A183228 -id:A183228 - cp's OEIS frontend

A183226 Sum of digits of (2^n) in base 5, also sum of digits of (10^n) in base 5.

1, 2, 4, 4, 4, 4, 8, 4, 4, 8, 12, 12, 12, 12, 8, 12, 16, 20, 20, 20, 16, 12, 20, 24, 28, 20, 32, 32, 24, 32, 40, 40, 32, 24, 28, 32, 32, 40, 28, 36, 36, 40, 44, 40, 36, 40, 36, 44, 44, 44, 44, 48, 52, 52, 48, 56, 40, 56, 68, 60, 52, 52, 48, 60, 56, 64, 60, 48, 56, 60, 60, 64, 60, 60, 60, 64, 52, 48, 64, 68, 56, 80, 80

Offset: 0

Washington Bomfim, Jan 01 2011

If i >= 2, a(n) mod 4 = 0. (Cf. A053824)

			a(9) = 8 because 10^9 = 4022000000000_5, and 2^9 = 512 = 4022_5.

Robert Israel, Table of n, a(n) for n = 0..10000

Cf. A055476, A173670, A183227, A183228.

a:= n-> add(i, i=convert (2^n, base, 5)):
seq(a(n), n=0..82);  # Alois P. Heinz, Jan 06 2011

Table[Plus@@IntegerDigits[2^n, 5], {n, 0, 49}] (* Either that one or this one *) Table[Plus@@IntegerDigits[10^n, 5], {n, 0, 49}] (* Alonso del Arte, Jan 06 2011 *)

\\  L is the list of the N digits of 2^n in quinary.
     \\ L[1] = a_0 , ..., L[N] = a_(N-1).
convert(n)={n=2^n; x=n; N=floor(log(n)/log(5))+1;
L = listcreate(N);
while(x, n=floor(n/5); r=x-5*n; listput(L, r); x=n; );
L; N};
for(n=0,100,convert(n);an=0;for(i=1,N,an+=L[i];); print1(an,", "));

t(n) = if(n<1, 0, if(n%5, t(n-1)+1, t(n/5)));
vector(200, n, n--; t(2^n)) \\ Altug Alkan, Oct 28 2015

A183227 a(n) is the base-5 digit sum of 10^n-1.

Original entry on oeis.org

0, 5, 11, 15, 19, 23, 31, 31, 35, 43, 51, 55, 59, 63, 63, 71, 79, 87, 91, 95, 95, 95, 107, 115, 123, 119, 135, 139, 135, 147, 159, 163, 159, 155, 163, 171, 175, 187, 179, 191, 195, 203, 211, 211, 211, 219, 219, 231, 235, 239

Offset: 0

Washington Bomfim, Jan 01 2011

			a(9) = 43 because 10^9 - 1 is written as 4021444444444_5, and 2^9 - 1 = 511 is written as 4021_5.

Cf. A183226, A183228.

A053824 := proc(n) add(d,d=convert(n,base,5)) ; end proc:
A183227 := proc(n) A053824(10^n-1) ; end proc: # R. J. Mathar, Jan 09 2011

\\L is a list of the N digits of 2^n - 1 in  quinary
convert(n)={n = 2^n - 1; x=n; N=floor(log(n)/log(5))+1;
L = listcreate(N);
while(x, n=floor(n/5); r=x-5*n; listput(L, r); x=n; );
L; N};
print1("0, "); for(n = 1,100, convert(n); s = 0; for(i = 1, N, s += L[i];); print1(s+4*n, ", "));

a(n) = A053824(10^n-1) = 4*n + A053824(2^n-1).

cp's OEIS Frontend

A183226 Sum of digits of (2^n) in base 5, also sum of digits of (10^n) in base 5.

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A183227 a(n) is the base-5 digit sum of 10^n-1.

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