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For n >= 1, a(n) is the least k >= 1 such that 1/d(k) + … + 1/d(k + n - 1) is an integer, d(i) = A000005(i).
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%I #13 Sep 27 2023 13:45:36

%S 1,2,1,25,54,7,53,65,6,22,51,49,343,209,416,624,17,18,338,410,1622,

%T 341,140,849,139,337,1939,338,849,4365,2565,6368,496,4366,132,8392,

%U 131,4453,128,4173,127,487,123,4437,492,122,3011,491,3724,4171,2637,1231,1631,12765,119

%N For n >= 1, a(n) is the least k >= 1 such that 1/d(k) + … + 1/d(k + n - 1) is an integer, d(i) = A000005(i).

%C Conjecture : The sum 1/d(k) + … + 1/d(k + n - 1) = C, C an integer, exists for all k >= 1, n >= 1.

%C Are there, for some fixed n >= 3, infinitely many k's such that 1/d(k) + … + 1/d(k + n - 1) is an integer ?

%e n = 3: 1/d(k) + 1/d(k + 1) + 1/d(k + 2) = C, C an integer, is valid for the least k = 1, thus a(3) = 1.

%e n = 4: 1/d(k) + 1/d(k + 1) + 1/d(k + 2) + 1/d(k + 3) = C, C an integer, is valid for the least k = 25, thus a(4) = 25.

%o (PARI) a(n) = my(k=1); while (denominator(sum(i=0, n-1, 1/numdiv(k+i))) != 1, k++); k; \\ _Michel Marcus_, Sep 27 2023

%Y Cf. A000005, A359056.

%K nonn

%O 1,2

%A _Ctibor O. Zizka_, Sep 27 2023

%E More terms from _Michel Marcus_, Sep 27 2023