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1 + Sum[ k^(n-1), {k,1,n}].
0

%I #2 Mar 31 2012 13:20:27

%S 2,4,15,101,980,12202,184821,3297457,67731334,1574304986,40851766527,

%T 1170684360925,36720042483592,1251308658130546,46034015337733481,

%U 1818399978159990977,76762718946972480010,3448810852242967123282

%N 1 + Sum[ k^(n-1), {k,1,n}].

%C Prime p divides a(p). Prime p divides a(p-2) for p>3. p^2 divides a(p-2) for prime p=7. p^2 divides a(p^2-2) for prime p except p=3. p^3 divides a(p^2-2) for prime p=7. p^3 divides a(p^3-2) for prime p>3. p^4 divides a(p^3-2) for prime p=7. p^4 divides a(p^4-2) for prime p>3. p^5 divides a(p^3-2) for prime p=7. It appears that p^k divides a(p^k-2) for prime p>3 and 7^(k+1) divides a(7^k-2) for integer k>0.

%F a(n) = 1 + Sum[ k^(n-1), {k,1,n}]. a(n) = 1 + A076015[n].

%t Table[(1+Sum[k^(n-1),{k,1,n}]),{n,1,23}]

%Y Cf. A076015.

%K nonn

%O 1,1

%A _Alexander Adamchuk_, Aug 04 2006