A033940 -id:A033940 - cp's OEIS frontend

A119910 Period 6: repeat [1, 3, 2, -1, -3, -2].

1, 3, 2, -1, -3, -2, 1, 3, 2, -1, -3, -2, 1, 3, 2, -1, -3, -2, 1, 3, 2, -1, -3, -2, 1, 3, 2, -1, -3, -2, 1, 3, 2, -1, -3, -2, 1, 3, 2, -1, -3, -2, 1, 3, 2, -1, -3, -2, 1, 3, 2, -1, -3, -2, 1, 3, 2, -1, -3, -2, 1, 3, 2, -1, -3, -2, 1, 3, 2, -1, -3, -2, 1, 3, 2, -1, -3, -2, 1, 3, 2, -1, -3, -2

Offset: 1

Kartikeya Shandilya (kartikeya.shandilya(AT)gmail.com), May 28 2006

Take any of term, multiply it to units place digit of any taken no. then save the product, then take the next term of this sequence, multiply it to the next place digit of the taken no., add the product to previous one and save it, then take the next term of the sequence, multiply it to the next place digit of the taken no. and add it to the previous sum, keep on doing this until all the digits of the taken no. are done, now if the calculated sum is divisible by 7, then the initial number taken must also be completely divisible by seven, otherwise not.
Can be converted into the sequence "10^n mod 7", 1) 1,3,2,6,4,5,1,3,2,6,4,5,1,3,2,6,4,5,1,3,2,6,4,5 .... 2) -6,-4,-5,6,4,5,-6,-4,-5,6,4,5,-6,-4,-5,6,4,5 ... 3) -6,-4,-5,-1,-3,-2,-6,-4,-5,-1,-3,-2,-6,-4,-5,-1,-3,-2 ...
Many variations can be made by adding or subtracting 7 from any term of the previous sequences. Still the divisibility rule will be valid.
Nonsimple continued fraction of (6+2*sqrt(2))/7 = 1.26120387... - R. J. Mathar, Mar 08 2012

			a(32)=?: 32%7=4, therefore a(32)=-1.
Let us test the divisibility of 342 with the series:
Take 1 from the sequence, multiply it by 2, the product is 2,
take 3 from the sequence, multiply it by 4, the product is 12,
take 2 from the sequence, multiply it by 3, the product is 6,
the sum of the products is 2 + 12 + 6 = 20,
because 20 is not divisible by 7, therefore 342 will also not be.

Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (1,-1).

Cf. A010892, A033940.

&cat[[1, 3, 2, -1, -3, -2]^^20]; // Wesley Ivan Hurt, Jun 19 2016

A119910:=n->[1, 3, 2, -1, -3, -2][(n mod 6)+1]: seq(A119910(n), n=0..100); # Wesley Ivan Hurt, Jun 19 2016

PadRight[{}, 100, {1, 3, 2, -1, -3, -2}] (* Wesley Ivan Hurt, Jun 19 2016 *)

From R. J. Mathar, Feb 08 2008: (Start)
O.g.f.: 2 + (3*x-2)/(x^2-x+1).
a(n) = 3*A010892(n-1) - 2*A010892(n).
a(n) = -a(n-3) for n>3. (End)
a(n) = a(n-1) - a(n-2) for n>2. - Philippe Deléham, Nov 16 2008
a(n) = (4*sqrt(3)*sin(n*Pi/3) - 6*cos(n*Pi/3))/3. - Wesley Ivan Hurt, Jun 19 2016

New name from Omar E. Pol, Oct 31 2013

A201909 Irregular triangle of 3^k mod prime(n).

Original entry on oeis.org

1, 0, 1, 3, 4, 2, 1, 3, 2, 6, 4, 5, 1, 3, 9, 5, 4, 1, 3, 9, 1, 3, 9, 10, 13, 5, 15, 11, 16, 14, 8, 7, 4, 12, 2, 6, 1, 3, 9, 8, 5, 15, 7, 2, 6, 18, 16, 10, 11, 14, 4, 12, 17, 13, 1, 3, 9, 4, 12, 13, 16, 2, 6, 18, 8, 1, 3, 9, 27, 23, 11, 4, 12, 7, 21, 5, 15

Offset: 1

T. D. Noe, Dec 07 2011

The row lengths are in A062117. Except for the second row, the first term of each row is 1. Many sequences are in this one: starting at A036119 (mod 17) and A070341 (mod 11).

			The first 9 rows are:
  1
  0
  1, 3, 4,  2
  1, 3, 2,  6,  4,  5
  1, 3, 9,  5,  4
  1, 3, 9
  1, 3, 9, 10, 13,  5, 15, 11, 16, 14,  8,  7,  4, 12, 2,  6
  1, 3, 9,  8,  5, 15,  7,  2,  6, 18, 16, 10, 11, 14, 4, 12, 17, 13
  1, 3, 9,  4, 12, 13, 16,  2,  6, 18,  8

T. D. Noe, Rows n = 1..60. flattened

Cf. A062117, A201908 (2^k), A201910 (5^k), A201911 (7^k).

Cf. A070352 (5), A033940 (7), A070341 (11), A168399 (13), A036119 (17), A070342 (19), A070356 (23), A070344 (29), A036123 (31), A070346 (37), A070361 (41), A036126 (43), A070364 (47), A036134 (79), A036136 (89), A036142 (113), A036143 (127), A036145 (137), A036158 (199), A036160 (223).

P:=Filtered([1..350],IsPrime);;
R:=List([1..Length(P)],n->OrderMod(7,P[n]));;
Flat(Concatenation([1,1,1,2,4,3,0],List([5..10],n->List([0..R[n]-1],k->PowerMod(7,k,P[n]))))); # Muniru A Asiru, Feb 01 2019

nn = 10; p = 3; t = p^Range[0,Prime[nn]]; Flatten[Table[If[Mod[n, p] == 0, {0}, tm = Mod[t, n]; len = Position[tm, 1, 1, 2][[-1,1]]; Take[tm, len-1]], {n, Prime[Range[nn]]}]]

A271350 a(n) = 3^n mod 83.

Original entry on oeis.org

1, 3, 9, 27, 81, 77, 65, 29, 4, 12, 36, 25, 75, 59, 11, 33, 16, 48, 61, 17, 51, 70, 44, 49, 64, 26, 78, 68, 38, 31, 10, 30, 7, 21, 63, 23, 69, 41, 40, 37, 28, 1, 3, 9, 27, 81, 77, 65, 29, 4, 12, 36, 25, 75, 59, 11, 33, 16, 48, 61, 17, 51, 70, 44, 49, 64, 26

Offset: 0

Vincenzo Librandi, Apr 05 2016

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
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, 1).

Cf. similar sequences of the type 3^n mod p, where p is a prime: A070352 (5), A033940 (7), A070341 (11), A168399 (13), A036119 (17), A070342 (19), A070356 (23), A070344 (29), A036123 (31), A070346 (37), A070361 (41), A036126 (43), A070364 (47), A036134 (79), this sequence (83), A036136 (89), A036142 (113), A036143 (127), A271351 (131), A036145 (137), A036158 (199), A271352 (211), A036160 (223).

Magma
```
[Modexp(3, n, 83): n in [0..100]];
```
Mathematica
```
PowerMod[3, Range[0, 100], 83]
```

a(n) = lift(Mod(3, 83)^n); \\ Altug Alkan, Apr 05 2016

a(n) = a(n-41).

A155751 A variation on 10^n mod 17.

Original entry on oeis.org

1, -7, -2, -3, 4, 6, -8, 5, -1, 7, 2, 3, -4, -6, 8, -5, 1, -7, -2, -3, 4, 6, -8, 5, -1, 7, 2, 3, -4, -6, 8, -5, 1, -7, -2, -3, 4, 6, -8, 5, -1, 7, 2, 3, -4, -6, 8, -5, 1, -7, -2, -3, 4, 6, -8, 5, -1, 7, 2, 3, -4, -6, 8, -5, 1, -7, -2, -3, 4, 6, -8, 5, -1, 7, 2, 3, -4, -6, 8, -5

Offset: 0

Ferruccio Guidi (fguidi(AT)cs.unibo.it), Jan 26 2009, Feb 08 2009

This is 10^n mod 17, using values -8,-7,...,7,8 (instead of 0..16). - Don Reble, Sep 02 2017.
This sequence can be employed in a test for divisibility by 17 and works like A033940 works for 7.
The use of negative coefficients ensures the termination of the test because the modulus of the intermediate sum at each step of the test decreases strictly.
The test is successful if the final sum is 0.
The negative coefficients have the form (10^n mod 17) - 17 when 10^n mod 17 > 8.
Example: 9996 is divisible by 17 since |6*1 + 9*(-7) + 9*(-2) + 9*(-3)| = 102 and 2*1 + 0*(-7) + 1*(-2) = 0.

Cf. A033940, A119910, A117378.

a(n)= -a(n-8). G.f.:(1-7x-2x^2-3x^3+4x^4+6x^5-8x^6+5x^7)/(1+x^8). [From R. J. Mathar, Feb 13 2009]

A178247 Decimal expansion of (269+11*sqrt(1086))/490.

Original entry on oeis.org

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

Offset: 1

Klaus Brockhaus, May 24 2010

Continued fraction expansion of (269+11*sqrt(1086))/490 is A033940.

			(269+11*sqrt(1086))/490 = 1.28877480647397915221...

Cf. A178230 (decimal expansion of sqrt(1086)), A033940 (repeat 1, 3, 2, 6, 4, 5).

RealDigits[(269+11Sqrt[1086])/490,10,120][[1]] (* Harvey P. Dale, Mar 18 2022 *)

A155754 A variation on 10^n mod 19.

Original entry on oeis.org

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

Offset: 0

Ferruccio Guidi (fguidi(AT)cs.unibo.it), Jan 26 2009

This sequence can be employed in a test for divisibility by 19 and works like A033940 works for 7.
The use of negative coefficients ensures the termination of the test because the modulus of the intermediate sum at each step of the test decreases strictly.
The test is successful if the final sum is 0.
The negative coefficients have the form (10^n mod 19) - 19 when 10^n mod 19 > 9.
Example: 8284 is divisible by 19 since |4*1 + 8*(-9) + 2*5 + 8*(-7)| = 114 and 4*1 + 1*(-9) + 1*5 = 0.

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

Cf. A033940, A119910, A117378.

a(n) = -a(n-9). G.f.: (-2*x^8-4*x^7-8*x^6+3*x^5+6*x^4-7*x^3+5*x^2-9*x+1) / (x^9+1). [Colin Barker, Feb 14 2013]

A369562 Smallest positive n-digit number divisible by 7.

Original entry on oeis.org

7, 14, 105, 1001, 10003, 100002, 1000006, 10000004, 100000005, 1000000001, 10000000003, 100000000002, 1000000000006, 10000000000004, 100000000000005, 1000000000000001, 10000000000000003, 100000000000000002, 1000000000000000006, 10000000000000000004, 100000000000000000005

Offset: 1

J. Lowell, Jan 25 2024

The only semiprime terms are a(2) = 14 and a(n) such that (10^(n-1) + 3)/7 is a prime. - Jon E. Schoenfield, Jan 27 2024

			a(3) = 105 = 7*15.

Index entries for linear recurrences with constant coefficients, signature (11,-10,-1,11,-10).

Cf. A033940, A241217, A369613.

a[n_] := 10^(n - 1) + {6, 4, 5, 1, 3, 2}[[Mod[n, 6, 1]]]; Array[a, 30]
(* or *)
LinearRecurrence[{11, -10, -1, 11, -10}, {7, 14, 105, 1001, 10003, 100002}, 30] (* Amiram Eldar, Jan 27 2024 *)
Table[10^n+7-PowerMod[10,n,7],{n,0,20}] (* Harvey P. Dale, Jan 13 2025 *)

a(n) = (floor(10^(n-1)/7) + 1)*7.
a(n) = 10^(n-1) + A033940(n+2). - Amiram Eldar, Jan 27 2024
G.f.: 7*x*(1 - 9*x + 3*x^2 - x^3 - 3*x^4)/((1 - x)*(1 + x)*(1 - 10*x)*(1 - x + x^2)). - Stefano Spezia, Jan 28 2024

cp's OEIS Frontend

A119910 Period 6: repeat [1, 3, 2, -1, -3, -2].

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A201909 Irregular triangle of 3^k mod prime(n).

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A271350 a(n) = 3^n mod 83.

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A155751 A variation on 10^n mod 17.

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A178247 Decimal expansion of (269+11*sqrt(1086))/490.

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Mathematica

A155754 A variation on 10^n mod 19.

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A369562 Smallest positive n-digit number divisible by 7.

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