A298102 The first of five consecutive integers the sum of which is equal to the sum of five consecutive prime numbers.
77, 279, 293, 327, 347, 353, 401, 437, 509, 641, 675, 683, 785, 803, 839, 885, 947, 961, 1169, 1177, 1193, 1239, 1325, 1337, 1395, 1433, 1461, 1501, 1545, 1639, 1683, 1715, 1731, 1777, 1809, 1915, 1955, 1989, 2031, 2059, 2139, 2145, 2345, 2387, 2393, 2431
Offset: 1
Keywords
Examples
77 is in the sequence because 77+78+79+80+81 = 395 = 71+73+79+83+89.
Links
- Colin Barker, Table of n, a(n) for n = 1..1000
Programs
-
Mathematica
p = {2, 3, 5, 7, 11}; lst = {}; While[p[[1]] < 3001, t = Plus @@ p; If[Mod[t, 10] == 5, AppendTo[lst, (t - 10)/5]]; p = Join[Rest@p, {NextPrime[p[[-1]]]}]]; lst (* Robert G. Wilson v, Jan 14 2018 *) Select[(#-10)/5&/@(Total/@Partition[Prime[Range[400]],5,1]),IntegerQ] (* Harvey P. Dale, Jun 22 2019 *) -
PARI
L=List(); forprime(p=2, 2500, q=nextprime(p+1); r=nextprime(q+1); s=nextprime(r+1); t=nextprime(s+1); u=p+q+r+s+t; if((u-10)%5==0, listput(L, (u-10)\5))); Vec(L)
-
PARI
upto(n) = my(res = List(), pr = primes(5), s = vecsum(pr)); while(pr[5] < n, if(s == 5 * pr[3], listput(res, pr[1])); lp = nextprime(pr[5] + 1); s += (lp - pr[1]); for(i = 1, 4, pr[i] = pr[i+1]); pr[5] = lp); res \\ David A. Corneth, Jan 12 2018
Extensions
New name by David A. Corneth, Jan 12 2018
Comments