cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-6 of 6 results.

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

Views

Author

Anthony Sand, May 01 2014

Keywords

Comments

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

Examples

			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).
		

Crossrefs

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.

Programs

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

Formula

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

Extensions

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

Views

Author

Anthony Sand, May 01 2014

Keywords

Examples

			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).
		

Crossrefs

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

Programs

  • PARI
    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

Formula

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

Views

Author

Anthony Sand, May 01 2014

Keywords

Examples

			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.
		

Crossrefs

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

Formula

This is the increasingly ordered set of numbers
S:= {odd primes: prime(m) = sum_{i=1..m-1} A239619(i) + A053735(m-1)}.

Extensions

Edited. - Wolfdieter Lang, May 19 2014

A242478 Primes p such that, in base 17, p = the cumulative sum of the digit-mult(digit-sum(prime)) of each prime < p.

Original entry on oeis.org

5, 57839, 58013, 105683, 160367, 926899, 926983, 927007, 928819, 963121, 963223, 2329777, 2384821, 2384881, 3228713, 3228751, 3229081, 3229097, 3246653, 3259547, 7327781, 7339447
Offset: 1

Views

Author

Anthony Sand, May 16 2014

Keywords

Examples

			5 = digit-mult(digit-sum(2)) + digit-mult(digit-sum(3)). 57839 = digit-mult(digit-sum(2)) + digit-mult(digit-sum(3)) + ...  digit-mult(digit-sum(BD1C)) = digit-mult(2) + digit-mult(3) + ... digit-mult(23) = 2 + 3 + ... 2*3. Note that BD1C and 23 in base 17 = 57829 and 37 in base 10.
		

Crossrefs

Cf. A240886.

Formula

The function digit-mult(n) multiplies all digits d of n, where d > 0. For example, digit-mult(1230) = 1 * 2 * 3 = 6. Therefore, in base 17, digit-mult(digit-sum(9999)) = digit-mult(22) = 2 * 2 = 4 (22 in base 17 = 36 in base 10).

A242479 Primes p such that, in base 17, p = the cumulative sum of the digit-mult(digit-sum(prime)) of each prime <= p.

Original entry on oeis.org

105701, 160309, 927137, 927149, 964973, 2329081, 2329097, 2329549, 2384587, 3228733, 3237527, 3242851, 7338377, 7338431, 7338557, 7338719
Offset: 1

Views

Author

Anthony Sand, May 16 2014

Keywords

Examples

			105701 = digit-mult(digit-sum(2)) + digit-mult(digit-sum(3)) + ... digit-mult(digit-sum(148CC)) = digit-mult(2) + digit-mult(3) + ... digit-mult(23) = 2 + 3 + ... 2*3. Note that 148CC and 23 in base 17 = 105701 and 37 in base 10.
		

Crossrefs

Cf. A240886.

Formula

The function digit-mult(n) multiplies all digits d of n, where d > 0. For example, digit-mult(1230) = 1 * 2 * 3 = 6. Therefore, in base 17, digit-mult(digit-sum(9999)) = digit-mult(22) = 2 * 2 = 4 (22 in base 17 = 36 in base 10).

A242589 Primes p such that p = the cumulative sum of the digit-sum in base 15 of the digit-product in base 4 of each prime < p.

Original entry on oeis.org

5, 19, 37, 43, 97, 107, 6091, 6389, 7121, 21727, 147107, 148151, 148279, 148429, 148469, 172877, 173209, 173741, 2621387, 5642293, 5642321, 8932771, 8981827, 8981879, 9094979, 9095089, 9997783, 10010687, 10010789, 10037749, 10144523, 40179929, 40365217, 40379077, 40379197, 40386811, 40612933
Offset: 1

Views

Author

Anthony Sand, May 20 2014

Keywords

Examples

			5 = digit-sum(digit-mult(2,b=4),b=15) + sum(mult(3,b=4),b=15) = 2 + 3.
19 = digit-sum(digit-mult(2,b=4),b=15) + sum(mult(3,b=4),b=15) + sum(mult(11,b=4),b=15) + sum(mult(13,b=4),b=15) + sum(mult(23,b=4),b=15) + sum(mult(31,b=4),b=15) + sum(mult(101,b=4),b=15) = 2 + 3 + 1 + 3 + 6 + 3 + 1.
		

Crossrefs

Cf. A240886 (similar sequence with digit sums in base 3).

Formula

sum = sum + digit-sum(digit-mult(prime,base=4),base=15). The function digit-mult(n) multiplies all digits d of n, where d > 0. For example, digit-mult(1230) = 1 * 2 * 3 = 6. Therefore, the digit-sum in base 15 of the digit-mult(333) in base 4 = digit-sum(3 * 3 * 3) = digit-sum(1C) = 1 + C = 13. (1C in base 15 = 27 in base 10).
Showing 1-6 of 6 results.