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 A229784 a(n) = (1^1^1 + 2^2^2 . . . + n^n^n) mod 10. 1
 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 (list; graph; refs; listen; history; text; internal format)
 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 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

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Last modified March 29 09:20 EDT 2023. Contains 361598 sequences. (Running on oeis4.)