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Sum of digits of n in bases 2 and 5.
2

%I #14 Jul 28 2023 04:05:10

%S 0,2,3,5,5,3,4,6,5,7,4,6,6,8,9,7,5,7,8,10,6,8,9,11,10,4,5,7,7,9,6,8,5,

%T 7,8,6,6,8,9,11,6,8,9,11,11,9,10,12,10,12,5,7,7,9,10,8,7,9,10,12,8,10,

%U 11,13,9,7,8,10,10,12,9,11,10,12,13,7,7,9,10,12,6,8,9,11,11,9,10,12,11,13

%N Sum of digits of n in bases 2 and 5.

%C Let n = Sum(c(k)*2^k), c(k) = 0,1, be the binary form of n, n = Sum(d(k)*5^k), d(k) = 0,1,2,3,4 the base 5 form; then a(n) = Sum(c(k)+d(k)).

%C a(n) mod 2 = doubled Thue-Morse sequence A095190.

%H Amiram Eldar, <a href="/A096289/b096289.txt">Table of n, a(n) for n = 0..10000</a>

%F a(n) = A000120(n) + A053824(n). - _Amiram Eldar_, Jul 28 2023

%e n=13: 13=1*2^3+1*2^2+1*2^0, 1+1+1=3, 13=2*5^1+3*5^0, 2+3=5, so a(13)=3+5=8.

%t a[n_] := Total[Flatten@ IntegerDigits[n, {2, 5}]]; Array[a, 100, 0] (* _Amiram Eldar_, Jul 28 2023 *)

%Y Cf. A000120, A010059, A010060, A053824, A095190, A096288.

%K easy,nonn,base

%O 0,2

%A _Miklos Kristof_, Peter Boros, Jun 24 2004