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%I #27 Feb 04 2023 20:47:27
%S 1,13,313,1359,245895,131186351,468729047,1830140937
%N Numbers k such that k divides Sum_{i=1..k} 10^(1 + floor(log_10(p(i)))) - 1 - p(i), where p(i) is the i-th prime number.
%C Alternative Name : The arithmetic mean of the first k 9's complements of primes is an integer.
%F k: (Sum_{i=1..k} 10^(1 + floor(log_10(p(i)))) - 1 - p(i)) / k = c, c an integer.
%e k = 13: first 13 prime numbers are {2,3,5,7,11,13,17,19,23,29,31,37,41}, their 9's complements are {7,6,4,2,88,86,82,80,76,70,68,62,58} and (7 + 6 + ... + 62 + 58) / 13 = 53, thus 13 is a term.
%t s = 0; p = 2; pow = 10; seq = {}; Do[s += pow - 1 - p; If[Divisible[s, k], AppendTo[seq, k]]; p = NextPrime[p]; If[p > pow, pow *= 10], {k, 1, 250000}]; seq (* _Amiram Eldar_, Feb 04 2023 *)
%Y Cf. A000040, A045345, A055642, A061601.
%K nonn,more
%O 1,2
%A _Ctibor O. Zizka_, Feb 04 2023
%E a(5)-a(8) from _Amiram Eldar_, Feb 04 2023
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