A239619 -id:A239619 - cp's OEIS frontend

A239690 Base 4 sum of digits of prime(n).

2, 3, 2, 4, 5, 4, 2, 4, 5, 5, 7, 4, 5, 7, 8, 5, 8, 7, 4, 5, 4, 7, 5, 5, 4, 5, 7, 8, 7, 5, 10, 5, 5, 7, 5, 7, 7, 7, 8, 8, 8, 7, 11, 4, 5, 7, 7, 10, 8, 7, 8, 11, 7, 11, 2, 5, 5, 7, 4, 5, 7, 5, 7, 8, 7, 8, 7, 4, 8, 7, 5, 8, 10, 7, 10, 11, 5, 7, 5, 7, 8, 7, 11, 7

Offset: 1

Tom Edgar, Mar 24 2014

a(n) is the rank of prime(n) in the base-4 dominance order on the natural numbers.

			The sixth prime is 13, 13 in base 4 is (3,1) so a(6)=3+1=4.

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Tyler Ball and Daniel Juda, Dominance over N, Rose-Hulman Undergraduate Mathematics Journal, Vol. 13, No. 2, Fall 2013.

Cf. A007605, A014499, A053737, A239619.

Cf. A000040, A007090.

a239690 = a053737 . a000040  -- Reinhard Zumkeller, Mar 20 2015

[&+Intseq(NthPrime(n),4): n in [1..100]]; // Vincenzo Librandi, Mar 25 2014

Table[Plus @@ IntegerDigits[Prime[n], 4], {n, 1, 100}] (* Vincenzo Librandi, Mar 25 2014 *)

[sum(i.digits(base=4)) for i in primes_first_n(200)]

a(n) = A053737(A000040(n)).

A240886 Primes p equal to the sum in base-3 of the digits of all primes < p.

Original entry on oeis.org

23, 31, 47, 59, 695689, 698471, 883517, 992609, 992737, 993037, 1314239, 1324361, 1324571, 1326511, 1327289, 1766291, 3174029

Offset: 1

Anthony Sand, Apr 14 2014

Conjecture: this sequence is finite and all terms are shown. - Robert G. Wilson v, Jul 27 2014

The sum of the digits in base three of all primes < 10^10 is 9694409092. - Robert G. Wilson v, Jul 27 2014

			For example, 23 = digit-sum(primes < 23, base=3) = sum(2) + sum(1,0) + sum(1,2) + sum(2,1) + sum(1,0,2) + sum(1,1,1) + sum(1,2,2) + sum(2,0,1).

Cf. A239619 (Base 3 sum of digits of prime(n)).

p = 2; s = 0; lst = {}; While[p < 3200000, If[s == p, AppendTo[lst, p]; Print[p]]; s = s + Total@ IntegerDigits[p, 3]; p = NextPrime[p]] (* Robert G. Wilson v, Jul 27 2014 *)

sdt(n) = my(d = digits(n, 3)); sum(i=1, #d, d[i]);
lista(nn) = {sp = 0; forprime(p=1, nn, if (p == sp, print1(p, ", ")); sp += sdt(p););} \\ Michel Marcus, May 02 2014

prime(n) such that, using base 3, prime(n) = sum_{1..n-1} A239619(i).

A241897 Primes p equal to the sum in base 3 of the digits of all primes < p - digit-sum of the index of prime p(i-1).

Original entry on oeis.org

67, 71, 97, 101, 149, 223, 656267, 697511, 697951, 698447, 699493, 700277, 715373, 883963, 888203, 888211, 992021, 992183, 992891, 993241, 994181, 1155607, 1155829, 1308121, 1308649, 1310093, 1313083, 1317409, 1320061, 1320157, 1320379, 1322521, 1322591

Offset: 1

Anthony Sand, May 01 2014

There are no further solutions beyond a(46)=4539541 up to at least 10^10. - Andrew Howroyd, Mar 02 2018

			67 = digit-sum(2..61,b=3) - digit-sum(index(61),b=3) = sum(2) + sum(1,0) + sum(1,2) + sum(2,1) + sum(1,0,2) + sum(1,1,1) + sum(1,2,2) + sum(2,0,1) + sum(2,1,2) + sum(1,0,0,2) + sum(1,0,1,1) + sum(1,1,0,1) + sum(1,1,1,2) + sum(1,1,2,1) + sum(1,2,0,2) + sum(1,2,2,2) + sum(2,0,1,2) + sum(2,0,2,1) - digit-sum(200).

Anthony Sand, Table of n, a(n) for n = 1..46

A240886. Primes p equal to the digit-sum in base 3 of all primes < p. A168161. Primes p which are equal to the sum of the binary digits in all primes <= p.

seq(maxp)={my(p=1,L=List(),s=0,k=0); while(pAndrew Howroyd, Mar 01 2018

prime(n) such that, using base 3, prime(n) = sum_{1..n-1} A239619(i) - sum_{index(n-1)}

a(29)-a(33) from Andrew Howroyd, Mar 02 2018

A241895 Primes p equal to the sum in base 3 of the digits of all primes <= p.

Original entry on oeis.org

3, 37, 695663, 695881, 1308731, 1308757, 1313153, 1314301, 1326097, 1766227, 3204779, 14328191

Offset: 1

Anthony Sand, May 01 2014

			3 = digit-sum(primes <= 3,base=3) = sum(2) + sum(1,0). 37 = digit-sum(primes <= 37,base=3) = sum(2) + sum(1,0) + sum(1,2) + sum(2,1) + sum(1,0,2) + sum(1,1,1) + sum(1,2,2) + sum(2,0,1) + sum(2,1,2) + sum(1,0,0,2) + sum(1,0,1,1) + sum(1,1,0,1).

Cf. A168161 (similar in base 2), A240886 (similar but excluding p from the sum).

sdt(n) = my(d = digits(n, 3)); sum(i=1, #d, d[i]);
lista(nn) = {sp = 0; forprime(p=1, nn, sp += sdt(p); if (p == sp, print1(p, ", ")););} \\ Michel Marcus, May 02 2014

prime(n) such that, using base 3, prime(n) = sum_{1..n} A239619(i).

A241896 Increasingly ordered odd primes p(m) with p(m) = (sum of the digits of all primes p(i) in base 3 for i=1, 2, ..., m-1) + (sum of digits of m-1 in base 3).

Original entry on oeis.org

3, 5, 7, 11, 17, 29, 37, 695641, 695687, 695749, 695881, 699943, 700199, 715457, 883433, 883451, 883471, 883621, 992111, 992357, 992591, 993683, 1308563, 1309999, 1310041, 1310359, 1310993, 1313161, 1314191, 1314377, 1317271, 1324567, 1326097, 1326109, 1326649, 1760113, 1760287, 1766509, 1766537, 3173761, 3204779, 3204827, 4539191

Offset: 1

Anthony Sand, May 01 2014

			prime(2) = 3  = A239619(1) + A053735(1) = 2 + 1. This is a(1) because it is the smallest odd prime from the defined set S.
prime(7) = 17 = sum_{i=1..6} A239619(i) + A053735(6) = (2 + 1 + 3 + 3 + 3 + 3) + 2 = 17. This is a(5) because it is the fifth smallest odd prime from the set S.
prime(6) = 13 is not a member of this sequence because (2 + 1 + 3 + 3 + 3) + 3 = 15 which is not equal 13, hence prime(6) is not a member of the set S.

CF. A240886 (similar sequence with digit-sums), A168161 (similar sequence but in binary). A053735, A239619.

This is the increasingly ordered set of numbers

S:= {odd primes: prime(m) = sum_{i=1..m-1} A239619(i) + A053735(m-1)}.

Edited. - Wolfdieter Lang, May 19 2014

A331112 Sum of the digits of the n-th prime number in balanced ternary.

Original entry on oeis.org

0, 1, -1, 1, 1, 3, -1, 1, -1, 1, 3, 3, -3, -1, -1, -1, -1, 1, 3, -1, 1, 1, 1, 1, 1, 1, 3, 1, 3, 1, -1, -3, -1, 1, -3, -1, 1, 1, -1, 1, -1, 1, 1, 3, 1, 3, 1, 1, 1, 3, -1, -1, 1, 1, -1, 1, 1, 3, 3, 3, 5, -1, 3, -1, 1, 1, 3, 5, 1, 3, 3, 3, -3, -1, -1, -3, -1, -1, -3, 1, -3, -1, -1, 1, 1, 1, -3, -1, -1, 1, -1, -1, 1, -1, 3, -1, -1, 1, 3, 1

Offset: 1

Thomas König, Jan 09 2020

			Using T for -1 and _bt as suffix for balanced ternary: 2_10 = 1T_bt, sum of digits is zero; 3_10 = 10_bt, sum of digits is 1 and 5_10 = 1TT, sum of digits = -1.

Wikipedia, Balanced ternary

See A007605 (sum of digits of primes in base 10); A239619 (sum of digits of primes in base 3).

Cf. A000040, A065363, A117966.

#include 
#include 
#define N 1000 /* Largest prime considered - 1  */
char x[N];
int main()
{
  int i, n, v, s, r;
  for (i=4; i

b:= proc(n) `if`(n=0, 0, (d-> `if`(d=2,
      b(q+1)-1, d+b(q)))(irem(n, 3, 'q')))
    end:
a:= n-> b(ithprime(n)):
seq(a(n), n=1..100);  # Alois P. Heinz, Jan 09 2020

a(n) = A065363(A000040(n)). - Alois P. Heinz, Jan 09 2020

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A239690 Base 4 sum of digits of prime(n).

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A240886 Primes p equal to the sum in base-3 of the digits of all primes < p.

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A241897 Primes p equal to the sum in base 3 of the digits of all primes < p - digit-sum of the index of prime p(i-1).

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A241895 Primes p equal to the sum in base 3 of the digits of all primes <= p.

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A241896 Increasingly ordered odd primes p(m) with p(m) = (sum of the digits of all primes p(i) in base 3 for i=1, 2, ..., m-1) + (sum of digits of m-1 in base 3).

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A331112 Sum of the digits of the n-th prime number in balanced ternary.

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