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 1-5 of 5 results.

A277487 a(n) = number of primes encountered before reaching (n^2)-1 when starting from k = ((n+1)^2)-1 and iterating map k -> k - A002828(k).

Original entry on oeis.org

1, 0, 1, 0, 0, 1, 2, 1, 1, 0, 2, 1, 2, 0, 3, 2, 0, 3, 0, 2, 0, 1, 4, 2, 3, 2, 4, 2, 0, 3, 3, 2, 5, 3, 4, 3, 3, 3, 2, 4, 2, 2, 4, 3, 3, 3, 6, 3, 1, 3, 4, 2, 6, 3, 3, 2, 5, 5, 5, 5, 4, 3, 7, 4, 4, 6, 4, 2, 4, 6, 5, 5, 5, 4, 7, 4, 4, 7, 4, 0, 5, 6, 7, 4, 4, 9, 4, 5, 2, 6, 6, 7, 11, 3, 6, 4, 9, 5, 7, 7, 7, 6, 8, 8, 7, 6, 4, 6, 5, 7, 8, 5, 9, 8, 8, 5, 12, 7, 5, 6
Offset: 1

Views

Author

Antti Karttunen, Nov 08 2016

Keywords

Comments

Number of primes on row n of A276574, after the initial zero-row.
Note how for the most n in range 1..10000, a(n) < A277486(n), even though for the most n in the same range A277890(n) < A277891(n). In range n=1..10000, there are only 209 cases where a(n) >= A277486(n).
On the other hand, when a(n) is compared to A277488(n), there is no such marked bias.

Examples

			For n=3, starting from k = ((3+1)^2)-1, and iterating k -> A255131(k), yields 15 -> 11 -> 8, where the iteration stops as the next lower number one less than a square has been reached. Of these numbers only 11 is a prime, thus a(3) = 1.

Links

Crossrefs

Cf. A000290, A002828, A010052, A276574, A277486, A277488, A277890, A277891, A278167 (partial sums).
Cf. A277888.

Programs

  • PARI
    istwo(n:int)=my(f); if(n<3, return(n>=0); ); f=factor(n>>valuation(n, 2)); for(i=1, #f[, 1], if(bitand(f[i, 2], 1)==1&&bitand(f[i, 1], 3)==3, return(0))); 1
isthree(n:int)=my(tmp=valuation(n, 2)); bitand(tmp, 1)||bitand(n>>tmp, 7)!=7
A002828(n)=if(issquare(n), !!n, if(istwo(n), 2, 4-isthree(n))) \\ From Charles R Greathouse IV, Jul 19 2011
A277487(n) = { my(orgk = ((n+1)^2)-1); my(k = orgk, s = 0); while(((k == orgk) || !issquare(1+k)), s = s + if(isprime(k),1,0); k = k - A002828(k)); s; };
for(n=1, 10000, write("b277487.txt", n, " ", A277487(n)));
  • Scheme
    (define (A277487 n) (let ((org_k (- (A000290 (+ 1 n)) 1))) (let loop ((k org_k) (s 0)) (if (and (< k org_k) (= 1 (A010052 (+ 1 k)))) s (loop (- k (A002828 k)) (+ s (A010051 k)))))))

Formula

a(n) <= A277891(n).

A277486 a(n) = number of integers one more than a prime encountered before reaching (n^2)-1 when starting from k = ((n+1)^2)-1 and iterating map k -> k - A002828(k).

Original entry on oeis.org

1, 2, 0, 2, 2, 2, 0, 2, 1, 2, 1, 3, 1, 3, 1, 3, 3, 2, 3, 3, 5, 4, 1, 4, 3, 4, 2, 4, 4, 2, 4, 4, 4, 3, 3, 4, 3, 4, 5, 5, 5, 4, 4, 6, 6, 3, 3, 9, 4, 5, 6, 9, 4, 6, 4, 4, 8, 6, 5, 7, 5, 9, 5, 5, 7, 8, 6, 11, 5, 9, 4, 7, 9, 9, 6, 10, 5, 5, 17, 4, 10, 9, 10, 7, 3, 3, 10, 8, 7, 10, 6, 9, 5, 10, 10, 10, 8, 11, 6, 9, 10, 7, 7, 7, 7, 12, 9, 11, 13, 9, 12, 6, 10, 9, 6
Offset: 1

Views

Author

Antti Karttunen, Nov 08 2016

Keywords

Examples

			For n=6, we start iterating from k = ((6+1)^2)-1 = 48, and then 48 - A002828(48) = 45, 45 - A002828(45) = 43, 43 - A002828(43) = 40, 40 - A002828(40) = 38, and 38 - A002828(38) = 35 (which is 6^2 - 1), and when we subtract one from each, only 47 and 37 are primes, thus a(6) = 2.
For n=7, we start iterating from k = ((7+1)^2)-1 = 63, and 63 -> 59, 59 -> 56, 56 -> 53, 53 -> 51, 51 -> 48 (which is 7^2 - 1), and subtracting one from each of 63, 59, 56, 53 and 51, doesn't yield a prime for any, thus a(7)=0. (Note that even though 48-1 = 47 is a prime, it is not included in the count for n=7).

Links

Crossrefs

Cf. A000290, A010051, A010052, A277487, A277488, A277890, A278166 (partial sums).

Programs

  • PARI
    istwo(n:int)=my(f); if(n<3, return(n>=0); ); f=factor(n>>valuation(n, 2)); for(i=1, #f[, 1], if(bitand(f[i, 2], 1)==1&&bitand(f[i, 1], 3)==3, return(0))); 1
isthree(n:int)=my(tmp=valuation(n, 2)); bitand(tmp, 1)||bitand(n>>tmp, 7)!=7
A002828(n)=if(issquare(n), !!n, if(istwo(n), 2, 4-isthree(n))) \\ From Charles R Greathouse IV, Jul 19 2011
A277486(n) = { my(orgk = ((n+1)^2)-1); my(k = orgk, s = 0); while(((k == orgk) || !issquare(1+k)), s = s + if(isprime(k-1),1,0); k = k - A002828(k)); s; };
for(n=1, 10000, write("b277486.txt", n, " ", A277486(n)));
  • Scheme
    (define (A277486 n) (let ((org_k (- (A000290 (+ 1 n)) 1))) (let loop ((k org_k) (s 0)) (if (and (< k org_k) (= 1 (A010052 (+ 1 k)))) s (loop (- k (A002828 k)) (+ s (A010051 (+ -1 k))))))))

Formula

For n >= 2, a(n) <= A277890(n).

A277890 Number of even numbers encountered before (n^2)-1 is reached when starting from k = ((n+1)^2)-1 and iterating map k -> k - A002828(k).

Original entry on oeis.org

0, 2, 0, 3, 2, 3, 1, 5, 3, 4, 4, 6, 3, 5, 3, 7, 8, 8, 6, 8, 9, 10, 6, 8, 10, 10, 7, 11, 10, 13, 11, 12, 12, 14, 10, 13, 12, 13, 14, 15, 13, 15, 15, 18, 18, 16, 15, 17, 21, 18, 18, 18, 19, 20, 16, 21, 20, 20, 22, 20, 23, 20, 22, 23, 21, 23, 23, 27, 25, 24, 22, 28, 22, 27, 24, 26, 25, 25, 29, 29, 28, 26, 30, 31, 28, 28, 31, 30, 32, 33, 27, 32, 34, 34, 30, 33, 33
Offset: 1

Views

Author

Antti Karttunen, Nov 08 2016

Keywords

Comments

The starting point ((n+1)^2)-1 of the iteration is included if it is even, but the ending point (n^2)-1 is never included in the count.
a(n) = number of even numbers on row n of A276574, after the initial zero-row.
See also comments in A277891.

Examples

			For n=6, we start iterating from k = ((6+1)^2)-1 = 48, and then 48 - A002828(48) = 45, 45 - A002828(45) = 43, 43 - A002828(43) = 40, 40 - A002828(40) = 38, and 38 - A002828(38) = 35 (which is 6^2 - 1), and three of these numbers are even, thus a(6) = 3.

Links

Crossrefs

Cf. A000035, A000290, A002828, A010052, A260734, A276573, A276574, A277486, A277488, A277889, A277891.

Programs

  • PARI
    istwo(n:int)=my(f); if(n<3, return(n>=0); ); f=factor(n>>valuation(n, 2)); for(i=1, #f[, 1], if(bitand(f[i, 2], 1)==1&&bitand(f[i, 1], 3)==3, return(0))); 1
isthree(n:int)=my(tmp=valuation(n, 2)); bitand(tmp, 1)||bitand(n>>tmp, 7)!=7
A002828(n)=if(issquare(n), !!n, if(istwo(n), 2, 4-isthree(n))) \\ From Charles R Greathouse IV, Jul 19 2011
A277890(n) = { my(orgk = ((n+1)^2)-1); my(k = orgk, s = 0); while(((k == orgk) || !issquare(1+k)), s = s + (1-(k%2)); k = k - A002828(k)); s; };
for(n=1, 10000, write("b277890.txt", n, " ", A277890(n)));
  • Scheme
    (define (A277890 n) (let ((org_k (- (A000290 (+ 1 n)) 1))) (let loop ((k org_k) (s 0)) (if (and (< k org_k) (= 1 (A010052 (+ 1 k)))) s (loop (- k (A002828 k)) (+ s (- 1 (A000035 k))))))))

Formula

a(n) + A277891(n) = A260734(n).
For n >= 2, a(n) >= A277486(n).
a(n) >= A277488(n).

A278168 a(n) = number of integers one less than a prime encountered before reaching 0 when starting from k = ((n+1)^2)-1 and iterating map k -> k - A002828(k).

Original entry on oeis.org

0, 1, 1, 3, 4, 5, 5, 8, 10, 13, 15, 16, 17, 19, 20, 23, 25, 28, 29, 31, 35, 39, 40, 42, 45, 47, 49, 52, 56, 59, 62, 66, 69, 73, 76, 78, 82, 87, 92, 96, 100, 103, 107, 112, 116, 120, 123, 127, 133, 137, 143, 151, 155, 159, 162, 167, 174, 177, 184, 186, 192, 198, 202, 209, 216, 220, 225, 232, 236, 244, 250, 254, 258, 261, 267, 278, 282, 287, 292, 301
Offset: 1

Views

Author

Antti Karttunen, Nov 13 2016

Keywords

Examples

			For n=4, starting from k = ((4+1)^2)-1, and iterating k -> A255131(k), yields 24 -> 21 -> 18 -> 16 -> 15 -> 11 -> 8 -> 6 -> 3 before 0 is reached. Subtracting one from each gives [25, 22, 19, 17, 16, 12, 9, 7, 4], of which only 19, 17, and 7 are primes, thus a(4) = 3.

Links

Crossrefs

Partial sums of A277488.
Cf. A002828, A255131, A278166, A278167.

Programs

Formula

a(1) = A277488(1); for n > 1, a(n) = A277488(n) + a(n-1).

A278495 a(n) = number of primes in range [n^2, (n+1)^2] that are leaves in "the least squares beanstalk" tree.

Original entry on oeis.org

1, 2, 1, 2, 2, 2, 1, 1, 2, 4, 1, 2, 1, 3, 2, 4, 3, 3, 3, 5, 3, 2, 2, 4, 4, 4, 4, 3, 4, 4, 4, 4, 2, 3, 3, 2, 4, 2, 5, 4, 6, 3, 5, 4, 5, 5, 4, 6, 3, 3, 6, 8, 4, 5, 3, 5, 5, 5, 4, 6, 6, 7, 5, 5, 7, 6, 8, 8, 8, 8, 5, 5, 5, 8, 7, 7, 7, 3, 13, 5, 8, 6, 8, 7, 8, 5, 14, 7, 8, 8, 10, 7, 5, 8, 6, 7, 6, 9, 4, 10, 4, 9, 8, 6, 8, 8, 8, 6, 10, 11, 13, 9
Offset: 1

Views

Author

Antti Karttunen, Nov 25 2016

Keywords

Comments

Number of terms of A278494 in range [n^2, (n+1)^2], where A278494 are primes p for which there does not exist any such integer k that k - A002828(k) = p.
In other words, number of primes p in range [n^2, (n+1)^2] for which (A002828(1+p) <> 1) and (A002828(2+p) <> 2) and (A002828(3+p) <> 3) and (A002828(4+p) <> 4).
Conjecture: a(n) > 0 for all n >= 1.
Similar guesses are easy to make but hard to prove. I also conjecture that A277487(n) > 0 for all n > 80, and that both A277486(n) > 0 and A277488(n) > 0 for all n > 7. If any of these claims were proved true, it would imply the proof of Legendre's conjecture as well. See also comments in A014085 and sequences A277888 & A278487.

Links

Crossrefs

Cf. A000290, A002828, A010051, A010052, A014085 (an upper bound), A278216, A278494 (primes that are counted), A278496.
Cf. also A277486, A277487, A277488.

Programs

  • PARI
    istwo(n:int)=my(f); if(n<3, return(n>=0); ); f=factor(n>>valuation(n, 2)); for(i=1, #f[, 1], if(bitand(f[i, 2], 1)==1&&bitand(f[i, 1], 3)==3, return(0))); 1
isthree(n:int)=my(tmp=valuation(n, 2)); bitand(tmp, 1)||bitand(n>>tmp, 7)!=7
A002828(n)=if(issquare(n), !!n, if(istwo(n), 2, 4-isthree(n))) \\ From Charles R Greathouse IV, Jul 19 2011
A278495(n) = { my(s = 0); for(k=(n^2),(n+1)^2, if((isprime(k) && (A002828(1+k) <> 1) && (A002828(2+k) <> 2) && (A002828(3+k) <> 3) && (A002828(4+k) <> 4)),s = s+1) ); s; };
for(n=1, 10000, write("b278495.txt", n, " ", A278495(n)));
  • Scheme
    (define (A278495 n) (let loop ((k (+ -1 (A000290 (+ 1 n)))) (s 0)) (if (= 1 (A010052 k)) s (loop (- k 1) (+ s (* (A010051 k) (if (zero? (A278216 k)) 1 0)))))))

Formula

For all n >= 1, a(n) <= A014085(n).
Showing 1-5 of 5 results.

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