A378270 Number of partitions of 1 into {1/1^2, 1/2^2, 1/3^2, ..., 1/n^2}.

1, 2, 3, 7, 8, 58, 59, 259, 664, 3427, 3428, 73351, 73352, 298785, 7060868, 43070304, 43070305, 901194373, 901194374, 32808600352, 1204438945226

Offset

1,2

Comments

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)

Links

Table of n, a(n) for n=1..21.

Index entries for sequences related to Egyptian fractions

Formula

a(p) = a(p-1) + 1 for prime p. - David A. Corneth, Nov 22 2024

Example

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.
From David A. Corneth, Nov 24 2024: (Start)
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)

Crossrefs

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

Sequence in context: A068342 A038034 A103173 * A098863 A053960 A114056

Adjacent sequences: A378267 A378268 A378269 * A378271 A378272 A378273

Keyword

nonn,more,new

Author

Ilya Gutkovskiy, Nov 21 2024

Extensions

a(12)-a(21) from David A. Corneth, Nov 22 2024

Status

approved