100][000100][001000][010000][0000000][0110][00010][00100][01000][000000][0010][0100][00000][010][0000][000][00][0]]
The encoding of the Fibonacci codes start:
[0] [0]
[1] [1]
[2] [3, 1]
[3] [7, 3, 1]
[4] [9, 15, 7, 3, 1]
[5] [19, 17, 31, 9, 15, 7, 3, 1]
[6] [21, 39, 35, 33, 63, 19, 17, 31, 9, 15, 7, 3, 1]
[7] [43, 41, 79, 37, 71, 67, 65, 127, 21, 39, 35, 33, 63, 19, 17, 31, 9, 15, 7, 3, 1]

Cf. A309896, A000045, A001924.

@cached_function
def FibonacciCodes(n):
    if n == 0 : return [[]]
    if n == 1 : return [[0]]
    A = [c.conjugate() for c in Compositions(n) if not(1 in c)]
    B = [[i-1 for i in a] for a in A]
    return B + FibonacciCodes(n-1)
def A327990row(n):
    FC = FibonacciCodes(n)
    B = lambda C: sum((c+1)*2^i for (i, c) in enumerate(C))
    return [B(c) for c in FC]
for n in (0..6): print(A327990row(n))

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

A327991 The complementary Fibonacci codes. Irregular triangle T(n, k) with n >= 0 and 0 <= k < A000045(n+1).

Original entry on oeis.org

1, 2, 2, 6, 2, 6, 30, 2, 6, 30, 10, 210, 2, 6, 30, 10, 210, 42, 70, 2310, 2, 6, 30, 10, 210, 42, 70, 2310, 14, 330, 462, 770, 30030, 2, 6, 30, 10, 210, 42, 70, 2310, 14, 330, 462, 770, 30030, 66, 110, 2730, 154, 4290, 6006, 10010, 510510

Offset: 0

Peter Luschny, Oct 09 2019

The complementary Fibonacci codes are binary strings enumerated in an irregular triangle CF(n, k). The first few are shown below in the Example section. The complementary Fibonacci codes are the bitwise complements of the Fibonacci codes described in A327990, in ascending order.
The complementary Fibonacci codes are represented here through
    T(n, k) = Product_{j=0..m} p(j)^c(j),
where p(j) is the j-th prime number, c = CF(n, k) and m = length(CF(n, k)).

			The complementary Fibonacci codes start:
[0] [[]]
[1] [[1]]
[2] [[1][11]]
[3] [[1][11][111]]
[4] [[1][11][111][101][1111]]
[5] [[1][11][111][101][1111][1101][1011][11111]]
[6] [[1][11][111][101][1111][1101][1011][11111][1001][11101][11011][10111][111111]]
[7] [[1][11][111][101][1111][1101][1011][11111][1001][11101][11011][10111][111111] [11001][10101][111101][10011][111011][110111][101111][1111111]]
The representation of the complementary Fibonacci codes starts:
[0] [1]
[1] [2]
[2] [2, 6]
[3] [2, 6, 30]
[4] [2, 6, 30, 10, 210]
[5] [2, 6, 30, 10, 210, 42, 70, 2310]
[6] [2, 6, 30, 10, 210, 42, 70, 2310, 14, 330, 462, 770, 30030]
[7] [2, 6, 30, 10, 210, 42, 70, 2310, 14, 330, 462, 770, 30030, 66, 110, 2730, 154, 4290, 6006, 10010, 510510]

The diagonal is A002110 (primorial numbers).

Cf. A309896, A327990, A000045, A001924.

@cached_function
def FibonacciCodes(n):
    if n == 0 : return [[]]
    if n == 1 : return [[1]]
    A = [c.conjugate() for c in Compositions(n) if not(1 in c)]
    return FibonacciCodes(n-1) + [[2-i for i in a] for a in A]
def A327991row(n):
    P = Primes()
    M = lambda C: mul(P[i]^c for (i, c) in enumerate(C))
    return [M(c) for c in FibonacciCodes(n)]
for n in (0..7): print(A327991row(n))

A335184 a(n) is the number of subsets of {1,2,...,n} with at least two elements and the difference between successive elements at least 6.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 3, 6, 10, 15, 21, 29, 40, 55, 75, 101, 134, 176, 230, 300, 391, 509, 661, 856, 1106, 1427, 1840, 2372, 3057, 3938, 5070, 6524, 8392, 10793, 13880, 17849, 22951, 29508, 37934, 48762, 62678, 80564, 103553, 133100, 171074, 219877, 282597, 363204, 466801, 599946, 771066, 990990

Offset: 0

Enrique Navarrete, May 25 2020

nonn

For n >= 6 the sequence contains the triangular numbers; for n >= 12 we have to add the tetrahedral numbers; for n >= 18 we have to add the numbers binomial(n,4) (starting with 0,1,5,...); for n >= 24 we have to add the numbers binomial(n,5) (starting with 0,1,6,..); in general, for n >= 6*k we have to add to the sequence the numbers binomial(n, k+1), k >= 1.
For example, a(26) = 1106 = 210+560+330+6, where 210 is a triangular number, 560 is a tetrahedral number, 330 is a number binomial(n,4) and 6 is a number binomial(m,5) (with the proper n, m due to shifts in names of the sequences).
The sequence counts sets with more than 2 elements such as {1,7,14}, {1,8,14}, {2,8,14,20}, etc. The first 3-element set is {1,7,13}, the first 4-element set is {1,7,13,19}, etc. Every time a larger set needs to be counted is when we have to add a term binomial(n, k+1).

			a(11) = 15 and the 15 subsets of {1,2,...11} with at least two elements and whose difference between successive elements is at least 6 are: {1,7}, {1,8}, {1,9}, {1,10}, {1,11}, {2,8}, {2,9}, {2,10}, {2,11}, {3,9}, {3,10}, {3,11}, {4,10}, {4,11}, {5,11}.

Index entries for linear recurrences with constant coefficients, signature (3,-3,1,0,0,1,-2,1).

Similar sequences with minimum difference 1..5 are A000295, A001924, A050228, A145131, A330910.

With[{k = 6}, Array[Count[Subsets[Range[# + k], {2, # + k}], ?(AllTrue[Differences@ #, # >= k &] &)] &, 16]] (* _Michael De Vlieger, Jun 26 2020 *)
LinearRecurrence[{3,-3,1,0,0,1,-2,1},{0,0,0,0,0,0,0,1},60] (* Harvey P. Dale, Nov 22 2022 *)

a(n) = {my(d=6); sum(k=0, (n-1)\d, binomial(n-d*k+k+1, k+2))} \\ Andrew Howroyd, Aug 11 2020

a(n) = Sum_{k=0..floor((n-1)/6)} binomial(n-6*k+k+1, k+2). - Andrew Howroyd, Aug 11 2020
From Colin Barker, May 26 2020: (Start)
G.f.: x^7 / ((1 - x)^2*(1 - x - x^6)).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + a(n-6) - 2*a(n-7) + a(n-8) for n>=8.
(End)

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A327990 The Fibonacci Codes. Irregular triangle T(n, k) with n >= 0 and 0 <= k < A000045(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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A327991 The complementary Fibonacci codes. Irregular triangle T(n, k) with n >= 0 and 0 <= k < A000045(n+1).

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SageMath

A335184 a(n) is the number of subsets of {1,2,...,n} with at least two elements and the difference between successive elements at least 6.

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