A349904 -id:A349904 - cp's OEIS frontend

A349977 Inverse Euler transform of the classical tribonacci numbers.

0, 1, 1, 1, 3, 4, 8, 13, 23, 38, 68, 114, 201, 343, 600, 1037, 1817, 3157, 5543, 9692, 17047, 29952, 52828, 93157, 164743, 291459, 516679, 916626, 1628684, 2896261, 5156925, 9189769, 16393652, 29268223, 52300907, 93529331, 167390342, 299787639, 537281476

Offset: 1

Peter Luschny, Dec 07 2021

nonn

The classical tribonacci numbers are defined a(n) = a(n-1) + a(n-2) + a(n-3) for n >= 3 with a(0) = 0 and a(1) = a(2) = 1.

See A349904 for the analogous sequence for the shifted tribonacci numbers A000073.

Cf. A349904, A000073.

(* EulerInvTransform is defined in A022562. *)
EulerInvTransform[LinearRecurrence[{1, 1, 1}, {0, 1, 1}, 40]]

# After the Maple program of Alois P. Heinz in A349904.
from functools import cache
from math import comb
def euler_inv_trans(a: callable, len: int):
    @cache
    def h(n: int, k: int):
        if n == 0: return 1
        if k <  1: return 0
        bk = b(k)
        R = range(int(bk == 0), 1 + n // k)
        return sum(comb(bk + j - 1, j) * h(n - k * j, k - 1) for j in R)
    @cache
    def b(n: int): return a(n - 1) - h(n, n - 1)
    return [b(n) for n in range(1, len)]
@cache
def tribonacci(n: int):
    return sum(tribonacci(n - j - 1) for j in range(3)) if n >= 3 else min(n, 1)
print(euler_inv_trans(tribonacci, 40))

A376789 Table read by antidiagonals: T(n,k) is the number of Lyndon words of length k on the alphabet {0,1} whose prefix is the bitwise complement of the binary expansion of n with n >= 1 and k >= 1.

Original entry on oeis.org

1, 1, 0, 2, 1, 0, 3, 1, 0, 0, 6, 1, 1, 0, 0, 9, 2, 2, 1, 0, 0, 18, 2, 4, 1, 0, 0, 0, 30, 4, 7, 1, 0, 1, 0, 0, 56, 5, 14, 1, 1, 1, 0, 0, 0, 99, 8, 25, 2, 1, 2, 1, 0, 0, 0, 186, 11, 48, 2, 2, 3, 2, 1, 0, 0, 0, 335, 18, 88, 3, 3, 6, 4, 1, 0, 0, 0, 0

Offset: 1

Peter Kagey, Oct 04 2024

T(n,k) = 0 if n is in A366195.
Row 1 is A059966.
Row 2 is A006206 for n > 1.
Row 3 is A065491 for n > 2.
Row 4 is A065417.
Row 6 is A349904.

			Table begins
n\k| 1  2  3  4  5  6   7   8   9  10   11   12
---+-------------------------------------------
 1 | 1, 1, 2, 3, 6, 9, 18, 30, 56, 99, 186, 335
 2 | 0, 1, 1, 1, 2, 2,  4,  5,  8, 11,  18,  25
 3 | 0, 0, 1, 2, 4, 7, 14, 25, 48, 88, 168, 310
 4 | 0, 0, 1, 1, 1, 1,  2,  2,  3,  4,   6,   7
 5 | 0, 0, 0, 0, 1, 1,  2,  3,  5,  7,  12,  18
 6 | 0, 0, 1, 1, 2, 3,  6, 10, 18, 31,  56,  96
 7 | 0, 0, 0, 1, 2, 4,  8, 15, 30, 57, 112, 214
 8 | 0, 0, 0, 1, 1, 1,  1,  1,  2,  2,   3,   3
 9 | 0, 0, 0, 0, 0, 0,  1,  1,  1,  2,   3,   4
10 | 0, 0, 0, 0, 1, 1,  2,  3,  5,  7,  12,  18
11 | 0, 0, 0, 0, 0, 0,  0,  0,  0,  0,   0,   0
12 | 0, 0, 0, 1, 1, 2,  3,  5,  9, 15,  26,  43
T(6,5) = 2 because 6 is 110 in base 2, its bitwise complement is 001, and there are T(6,5) = 2 length-5 Lyndon words that begin with 001: 00101 and 00111.

Cf. A059966, A006206, A065491, A065417, A349904.

Cf. A365746.

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A349977 Inverse Euler transform of the classical tribonacci numbers.

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