A001370 -id:A001370 - cp's OEIS frontend

A261009 Write 2^n in base 3, add up the "digits".

1, 2, 2, 4, 4, 4, 4, 6, 4, 8, 8, 10, 10, 8, 10, 16, 12, 14, 12, 16, 14, 18, 16, 12, 10, 12, 14, 20, 20, 22, 24, 26, 24, 22, 22, 22, 18, 20, 26, 28, 28, 28, 26, 30, 30, 30, 26, 26, 26, 32, 38, 40, 38, 38, 28, 34, 40, 42, 38, 40, 46, 40, 38, 42, 48, 44, 42, 40, 42, 48, 48, 44

Offset: 0

N. J. A. Sloane, Aug 14 2015

Comment from Jean-Paul Allouche, Oct 25 2015: As mentioned by Holdum et al. (2015) the following problem, cited in "Concrete Mathematics" by Graham, Knuth, and Patashnik (1994), is still open: prove that for all n > 256, binomial(2n,n) is either divisible by 4 or by 9 (cf. A000984). This can be easily reduced to show that, for all k >= 9, 2*a(k) - a(k+1) >= 4. This has been proved up to huge values of k (Holdum et al. mention k = 10^{13}).

For additional information about the divisibility of binomial(2n,n) by squares see the comments and references in A000984, - N. J. A. Sloane, Oct 29 2015

			2^7 = 128_10 = 11202_3, so a(7) = 1+1+2+0+2 = 6.

Giovanni Resta, Table of n, a(n) for n = 0..10000
Cernenoks J., Iraids J., Opmanis M., Opmanis R., Podnieks K., Integer complexity: experimental and analytical results II, arXiv:1409.0446 [math.NT] (September 2014)
Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete Math., 2n-d ed.; Addison-Wesley, 1994
Sebastian Tim Holdum, Frederik Ravn Klausen, Peter Michael Reichstein Rasmussen, Powers in prime bases and a problem on central coefficients, Integers 15 (2015), #A43
K. Podnieks, Digits of pi: limits to the seeming randomness, arXiv:1411.3911 [math.NT], 2014.

Cf. A000079, A000984, A053735, A007089.

Sum of digits of k^n in base b for various pairs (k,b): A001370 (2,10), A011754 (3,2), A261009 (2,3), A261010 (5,3).

a261009 = a053735 . a000079  -- Reinhard Zumkeller, Aug 14 2015

S:=n->add(i,i in convert(2^n,base,3)); [seq(S(n),n=0..100)];

Table[Total@ IntegerDigits[2^n, 3], {n, 0, 100}] (* Giovanni Resta, Aug 14 2015 *)

a(n) = vecsum(digits(2^n, 3)); \\ Michel Marcus, Aug 14 2015

a(n) = A053735(A000079(n)). - Michel Marcus, Aug 14 2015

A175512 (Digit sum of 7^n) mod n.

Original entry on oeis.org

0, 1, 1, 3, 2, 4, 4, 7, 1, 3, 5, 1, 0, 2, 4, 4, 7, 1, 3, 16, 19, 9, 16, 1, 22, 8, 10, 21, 5, 28, 22, 7, 28, 6, 17, 28, 4, 16, 19, 22, 25, 10, 40, 43, 28, 3, 34, 40, 31, 34, 13, 40, 8, 1, 49, 43, 10, 9, 34, 19, 22, 8, 19, 40, 52, 64, 0, 66, 37, 49, 52, 55, 58, 70, 37, 49, 11, 64, 6, 25, 28, 67, 43, 55, 40, 20, 64, 13, 8, 28, 49, 34, 82, 0, 79, 82, 85, 61, 73, 67

Offset: 1

N. J. A. Sloane, Dec 03 2010

Cf. A000420, A066003, A067863; A000079, A001370, A175434, A175169.

Table[Mod[Plus@@IntegerDigits[7^n],n],{n,150}] (* Harvey P. Dale, Dec 16 2010 *)

A066538 Sum of the digits of the n-th Mersenne prime (A000668).

Original entry on oeis.org

3, 7, 4, 10, 19, 13, 28, 46, 73, 112, 139, 154, 697, 847, 1675, 3106, 3106, 4258, 5755, 5950, 13216, 13693, 14980, 27202, 28939, 31339, 60337, 116455, 149365, 179488, 291745, 1026544, 1163443, 1704376, 1893388, 4038358, 4092673, 9440671, 18243946, 28445131, 32580433, 35170384, 41201947, 44142151, 50349694, 57766339, 58416637

Offset: 1

Robert G. Wilson v, Jan 06 2002

From Gord Palameta, Jul 21 2018: (Start)
a(38) and a(39) were calculated by Enoch Haga, Sep 07 1999 and Dec 17 2001; a(40) through a(42) were calculated by Andrew Rupinski, Mar 12 2005. (See the Carlos Rivera link.)
It appears that asymptotically a(n)/A000043(n) = 9*log_10(2)/2. (End)

Amiram Eldar, Table of n, a(n) for n = 1..48 (terms 1..47 from Gord Palameta)
Carlos Rivera, Puzzle 74.- SOD(2^6972593-1), The Prime Puzzles & Problems Connection.

Cf. A000043, A000668, A007953, A001348.
Subsequence of: A007953, A007605.
Cf. A001370 (sum of digits of 2^n).

ep = {the exponents from A000043}; a = {}; Do[ a = Append[a, Apply[ Plus, IntegerDigits[ 2^ep[[n]] - 1]]], {n, 1, 47} ]; a
(* Second program: *)
Array[Total@ IntegerDigits[2^MersennePrimeExponent@ # - 1] &, 45] (* Michael De Vlieger, Jul 22 2018 *)

a(n) = A007953(A000668(n)). - Amiram Eldar, Oct 16 2024

Definition corrected by Omar E. Pol, Apr 01 2008

a(38)-a(47) from Gord Palameta, Jul 21 2018

A112435 Next term is the sum of the last 10 digits in the sequence, beginning with a(10) = 4.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 4, 8, 16, 23, 28, 38, 45, 42, 41, 41, 36, 34, 32, 31, 30, 28, 29, 33, 34, 37, 44, 42, 37, 41, 39, 41, 38, 43, 40, 39, 39, 46, 45, 47, 54, 51, 45, 44, 43, 39, 42, 42, 39, 43, 43, 38, 43, 44, 40, 37, 40, 33, 32, 29, 36, 35, 39, 45, 49, 51, 48, 52, 47

Offset: 1

Alexandre Wajnberg, Dec 11 2005

Variation on Angelini's A112395. The sequence cycles at a(16)=38 and the loop has 312 terms. It is exactly the same loop as in A112433 "Next term is the sum of the last 10 digits in the sequence, beginning with a(10) = 2" and the first term of this 4-loop is the first term of that 2-loop . Computed by Gilles Sadowski.

			a(16)=38 because 4 + 4 + 8 + 1+6 +2+3 + 2+8 = 38

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A112395, A004207, A065075, A001370.

nxt[{t_,a_}]:={Take[Join[t,IntegerDigits[Total[t]]],-10],Total[t]}; Join[ {0,0,0,0,0,0,0,0,0},NestList[nxt,{{0,0,0,0,0,0,0,0,0,4},4},80][[All,2]]] (* Harvey P. Dale, Jun 22 2019 *)

A186775 Numbers k such that digitsum(2^k) > digitsum(2^(k+1)).

Original entry on oeis.org

3, 4, 8, 9, 15, 16, 20, 21, 23, 24, 26, 28, 29, 33, 34, 36, 39, 40, 41, 46, 48, 51, 52, 55, 56, 57, 60, 63, 64, 67, 68, 69, 74, 75, 76, 77, 80, 82, 83, 85, 86, 88, 91, 92, 94, 95, 97, 98, 100, 102, 106, 108, 112, 113, 116, 118, 121, 124, 126

Offset: 1

Thomas Nordhaus, Feb 26 2011

If 2^k and 2^(k+1) acted like random numbers of their size, the probability that k would be in the sequence would be 1/2 + O(1/k). So very possibly a(n) ~ 2n. - Charles R Greathouse IV, Aug 08 2022

			3 is in the sequence because digitsum(2^3) = 8 > 7 = digitsum(2^4).

J.W.L. (Jan) Eerland, Table of n, a(n) for n = 1..49688

Cf. A001370.

DeleteCases[Table[If[Total[Total[IntegerDigits[2^n]]]>Total[IntegerDigits[2^(n+1)]],n,k],{n,0, 10^5}],k] (* J.W.L. (Jan) Eerland, Aug 08 2022 *)

from itertools import count, islice, pairwise
def ds2(n): return sum(map(int, str(1< t[1])
print(list(islice(agen(), 60))) # Michael S. Branicky, Aug 08 2022

def is_A186775(n): return sum((2^n).digits()) > sum((2^(n+1)).digits()) # D. S. McNeil, Feb 27 2011

A253298 Digital root for the following sequences, F(4n)/F(4); F(12n)/F(12); F(20*n)/F(20), where the pattern increases by 8, ad infinitum, with the Fibonacci numbers F = A000045.

Original entry on oeis.org

1, 7, 3, 5, 5, 3, 7, 1, 9, 8, 2, 6, 4, 4, 6, 2, 8, 9, 1, 7, 3, 5, 5, 3, 7, 1, 9, 8, 2, 6, 4, 4, 6, 2, 8, 9, 1, 7, 3, 5, 5, 3, 7, 1, 9, 8, 2, 6, 4, 4, 6, 2, 8, 9, 1, 7, 3, 5, 5, 3, 7, 1, 9, 8, 2, 6, 4, 4, 6, 2, 8, 9

Offset: 1

Peter M. Chema, Dec 30 2014

Cyclical and palindromic in two parts with periodicity 18: {1, 7, 3, 5, 5, 3, 7, 1, 9, 8, 2, 6, 4, 4, 6, 2, 8, 9}.
Digital root of the period is 9, its mean and median is 5, and its product is (9!)^2.
See A253368 for the initial motivation for this sequence.
From Peter M. Chema, Jul 04 2016: (Start)
A composite of three respective digital root sequences in alternation: a "halving sequence" of 1, 5, 7, 8, 4, 2, a "doubling sequence" of 7, 5, 1, 2, 4, 8, and a three-six-nine circuit of 3, 3, 9, 6, 6, 9.
Also the digital root of A000045(4n)/3 or A004187(n).  In general terms, sequences defined by Fib(x*n)/ Fib(x) where x=(8*a-4) all share the same digital root (e.g., F(4*n)/F(4); F(12*n)/F(12); F(20*n)/F(20); F(28*n)/F(28); F(36*n)/F(36), etc.) (End)

Tom Barnett, Phi VBM Tori Array, YouTube video (see first two minutes).
Index entries for linear recurrences with constant coefficients, signature (1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1,-1,1).

Cf. A010888, A253368, A070366 and A001370.

f[n_] := Mod[ Fibonacci[ 12n]/144, 9]; Array[f, 5*18] (* Robert G. Wilson v, Jan 23 2015 *)
LinearRecurrence[{1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1, -1, 1},{1, 7, 3, 5, 5, 3, 7, 1, 9, 8, 2, 6, 4, 4, 6, 2, 8},72] (* Ray Chandler, Aug 12 2015 *)

a(n) = A010888(A253368(n)).

G.f.: x*(1 + 7*x + 3*x^2 + 5*x^3 + 5*x^4 + 3*x^5 + 7*x^6 + x^7 + 9*x^8 + 8*x^9 + 2*x^10 + 6*x^11 + 4*x^12 + 4*x^13 + 6*x^14 + 2*x^15 + 8*x^16 + 9*x^17)/(1 - x^18). - Vincenzo Librandi, Mar 28 2016

Edited. Numbers and name changed to fit A253368. Formula adapted. Cross reference added. - Wolfdieter Lang, Jan 28 2015

Name generalized by Peter M. Chema, Jul 04 2016

A261010 Write 5^n in base 3, add up the "digits".

Original entry on oeis.org

1, 3, 5, 7, 7, 9, 9, 13, 15, 13, 13, 17, 19, 21, 21, 27, 25, 25, 25, 23, 27, 33, 31, 39, 35, 45, 37, 57, 45, 47, 45, 45, 53, 47, 55, 51, 57, 59, 67, 67, 69, 65, 67, 65, 71, 79, 71, 65, 67, 75, 65, 71, 73, 83, 69, 79, 81, 85, 79, 89, 87, 95, 89, 85, 97, 99, 93, 101, 107

Offset: 0

N. J. A. Sloane, Aug 14 2015

Chai Wah Wu, Table of n, a(n) for n = 0..10000
Cernenoks J., Iraids J., Opmanis M., Opmanis R., Podnieks K., Integer complexity: experimental and analytical results II, arXiv:1409.0446 [math.NT] (September 2014)
K. Podnieks, Digits of pi: limits to the seeming randomness, arXiv:1411.3911 [math.NT], 2014.

Sum of digits of k^n in base b for various pairs (k,b): A001370 (2,10), A011754 (3,2), A261009 (2,3), A261010 (5,3).

S:=n->add(i,i in convert(5^n,base,3)); [seq(S(n),n=0..100)];

def digits(n, b=10): # digits of n in base 2 <= b <= 62
    x, y = n, ''
    while x >= b:
        x, r = divmod(x,b)
        y += str(r) if r < 10 else (chr(r+87) if r < 36 else chr(r+29))
    y += str(x) if x < 10 else (chr(x+87) if x < 36 else chr(x+29))
    return y[::-1]
def A261010(n):
    return sum([int(d) for d in digits(5**n,3)]) # Chai Wah Wu, Aug 14 2015

A071906 Sum of digits of 2^n (mod 2).

Original entry on oeis.org

1, 0, 0, 0, 1, 1, 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1

Offset: 0

Santi Spadaro, Jun 13 2002

G. C. Greubel, Table of n, a(n) for n = 0..10000

Cf. A001370.

f[n_] := Mod[Plus @@ IntegerDigits[2^n], 2]; Table[ f@n, {n, 0, 104}] (* Robert G. Wilson v, May 04 2009 *)

a(n) = sumdigits(2^n) % 2; \\ Michel Marcus, Apr 20 2017

A112437 Next term is the sum of the last 10 digits in the sequence, beginning with a(10) = 6.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 0, 6, 6, 12, 15, 21, 24, 30, 21, 21, 18, 24, 24, 27, 33, 36, 36, 39, 45, 45, 48, 51, 48, 48, 51, 48, 48, 54, 51, 45, 48, 48, 48, 51, 51, 48, 48, 48, 48, 54, 57, 57, 57, 57, 57, 60, 54, 51, 45, 42, 36, 39, 42, 42, 39, 45, 45, 42, 42, 42, 36, 36, 36, 39, 45

Offset: 1

Alexandre Wajnberg, Dec 11 2005

Variation on Angelini's A112395. The sequence cycles at a(24)=36 and the loop has 104 terms. It is the same loop as in A112434 "Next term is the sum of the last 10 digits in the sequence, beginning with a(10) = 3"; the first term of this 6-loop is the 47th term of that 3-loop and the first term of that 3-loop is the 59th of this 6-loop. Computed by Gilles Sadowski.

			a(24)=36 because 1+8 + 2+4 + 2+4 + 2+7 + 3+3 = 36

Cf. A112395, A004207, A065075, A001370.

A293011 a(n) is the smallest positive k such that f(k) = n*g(k) where f = A007953 and g = A000120, or 0 if no such k exists.

Original entry on oeis.org

1, 2, 6, 4, 32, 48, 16, 8, 288, 64, 128, 8196, 256, 2048, 16896, 278528, 2097664, 589824, 4096, 8192, 8388609, 16384, 536870944, 268435488, 65536, 32768, 268959744, 17179869440, 524288, 4294967298, 1048576, 8589934594, 8589934596

Offset: 1

Altug Alkan, Sep 28 2017

Conjecture: a(n) > 0 for all n >= 1.
Numbers n such that a(n) is not a power of 2 are 3, 6, 9, 12, 15, 16, 17, 18, ...
a(21) = 8388609 = 2^23 + 1 is the second odd term of this sequence after a(1) = 1.
Smallest n such that a(n + 1) = a(n) + 2 is 32 and a(32) = 2*(2^32 + 1).
For n <= 170, A000120(a(n)) <= 2. - Robert Israel, Nov 22 2019

			a(9) = 288 = 2^8 + 2^5 because A007953(288) = 2 + 8 + 8 = 18, 18 / 2 = 9 and 288 is the least number with this property.

Robert Israel, Table of n, a(n) for n = 1..170
Robert Israel, Color-coded logarithmic plot

Cf. A000120, A001370, A007953.

# This code returns a(n) if A000120(a(n)) <= 3 and it can prove that no
# smaller number with A000120 >= 4 can have A007953 large enough. If it
# can't prove that, it returns FAIL.
sdd:= n -> convert(convert(n,base,10),`+`):
g:= proc(n) local found, k1, k2, k3, x, y, m,bd;
  found:= false;
  for k1 from 1 while not found do
    for k2 from 0 to k1-1 do
      x:= 2^k1 + 2^k2;
      if sdd(x) = 2*n then found:= true; break fi
  od od;
  for k1 from 0 to ilog2(x) do
    if sdd(2^k1) = n then x:= 2^k1; break fi
  od;
  m:= ilog10(x);
  bd:= floor(x/10^m)+9*m;
  if bd <= 3*n then return x fi;
  found:= false;
  for k1 from 2 to ilog2(x) while not found do
    for k2 from 1 to k1-1 while not found do
      for k3 from 0 to k2-1 do
         y:= 2^k1 + 2^k2 + 2^k3;
         if y > x or sdd(y) = 3*n then found:= true; break fi;
  od od od;
  if found then x:= min(x,y) fi;
  bd:= floor(x/10^m)+9*m;
  if bd <= 4*n then x else FAIL fi;
end proc:
map(g, [$1..50]); # Robert Israel, Nov 22 2019

a(n) = {my(k=1); while ((hammingweight(k))*n != sumdigits(k), k++); k; }

cp's OEIS Frontend

A261009 Write 2^n in base 3, add up the "digits".

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A175512 (Digit sum of 7^n) mod n.

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A066538 Sum of the digits of the n-th Mersenne prime (A000668).

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A112435 Next term is the sum of the last 10 digits in the sequence, beginning with a(10) = 4.

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A186775 Numbers k such that digitsum(2^k) > digitsum(2^(k+1)).

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Sage

A253298 Digital root for the following sequences, F(4*n)/F(4); F(12*n)/F(12); F(20*n)/F(20), where the pattern increases by 8, ad infinitum, with the Fibonacci numbers F = A000045.

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A261010 Write 5^n in base 3, add up the "digits".

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Maple

Python

A071906 Sum of digits of 2^n (mod 2).

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A253298 Digital root for the following sequences, F(4n)/F(4); F(12n)/F(12); F(20*n)/F(20), where the pattern increases by 8, ad infinitum, with the Fibonacci numbers F = A000045.