A177955 Partial sums of A045542.

3, 10, 18, 33, 57, 83, 114, 149, 197, 260, 340, 439, 559, 683, 810, 953, 1121, 1316, 1531, 1755, 1997, 2252, 2540, 2863, 3205, 3565, 3964, 4404, 4887, 5398, 5926, 6501, 7125, 7800, 8528, 9311, 10151, 11050, 12010, 13009, 14032, 15120, 16275, 17499

Offset: 1

Author

Jonathan Vos Post, May 16 2010

Keywords

• easy
• nonn

Comments

Partial sums of sub-perfect powers: perfect powers (squares, cubes, etc.) minus 1. The subsequence of primes in the partial sum begins: 3, 83, 149, 197, 439, 683, 953, 1531, 1997, 9311, 10151, 13009. The subsequence of subperfect powers in the partial sum (numbers n such that n-1 is a perfect power) begins: 10 (because 10-1=9=3^2), 197 because 197-1=196=2^2 * 7^2.

Examples

			a(40) = 3 + 7 + 8 + 15 + 24 + 26 + 31 + 35 + 48 + 63 + 80 + 99 + 120 + 124 + 127 + 143 + 168 + 195 + 215 + 224 + 242 + 255 + 288 + 323 + 342 + 360 + 399 + 440 + 483 + 511 + 528 + 575 + 624 + 675 + 728 + 783 + 840 + 899 + 960 + 999 = 13009 is prime.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A001597, A045542.

Programs

Maple

N:= 10^4:
P:= sort(convert({seq(seq(i^p-1, p=2..floor(log[i](N))),i=2..isqrt(N))},list)):
ListTools:-PartialSums(P); # Robert Israel, Jul 06 2017

Formula

a(n) = SUM[i=1..n] A045542(i) = SUM[i=1..n] (A001597(i+1) - 1) = (SUM[i=1..n] A001597(i+1)) - n.