Haris Ziko has authored 3 sequences.
A330822
a(n) = digsum(2^a(n-1)) with a(0) = 0.
Original entry on oeis.org
0, 1, 2, 4, 7, 11, 14, 22, 25, 29, 41, 50, 76, 106, 142, 214, 313, 380, 508, 691, 875, 1184, 1687, 2243, 3164, 4126, 5578, 7855, 10676, 13981, 18659, 25421, 34277, 46409, 63023, 85658, 116248, 157660, 213892, 290554, 393145, 532838, 723451, 981020, 1328449, 1799363
Offset: 0
Starting with a(0) = 0.
a(1) = digsum(2^0) = digsum(1) = 1.
a(2) = digsum(2^1) = digsum(2) = 2.
a(3) = digsum(2^2) = digsum(4) = 4.
a(4) = digsum(2^4) = digsum(16) = 7.
etc.
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a:=[0,1]; [n le 2 select a[n] else &+Intseq(2^Self(n-1)):n in [1..50]]; // Marius A. Burtea, Jan 03 2020
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a[0] = 0; a[n_] := a[n] = Total @ IntegerDigits[2^a[n-1]]; Array[a, 45, 0] (* Amiram Eldar, Jan 03 2020 *)
NestList[Total[IntegerDigits[2^#]]&,0,50] (* Harvey P. Dale, Jul 19 2025 *)
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a(n)={my(x=0); for(i=0, n-1, x=sumdigits(2^x)); x};
lista(nn)={my(x=0, v=vector(nn+1)); v[1]=0; for(i=1, nn, x=sumdigits(2^x); v[i+1]=x); v};
A329814
The smallest base b where the sum of the digits for the number n in the base b is the smallest, with 1 < b < n and a(1) = a(2) = 1.
Original entry on oeis.org
1, 1, 2, 2, 2, 2, 6, 2, 3, 2, 10, 2, 12, 7, 14, 2, 2, 2, 18, 2, 20, 11, 22, 2, 5, 5, 3, 3, 28, 3, 30, 2, 2, 2, 34, 6, 6, 19, 38, 2, 40, 6, 42, 22, 44, 23, 46, 2, 7, 5, 50, 26, 52, 3, 54, 7, 56, 29, 58, 30, 60, 31, 62, 2, 2, 2, 66, 2, 68, 35, 70, 2, 72, 37, 74
Offset: 1
For n = 5:
n in base 2 = [1, 0, 1] -> digitSum(5, 2) = 2.
n in base 3 = [1, 2] -> digitSum(5, 3) = 3.
n in base 4 = [1, 1] -> digitSum(5, 4) = 2.
Base 2 has the smallest sum of the digits for n = 5 ->
therefore a(5) = 2.
For n = 7:
n in base 2 = [1, 1, 1] -> digitSum(7, 2) = 3.
n in base 3 = [2, 1] -> digitSum(7, 3) = 3.
n in base 4 = [1, 3] -> digitSum(7, 4) = 4.
n in base 5 = [1, 2] -> digitSum(7, 5) = 3.
n in base 6 = [1, 1] -> digitSum(7, 6) = 2.
Base 6 has the smallest sum of the digits for n = 7 ->
therefore a(7) = 6.
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f:= proc(n) local F, t,d,bmin,s,r,b;
F:= ifactors(n)[2];
d:= igcd(seq(t[2],t=F));
if d > 1 then return mul(t[1]^(t[2]/d),t=F) fi;
F:= ifactors(n-1)[2];
d:= igcd(seq(t[2],t=F));
if d=1 then bmin:= n-1 else bmin:= mul(t[1]^(t[2]/d),t=F) fi;
for s in numtheory:-divisors(n) do
r:= n/s-1;
F:= ifactors(s)[2];
d:= igcd(seq(t[2],t=F));
b:= mul(t[1]^(t[2]/d),t=F);
if b < bmin and r = b^padic:-ordp(r,b) then bmin:= b fi
od;
bmin;
end proc:
map(f, [$1..100]); # Robert Israel, Dec 05 2019
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a[n_] := Block[{b=1, r=n, t}, Do[t = Plus @@ IntegerDigits[n, i]; If[t < r, r=t; b=i], {i, 2, n-1}]; b]; Array[a, 75] (* Giovanni Resta, Nov 22 2019 *)
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a(n)={my(best_b=1, best_dig_sum=n); if(n>1, for(b=2, n-1, dig_sum=sumdigits(n, b); if(best_dig_sum>dig_sum, best_dig_sum=dig_sum; best_b=b))); best_b};
A308828
Number of sequences that include all residues modulo n starting with x_0 = 0 and then x_i given recursively by x_{i+1} = a * x_i + c (mod n) where a and c are integers in the interval [0..n-1].
Original entry on oeis.org
1, 1, 2, 2, 4, 2, 6, 8, 18, 4, 10, 4, 12, 6, 8, 32, 16, 18, 18, 8, 12, 10, 22, 16, 100, 12, 162, 12, 28, 8, 30, 128, 20, 16, 24, 36, 36, 18, 24, 32, 40, 12, 42, 20, 72, 22, 46, 64, 294, 100, 32, 24, 52, 162, 40, 48, 36, 28, 58, 16, 60, 30, 108, 512, 48, 20, 66, 32, 44, 24
Offset: 1
For n = 1:
a = 0, c = 0: [0];
#cycles = 1 -> a(1) = 1.
For n = 5:
a = 1, c = 1: [0, 1, 2, 3, 4];
a = 1, c = 2: [0, 2, 4, 1, 3];
a = 1, c = 3: [0, 3, 1, 4, 2];
a = 1, c = 4: [0, 4, 3, 2, 1];
#cycles = 4 -> a(5) = 4.
For n = 8:
a = 1, c = 1: [0, 1, 2, 3, 4, 5, 6, 7];
a = 1, c = 3: [0, 3, 6, 1, 4, 7, 2, 5];
a = 1, c = 5: [0, 5, 2, 7, 4, 1, 6, 3];
a = 1, c = 7: [0, 7, 6, 5, 4, 3, 2, 1];
a = 5, c = 1: [0, 1, 6, 7, 4, 5, 2, 3];
a = 5, c = 3: [0, 3, 2, 5, 4, 7, 6, 1];
a = 5, c = 5: [0, 5, 6, 3, 4, 1, 2, 7];
a = 5, c = 7: [0, 7, 2, 1, 4, 3, 6, 5];
#cycles = 8 -> a(8) = 8.
- D. E. Knuth, The Art of Computer Programming, Vol. 3, Random Numbers, Section 3.2.1.2, p. 16.
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checkFullSet(v,n)={my(v2=vector(n), unique=1); for(i=1, n, my(j=v[i]+1); if(v2[j]==1, unique=0; break, v2[j]=1;);); unique;};
doCycle(a,c,m)={my(v_=vector(m), x=c); v_[1]=c; for(i=1, m-1, v_[i+1]=(a*v_[i]+c)%m;); v_;};
getCycles(m)={my(L=List()); for(a=0, m-1, for(c=0, m-1, my(v1=doCycle(a,c,m)); if(checkFullSet(v1,m), listput(L, v1)););); Mat(Col(L))};
a(n)={my(M=getCycles(n)); matsize(M)[1]};
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