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A229784
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a(n) = (1^1^1 + 2^2^2 . . . + n^n^n) mod 10.
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1
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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
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OFFSET
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0,3
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COMMENTS
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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
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LINKS
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Table of n, a(n) for n=0..110.
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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Cf. A120962, A185353.
Sequence in context: A329091 A306398 A093825 * A091494 A021139 A020790
Adjacent sequences: A229781 A229782 A229783 * A229785 A229786 A229787
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KEYWORD
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nonn,base,easy
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AUTHOR
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José María Grau Ribas, Sep 29 2013
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EXTENSIONS
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a(0)=0 prepended by Alois P. Heinz, Jun 17 2022
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STATUS
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approved
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