A000042 -id:A000042 - cp's OEIS frontend

A120449 Array by antidiagonals of all primitive Orloj clock striking sequences.

1, 11, 1, 111, 2, 1, 1111, 12, 2, 1, 11111, 121, 21, 2, 1, 111111, 212, 22, 3, 2, 1, 1111111, 1212, 122, 112, 3, 2, 1, 11111111, 12121, 1221, 311, 31, 12, 11, 1, 111111111, 21212, 2212, 231, 23, 4, 3, 2, 1, 1111111111, 121212, 21221, 1231, 312

Offset: 1

Franklin T. Adams-Watters, Jul 19 2006

This is the sequences of strikes at each hour, represented by concatenation of the digits. The repeating pattern for each row is in A118382. This table eventually contains non-decimal digits. Row 47 is the first one containing a non-decimal digit.

			The table starts:
1 11 111 1111 11111 111111 ...
1 2 12 121 212 1212 ...
1 2 21 22 122 1221 ...
1 2 3 112 311 231 ...
1 2 3 31 23 312 ...

Cf. A118382, A117510, rows A000042, A028359, A028355.

A245323 a(n) = F(6n-3)(L(2*n-1)+1), where F = A000045 are the Fibonacci and L = A000032 are the Lucas numbers.

Original entry on oeis.org

4, 170, 7320, 328380, 15124186, 704915600, 33014404692, 1549142827050, 72743819556328, 3416820019114700, 160507201018772634, 7540231471940495520, 354226959651753624100, 16641065639596669234730, 781774759322033582085816, 36726752905662141638238300

Offset: 1

Rémi Schulz, Jul 18 2014

Let n = 2*m-1 where m is 1,2,3,...; then a(m) = F(3*n)*(L(n)+1) with F(n) a Fibonacci number and L(n) a Lucas number (A000032); also a(n) = F(n)*(L(n)^3+L(n)^2+L(n)+1) which is a repdigit in base L(n), made of four digits F(n).
For n = 1, unary representation must be used to give a repdigit, then 4(10) = 1111(1).
For n = 3, 170(10) = 2222(4).
For n = 5, 7320(10) = 5555(11).
Starting from n = 5, the repdigit is the 4th term of a Fibonacci-type sequence of 5 palindromes.
For example this sequence for n=5 in base 11 is: 1331, 2112, 3443, 5555, 8998 which are the 5 2-digit Fibonacci numbers in base 10 concatenated with their reversed forms.
If the repdigit F(3*n)*(L(n)+1) is the most noticeable result in base L(n), there are other recurrences; naming f the digit F(n) and g the digit F(n)*2, we find
F(0*n)*(L(n)+1) = 0
F(1*n)*(L(n)+1) = ff
F(2*n)*(L(n)+1) = ff0
F(3*n)*(L(n)+1) = ffff
F(4*n)*(L(n)+1) = ffgg0
The base L(n) gives a visual aspect to the formula
F(n*k) = L(n)*F(n*(k-1)) + F(n*(k-2)) with n odd, which is a particular case of the general formula for any integers n,k,m:
F(n*(k)+m) = L(n) * F(n*(k-1)+m) - (-1)^n * F(n*(k-2)+m)

			Example: for n = 5, F(15) = 610, L(5) = 11, then a(5) = 610*12 = 7320 which is 5555 in base 11; F(5) = 5.

Index entries for linear recurrences with constant coefficients, signature (72,-1304,6066,-1304,72,-1).

Cf. A000045, A000032, A000042.

LinearRecurrence[{72,-1304,6066,-1304,72,-1}, {4,170,7320,328380,15124186,704915600},30] (* Harvey P. Dale, Aug 26 2014 *)
Table[Fibonacci[6 n - 3] (LucasL[2 n - 1] + 1), {n, 16}] (* Michael De Vlieger, Oct 21 2016 *)

vector(50, m, fibonacci(6*m-3)*(lucas(2*m-1)+1)) \\ Colin Barker, Jul 18 2014

a((n-1)/2) = F(3*n)*(L(n)+1) for any positive odd n. [Corrected by M. F. Hasler, Oct 20 2016]
a(n) = F(n)*(L(n)^3+L(n)^2+L(n)+1).
G.f.: 2*x*(51*x^4-622*x^3+148*x^2-59*x+2) / ((x^2-47*x+1)*(x^2-18*x+1)*(x^2-7*x+1)). - Colin Barker, Jul 18 2014

More terms from Colin Barker, Jul 18 2014

Edited and partially corrected by M. F. Hasler, Oct 09 and Oct 20 2016

A335012 Number of different remainders when the first n terms of 1, 11, 111, 1111, ... are divided by n.

Original entry on oeis.org

1, 1, 3, 2, 1, 3, 6, 3, 9, 1, 2, 4, 6, 6, 3, 4, 16, 9, 18, 2, 6, 2, 22, 5, 2, 6, 27, 7, 28, 3, 15, 5, 6, 16, 6, 10, 3, 18, 6, 3, 5, 6, 21, 3, 9, 22, 46, 6, 42, 2, 48, 7, 13, 27, 2, 8, 18, 28, 58, 4, 60, 15, 18, 6, 6, 6, 33, 17, 66, 6, 35, 11, 8, 3, 4, 19, 6, 6, 13, 4, 81

Offset: 1

Sen-Peng Eu, May 19 2020

a(n) = n if and only if n is a power of 3.
Conjecture: a(n) = n-1 if and only if n is a long period prime (A006883), that is, n is a prime and the decimal expansion of 1/n has period n-1.
If gcd(n,30) = 1 then a(n) = A084680(n). - Robert Israel, Jun 25 2020

			a(4) = 2 since when 1, 11, 111, 1111 are divided by 4 the remainders are 1, 3, 3, 3, two different numbers.
a(6) = 3 since when 1, 11, 111, 1111, 11111, 111111 are divided by 6 the remainders are 1, 5, 3, 1, 5, 3, three different numbers.

Cf. A000042, A006883, A084680.

with(ListTools): a := proc (n) return add(10^i, i = 0 .. n-1) end proc: r := proc (n) return seq(`mod`(a(i), n), i = 1 .. n) end proc: seq(nops(MakeUnique([r(n)])), n = 1 .. 243);
# Simpler:
f:= n -> nops({seq(((10^i-1)/9) mod n,i=1..n)}):
map(f, [$1..100]); # Robert Israel, Jun 25 2020

Table[Length@ Union@ Array[Mod[(10^# - 1)/9, n] &, n], {n, 81}] (* Michael De Vlieger, Jun 28 2020 *)

a(n) = #Set(vector(n, k, (10^k-1)/9) % n); \\ Michel Marcus, Jun 15 2020

A346821 Lexicographically earliest sequence of distinct terms > 0 such that the concatenation of three successive terms form a palindrome in base 10.

Original entry on oeis.org

1, 10, 11, 101, 10111, 101101, 11101, 11101101, 1110111, 10110111, 1011011110111, 10110111101101, 1110111101101, 111011110110111101101, 11101111011011110111, 101101111011011110111, 1011011110110111101111011011110111, 10110111101101111011110110111101101, 1110111101101111011110110111101101

Offset: 1

Eric Angelini and Carole Dubois, Aug 30 2021

This sequence shows only 0's and 1's. The lexicographically earliest sequence of distinct terms > 0 such that the concatenation of two successive terms form a palindrome in base 10 is A000042.

			a(1) = 1, a(2) = 10, a(3) = 11 form the palindrome 11011 when concatenated;
a(2) = 10, a(3) = 11, a(4) = 101 form the palindrome 1011101 when concatenated;
a(3) = 11, a(4) = 101, a(5) = 10111 form the palindrome 1110110111 when concatenated;
a(4) = 101, a(5) = 10111, a(6) = 101101 form the palindrome 10110111101101 when concatenated; etc.

Cf. A000042.

cp's OEIS Frontend

A120449 Array by antidiagonals of all primitive Orloj clock striking sequences.

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A245323 a(n) = F(6n-3)(L(2*n-1)+1), where F = A000045 are the Fibonacci and L = A000032 are the Lucas numbers.

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A335012 Number of different remainders when the first n terms of 1, 11, 111, 1111, ... are divided by n.

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Maple

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PARI

A346821 Lexicographically earliest sequence of distinct terms > 0 such that the concatenation of three successive terms form a palindrome in base 10.

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Author

Keywords

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cp's OEIS Frontend

A120449 Array by antidiagonals of all primitive Orloj clock striking sequences.

Views

Author

Keywords

Comments

Examples

Crossrefs

A245323 a(n) = F(6*n-3)*(L(2*n-1)+1), where F = A000045 are the Fibonacci and L = A000032 are the Lucas numbers.

Views

Author

Keywords

Comments

Examples

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Programs

Mathematica

PARI

Formula

Extensions

A335012 Number of different remainders when the first n terms of 1, 11, 111, 1111, ... are divided by n.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Maple

Mathematica

PARI

A346821 Lexicographically earliest sequence of distinct terms > 0 such that the concatenation of three successive terms form a palindrome in base 10.

Views

Author

Keywords

Comments

Examples

Crossrefs

A245323 a(n) = F(6n-3)(L(2*n-1)+1), where F = A000045 are the Fibonacci and L = A000032 are the Lucas numbers.