Anton Joha - cp's OEIS frontend

A181761 Start with 18; write down the sum of its digits; add last two terms; repeat.

18, 9, 27, 9, 36, 9, 45, 9, 54, 9, 63, 9, 72, 9, 81, 9, 90, 9, 99, 18, 117, 9, 126, 9, 135, 9, 144, 9, 153, 9, 162, 9, 171, 9, 180, 9, 189, 18, 207, 9, 216, 9, 225, 9, 234, 9, 243, 9, 252, 9, 261, 9, 270, 9, 279, 18, 297, 18, 315, 9, 324, 9, 333, 9, 342, 9

Offset: 1

Anton Joha, Nov 14 2010

			a(1)=18; a(2)=1+8=9; a(3)=18+9=27, etc.

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A129888.

NestList[{Total[#],Total[IntegerDigits[Total[#]]]}&,{18,9},40]// Flatten (* Harvey P. Dale, Nov 10 2016 *)

a(2n+1) = A016096(n+2). - Michel Marcus, Oct 31 2014

A119707 Number of distinct primes appearing in all partitions of n into prime parts.

Original entry on oeis.org

0, 1, 1, 1, 3, 2, 4, 3, 4, 4, 5, 4, 6, 5, 6, 6, 7, 6, 8, 7, 8, 8, 9, 8, 9, 9, 9, 9, 10, 9, 11, 10, 11, 11, 11, 11, 12, 11, 12, 12, 13, 12, 14, 13, 14, 14, 15, 14, 15, 15, 15, 15, 16, 15, 16, 16, 16, 16, 17, 16, 18, 17, 18, 18, 18, 18, 19, 18, 19, 19, 20, 19, 21, 20, 21, 21, 21, 21, 22

Offset: 1

Anton Joha, Jun 10 2006

nonn

			There is only 1 distinct prime number involved in the partitions of 4, namely 2 (in 2+2 = 4). The partition 3+1 does not count, as 1 is not a prime. So a(4)= 1.
There are 3 distinct primes involved in the partitions of 5 = 2+3, so a(5) = 3.

Cf. A000720.

f[n_] := If[OddQ@n, If[n == 3, 1, PrimePi@n], If[n == 2, 1, PrimePi[n - 2]]]; Array[f, 80] (* Robert G. Wilson v *)

When n = odd and >=5 then a(n) = pi(n) = A000720(n). When n = even and >=4 then a(n) = pi(n-2) = A000720(n-2)

Edited and extended by Robert G. Wilson v, Jun 15 2006

A120353 Sum of 5 consecutive powers of 3, starting with a power of 9.

Original entry on oeis.org

121, 1089, 9801, 88209, 793881, 7144929, 64304361, 578739249, 5208653241, 46877879169, 421900912521, 3797108212689, 34173973914201, 307565765227809, 2768091887050281, 24912826983452529, 224215442851072761, 2017938985659654849, 18161450870936893641

Offset: 1

Anton Joha, Jun 25 2006

Always a square.

			a(2) = 3^2 + 3^3 + 3^4 + 3^5 + 3^6 = 121*(3^2) = 1089.

Vincenzo Librandi, Table of n, a(n) for n = 1..300
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (9).

[121*3^(2*n-2): n in [1..30]]; // Vincenzo Librandi, Jun 10 2011

Total/@Select[Partition[3^Range[0,60],5,1],IntegerQ[Log[9, First[#]]]&] (* or *) Table[121 3^(2n-2),{n,30}] (* Harvey P. Dale, May 03 2011 *)

a(n)=121*9^(n-1) \\ Charles R Greathouse IV, Jul 11 2016

a(n) = 121*3^(2n-2).

Corrected and extended by Harvey P. Dale, May 03 2011

A120354 a(n) = 11*3^n.

Original entry on oeis.org

11, 33, 99, 297, 891, 2673, 8019, 24057, 72171, 216513, 649539, 1948617, 5845851, 17537553, 52612659, 157837977, 473513931, 1420541793, 4261625379, 12784876137, 38354628411, 115063885233, 345191655699, 1035574967097, 3106724901291, 9320174703873

Offset: 0

Anton Joha, Jun 25 2006

Square root of A120353.

Vincenzo Librandi, Table of n, a(n) for n = 0..300
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (3).

Cf. A000244, A120353.

[11*3^n: n in [0..30]]; // Vincenzo Librandi, Jun 10 2011

11*3^Range[0, 60] (* Vladimir Joseph Stephan Orlovsky, Jun 09 2011 *)
NestList[3#&,11,30] (* Harvey P. Dale, Mar 04 2013 *)

a(n)=11*3^n \\ Charles R Greathouse IV, Sep 04 2013

From Philippe Deléham, Nov 23 2008: (Start)
a(n) = 3*a(n-1), n > 0; a(0)=11.
G.f.: 11/(1-3*x). (End)
a(n) = 11*A000244(n). - Michel Marcus, Sep 04 2013
E.g.f.: 11*exp(3*x). - Elmo R. Oliveira, Aug 16 2024

Better definition from Tanya Khovanova, Jan 20 2007

A059324 Numbers n such that 6n + 5 is composite.

Original entry on oeis.org

5, 10, 12, 15, 19, 20, 23, 25, 26, 30, 33, 34, 35, 36, 40, 45, 47, 49, 50, 53, 54, 55, 56, 60, 61, 62, 65, 67, 68, 70, 72, 75, 78, 80, 82, 85, 87, 88, 89, 90, 91, 95, 96, 100, 101, 103, 104, 105, 110, 111, 114, 115, 117, 118, 120, 121, 122, 124, 125, 127, 129, 130

Offset: 1

Anton Joha, Jan 26 2001

Conjecture: There exists no pair of primes (p, q > p^2) such that q - p^2 = 6*n - 4 (see A138479). - Philippe LALLOUET (philip.lallouet(AT)orange.fr), Mar 20 2008

			a(3) = 12 because 6*12 + 5 = 77 is composite.

G. C. Greubel, Table of n, a(n) for n = 1..5000

Complement of A059325.

Cf. A138479.

Select[Range[200],!PrimeQ[6#+5]&]  (* Harvey P. Dale, Mar 13 2011 *)

isok(n) = ! isprime(6*n+5); \\ Michel Marcus, Jan 06 2017

a(n) = A046953(n-1) - 1.

More terms from Henry Bottomley, Jan 29 2001

A059482 a(0)=1, a(n) = a(n-1) + 8*10^(n-1).

Original entry on oeis.org

1, 9, 89, 889, 8889, 88889, 888889, 8888889, 88888889, 888888889, 8888888889, 88888888889, 888888888889, 8888888888889, 88888888888889, 888888888888889, 8888888888888889, 88888888888888889, 888888888888888889, 8888888888888888889, 88888888888888888889, 888888888888888888889

Offset: 0

Anton Joha, Feb 04 2001

Related to the sum of Fibonacci-variants: Sum of the (Fibonacci numbers)/(10^n) = 0/(10^1) + 1/(10^2) + 1/(10^3) + 2/(10^4) + 3/(10^5) + 5/(10^6) + ... = 1/89. Sum of the (tribonacci numbers)/(10^(n+1)) = 1/889. Sum of the (tetranacci numbers)/(10^(n+2)) = 1/8889, etc. The denominators of those sums is the current sequence. The first one is of course 0.11111111111... = 1/9. - partially edited by Michel Marcus, Jan 27 2014
Let A be the Hessenberg matrix of the order n, defined by: A[1,j]=1, A[i,i]:=12, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n-1)=(-1)^(n-1)*charpoly(A,2). - Milan Janjic, Feb 21 2010
Except for the initial term, these are the 9-automorphic numbers ending in 9. - Eric M. Schmidt, Aug 17 2012

			a(3) = (10^3)*(1000/1125) + (1/9) = (8000/9) + (1/9) = 889.

Harry J. Smith, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (11,-10).

Cf. A002282.

Table[(8*10^n+1)/9, {n,0,50}] (* G. C. Greubel, May 15 2017 *)

{ a=1/5; for (n = 0, 200, a+=8*10^(n - 1); write("b059482.txt", n, " ", a); ) } \\ Harry J. Smith, Jun 27 2009

def a(n): return (8*10**n+1)//9 # Martin Gergov, Oct 20 2022

a(n) = (10^n)*(1000/1125) + (1/9).
a(n) = A002282(n) + 1 = (8*10^n + 1)/9.
a(n) = 10*a(n-1) - 1 with n > 0, a(0)=1. - Vincenzo Librandi, Aug 07 2010
G.f.: -(2*x-1) / ((x-1)*(10*x-1)). - Colin Barker, Feb 02 2013
a(n) = 10^n - Sum_{i=0..n-1} 10^i for n > 0. - Bruno Berselli, Jun 20 2013
E.g.f.: exp(x)*(1 + 8*exp(9*x))/9. - Stefano Spezia, Oct 25 2023

More terms from Henry Bottomley, Feb 05 2001

A059325 Numbers n such that 6n + 5 is prime.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 7, 8, 9, 11, 13, 14, 16, 17, 18, 21, 22, 24, 27, 28, 29, 31, 32, 37, 38, 39, 41, 42, 43, 44, 46, 48, 51, 52, 57, 58, 59, 63, 64, 66, 69, 71, 73, 74, 76, 77, 79, 81, 83, 84, 86, 92, 93, 94, 97, 98, 99, 102, 106, 107, 108, 109, 112, 113, 116, 119, 123, 126

Offset: 1

Anton Joha, Jan 26 2001

			a(4)=3 because 6*3 + 1 = 19 is prime.

G. C. Greubel, Table of n, a(n) for n = 1..5000

Complement of A059324.

[n: n in [0..200]| IsPrime(6*n+5)] // Vincenzo Librandi, Aug 07 2010

Select[Range[0, 1000], PrimeQ[6 # + 5] &] (* G. C. Greubel, Jan 06 2017 *)

is(n)=isprime(6*n+5) \\ Charles R Greathouse IV, Feb 17 2017

More terms from Henry Bottomley, Jan 29 2001

A059425 Primes of form n^2 + 19n + 17.

Original entry on oeis.org

17, 37, 59, 83, 109, 137, 167, 199, 233, 269, 307, 347, 389, 433, 479, 577, 683, 739, 797, 857, 919, 983, 1049, 1117, 1187, 1259, 1409, 1487, 1567, 1733, 1907, 1997, 2089, 2377, 2477, 2579, 2683, 2789, 2897, 3119, 3467, 3709, 3833, 4217, 4349, 4483

Offset: 1

Anton Joha, Jan 31 2001

nonn

0<=n<=14 gives primes.

			a(3) = 83 = 3^2 + 19*3 + 17 is in the sequence because it is prime. a(15)=527 is not because 527 = 17*31.

with(numtheory): for n from 0 to 300 do if isprime(n^2 + 19*n + 17) then printf(`%d,`,n^2 + 19*n + 17) fi; od:

Select[Table[n^2+19n+17,{n,0,60}],PrimeQ] (* Harvey P. Dale, Jun 21 2011 *)

More terms from James Sellers, Feb 03 2001

cp's OEIS Frontend

User: Anton Joha

A181761 Start with 18; write down the sum of its digits; add last two terms; repeat.

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A119707 Number of distinct primes appearing in all partitions of n into prime parts.

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A120353 Sum of 5 consecutive powers of 3, starting with a power of 9.

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A120354 a(n) = 11*3^n.

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A059324 Numbers n such that 6n + 5 is composite.

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A059482 a(0)=1, a(n) = a(n-1) + 8*10^(n-1).

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A059325 Numbers n such that 6n + 5 is prime.

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