|
|
A167790
|
|
a(n) is the index k of k-th prime p(k) in the smallest sum s(k)=2+3+...+p(k)=t*p(n) of first k primes where t is a true divisor and first occurrence of factor p(n) (n=1,2,3,...)
|
|
2
|
|
|
3, 10, 3, 5, 8, 49, 13, 23, 23, 7, 39, 29, 15, 10, 39, 25, 30, 151, 38, 19, 139, 27, 174, 21, 287, 422, 240, 24, 94, 22, 16, 173, 861, 231, 143, 140, 213, 902, 18, 134, 143, 310, 70, 58, 295, 550, 237, 210, 229, 57, 221, 271, 194, 540, 145, 718, 116, 184, 90, 71, 168
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
It is conjectured that the sequence is infinite
If t is not restricted to nontrivial divisors, the sequence becomes A111287. - R. J. Mathar, Nov 17 2009
|
|
REFERENCES
|
Richard E. Crandall and Carl Pomerance, Prime Numbers, Springer 2005
Leonard E. Dickson, History of the Theory of numbers, vol. I, Dover Publications 2005)
Paulo Ribenboim, The New Book of Prime Number Records, Springer 1996
|
|
LINKS
|
|
|
FORMULA
|
a(n) = min[2+3+...+p(k)/t], where the minimum is taken with respect to k, the denominator t > 1 is an integer divisor of numerator s(k)=2+3+...+p(k).
|
|
EXAMPLE
|
(1) s(5)=2+3+5+7+11=28=2^2*7=4*p(4) gives a(4)=5 as first occurrence of prime factor p(4)=7;
(2) s(8)=2+3+5+7+11+13+17+19=77=7*11=7*p(5) gives a(5)=8 as first occurrence of prime factor p(5)=11;
(3) s(422)=2+3+5+...+2917=570145= 5 * 101 * 1129=5645*p(26) gives a(26)=422 and demonstrates the numerical difficulties.
|
|
CROSSREFS
|
Cf. A007504 (sum of first n primes).
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
Eva-Maria Zschorn (e-m.zschorn(AT)zaschendorf.km3.de), Nov 12 2009, Nov 13 2009
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|