A229784 a(n) = (1^1^1 + 2^2^2 . . . + n^n^n) mod 10.

0, 1, 7, 4, 0, 5, 1, 4, 0, 9, 9, 0, 6, 9, 5, 0, 6, 3, 9, 8, 8, 9, 5, 2, 8, 3, 9, 2, 8, 7, 7, 8, 4, 7, 3, 8, 4, 1, 7, 6, 6, 7, 3, 0, 6, 1, 7, 0, 6, 5, 5, 6, 2, 5, 1, 6, 2, 9, 5, 4, 4, 5, 1, 8, 4, 9, 5, 8, 4, 3, 3, 4, 0, 3, 9, 4, 0, 7, 3, 2, 2, 3, 9, 6, 2, 7, 3, 6, 2, 1, 1, 2, 8, 1, 7, 2, 8, 5, 1, 0, 0, 1, 7, 4, 0, 5, 1, 4, 0, 9, 9

Offset

0,3

Comments

The last digit of 1^1^1 + 2^2^2 +...+ n^n^n, which has period 100.

Sum of A120962 mod 10. - T. D. Noe, Sep 30 2013

Links

Ray Chandler, Table of n, a(n) for n = 0..999

Index entries for linear recurrences with constant coefficients, signature (1, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1, -1, 1).

Mathematica

Table[Mod[Sum[PowerMod[i, i^i, 10], {i, 1, n}], 10], {n, 200}]
Mod[Accumulate[Table[PowerMod[i, i^i, 10], {i, 100}]], 10] (* T. D. Noe, Sep 30 2013 *)

Prog

(PARI) a(n)=lift(sum(k=1, n%100, Mod(k, 10)^k^k)) \\ Charles R Greathouse IV, Dec 27 2013
(Python)
from itertools import count, accumulate, islice
def A229784_gen(): # generator of terms
    yield from accumulate((pow(k, k**k, 10) for k in count(1)), func=lambda x, y:(x+y)%10)
A229784_list = list(islice(A229784_gen(), 40)) # Chai Wah Wu, Jun 17 2022

Crossrefs

Cf. A120962, A185353.

Sequence in context: A329091 A306398 A093825 * A091494 A021139 A020790

Adjacent sequences: A229781 A229782 A229783 * A229785 A229786 A229787

Keyword

nonn,base,easy

Author

José María Grau Ribas, Sep 29 2013

Extensions

a(0)=0 prepended by Alois P. Heinz, Jun 17 2022

Status

approved