OFFSET
1,1
COMMENTS
It doesn't matter if the primes are included in the sums or excluded as long as the symmetry is taken into account (d(prime) is always 2).
LINKS
Karl-Heinz Hofmann, Table of n, a(n) for n = 1..10000
EXAMPLE
For n=1, the primes are a(1) = p1 = 105373, p2 = 105379, p3 = 105389, and p4 = 105397; the composites between them have the following numbers of divisors:
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Between p1 and p2 Between p2 and p3 Between p3 and p4
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d(105380) = 24
+ d(105381) = 10
+ d(105382) = 4 d(105390) = 24
+ d(105383) = 4 + d(105391) = 12
d(105374) = 16 + d(105384) = 16 + d(105392) = 20
+ d(105375) = 16 + d(105385) = 8 + d(105393) = 12
+ d(105376) = 24 + d(105386) = 16 + d(105394) = 4
+ d(105377) = 4 + d(105387) = 4 + d(105395) = 8
+ d(105378) = 32 + d(105388) = 6 + d(105396) = 12
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sum = 92 sum = 92 sum = 92
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Sums are equal. Thus 105373 is a term.
MATHEMATICA
f[p_, i_, j_] := Sum[DivisorSigma[0, k], {k, p[[i]], p[[j]]}]; Select[Partition[Prime[Range[3*10^5]], 4, 1], f[#, 1, 2] == f[#, 2, 3] == f[#, 3, 4] &][[;; , 1]] (* Amiram Eldar, May 15 2022 *)
PROG
(Python)
from sympy import sieve as p, divisor_count as d
def sum_of_taus(a, b): return sum(d(ab) for ab in range(a+1, b))
print([p[k] for k in range (1, 100000) if sum_of_taus(p[k], p[k+1]) == sum_of_taus(p[k+1], p[k+2]) == sum_of_taus(p[k+2], p[k+3])])
(PARI) a353553(limit) = {my(p1=3, p2=5, p3=7, ns1=sum(k=p1+1, p2-1, numdiv(k)), ns2=sum(k=p2+1, p3-1, numdiv(k)), ns3); forprime(p4=11, limit, ns3=sum(k=p3+1, p4-1, numdiv(k)); if(ns3==ns2 && ns3==ns1, print1(p1, ", ")); p1=p2; p2=p3; p3=p4; ns1=ns2; ns2=ns3)};
a353553(13000000) \\ Hugo Pfoertner, May 17 2022
CROSSREFS
KEYWORD
nonn
AUTHOR
Karl-Heinz Hofmann, May 15 2022
STATUS
approved