

A104477


Number of successive squarefree intervals between primes.


2



1, 0, 1, 0, 1, 0, 2, 0, 1, 0, 3, 0, 2, 0, 3, 0, 2, 0, 4, 0, 3, 0, 4, 0, 4, 0, 3, 0, 5, 0, 6, 0, 4, 0, 5, 0, 5, 0, 6, 0, 6, 0, 6, 0, 5, 0, 8, 0, 7, 0, 6, 0, 7, 0, 8, 0, 7, 0, 7, 0, 9, 0, 8, 0, 9, 0, 8, 0, 9, 0, 8, 0, 8, 0, 11, 0, 10, 0, 11, 0, 10, 0, 8, 0, 11, 0, 10, 0, 12, 0, 9, 0, 12, 0, 14, 0, 9, 0, 10, 0
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OFFSET

1,7


COMMENTS

Find the number (the "run length") of successive intervals [p, p'=nextprime(p)] (followed by [p', p''], then [p'', p'''] etc.) which do not contain a square. When a square (n+1)^2 is found in such an interval, this will result in a term a(2n) = 0, preceded by a(2n1) = the number of intervals of primes counted before reaching that square, i.e., between n^2 and (n+1)^2.  M. F. Hasler, Oct 01 2018


LINKS

Vincenzo Librandi, Table of n, a(n) for n = 1..10000


FORMULA

a(2n) = 0: this is the interval from the greatest prime less than the (n+1)th square, through that square and up to the least prime greater than that square.  Robert G. Wilson v, Apr 23 2005
a(2n1) = the difference between the indices of the greatest prime less than (n+1)^2 and the least prime greater than n^2.  Robert G. Wilson v, Apr 23 2005
a(2n1) = A014085(n)  1 = primepi((n+1)^2  primepi(n^2)  1.  M. F. Hasler, Oct 01 2018


EXAMPLE

a(1)=1 because the first interval between primes (2 to 3) is free of squares.
a(2)=0 because there is a square between 3 and 5.
a(7)=2 because there are two successive squarefree intervals: 17 to 19; and 19 to 23.
a(8)=0 because between 23 and 29 there is a square: 25.


MATHEMATICA

NextPrim[n_] := Block[{k = n + 1}, While[ !PrimeQ[k], k++ ]; k]; PrevPrim[n_] := Block[{k = n  1}, While[ !PrimeQ[k], k ]; k]; f[n_] := If[ EvenQ[n], 0, PrimePi[ PrevPrim[(n + 3)^2/4]]  PrimePi[ NextPrim[(n + 1)^2/4]]]; Table[ f[n], {n, 100}] (* Robert G. Wilson v, Apr 23 2005 *)


PROG

(PARI) p=2; c=0; forprime(np=p+1, 1e4, if( sqrtint(p) < sqrtint(np), print1(c", ", c=0, ", "), c++); p=np) \\ For illustrative purpose. Better:
A104477(n)=if(bittest(n, 0), primepi((1+n\/=2)^2)primepi(n^2)1, 0) \\ M. F. Hasler, Oct 01 2018


CROSSREFS

Cf. A061265, A031265, A104481.
Equals A014085  1 without the initial term, interleaved with 0's.
Sequence in context: A112554 A120616 A108044 * A224928 A308066 A052173
Adjacent sequences: A104474 A104475 A104476 * A104478 A104479 A104480


KEYWORD

easy,nonn


AUTHOR

Alexandre Wajnberg, Apr 18 2005


EXTENSIONS

More terms from Robert G. Wilson v, Apr 23 2005
Offset corrected by M. F. Hasler, Oct 01 2018


STATUS

approved



