A378270 - OEIS

%I #21 Dec 08 2024 17:15:15

%S 1,2,3,7,8,58,59,259,664,3427,3428,73351,73352,298785,7060868,

%T 43070304,43070305,901194373,901194374,32808600352,1204438945226

%N Number of partitions of 1 into {1/1^2, 1/2^2, 1/3^2, ..., 1/n^2}.

%C From _David A. Corneth_, Nov 24 2024: (Start) Primes n /2 < p <= n occur in exactly one solution namely (p^2) * (1/p^2). Proof If the numerator k of k/p^2 is less then p^2 then p divides the denominator of the sum of the Egyptian fractions as p divides no other number <= n. But the goal is have sum 1 i.e. denominator 1 so p cannot be a divisor of the denominator. Consequently this can be reduced to "Number of partitions of 1 into {1/1^2, 1/2^2, ..., 1/(n/2)^2, ..., 1/n^2}" plus number of primes n/2 < p <= n. The denominators for the first part can be cleared turning this into a partitioning problem of positive integers. (End)

%H <a href="/index/Ed#Egypt">Index entries for sequences related to Egyptian fractions</a>

%F a(p) = a(p-1) + 1 for prime p. - _David A. Corneth_, Nov 22 2024

%e a(4) = 7 because we have 16 * (1/16) = 12 * (1/16) + 1/4 = 8 * (1/16) + 2 * (1/4) = 4 * (1/16) + 3 * (1/4) = 9 * (1/9) = 4 * (1/4) = 1.

%e From _David A. Corneth_, Nov 24 2024: (Start)

%e To find a(12) we can rewrite the problem as "Number of partitions of 1 into {1/1^2, 1/2^2, 1/3^2, 1/4^2, 1/5^2, 1/6^2, 1/8^2, 1/9^2, 1/10^2, 1/12^2} + |{7, 11}|". The lcm of (1^2, 2^2, 3^2, 4^2, 5^2, 6^2, 8^2, 9^2, 10^2, 12^2) is 129600. So this comes a partition problem of (number of partitions of 129600 into parts 129600, 32400, 14400, 8100, 5184, 3600, 2025, 1600, 1296, 900) + |{7, 11}|. (End)

%Y Cf. A000290, A020473, A038034, A348625, A378271.

%K nonn,more,new

%O 1,2

%A _Ilya Gutkovskiy_, Nov 21 2024

%E a(12)-a(21) from _David A. Corneth_, Nov 22 2024