|
|
A257077
|
|
a(n) = prime(n)-prime(1)-prime(2)-...-prime(k), while the result > 0.
|
|
1
|
|
|
2, 1, 3, 2, 1, 3, 7, 2, 6, 1, 3, 9, 13, 2, 6, 12, 1, 3, 9, 13, 15, 2, 6, 12, 20, 1, 3, 7, 9, 13, 27, 2, 8, 10, 20, 22, 28, 3, 7, 13, 19, 21, 31, 33, 37, 2, 14, 26, 30, 32, 36, 1, 3, 13, 19, 25, 31, 33, 39, 43, 2, 12, 26, 30, 32, 36, 3, 9, 19, 21, 25, 31, 39
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
It appears that a(n) = n occurs only for n=3, 7, 13. It also appears that a(n+1) is never equal to a(n).
The list of indices such that a(n)=1 correspond to the primes in A053845. - Michel Marcus, Apr 16 2015
|
|
LINKS
|
|
|
FORMULA
|
|
|
EXAMPLE
|
a(1) = 2, since there is no previous prime.
a(2) = 1, since 3 - 2 = 1.
a(3) = 3, since 5 - 2 = 3.
a(4) = 2, since 7 - 2 - 3 = 2.
a(5) = 1, since 11 - 2 - 3 - 5 = 1.
a(6) = 3, since 13 - 2 - 3 - 5 = 3.
a(13) = 13, since 41 - 2 - 3 - 5 - 7 - 11 = 13.
|
|
MATHEMATICA
|
lst = {}; i = 1; While[i <= 1000, x = Prime[i]; k = 1; While[x > 0, x -= Prime[k]; k++]; x += Prime[k - 1]; AppendTo[lst, x]; i++]; lst
|
|
PROG
|
(PARI) a(n) = {s = prime(n); k = 1; while ((ns = (s - prime(k))) > 0, s = ns; k++); s; } \\ Michel Marcus, Apr 16 2015
(PARI) s=0; q=2; forprime(p=2, 10, if(s+q>p, s+=q; q=nextprime(q+1)); print1(p-s", ")) \\ Charles R Greathouse IV, Apr 22 2015
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,easy
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|