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Sum of digits of 4^n.
13

%I #29 Dec 19 2020 10:32:42

%S 1,4,7,10,13,7,19,22,25,19,31,25,37,40,43,37,58,61,64,67,61,46,58,70,

%T 73,76,79,82,85,70,82,85,88,109,103,70,109,130,106,100,112,124,118,

%U 112,115,118,139,151,127,112,115,118,121,142,145,121,160

%N Sum of digits of 4^n.

%H Harry J. Smith, <a href="/A065713/b065713.txt">Table of n, a(n) for n = 0..1000</a>

%F a(n) = A007953(A000302(n)). - _Michel Marcus_, Nov 01 2013 [corrected by _Georg Fischer_, Dec 19 2020]

%F a(n) = A001370(2n). Results given there imply a(n) > log_4(n) + 1/2, n > 0, but we can conjecture & expect a(n) ~ 9*log_10(2)*n. - _M. F. Hasler_, May 18 2017

%t Table[Total[IntegerDigits[4^n]], {n, 0, 60}] (* _Vincenzo Librandi_, Oct 08 2013 *)

%o (PARI) SumD(x)= { local(s=0); while (x>9, s+=x-10*(x\10); x\=10); return(s + x) } { for (n=0, 1000, a=SumD(4^n); write("b065713.txt", n, " ", a) ) } \\ _Harry J. Smith_, Oct 27 2009

%o (PARI) a065713(n)=sumdigits(4^n); \\ _Michel Marcus_, Nov 01 2013

%Y Cf. A000302, A007953.

%Y Cf. sum of digits of k^n: A001370 (k=2), A004166 (k=3), this sequence (k=4), A066001 (k=5), A066002 (k=6), A066003(k=7), A066004 (k=8), A065999 (k=9), A066005 (k=11), A066006 (k=12), A175527 (k=13).

%K nonn,base

%O 0,2

%A _N. J. A. Sloane_, Dec 11 2001