A062397 -id:A062397 - cp's OEIS frontend

A034524 a(n) = 11^n + 1.

2, 12, 122, 1332, 14642, 161052, 1771562, 19487172, 214358882, 2357947692, 25937424602, 285311670612, 3138428376722, 34522712143932, 379749833583242, 4177248169415652, 45949729863572162, 505447028499293772

Offset: 0

N. J. A. Sloane

T. D. Noe, Table of n, a(n) for n = 0..200
Index entries for linear recurrences with constant coefficients, signature (12,-11).

Sequences of the form m^n + 1: A000012 (m=0), A007395 (m=1), A000051 (m=2), A034472 (m=3), A052539 (m=4), A034474 (m=5), A062394 (m=6), A034491 (m=7), A062395 (m=8), A062396 (m=9), A062397 (m=10), this sequence (m=11), A178248 (m=12), A141012 (m=13), A228081 (m=64).

Cf. A001020.

[11^n +1: n in [0..30]]; // G. C. Greubel, Mar 11 2023

LinearRecurrence[{12,-11},{2,12},18] (* Ray Chandler, Aug 26 2015 *)
11^Range[0,30]+1 (* G. C. Greubel, Mar 11 2023 *)

a(n)=11^n+1 \\ Charles R Greathouse IV, Sep 24 2015

[sigma(11,n)for n in range(0,18)] # - Zerinvary Lajos, Jun 04 2009

From Mohammad K. Azarian, Jan 02 2009: (Start)
G.f.: 1/(1-x) + 1/(1-11*x).
E.g.f.: exp(x) + exp(11*x). (End)
From G. C. Greubel, Mar 11 2023: (Start)
a(n) = 11*a(n-1) - 10.
a(n) = A001020(n) + 1. (End)

A178248 a(n) = 12^n + 1.

Original entry on oeis.org

2, 13, 145, 1729, 20737, 248833, 2985985, 35831809, 429981697, 5159780353, 61917364225, 743008370689, 8916100448257, 106993205379073, 1283918464548865, 15407021574586369, 184884258895036417, 2218611106740436993

Offset: 0

Stuart Clary, Dec 20 2010

Prime factors of a(n) are in the Cunningham Project.

			a(3) = 12^3 + 1 = 1729.

Vincenzo Librandi, Table of n, a(n) for n = 0..200
S. S. Wagstaff, Jr., The Cunningham Project.
Index entries for linear recurrences with constant coefficients, signature (13,-12).

Cf. A000051, A001021, A007689, A016125, A024140, A034472, A034474, A034491, A034524, A052539, A062394, A062395, A062396, A062397, A063376, A063481, A074600-A074624, A228081.

[12^n + 1: n in [0..20]]; // Vincenzo Librandi, May 02 2011

nmax = 17; 12^Range[0, nmax] + 1
LinearRecurrence[{13,-12},{2,13},20] (* Harvey P. Dale, Nov 02 2015 *)

vector(21,n,12^(n-1)+1) \\ Charles R Greathouse IV, May 02 2011

O.g.f.: (2-13*x)/(1-13*x+12*x^2) = (2-13*x)/((1-x)(1-12*x)).
E.g.f.: exp(x) + exp(12*x). - Stefano Spezia, Mar 20 2023
From Elmo R. Oliveira, Dec 15 2023: (Start)
a(n) = 12*a(n-1) - 11 for n>0.
a(n) = 13*a(n-1) - 12*a(n-2) for n>1.
a(n) = A001021(n)+1 = A024140(n)+2.
a(n) = (11*A016125(n) + 13)/12. (End)

A047855 a(n) = A047848(7,n).

Original entry on oeis.org

1, 2, 12, 112, 1112, 11112, 111112, 1111112, 11111112, 111111112, 1111111112, 11111111112, 111111111112, 1111111111112, 11111111111112, 111111111111112, 1111111111111112, 11111111111111112, 111111111111111112, 1111111111111111112, 11111111111111111112, 111111111111111111112

Offset: 0

Clark Kimberling

n-th difference of a(n), a(n-1), ..., a(0) is A001019(n-1) for n >= 1.
Range of A164898, apart from first term. - Reinhard Zumkeller, Aug 30 2009
a(n) is the number of integers less than or equal to 10^n, whose initial digit is 1. - Michel Marcus, Jul 04 2019
a(n) is 2^n represented in bijective base-2 numeration. - Alois P. Heinz, Aug 26 2019
This sequence proves both A028842 (numbers with prime product of digits) and A028843 (numbers with prime iterated product of digits) are infinite. Proof: Suppose either of those sequences is finite. Label as omega the supposed last term. Compute n = ceiling(log_10 omega) + 1. Then a(n) > omega. The product of digits of a(n) is 2, contradicting the assumption that omega is the final term of either A028842 or A028843. - Alonso del Arte, Apr 14 2020
For n >= 2, the concatenation of a(n) with 8*a(n) equals (3*R_n+3)^2, where R_n = A002275(n) is the repunit with n 1's; hence this sequence, except for {1,2}, is a subsequence of A115549. - Bernard Schott, Apr 30 2022

Ivan Panchenko, Table of n, a(n) for n = 0..200
Wikipedia, Bijective numeration.
Index entries for linear recurrences with constant coefficients, signature (11,-10).

Cf. A001019, A002275, A028842, A028843, A047848, A115549, A164898.

Cf. A002281, A062397.

[(10^n + 8)/9: n in [0..40]]; // G. C. Greubel, Jan 11 2025

a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=10*a[n-1]+1 od: seq(a[n]+1, n=0..18); # Zerinvary Lajos, Mar 20 2008

Join[{1}, Table[FromDigits[PadLeft[{2}, n, 1]], {n, 30}]] (* Harvey P. Dale, Apr 17 2013 *)
(10^Range[0, 29] + 8)/9 (* Alonso del Arte, Apr 12 2020 *)

a(n)=if(n==0,1,if(n==1,2,11*a(n-1)-10*a(n-2)))
for(i=0,10,print1(a(i),",")) \\ Lambert Klasen, Jan 28 2005

def A047855(n): return (pow(10,n) +8)//9
print([A047855(n) for n in range(41)]) # G. C. Greubel, Jan 11 2025

[gaussian_binomial(n,1,10)+1 for n in range(17)] # Zerinvary Lajos, May 29 2009

(List.fill(20)(10: BigInt)).scanLeft(1: BigInt)( * ).map(n => (n + 8)/9) // Alonso del Arte, Apr 12 2020

a(n) = (10^n + 8)/9. - Ralf Stephan, Feb 14 2004
a(0) = 1, a(1) = 2, a(n) = 11*a(n-1) - 10*a(n-2) for n > 1. - Lambert Klasen (lambert.klasen(AT)gmx.net), Jan 28 2005
G.f.: (1 - 9*x)/(1 - 11*x + 10*x^2). - Philippe Deléham, Oct 05 2009
a(n) = 10*a(n-1) - 8 (with a(0) = 1). - Vincenzo Librandi, Aug 06 2010
From Elmo R. Oliveira, Apr 03 2025: (Start)
E.g.f.: exp(x)*(8 + exp(9*x))/9.
a(n) = (A062397(n) - A002281(n))/2. (End)

More terms from Harvey P. Dale, Apr 17 2013

A069283 a(n) = -1 + number of odd divisors of n.

Original entry on oeis.org

0, 0, 0, 1, 0, 1, 1, 1, 0, 2, 1, 1, 1, 1, 1, 3, 0, 1, 2, 1, 1, 3, 1, 1, 1, 2, 1, 3, 1, 1, 3, 1, 0, 3, 1, 3, 2, 1, 1, 3, 1, 1, 3, 1, 1, 5, 1, 1, 1, 2, 2, 3, 1, 1, 3, 3, 1, 3, 1, 1, 3, 1, 1, 5, 0, 3, 3, 1, 1, 3, 3, 1, 2, 1, 1, 5, 1, 3, 3, 1, 1, 4, 1, 1, 3, 3, 1, 3, 1, 1, 5, 3, 1, 3, 1, 3, 1, 1, 2, 5, 2

Offset: 0

Reinhard Zumkeller, Mar 13 2002

Number of nontrivial ways to write n as sum of at least 2 consecutive integers. That is, we are not counting the trivial solution n=n. E.g., a(9)=2 because 9 = 4 + 5 and 9 = 2 + 3 + 4. a(8)=0 because there are no integers m and k such that m + (m+1) + ... + (m+k-1) = 8 apart from k=1, m=8. - Alfred Heiligenbrunner, Jun 07 2004
Also number of sums of sequences of consecutive positive integers excluding sequences of length 1 (e.g., 9 = 2+3+4 or 4+5 so a(9)=2). (Useful for cribbage players.) - Michael Gilleland, Dec 29 2002
Let M be any positive integer. Then a(n) = number of proper divisors of M^n + 1 of the form M^k + 1.
This sequence gives the distinct differences of triangular numbers Ti giving n : n = Ti - Tj; none if n = 2^k. If factor a = n or a > (n/a - 1)/2 : i = n/a + (a - 1)/2; j = n/a - (a+1)/2. Else : i = n/2a + (2a - 1)/2; j = n/2a - (2a - 1)/2. Examples: 7 is prime; 7 = T4 - T2 = (1 + 2 + 3 + 4) - (1 + 2) (a = 7; n/a = 1). The odd factors of 35 are 35, 7 and 5; 35 = T18 - T16 (a = 35) = T8 - T1 (a = 7) = T5 - T7 (a = 5). 144 = T20 - T11 (a = 9) = T49 - T46 (a = 3). - M. Dauchez (mdzzdm(AT)yahoo.fr), Oct 31 2005
Also number of partitions of n into the form 1 + 2 + ...( k - 1) + k + k + ... + k for some k >= 2. Example: a(9) = 2 because we have [2, 2, 2, 2, 1] and [3, 3, 2, 1]. - Emeric Deutsch, Mar 04 2006
a(n) is the number of nontrivial runsum representations of n, and is also known as the politeness of n. - Ant King, Nov 20 2010
Also number of nonpowers of 2 dividing n, divided by the number of powers of 2 dividing n, n > 0. - Omar E. Pol, Aug 24 2019
a(n) only depends on the prime signature of A000265(n). - David A. Corneth, May 30 2020, corrected by Charles R Greathouse IV, Oct 31 2021

			a(14) = 1 because the divisors of 14 are 1, 2, 7, 14, and of these, two are odd, 1 and 7, and -1 + 2 = 1.
a(15) = 3 because the divisors of 15 are 1, 3, 5, 15, and of these, all four are odd, and -1 + 4 = 3.
a(16) = 0 because 16 has only one odd divisor, and -1 + 1 = 0.
Using Ant King's formula: a(90) = 5 as 90 = 2^1 * 3^2 * 5^1, so a(90) = (1 + 2) * (1 + 1) - 1 = 5. - _Giovanni Ciriani_, Jan 12 2013
x^3 + x^5 + x^6 + x^7 + 2*x^9 + x^10 + x^11 + x^12 + x^13 + x^14 + ...
a(120) = 3 as the odd divisors of 120 are the odd divisors of 15 as 120 = 15*2^3. 15 has 4 odd divisors so that gives a(120) = 4 - 1 = 3. - _David A. Corneth_, May 30 2020

Ronald L. Graham, Donald E. Knuth and Oren Patashnik, Concrete Mathematics, 2nd ed., Addison-Wesley, 1994, see exercise 2.30 on p. 65.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Tom M. Apostol, Sums of Consecutive Positive Integers, The Mathematical Gazette, Vol. 87, No. 508, (March 2003), pp. 98-101.
Alfred Heiligenbrunner, Sum of adjacent numbers (in German).
Henri Picciotto, Staircases.
Wikipedia, Polite Number.

Cf. A000265, A001227, A001620, A062397, A057934, A138591, A182469.

Cf. A095808 (sums of ascending and descending consecutive integers).

a069283 0 = 0
a069283 n = length $ tail $ a182469_row n
-- Reinhard Zumkeller, May 01 2012

[0] cat [-1 + #[d:d in Divisors(n)| IsOdd(d)]:n in [1..100]]; // Marius A. Burtea, Aug 24 2019

g:=sum(x^(k*(k+1)/2)/(1-x^k),k=2..20): gser:=series(g,x=0,115): seq(coeff(gser,x,n),n=0..100); # Emeric Deutsch, Mar 04 2006
A069283 := proc(n)
    A001227(n)-1 ;
end proc: # R. J. Mathar, Jun 18 2015

g[n_] := Module[{dL = Divisors[2n], dP}, dP = Transpose[{dL, 2n/dL}]; Select[dP, ((1 < #[[1]] < #[[2]]) && (Mod[ #[[1]] - #[[2]], 2] == 1)) &] ]; Table[Length[g[n]], {n, 1, 100}]
Table[Length[Select[Divisors[k], OddQ[#] &]] - 1, {k, 100}] (* Ant King, Nov 20 2010 *)
Join[{0}, Times @@@ (#[[All, 2]] & /@ Replace[FactorInteger[Range[2, 50]], {2, a_} -> {2, 0}, Infinity] + 1) - 1] (* Horst H. Manninger, Oct 30 2021 *)

{a(n) = if( n<1, 0, sumdiv( n, d, d%2) - 1)} /* Michael Somos, Aug 07 2013 */

a(n) = numdiv(n >> valuation(n, 2)) - 1 \\ David A. Corneth, May 30 2020

from sympy import divisor_count
def A069283(n): return divisor_count(n>>(~n&n-1).bit_length())-1 if n else 0 # Chai Wah Wu, Jul 16 2022

a(n) = 0 if and only if n = 2^k.
a(n) = A001227(n)-1.
a(n) = 1 if and only if n = 2^k * p where k >= 0 and p is an odd prime. - Ant King, Nov 20 2010
G.f.: sum(k>=2, x^(k(k + 1)/2)/(1 - x^k) ). - Emeric Deutsch, Mar 04 2006
If n = 2^k p1^b1 p2^b2 ... pr^br, then a(n) = (1 + b1)(1 + b2) ... (1 + br) - 1. - Ant King, Nov 20 2010
Dirichlet g.f.: (zeta(s)*(1-1/2^s) - 1)*zeta(s). - Geoffrey Critzer, Feb 15 2015
a(n) = (A000005(n) - A001511(n))/A001511(n) = A326987(n)/A001511(n), with n > 0 in both formulas. - Omar E. Pol, Aug 24 2019
G.f.: Sum_{k>=1} x^(3*k) / (1 - x^(2*k)). - Ilya Gutkovskiy, May 30 2020
From David A. Corneth, May 30 2020: (Start)
a(2*n) = a(n).
a(n) = A001227(A000265(n)) - 1. (End)
Sum_{k=1..n} a(k) ~ n*log(n)/2 + (gamma + log(2)/2 - 3/2)*n, where gamma is Euler's constant (A001620). - Amiram Eldar, Dec 01 2023

Edited by Vladeta Jovovic, Mar 25 2002

A119704 a(n) = number of distinct prime factors of 10^n+1 = omega(10^n+1).

Original entry on oeis.org

1, 1, 1, 3, 2, 2, 2, 2, 2, 5, 3, 4, 3, 3, 4, 7, 5, 4, 3, 2, 4, 7, 4, 5, 3, 5, 3, 7, 4, 3, 7, 2, 4, 8, 4, 5, 6, 4, 3, 9, 4, 3, 7, 4, 4, 12, 4, 4, 9, 4, 7, 8, 4, 2, 6, 9, 5, 6, 5, 4, 6, 3, 3, 11, 3, 6, 8, 2, 4, 10, 11, 3, 5, 4, 7, 11, 6, 11, 7, 4, 9, 11, 3, 7, 8, 8, 3, 8, 4, 4, 11, 6, 4, 8, 4, 6, 8, 4

Offset: 0

Lekraj Beedassy, Jun 09 2006

nonn

			a(1) = number of distinct prime factors of 11 = 1.
a(3) = number of distinct prime factors of 1001 = 3.
a(11) = 4 because 10^11+1 = 11*11*23*4093*8779 has 4 distinct factors.

Max Alekseyev, Table of n, a(n) for n = 0..331 [First 309 terms from Ray Chandler, from the Kamada link]
Makoto Kamada, Factorizations of 100...001.

Cf. A001221, A003021, A038371, A057934, A062397.

Table[Length[FactorInteger[10^n + 1]], {n, 0, 50}] (* Stefan Steinerberger, Jun 13 2006 *)
PrimeNu[10^Range[0,100]+1] (* The program will take some time to run *) (* Harvey P. Dale, Aug 27 2019 *)

a(n) = A001221(A062397(n)). - Ray Chandler, May 02 2017

More terms from Don Reble, Jun 13 2006

A003021 Largest prime factor of 10^n + 1.

Original entry on oeis.org

2, 11, 101, 13, 137, 9091, 9901, 909091, 5882353, 52579, 27961, 8779, 99990001, 1058313049, 121499449, 9091, 69857, 21993833369, 999999000001, 909090909090909091, 5964848081, 909091, 1056689261, 549797184491917

Offset: 0

N. J. A. Sloane

J. Brillhart et al., Factorizations of b^n +- 1. Contemporary Mathematics, Vol. 22, Amer. Math. Soc., Providence, RI, 2nd edition, 1985; and later supplements.
Ehrhard Behrends, Five-Minute Mathematics, translated by David Kramer. American Mathematical Society (2008) p. 7.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Table of n, a(n) for n = 0..331
J. Brillhart et al., Factorizations of b^n +- 1, Contemporary Mathematics, Vol. 22, Amer. Math. Soc., Providence, RI, 3rd edition, 2002.
T. Granlund, Factors of 10^n + 1.
Makoto Kamada, Factorizations of 100...001.
S. S. Wagstaff, Jr., The Cunningham Project.

Cf. A006530, A062397.

[Maximum(PrimeDivisors(10^n+1)): n in [0..40]]; // Vincenzo Librandi, Jul 12 2016

Table[FactorInteger[10^n + 1][[-1, 1]], {n, 0, 29}] (* Alonso del Arte, Oct 21 2011 *)

for(n=0, 1e2, p=factor(10^n+1)[omega(10^n+1), 1]; print1(p, ", ")) \\ Felix Fröhlich, Aug 13 2014

a(n) = A006530(A062397(n)). - Vincenzo Librandi, Jul 12 2016

More terms from Jason Earls, Jul 11 2001
Terms up to a(280) in b-file from D. S. McNeil, Oct 22 2011
a(281)-a(310) in b-file from Ray Chandler, May 02 2017
a(311)-a(331) in b-file from Max Alekseyev, Apr 24 2019, Feb 13 2020, May 13 2022

A033934 a(n) = (10^n + 1)^2.

Original entry on oeis.org

4, 121, 10201, 1002001, 100020001, 10000200001, 1000002000001, 100000020000001, 10000000200000001, 1000000002000000001, 100000000020000000001, 10000000000200000000001, 1000000000002000000000001, 100000000000020000000000001, 10000000000000200000000000001

Offset: 0

Jeff Burch

The members of this sequence are both perfect squares and palindromes. Therefore A002779 is an infinite sequence. - Ant King, Jun 26 2011

Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A002779 (palindromic squares), A000290 (squares), A002113 (palindromes).

Cf. A011557, A062397, A066138.

(10^Range[0,20]+1)^2 (* or *) LinearRecurrence[{111,-1110,1000},{4,121,10201},20] (* Harvey P. Dale, Feb 16 2016 *)

my(x='x+O('x^15)); Vec((1210*x^2-323*x+4)/(-1000*x^3+1110*x^2-111*x+1)) \\ Elmo R. Oliveira, Jul 04 2025

a(n) = A062397(n)^2 = A066138(n) + A011557(n).
From Elmo R. Oliveira, Jul 04 2025: (Start)
G.f.: (4 - 323*x + 1210*x^2)/((1-x)*(1-10*x)*(1-100*x)).
E.g.f.: exp(x)*(1 + 2*exp(9*x) + exp(99*x)).
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3). (End)

Better description from Henry Bottomley, Dec 07 2001

More terms from Harvey P. Dale, Feb 16 2016

A066138 a(n) = 10^(2*n) + 10^n + 1.

Original entry on oeis.org

3, 111, 10101, 1001001, 100010001, 10000100001, 1000001000001, 100000010000001, 10000000100000001, 1000000001000000001, 100000000010000000001, 10000000000100000000001, 1000000000001000000000001, 100000000000010000000000001, 10000000000000100000000000001, 1000000000000001000000000000001

Offset: 0

Henry Bottomley, Dec 07 2001

Palindromes whose digit sum is 3.
Essentially the same as A135577. - R. J. Mathar Apr 29 2008
From Peter Bala, Sep 25 2015: (Start)
For n >= 1, the simple continued fraction expansion of sqrt(a(n)) = [10^n; 1, 1, 2/3*(10^n - 1), 1, 1, 2*10^n, ...] has period 6. As n increases, the expansion has the large partial quotients 2/3*(10^n - 1) and 2*10^n.
For n >= 1, the continued fraction expansion of sqrt(a(2*n))/a(n) = [0; 1, 10^n - 1, 1, 1, 1/3*(10^n - 4), 1, 4, 1, 1/3*(10^n - 4), 1, 1, 10^n - 1, 2, 10^n - 1, ...] has pre-period of length 3 and period 12 beginning 1, 1, 1/3*(10^n - 4), ....  As n increases, the expansion has the large partial quotients 10^n - 1 and 1/3*(10^n - 4).
A theorem of Kuzmin in the measure theory of continued fractions says that large partial quotients are the exception in continued fraction expansions.
Empirically, we also see exceptionally large partial quotients in the continued fraction expansions of the m-th root of the numbers a(m*n), for m >= 3. For example, it appears that the continued fraction expansion of a(3*n)^(1/3), for n >= 2, begins [10^(2*n); 3*10^n - 1, 1, 0.5*10^(2*n) - 1, 1.44*10^n - 1, 1, ...]. Cf. A000533, A002283 and A168624. (End)

			From _Peter Bala_, Sep 25 2015: (Start)
Simple continued fraction expansions showing large partial quotients:
a(9)^(1/3) =[1000000; 2999, 1, 499999, 1439, 1, 2582643, 1, 1, 1, 2, 3, 3, ...].
a(20)^(1/4) = [10000000000; 39999999999, 1, 3999999999, 16949152542, 2, 1, 2, 6, 1, 4872106, 3, 9, 2, 3, ...].
a(25)^(1/5) = [10000000000; 4999999999999999, 1, 3333333332, 2, 1, 217391304347825, 2, 2, 1, 1, 1, 2, 1, 23980814, 1, 1, 1, 1, 1, 7, ...]. (End)

Harry J. Smith, Table of n, a(n) for n=0..100
Index entries for linear recurrences with constant coefficients, signature (111,-1110,1000).

Cf. A000533, A002283, A033934, A062397, A074992, A135577, A168624.

[10^(2*n) + 10^n + 1: n in [0..20]]; // Vincenzo Librandi, Sep 27 2015

Table[10^(2 n) + 10^n + 1, {n, 0, 15}] (* Michael De Vlieger, Sep 27 2015 *)
CoefficientList[Series[(3 - 222 x + 1110 x^2)/((1 - 100 x) (1 - 10 x) (1 - x)), {x, 0, 33}], x] (* Vincenzo Librandi, Sep 27 2015 *)

a(n) = { 10^(2*n) + 10^n + 1 } \\ Harry J. Smith, Feb 02 2010

Vec(-3*(370*x^2-74*x+1)/((x-1)*(10*x-1)*(100*x-1)) + O(x^20)) \\ Colin Barker, Sep 27 2015

A168624(n) = a(2*n)/a(n). - Peter Bala, Sep 24 2015
G.f.: (3 - 222*x + 1110*x^2)/((1 - 100*x)*(1 - 10*x)*(1 - x)). - Vincenzo Librandi, Sep 27 2015
From Colin Barker, Sep 27 2015: (Start)
a(n) = 111*a(n-1) - 1110*a(n-2) + 1000*a(n-3) for n > 2.
G.f.: -3*(370*x^2-74*x+1)/((x-1)*(10*x-1)*(100*x-1)). (End)
From Elmo R. Oliveira, Aug 27 2024: (Start)
E.g.f.: exp(x)*(exp(99*x) + exp(9*x) + 1).
a(n) = 3*A074992(n). (End)

Offset changed from 1 to 0 by Harry J. Smith, Feb 02 2010

More terms from Michael De Vlieger, Sep 27 2015

A038371 Smallest prime factor of 10^n + 1.

Original entry on oeis.org

2, 11, 101, 7, 73, 11, 101, 11, 17, 7, 101, 11, 73, 11, 29, 7, 353, 11, 101, 11, 73, 7, 89, 11, 17, 11, 101, 7, 73, 11, 61, 11, 19841, 7, 101, 11, 73, 11, 101, 7, 17, 11, 29, 11, 73, 7, 101, 11, 97, 11, 101, 7, 73, 11, 101, 11, 17, 7, 101, 11, 73, 11, 101, 7, 1265011073

Offset: 0

Miklos SZABO (mike(AT)ludens.elte.hu)

nonn

a(n) >= 7 for all n >= 1 since 10^n + 1 is then not divisible by 2, 3 or 5.
Record values are a({0, 1, 2, 16, 32, 64, ...}). - M. F. Hasler, Apr 04 2008
The record values (2, 11, 101, 353, 19841, 1265011073, ...) are also found in A185121 and A102050 (smallest prime factor of 10^2^n+1). - M. F. Hasler, Jun 28 2024

			a(12) = 73 as 10^12+1 = 1000000000001 = 73*137*99990001.

Ehrhard Behrends, Five-Minute Mathematics, translated by David Kramer. American Mathematical Society (2008) p. 7

Max Alekseyev, Table of n, a(n) for n = 0..1023 (terms 0..500 from M. F. Hasler)
Makoto Kamada, Factorizations of 100...001.

Cf. A020639 (least prime factor), A062397 (10^n + 1), A003021 (largest prime factor of 10^n + 1), A057934 (number of prime factors of 10^n + 1, with multiplicity), A119704 (as before, without multiplicity), A185121 (smallest prime factor of 10^2^n+1), A102050 (as before, but 1 if 10^2^n+1 is prime).

[Min(PrimeFactors(10^n+1)):n in[0..70]]; // Vincenzo Librandi, Nov 08 2018

Table[FactorInteger[10^n + 1][[1, 1]], {n, 0, 49}] (* Alonso del Arte, Oct 21 2011 *)

A038371(n)=A020639(10^n+1) \\ Much more efficient than the naive {factor(10^n+1)[1,1]}. - M. F. Hasler, Apr 04 2008, edited Jun 29 2024

a(n) = A020639(A062397(n)).
For odd n, a(n) <= 11 since every (base 10) palindrome of even length is divisible by 11. - M. F. Hasler, Apr 04 2008 [See below for more precise formula.]
More generally, for k >= 0 and n == 2^k (mod 2^(k+1)), a(n) <= A185121(k) = (11, 101, 73, 17, 353, ...). This follows from  x^{2q+1} + 1 = (x+1) Sum_{m=0..2q} (-x)^m, with x=10^2^k. - M. F. Hasler, Jul 30 2019
From M. F. Hasler, Jun 28 2024: (Start)
a(2k+1) = 7 iff k == 1 (mod 3), else 11. [Making the 2008 formula more precise.]
a(4k+2) = 29 iff k == 3 (mod 7), else = 61 if k == 7 (mod 15), else = 89 if k == 5 (mod 11), else 101.
a(8k+4) = 73 for all k >= 0.
a(16k+8) = 17 for all k >= 0.
a(32k+16) = 97 iff k==1 (mod 3), else 353.
a(64k+32) = 193 iff k==1 (mod 3), else 1217 if k==9 (mod 19), else 2753 if k==21 (mod 43), else 3137 if k==24 (mod 49), else 3329 if k==6 (mod 13), else 4481 if k==17 (mod 35), else 4673 if k==36 (mod 73), else 5953 if k==15 (mod 31), else 6529 if k==8 (mod 17), else 13633 if k==35 (mod 71), else 15937 if k==41 (mod 83), else 19841. (End)

More terms from Reinhard Zumkeller, Mar 12 2002

A366668 Sum of the divisors of 10^n+1.

Original entry on oeis.org

3, 12, 102, 1344, 10212, 109104, 1010004, 10909104, 105882372, 1413350400, 10102223208, 114737461440, 1021097900424, 10921790676000, 104844305394000, 1355394166984704, 10073631600468000, 110177492439680640, 1010002989998020008, 10909090909090909104

Offset: 0

Sean A. Irvine, Oct 15 2023

nonn

			a(3)=1344 because 10^3+1 has divisors {1, 7, 11, 13, 77, 91, 143, 1001}.

Max Alekseyev, Table of n, a(n) for n = 0..331

Cf. A000203, A003021, A057934, A062397, A102146, A119704, A344897, A366666.

a:=n->numtheory[sigma](10^n+1):
seq(a(n), n=0..100);

DivisorSigma[1, 10^Range[0,19] + 1] (* Paul F. Marrero Romero, Nov 12 2023 *)

a(n) = sigma(10^n+1) = A000203(A062397(n)).

cp's OEIS Frontend

A034524 a(n) = 11^n + 1.

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A178248 a(n) = 12^n + 1.

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A047855 a(n) = A047848(7,n).

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A069283 a(n) = -1 + number of odd divisors of n.

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A119704 a(n) = number of distinct prime factors of 10^n+1 = omega(10^n+1).

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A003021 Largest prime factor of 10^n + 1.

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