A003893 -id:A003893 - cp's OEIS frontend

A072709 Integers m such that the last digit of Fibonacci(m) is 7.

14, 16, 17, 23, 34, 37, 43, 56, 74, 76, 77, 83, 94, 97, 103, 116, 134, 136, 137, 143, 154, 157, 163, 176, 194, 196, 197, 203, 214, 217, 223, 236, 254, 256, 257, 263, 274, 277, 283, 296, 314, 316, 317, 323, 334, 337, 343, 356, 374, 376, 377, 383, 394, 397, 403

Offset: 1

Benoit Cloitre, Aug 07 2002

Sequence contains numbers of form (14+60k), (16+60k), (17+60k), (23+60k), (24+60k), (34+60k), (37+60k), (43+60k), (56+60k), with k>=0.

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

Cf. A000045, A003893.

Select[Range[500],Last[IntegerDigits[Fibonacci[#]]]==7&] (* or *) LinearRecurrence[{1,0,0,0,0,0,0,1,-1},{14,16,17,23,34,37,43,56,74},60] (* Harvey P. Dale, Sep 12 2014 *)

G.f.: x*(4*x^8+13*x^7+6*x^6+3*x^5+11*x^4+6*x^3+x^2+2*x+14) / (x^9-x^8-x+1). - Colin Barker, Jun 16 2013

A072711 Integers m such that the last digit of Fibonacci(m) is 9.

Original entry on oeis.org

11, 29, 31, 32, 38, 49, 52, 58, 71, 89, 91, 92, 98, 109, 112, 118, 131, 149, 151, 152, 158, 169, 172, 178, 191, 209, 211, 212, 218, 229, 232, 238, 251, 269, 271, 272, 278, 289, 292, 298, 311, 329, 331, 332, 338, 349, 352, 358, 371, 389, 391, 392, 398, 409

Offset: 1

Benoit Cloitre, Aug 07 2002

Sequence contains numbers of form (11+60k), (29+60k), (31+60k), (32+60k), (38+60k), (49+60k), (52+60k), (58+60k), with k>=0.

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

Cf. A000045, A003893.

Select[Range[500],Last[IntegerDigits[Fibonacci[#]]]==9&] (* or *) LinearRecurrence[ {1,0,0,0,0,0,0,1,-1},{11,29,31,32,38,49,52,58,71},60] (* Harvey P. Dale, Sep 06 2015 *)

G.f.: x*(2*x^8+6*x^7+3*x^6+11*x^5+6*x^4+x^3+2*x^2+18*x+11) / (x^9-x^8-x+1). - Colin Barker, Jun 16 2013

A137290 Fibonacci(n) mod 30.

Original entry on oeis.org

1, 1, 2, 3, 5, 8, 13, 21, 4, 25, 29, 24, 23, 17, 10, 27, 7, 4, 11, 15, 26, 11, 7, 18, 25, 13, 8, 21, 29, 20, 19, 9, 28, 7, 5, 12, 17, 29, 16, 15, 1, 16, 17, 3, 20, 23, 13, 6, 19, 25, 14, 9, 23, 2, 25, 27, 22, 19, 11, 0, 11, 11, 22, 3, 25, 28, 23, 21, 14, 5, 19, 24, 13, 7, 20, 27, 17, 14

Offset: 1

Aaron M. Churchill (churchil(AT)math.udel.edu), Mar 15 2008

nonn

Has period 120.

Michel Marcus, Table of n, a(n) for n = 1..130
Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1).

Cf. A000045, A003893, A007887, A079345, A105471.

Mod[Fibonacci[Range[80]],30] (* Harvey P. Dale, Sep 12 2022 *)

a(n) = fibonacci(n) % 30 \\ Michel Marcus, Jun 12 2013

A181190 Maximal length of chain-addition sequence mod 10 with window of size n.

Original entry on oeis.org

1, 60, 124, 1560, 4686, 1456, 18744, 585936, 4882810, 212784

Offset: 1

Alexander Dashevsky (atanvarnoalda(AT)gmail.com), Oct 10 2010

Chain addition mod 10 with window n: take an n-digit 'seed'. Take the sum of its digits mod 10 and append to the seed. Repeat with the last n digits of the string, until the seed appears again.
This sequence shows the lengths of the longest sequences for different window sizes.
a(1)-a(10) all occur for seed 1 (among others). If this is always true, the sequence continues: 406224, 12695306, 4272460934, 380859180, 122070312496, 518798826, 3433227539058. - Lars Blomberg, Feb 12 2013
Comment from Michel Lagneau, Jan 20 2017, edited by N. J. A. Sloane, Jan 24 2017: (Start)
If seed 1 is always as good as or better than any other, as seems likely, then this sequence has the following alternative description.
Consider the n initial terms of an infinite sequence S(k, n) of decimal digits given by 0, 0,..., 0, 1. The succeeding terms are given by the final digits in the sum of the n immediately preceding terms. The sequence lists the period of each sequence corresponding to n = 2, 3, ...
a(2) = period of A000045 mod 10 (Fibonacci numbers mod 10) = A001175(10).
a(3) = period of A000073 mod 10 (tribonacci numbers mod 10) = A046738(10).
a(4) = period of A000078 mod 10 (tetranacci numbers mod 10) = A106295(10).
a(5) = period of A001591 mod 10 (pentanacci numbers mod 10) = A106303(10).
a(6) = period of A001592 mod 10 (hexanacci numbers mod 10).
a(7) = period of A122189 mod 10 (heptanacci numbers mod 10).
a(8) = period of A079262 mod 10 (octanacci numbers mod 10).
a(4) = 1560 because the four initial terms 0, 0, 0, 1 => S(k, 4) = 0, 0, 0, 1, 1, 2, 4, 8, 5, 9, 6, 8, 8, 1, 3, 0, 2, 6, 1, 9, 8, ... (tetranacci numbers mod 10). This sequence is periodic with period 1560:
S(1560 + 1, 4) = S(1, 4) = 0,
S(1560 + 2, 4) = S(2, 4) = 0,
S(1560 + 3, 4) = S(3, 4) = 0,
S(1560 + 4, 4) = S(4, 4) = 1.
(End)

			For n=2, the longest sequence begins with '01' (among others):
01123583145943707741561785381909987527965167303369549325729101.
It is 60 digits long (not counting the second '01' at the end).
For n=3, one of the longest sequences begins again with '001':
00112473441944756893025746770415061742394699425184352079627546556679289964992013
48570291225960516297849144970639807524172091001 (124 digits long without the second '001').

Cf. A000045, A000073, A000078, A001591, A001592, A003893, A079262, A122189

a(8)-a(10) from Lars Blomberg, Feb 12 2013

A248740 a(n) = Fibonacci(n) mod 1000.

Original entry on oeis.org

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 597, 584, 181, 765, 946, 711, 657, 368, 25, 393, 418, 811, 229, 40, 269, 309, 578, 887, 465, 352, 817, 169, 986, 155, 141, 296, 437, 733, 170, 903, 73, 976, 49, 25, 74, 99, 173, 272

Offset: 0

Franz Vrabec, Oct 13 2014

nonn

The sequence is periodic with period 1500 = A001175(1000).
A number m of {0, 1, ..., 999} is not in the range of this sequence, iff m is congruent to 4 or 6 mod 8.
These numbers are the 250 = 1000 - A066853(1000) numbers of the set {4, 6, 12, 14, ..., 996, 998}. E.g., a Fibonacci number will never end in the digits '100'.

			a(17) = (a(16) + a(15)) mod 1000 = (987 + 610) mod 1000 = 1597 mod 1000 = 597.

Alois P. Heinz, Table of n, a(n) for n = 0..3000
Index entries for linear recurrences with constant coefficients, order 1500.

Cf. A000045, A003893, A105471.

[Fibonacci(n) mod 1000: n in [0..80]]; // Vincenzo Librandi, Oct 17 2014

a:= proc(n) option remember;
      `if`(n<2, n, irem(a(n-1)+a(n-2), 1000))
    end:
seq(a(n), n=0..60);  # Alois P. Heinz, Oct 18 2015

vector(100,n,fibonacci(n-1)%1000) \\ Derek Orr, Oct 17 2014

a(n) = (a(n-1) + a(n-2)) mod 1000 for n>1, a(0) = 0, a(1) = 1.

More terms from Vincenzo Librandi, Oct 17 2014

A256985 Define a sequence {b(i), i >= 0} by b(i) = 1 for 0 <= i <= n, thereafter b(i+1) = (b(i)+b(i-n)) mod 10; consider the sequence of n-tuples [b(i-n+1), b(i-n+2),...,b(i)]; this repeats with period a(n).

Original entry on oeis.org

60, 217, 520, 42, 196812, 2480437, 2232, 7128815, 1736327236, 124516392, 203450520, 40193528485, 14417724597564, 22856442972, 324145501174, 7946757, 193726348876699204, 206135768515040, 581179046630097612, 32289695739703771, 275114595439871720

Offset: 1

N. J. A. Sloane, Apr 26 2015

nonn

			For n=1 we get the Fibonacci sequence (starting 1,1,2,3,...), A000045, which read mod 10 repeats with period 60 (see A003893), so a(1)=60.  The full period in this case is:
[1, 1, 2, 3, 5, 8, 3, 1, 4, 5, 9, 4, 3, 7, 0, 7, 7, 4, 1, 5, 6, 1, 7, 8, 5, 3, 8, 1, 9, 0, 9, 9, 8, 7, 5, 2, 7, 9, 6, 5, 1, 6, 7, 3, 0, 3, 3, 6, 9, 5, 4, 9, 3, 2, 5, 7, 2, 9, 1, 0].

Hiroaki Yamanouchi, Table of n, a(n) for n = 1..105
Popular Computing (Calabasas, CA), Contest 9, Vol. 4 (No. 40, July 1976), scanned copy of page PC40-1.
Popular Computing (Calabasas, CA), Contest 9, Vol. 4 (No. 40, July 1976), scanned copy of page PC40-3.

Cf. A000045, A003893.

from itertools import accumulate # requires python 3.2 or higher
def A256985(n):
    ilist, k = [1]*(n+1), 1
    jlist = [d % 10 for d in accumulate(ilist)]
    jlist = [jlist[-1]]+ jlist[:-1]
    while ilist != jlist:
        k += 1
        jlist = [d % 10 for d in accumulate(jlist)]
        jlist = [jlist[-1]]+ jlist[:-1]
    return k # Chai Wah Wu, Apr 30 2015

a(6)-a(11) from Chai Wah Wu, Apr 30 2015

a(12)-a(21) from Hiroaki Yamanouchi, May 04 2015

A341414 a(n) = (Fibonacci(n)*Lucas(n)) mod 10.

Original entry on oeis.org

0, 1, 3, 8, 1, 5, 4, 7, 7, 4, 5, 1, 8, 3, 1, 0, 9, 7, 2, 9, 5, 6, 3, 3, 6, 5, 9, 2, 7, 9, 0, 1, 3, 8, 1, 5, 4, 7, 7, 4, 5, 1, 8, 3, 1, 0, 9, 7, 2, 9, 5, 6, 3, 3, 6, 5, 9, 2, 7, 9, 0, 1, 3, 8, 1, 5, 4, 7, 7, 4, 5, 1, 8, 3, 1, 0, 9, 7, 2, 9, 5, 6, 3, 3, 6, 5, 9, 2, 7, 9

Offset: 0

Jens Rasmussen, Feb 11 2021

Fibonacci starting with 0,1 and Lucas starting with 2,1.
Blocks of 30 numbers with 10 even and 20 uneven numbers.
Symmetric as a(7-i)=a(8+i) for i=1,2,...,6, and a(22-j)=a(23+j) for j=1..21.
Decimal expansion of 13801675776055042253380279/999000999000999000999000999. - Jianing Song, Apr 04 2021

			For n=5: a(5) = (Fibonacci(5)*Lucas(5)) mod 10 = (5*11) mod 10 = 55 mod 10 = 5.

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

Cf. A000032, A000045, A001906, A130893.

Bisection of A003893.

Table[Mod[Fibonacci@n*LucasL@n, 10], {n, 0, 100}] (* Giorgos Kalogeropoulos, Mar 31 2021 *)

a(n) = fibonacci(2*(n%30)) % 10 \\ Jianing Song, Apr 04 2021

a(n) = (Fibonacci(n)*Lucas(n)) mod 10 = Fibonacci(2*n) mod 10 using Binet's formula for Fibonacci and corresponding formula for Lucas.
a(n) = a(n-30).
a(n) = a(n-3) - a(n-6) + a(n-9) - a(n-12) + a(n-15) - a(n-18) + a(n-21) - a(n-24) + a(n-27).
a(n) = A003893(2*n).

A367556 Comma transform of the Fibonacci numbers.

Original entry on oeis.org

1, 11, 12, 23, 35, 58, 81, 32, 13, 45, 58, 91, 42, 33, 76, 9, 71, 72, 44, 16, 51, 61, 12, 74, 87, 51, 31, 83, 15, 98, 1, 92, 93, 85, 79, 51, 22, 73, 96, 61, 51, 12, 64, 77, 31, 1, 32, 34, 67, 91, 52, 43, 95, 38, 21, 52, 73, 25, 99, 11, 2, 14, 16, 21, 31, 52, 84

Offset: 0

Alois P. Heinz, Nov 22 2023

See A367360 for further information.

Alois P. Heinz, Table of n, a(n) for n = 0..10000

Cf. A000045, A003893, A008963, A138844, A367360.

F:= combinat[fibonacci]:
a:= n-> parse(cat(""||(F(n))[-1], ""||(F(n+1))[1])):
seq(a(n), n=0..92);

With[{nmax=100},Map[10Mod[#[[1]],10]+IntegerDigits[#[[2]]][[1]]&,Partition[Fibonacci[Range[0,nmax+1]],2,1]]] (* Paolo Xausa, Nov 24 2023 *)

from sympy import fibonacci
from itertools import islice, pairwise, count
def S(): yield from (fibonacci(i) for i in count(0))
def C(g): # generator of comma transform of sequence passed as a generator
    yield from (10*(t%10) + int(str(u)[0]) for t, u in pairwise(g))
def agen(): return C(S())
print(list(islice(agen(), 67))) # Michael S. Branicky, Jan 05 2024

a(n) = 10 * A003893(n) + A008963(n+1).

A080787 a(1)=a(2)=1; a(n) = a(n-1) + last decimal digit of a(n-2).

Original entry on oeis.org

1, 1, 2, 3, 5, 8, 13, 21, 24, 25, 29, 34, 43, 47, 50, 57, 57, 64, 71, 75, 76, 81, 87, 88, 95, 103, 108, 111, 119, 120, 129, 129, 138, 147, 155, 162, 167, 169, 176, 185, 191, 196, 197, 203, 210, 213, 213, 216, 219, 225, 234, 239, 243, 252, 255, 257, 262, 269, 271

Offset: 1

Benoit Cloitre, Mar 12 2003

a(n)=a(n-1)+a(n-2)(mod 10); for n>=3 a(n)-a(n-1)=A003893(n-2)=A000045(n-2)(mod 10)

A160137 Lodumo_10 of Fibonacci numbers.

Original entry on oeis.org

0, 1, 11, 2, 3, 5, 8, 13, 21, 4, 15, 9, 14, 23, 7, 10, 17, 27, 24, 31, 25, 6, 41, 37, 18, 35, 33, 28, 51, 19, 20, 29, 39, 38, 47, 45, 12, 57, 49, 16, 55, 61, 26, 67, 43, 30, 53, 63, 36, 59, 65, 34, 69, 73, 22, 75, 77, 32, 79, 71, 40, 81, 91, 42, 83, 85, 48, 93, 101, 44, 95, 89, 54

Offset: 0

Philippe Deléham, May 02 2009

Permutation of nonnegative integers.

Cf. A000045, A003893

a(n)=lod_10(A000045(n)).

cp's OEIS Frontend

A072709 Integers m such that the last digit of Fibonacci(m) is 7.

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A072711 Integers m such that the last digit of Fibonacci(m) is 9.

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A137290 Fibonacci(n) mod 30.

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A181190 Maximal length of chain-addition sequence mod 10 with window of size n.

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A248740 a(n) = Fibonacci(n) mod 1000.

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A256985 Define a sequence {b(i), i >= 0} by b(i) = 1 for 0 <= i <= n, thereafter b(i+1) = (b(i)+b(i-n)) mod 10; consider the sequence of n-tuples [b(i-n+1), b(i-n+2),...,b(i)]; this repeats with period a(n).

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A341414 a(n) = (Fibonacci(n)*Lucas(n)) mod 10.

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A367556 Comma transform of the Fibonacci numbers.

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