How do you find the number of primes in a sequence up to the first thousand terms?

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I have a sequence:



$ 7,19,61,193,577,1633,4417, ... $



and I'd like to find out how many primes and semiprimes there are in this sequence up to the first thousand terms? Any idea how I could do that quickly?



To form the sequence I did: $ 3(2)+1, 6(3)+1, 12(5)+1, 24(8)+1, 48(12)+1, 96(17)+1, ... $



i.e. : $ 3+3+1, 6+6+6+1, 12+12+12+12+12+1, ... $




sequences-and-series prime-numbers




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  • Well I took 3,3,1 and then 6,6,6,1, and then 12,12,12,12,12,1 etc. so the 2,3,5,12,17 are the number of times you multiply the 3, 6, 12, etc.
    – George Thomas
    Jul 22 at 23:24










  • If $t_n$ is the $n$th triangular number, then a formula for the sequence is $a_n = 3cdot 2^n-1(2+t_n-1)+1$ and we know that $t_n = n(n+1)/2.$
    – B. Goddard
    Jul 22 at 23:26











  • So this is a triangular sequence you're saying?
    – George Thomas
    Jul 22 at 23:29










  • No. The triangular numbers come into the formula. If you simplify my calculations you get $a_n = 3cdot 2^n-2(n^2-n+4) +1.$ It's easy to write a quick loop in Maple and get the list that Robert gets below.
    – B. Goddard
    Jul 22 at 23:33














up vote
-2
down vote

favorite












I have a sequence:



$ 7,19,61,193,577,1633,4417, ... $



and I'd like to find out how many primes and semiprimes there are in this sequence up to the first thousand terms? Any idea how I could do that quickly?



To form the sequence I did: $ 3(2)+1, 6(3)+1, 12(5)+1, 24(8)+1, 48(12)+1, 96(17)+1, ... $



i.e. : $ 3+3+1, 6+6+6+1, 12+12+12+12+12+1, ... $




sequences-and-series prime-numbers




share|cite|improve this question





















  • Well I took 3,3,1 and then 6,6,6,1, and then 12,12,12,12,12,1 etc. so the 2,3,5,12,17 are the number of times you multiply the 3, 6, 12, etc.
    – George Thomas
    Jul 22 at 23:24










  • If $t_n$ is the $n$th triangular number, then a formula for the sequence is $a_n = 3cdot 2^n-1(2+t_n-1)+1$ and we know that $t_n = n(n+1)/2.$
    – B. Goddard
    Jul 22 at 23:26











  • So this is a triangular sequence you're saying?
    – George Thomas
    Jul 22 at 23:29










  • No. The triangular numbers come into the formula. If you simplify my calculations you get $a_n = 3cdot 2^n-2(n^2-n+4) +1.$ It's easy to write a quick loop in Maple and get the list that Robert gets below.
    – B. Goddard
    Jul 22 at 23:33












up vote
-2
down vote

favorite









up vote
-2
down vote

favorite











I have a sequence:



$ 7,19,61,193,577,1633,4417, ... $



and I'd like to find out how many primes and semiprimes there are in this sequence up to the first thousand terms? Any idea how I could do that quickly?



To form the sequence I did: $ 3(2)+1, 6(3)+1, 12(5)+1, 24(8)+1, 48(12)+1, 96(17)+1, ... $



i.e. : $ 3+3+1, 6+6+6+1, 12+12+12+12+12+1, ... $




sequences-and-series prime-numbers




share|cite|improve this question













I have a sequence:



$ 7,19,61,193,577,1633,4417, ... $



and I'd like to find out how many primes and semiprimes there are in this sequence up to the first thousand terms? Any idea how I could do that quickly?



To form the sequence I did: $ 3(2)+1, 6(3)+1, 12(5)+1, 24(8)+1, 48(12)+1, 96(17)+1, ... $



i.e. : $ 3+3+1, 6+6+6+1, 12+12+12+12+12+1, ... $





sequences-and-series prime-numbers





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 22 at 23:27
























asked Jul 22 at 23:20









George Thomas

269416




269416











  • Well I took 3,3,1 and then 6,6,6,1, and then 12,12,12,12,12,1 etc. so the 2,3,5,12,17 are the number of times you multiply the 3, 6, 12, etc.
    – George Thomas
    Jul 22 at 23:24










  • If $t_n$ is the $n$th triangular number, then a formula for the sequence is $a_n = 3cdot 2^n-1(2+t_n-1)+1$ and we know that $t_n = n(n+1)/2.$
    – B. Goddard
    Jul 22 at 23:26











  • So this is a triangular sequence you're saying?
    – George Thomas
    Jul 22 at 23:29










  • No. The triangular numbers come into the formula. If you simplify my calculations you get $a_n = 3cdot 2^n-2(n^2-n+4) +1.$ It's easy to write a quick loop in Maple and get the list that Robert gets below.
    – B. Goddard
    Jul 22 at 23:33
















  • Well I took 3,3,1 and then 6,6,6,1, and then 12,12,12,12,12,1 etc. so the 2,3,5,12,17 are the number of times you multiply the 3, 6, 12, etc.
    – George Thomas
    Jul 22 at 23:24










  • If $t_n$ is the $n$th triangular number, then a formula for the sequence is $a_n = 3cdot 2^n-1(2+t_n-1)+1$ and we know that $t_n = n(n+1)/2.$
    – B. Goddard
    Jul 22 at 23:26











  • So this is a triangular sequence you're saying?
    – George Thomas
    Jul 22 at 23:29










  • No. The triangular numbers come into the formula. If you simplify my calculations you get $a_n = 3cdot 2^n-2(n^2-n+4) +1.$ It's easy to write a quick loop in Maple and get the list that Robert gets below.
    – B. Goddard
    Jul 22 at 23:33















Well I took 3,3,1 and then 6,6,6,1, and then 12,12,12,12,12,1 etc. so the 2,3,5,12,17 are the number of times you multiply the 3, 6, 12, etc.
– George Thomas
Jul 22 at 23:24




Well I took 3,3,1 and then 6,6,6,1, and then 12,12,12,12,12,1 etc. so the 2,3,5,12,17 are the number of times you multiply the 3, 6, 12, etc.
– George Thomas
Jul 22 at 23:24












If $t_n$ is the $n$th triangular number, then a formula for the sequence is $a_n = 3cdot 2^n-1(2+t_n-1)+1$ and we know that $t_n = n(n+1)/2.$
– B. Goddard
Jul 22 at 23:26





If $t_n$ is the $n$th triangular number, then a formula for the sequence is $a_n = 3cdot 2^n-1(2+t_n-1)+1$ and we know that $t_n = n(n+1)/2.$
– B. Goddard
Jul 22 at 23:26













So this is a triangular sequence you're saying?
– George Thomas
Jul 22 at 23:29




So this is a triangular sequence you're saying?
– George Thomas
Jul 22 at 23:29












No. The triangular numbers come into the formula. If you simplify my calculations you get $a_n = 3cdot 2^n-2(n^2-n+4) +1.$ It's easy to write a quick loop in Maple and get the list that Robert gets below.
– B. Goddard
Jul 22 at 23:33




No. The triangular numbers come into the formula. If you simplify my calculations you get $a_n = 3cdot 2^n-2(n^2-n+4) +1.$ It's easy to write a quick loop in Maple and get the list that Robert gets below.
– B. Goddard
Jul 22 at 23:33










1 Answer
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up vote
4
down vote



accepted










There are $24$ primes in the first $1000$ terms, namely terms number
$$1, 2, 3, 4, 5, 12, 21, 23, 24, 27, 46, 47, 48, 50, 54, 65, 68, 101, 159, 286, 483, 572, 802, 873$$



There's no shortcut, just generate the terms and test for primality.
Fortunately, that's quite fast. In Maple on my computer it took less than half a second.



Counting the semiprimes would be harder, I think, because I don't know of a way to test for semiprimes that's much faster than factoring, and the $1000$'th term has $307$ digits - no picnic to factor.






share|cite|improve this answer























  • I'm sorry I don't understand this
    – George Thomas
    Jul 22 at 23:31










  • What part don't you understand?
    – Robert Israel
    Jul 22 at 23:33










  • What are the numbers you listed?
    – George Thomas
    Jul 22 at 23:36










  • Oh are those the terms in the sequence that yield primes?
    – George Thomas
    Jul 22 at 23:39






  • 1




    Ah I see that it's the law of small numbers taking effect here
    – George Thomas
    Jul 22 at 23:56










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
4
down vote



accepted










There are $24$ primes in the first $1000$ terms, namely terms number
$$1, 2, 3, 4, 5, 12, 21, 23, 24, 27, 46, 47, 48, 50, 54, 65, 68, 101, 159, 286, 483, 572, 802, 873$$



There's no shortcut, just generate the terms and test for primality.
Fortunately, that's quite fast. In Maple on my computer it took less than half a second.



Counting the semiprimes would be harder, I think, because I don't know of a way to test for semiprimes that's much faster than factoring, and the $1000$'th term has $307$ digits - no picnic to factor.






share|cite|improve this answer























  • I'm sorry I don't understand this
    – George Thomas
    Jul 22 at 23:31










  • What part don't you understand?
    – Robert Israel
    Jul 22 at 23:33










  • What are the numbers you listed?
    – George Thomas
    Jul 22 at 23:36










  • Oh are those the terms in the sequence that yield primes?
    – George Thomas
    Jul 22 at 23:39






  • 1




    Ah I see that it's the law of small numbers taking effect here
    – George Thomas
    Jul 22 at 23:56














up vote
4
down vote



accepted










There are $24$ primes in the first $1000$ terms, namely terms number
$$1, 2, 3, 4, 5, 12, 21, 23, 24, 27, 46, 47, 48, 50, 54, 65, 68, 101, 159, 286, 483, 572, 802, 873$$



There's no shortcut, just generate the terms and test for primality.
Fortunately, that's quite fast. In Maple on my computer it took less than half a second.



Counting the semiprimes would be harder, I think, because I don't know of a way to test for semiprimes that's much faster than factoring, and the $1000$'th term has $307$ digits - no picnic to factor.






share|cite|improve this answer























  • I'm sorry I don't understand this
    – George Thomas
    Jul 22 at 23:31










  • What part don't you understand?
    – Robert Israel
    Jul 22 at 23:33










  • What are the numbers you listed?
    – George Thomas
    Jul 22 at 23:36










  • Oh are those the terms in the sequence that yield primes?
    – George Thomas
    Jul 22 at 23:39






  • 1




    Ah I see that it's the law of small numbers taking effect here
    – George Thomas
    Jul 22 at 23:56












up vote
4
down vote



accepted







up vote
4
down vote



accepted






There are $24$ primes in the first $1000$ terms, namely terms number
$$1, 2, 3, 4, 5, 12, 21, 23, 24, 27, 46, 47, 48, 50, 54, 65, 68, 101, 159, 286, 483, 572, 802, 873$$



There's no shortcut, just generate the terms and test for primality.
Fortunately, that's quite fast. In Maple on my computer it took less than half a second.



Counting the semiprimes would be harder, I think, because I don't know of a way to test for semiprimes that's much faster than factoring, and the $1000$'th term has $307$ digits - no picnic to factor.






share|cite|improve this answer















There are $24$ primes in the first $1000$ terms, namely terms number
$$1, 2, 3, 4, 5, 12, 21, 23, 24, 27, 46, 47, 48, 50, 54, 65, 68, 101, 159, 286, 483, 572, 802, 873$$



There's no shortcut, just generate the terms and test for primality.
Fortunately, that's quite fast. In Maple on my computer it took less than half a second.



Counting the semiprimes would be harder, I think, because I don't know of a way to test for semiprimes that's much faster than factoring, and the $1000$'th term has $307$ digits - no picnic to factor.







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Jul 22 at 23:37


























answered Jul 22 at 23:29









Robert Israel

304k22201441




304k22201441











  • I'm sorry I don't understand this
    – George Thomas
    Jul 22 at 23:31










  • What part don't you understand?
    – Robert Israel
    Jul 22 at 23:33










  • What are the numbers you listed?
    – George Thomas
    Jul 22 at 23:36










  • Oh are those the terms in the sequence that yield primes?
    – George Thomas
    Jul 22 at 23:39






  • 1




    Ah I see that it's the law of small numbers taking effect here
    – George Thomas
    Jul 22 at 23:56
















  • I'm sorry I don't understand this
    – George Thomas
    Jul 22 at 23:31










  • What part don't you understand?
    – Robert Israel
    Jul 22 at 23:33










  • What are the numbers you listed?
    – George Thomas
    Jul 22 at 23:36










  • Oh are those the terms in the sequence that yield primes?
    – George Thomas
    Jul 22 at 23:39






  • 1




    Ah I see that it's the law of small numbers taking effect here
    – George Thomas
    Jul 22 at 23:56















I'm sorry I don't understand this
– George Thomas
Jul 22 at 23:31




I'm sorry I don't understand this
– George Thomas
Jul 22 at 23:31












What part don't you understand?
– Robert Israel
Jul 22 at 23:33




What part don't you understand?
– Robert Israel
Jul 22 at 23:33












What are the numbers you listed?
– George Thomas
Jul 22 at 23:36




What are the numbers you listed?
– George Thomas
Jul 22 at 23:36












Oh are those the terms in the sequence that yield primes?
– George Thomas
Jul 22 at 23:39




Oh are those the terms in the sequence that yield primes?
– George Thomas
Jul 22 at 23:39




1




1




Ah I see that it's the law of small numbers taking effect here
– George Thomas
Jul 22 at 23:56




Ah I see that it's the law of small numbers taking effect here
– George Thomas
Jul 22 at 23:56












 

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