A119598 -id:A119598 - cp's OEIS frontend

Showing 1-10 of 10 results.

A065764 Sum of divisors of square numbers.

1, 7, 13, 31, 31, 91, 57, 127, 121, 217, 133, 403, 183, 399, 403, 511, 307, 847, 381, 961, 741, 931, 553, 1651, 781, 1281, 1093, 1767, 871, 2821, 993, 2047, 1729, 2149, 1767, 3751, 1407, 2667, 2379, 3937, 1723, 5187, 1893, 4123, 3751, 3871, 2257, 6643

Offset: 1

Labos Elemer, Nov 19 2001

Unlike A065765, the sums of divisors of squares give remainders r=1,3,5 modulo 6: sigma(4)==1, sigma(49)==3, sigma(2401)==5 (mod 6). See also A097022.
a(n) is also the number of ordered pairs of positive integers whose LCM is n, (see LeVeque). - Enrique Pérez Herrero, Aug 26 2013
Main diagonal of A319526. - Omar E. Pol, Sep 25 2018
Subsequence of primes is A023195 \ {3}; also, 31 is the only known prime to be twice in the data because 31 = sigma(16) = sigma(25) (see A119598 and Goormaghtigh conjecture link). - Bernard Schott, Jan 17 2021

W. J. LeVeque, Fundamentals of Number Theory, pp. 125 Problem 4, Dover NY 1996.

T. D. Noe, Table of n, a(n) for n=1..10000
Wikipedia, Goormaghtigh conjecture.

Cf. A000203, A000290, A028982, A319526.

a:=List([1..50],n->Sigma(n^2));; Print(a); # Muniru A Asiru, Jan 01 2019

[SumOfDivisors(n^2): n in [1..48]]; // Bruno Berselli, Apr 12 2011

with(numtheory): [sigma(n^2)$n=1..50]; # Muniru A Asiru, Jan 01 2019

Table[Plus@@Divisors[n^2], {n, 48}] (* Alonso del Arte, Feb 24 2012 *)
f[p_, e_] := (p^(2*e + 1) - 1)/(p - 1); a[1] = 1; a[n_] := Times @@ f @@@ FactorInteger[n]; Array[a, 50] (* Amiram Eldar, Sep 10 2020 *)

numlib::sigma(n^2)$ n=1..81 // Zerinvary Lajos, May 13 2008

a(n) = sigma(n^2); \\ Harry J. Smith, Oct 30 2009

from math import prod
from sympy import factorint
def A065764(n): return prod((p**((e<<1)+1)-1)//(p-1) for p,e in factorint(n).items()) # Chai Wah Wu, Oct 25 2023

[sigma(n^2,1)for n in range(1,49)] # Zerinvary Lajos, Jun 13 2009

a(n) = sigma(n^2) = A000203(A000290(n)).
Multiplicative with a(p^e) = (p^(2*e+1)-1)/(p-1). - Vladeta Jovovic, Dec 01 2001
Dirichlet g.f.: zeta(s)*zeta(s-1)*zeta(s-2)/zeta(2*s-2), inverse Mobius transform of A000082. - R. J. Mathar, Mar 06 2011
Dirichlet convolution of A001157 by the absolute terms of A055615. Also the Dirichlet convolution of A048250 by A000290. - R. J. Mathar, Apr 12 2011
a(n) = Sum_{d|n} d*Psi(d), where Psi is A001615. - Enrique Pérez Herrero, Feb 25 2012
a(n) >= (n+1) * sigma(n) - n, where sigma is A000203, equality holds if n is in A000961. - Enrique Pérez Herrero, Apr 21 2012
Sum_{k=1..n} a(k) ~ 5*Zeta(3) * n^3 / Pi^2. - Vaclav Kotesovec, Jan 30 2019
Sum_{k>=1} 1/a(k) = 1.3947708738535614499846243600124612760835313454790187655653356563282177118... - Vaclav Kotesovec, Sep 20 2020

A053696 Numbers that can be represented as a string of three or more 1's in a base >= 2.

Original entry on oeis.org

7, 13, 15, 21, 31, 40, 43, 57, 63, 73, 85, 91, 111, 121, 127, 133, 156, 157, 183, 211, 241, 255, 259, 273, 307, 341, 343, 364, 381, 400, 421, 463, 507, 511, 553, 585, 601, 651, 703, 757, 781, 813, 820, 871, 931, 993, 1023, 1057, 1093, 1111, 1123, 1191

Offset: 1

Henry Bottomley, Mar 23 2000

Numbers of the form (b^n-1)/(b-1) for n > 2 and b > 1. - T. D. Noe, Jun 07 2006
Numbers m that are nontrivial repunits for any base b >= 2. For k = 2 (I use k for the exponent since n is used as the index in a(n)) we get (b^k-1)/(b-1) = (b^2-1)/(b-1) = b+1, so every integer m >= 3 is a 2-digit repunit in base b = m-1. And for n = 1 (the 1-digit degenerate repunit) we get (b-1)/(b-1) = 1 for any base b >= 2. If we considered all k >= 1 we would get the sequence of all positive integers except 2 since it is the smallest uniform base used in positional representation (2 might be seen as the "repunit" in a nonpositional base representation such as the Roman numerals where 2 is expressed as II). - Daniel Forgues, Mar 01 2009
These repunits numbers belong to Brazilian numbers (A125134) (see Links: "Les nombres brésiliens" - section IV, p. 32). - Bernard Schott, Dec 18 2012
The Brazilian primes (A085104) belong to this sequence. - Bernard Schott, Dec 18 2012

			a(5) = 31 because 31 can be written as 111 base 5 (or indeed 11111 base 2).

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000 (first 1172 terms from T. D. Noe)
Bernard Schott, Les nombres brésiliens, Quadrature, no. 76, avril-juin 2010, pages 30-38; included here with permission from the editors of Quadrature.

Cf. A090503 (a subsequence), A119598 (numbers that are repunits in four or more bases), A125134, A085104.

Cf. A108348.

a053696 n = a053696_list !! (n-1)
a053696_list = filter ((> 1) . a088323) [2..]
-- Reinhard Zumkeller, Jan 22 2014, Nov 26 2013

N:= 10^4: # to get all terms <= N
V:= Vector(N):
for b from 2 while (b^3-1)/(b-1) <= N do
  inds:= [seq((b^k-1)/(b-1), k=3..ilog[b](N*(b-1)+1))];
  V[inds]:= 1;
od:
select(t -> V[t] = 1, [$1..N]); # Robert Israel, Dec 10 2015

fQ[n_] := Block[{d = Rest@ Divisors[n - 1]}, Length@ d > 2 && Length@ Select[ IntegerDigits[n, d], Union@# == {1} &] > 1]; Select[ Range@ 1200, fQ]
lim=1000; Union[Reap[Do[n=3; While[a=(b^n-1)/(b-1); a<=lim, Sow[a]; n++], {b, 2, Floor[Sqrt[lim]]}]][[2, 1]]]
Take[Union[Flatten[With[{l=Table[PadLeft[{},n,1],{n,3,100}]}, Table[ FromDigits[#,n]&/@l,{n,2,100}]]]],80] (* Harvey P. Dale, Oct 06 2011 *)

list(lim)=my(v=List(),e,t);for(b=2,sqrt(lim),e=3;while((t=(b^e-1)/(b-1))<=lim,listput(v,t);e++));vecsort(Vec(v),,8) \\ Charles R Greathouse IV, Oct 06 2011

list(lim)=my(v=List(),e,t);for(b=2,lim^(1/3),e=4;while((t=(b^e-1)/(b-1))<=lim,listput(v,t);e++));vecsort(concat(Vec(v), vector((sqrtint (lim\1*4-3)-3)\2,i,i^2+3*i+3)),,8) \\ Charles R Greathouse IV, May 30 2013

a(n) ~ n^2 since as n grows the density of repunits of degree 2 among all the repunits tends to 1. - Daniel Forgues, Dec 09 2008

A088323(a(n)) > 1. - Reinhard Zumkeller, Jan 22 2014

A060072 a(n) = (n^(n-1) - 1)/(n-1) for n>1, a(1) = 0.

Original entry on oeis.org

0, 1, 4, 21, 156, 1555, 19608, 299593, 5380840, 111111111, 2593742460, 67546215517, 1941507093540, 61054982558011, 2085209001813616, 76861433640456465, 3041324492229179280, 128583032925805678351, 5784852794328402307380, 275941052631578947368421

Offset: 1

Henry Bottomley, Feb 21 2001

nonn

(n-1)-digit repunits in base n written in decimal.

			a(10)=111111111; i.e., just nine 1's (converted from base 10 to decimal).

Harry J. Smith, Table of n, a(n) for n=1..200

Cf. A000169, A037205, A055869, A060073, A125118, A228275.

Cf. other sequences of generalized repunits, such as A053696, A055129, A031973, A125598, A173468, A023037, A119598, A085104, and A162861.

[0] cat [ (n^(n-1) -1)/(n-1) : n in [2..25]]; // G. C. Greubel, Aug 15 2022

Join[{0},Array[(#^(#-1)-1)/(#-1)&,20,2]] (* Harvey P. Dale, Jun 04 2013 *)

a(n) = if (n==1, 0, (n^(n - 1) - 1)/(n - 1)); \\ Harry J. Smith, Jul 01 2009

[0]+[(n^(n-1) -1)/(n-1) for n in (2..25)] # G. C. Greubel, Aug 15 2022

a(n+1) = Sum_{k=1..n} n^(k-1)*C(n, k). - Olivier Gérard, Jun 26 2001 [Corrected by Mathew Englander, Dec 15 2020]
a(n) = Sum_{j=2..n} n^(n-j). - Zerinvary Lajos, Sep 11 2006
a(n+1) = A125118(n,n). - Reinhard Zumkeller, Nov 21 2006
a(n) = Integral_{x=1/n..1} 1/x^n dx. - Francesco Daddi, Aug 01 2011
a(n) = A037205(n-1)/(n-1) = A060073(n)*(n-1) = A023037(n) - A000169(n).
a(n) = [x^n] x^2/((1 - x)*(1 - n*x)). - Ilya Gutkovskiy, Oct 04 2017
a(n) = 1 + A228275(n, n-2) for n >= 2. - Mathew Englander, Dec 14 2020

Name edited by Michel Marcus, Dec 14 2020

A167783 Numbers that are repdigits with length > 2 in more than one base.

Original entry on oeis.org

31, 63, 255, 273, 364, 511, 546, 728, 777, 931, 1023, 1365, 1464, 2730, 3280, 3549, 3783, 3906, 4095, 4557, 6560, 7566, 7812, 8191, 9114, 9331, 9841, 10507, 11349, 11718, 13671, 14043, 14763, 15132, 15624, 16383, 18291, 18662, 18915, 19608, 19682, 21845, 22351, 22698

Offset: 1

Andrew Weimholt, Nov 12 2009

Definition requires "length > 2" because all numbers n > 2 are trivially represented as "11" in base n-1.
From Daniel Forgues, Nov 13 2009: (Start)
0 = 00 = 000 = 0000 = 00000 = 000000 = 0000000 = 00000000 = ... in any positional number representation (includes fixed base radix b > 1, mixed base radix with each b_i > 1, i >= 0, such as factorial and primorial based radix...)
The sequence definition should be read as:
Nonnegative integers that are repdigits with length > 2 in more than one fixed base radix b > 1.
Considering all fixed and mixed base radix would include many more nonnegative integers (but not the integers 1 to 6) which are repdigits with length > 2 in more than one radix. (End)
From Bernard Schott, Aug 08 2017: (Start)
In this sequence data, the first number which is repdigit, with length > 2, in more than two bases is the twelfth Mersenne number 4095 with four Brazilian representations: M_12 = 4095 = 111111111111_2 = 333333_4 = 7777_8 = (15 15 15)_16.
The Mersenne number M_15 is the first number which is repdigit in exactly three bases with M_15 = 32767 = 111111111111111_2 = 77777_8 = (31 31 31)_32.
Only two numbers are repunits in more than one base: the Mersenne primes 31 and 8191 (Examples and A119598).
Some numbers are once repunit and once multiple of a Brazilian prime such that Mersenne number M_9 = 511 = 7 * 73 = 111111111_2 = 7 * 111_8 = 777_8.
Some numbers are once repunit and once multiple of a composite repunit such that Mersenne number M_6 = 63 = 3 * 21 = 111111_2 = 3 * 111_4 = 333_4.
Some numbers are repdigits in two different bases: 546 = 666_9 = 222_16. (End)

			31 is in the list because 31 = 11111_2 = 111_5;
8191 = 1111111111111_2 = 111_90;
10507 = {19 19 19}_23 = 111_102.

David Trimas, Table of n, a(n) for n = 1..8424 (first 158 terms from Michel Marcus)
David Trimas, Wolfram Cloud Implementation of A167783
Wolfram Demonstrations Project, Mixed Radix Number Representations

Cf. A167782 (numbers that are repdigits with length > 2 in some base).
Cf. A010785 (repdigits (base 10)).
Cf. A053696 (numbers which are repunits in some base).
Cf. A158235 (numbers n whose square is a repdigit in some base < n).
Cf. A290869 (Numbers that are repdigits with length > 2 in more than two bases).

Select[Range[550], Function[n, 1 < Count[Range[2, n - 1], ?(And[Length@ DeleteCases[#, 0] == 1, Union[#][[2]] > 2] &@ DigitCount[n, #] &)]]] (* _Michael De Vlieger, Aug 09 2017 *)

isok(n)=my(nb = 0); for (b=2, n-1, d = digits(n, b); if ((#d > 2) && (#Set(d) == 1), nb++); if (nb > 1, return (1));); return (0); \\ Michel Marcus, Aug 08 2017

a(41)-a(44) from Bernard Schott, Aug 08 2017

A326381 Numbers m such that beta(m) = tau(m)/2 + 1 where beta(m) is the number of Brazilian representations of m and tau(m) is the number of divisors of m.

Original entry on oeis.org

31, 63, 255, 273, 364, 511, 546, 728, 777, 931, 1023, 1365, 1464, 2730, 3280, 3549, 3783, 4557, 6560, 7566, 7812, 8191, 9114, 9331, 9841, 10507, 11349, 11718, 13671, 14043, 14763, 15132, 15624, 16383, 18291, 18662, 18915, 19608, 19682, 21845, 22351, 22698

Offset: 1

Bernard Schott, Jul 07 2019

As tau(m) = 2 * (beta(m) - 1), the terms of this sequence are not squares.
There are 3 subsequences which realize a partition of this sequence (see examples):
1) Non-oblong composites which have exactly two Brazilian representations with three digits or more, they form A326388.
2) Oblong numbers that have exactly three Brazilian representations with three digits or more; thanks to Michel Marcus, who found the smallest, 641431602. These oblong integers are a subsequence of A290869 and A309062.
3) The two Brazilian primes 31 and 8191 of the Goormaghtigh conjecture (A119598) for which beta(p) = tau(p)/2 + 1 = 2.

			One example for each type:
1) 63 = 111111_2 = 333_4 = 77_8 = 33_20 with tau(63) = 6 and beta(63) = 4.
2) 641431602 = 25326 * 25327 is oblong with tau(641431602) = 256. The three Brazilian representations with three digits or more of 641431602 are 999999_37 = (342,342,342)_1369 = (54,54,54)_3446, so beta"(641431602) = 3 and beta(641431602) = tau(641431602)/2 + 1 = 129.
3) 31 = 11111_2 = 111_5 and 8191 = 1111111111111_2 = 11_90 with beta(p) = tau(p)/2 + 1 = 2.

Wikipedia, Goormaghtigh conjecture.
Index entries for sequences related to Brazilian numbers.

Cf. A000005 (tau), A220136 (beta).
Cf. A119598 (Goormaghtigh conjecture).
Subsequence of A167783.
Cf. A326378 (tau(m)/2 - 2), A326379 (tau(m)/2 - 1), A326380 (tau(m)/2), A326382 (tau(m)/2 + 2), A326383 (tau(m)/2 + 3).

beta(n) = sum(i=2, n-2, #vecsort(digits(n, i), , 8)==1); \\ A220136
isok(n) = beta(n) == numdiv(n)/2 + 1; \\ Michel Marcus, Jul 08 2019

A088323 Number of numbers b>1 such that n is a repunit in base b representation.

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 2

Reinhard Zumkeller, Nov 06 2003

is a(n) < 4 ?;
n>2: a(n) > 0 as n = (n-1)^1 + (n-1)^0.
a(A119598(n)) > 3; a(A053696(n)) > 2; a(A085104(n)) > 2. - Reinhard Zumkeller, Jan 22 2014

			a(31)=3: 31 = 2^4+2^3+2^2+2^1+2^0 = 5^2+5^1+5^0 = 30^1+30^0.

Reinhard Zumkeller, Table of n, a(n) for n = 2..10000
Eric Weisstein's World of Mathematics, Repunit

Cf. A000225, A003462, A002275, A068953.

a088323 n = sum $ map (f n) [2 .. n-1] where
   f x b = if x == 0 then 1 else if d /= 1 then 0 else f x' b
                                 where (x',d) = divMod x b
-- Reinhard Zumkeller, Jan 22 2014

Example corrected by Reinhard Zumkeller, Jan 22 2014

A290015 Brazilian numbers which have exactly two Brazilian representations.

Original entry on oeis.org

15, 18, 21, 26, 28, 30, 31, 32, 44, 45, 50, 52, 56, 57, 62, 64, 68, 75, 76, 85, 86, 91, 92, 93, 98, 99, 110, 111, 116, 117, 129, 133, 146, 147, 148, 153, 164, 175, 183, 188, 207, 212, 215, 219, 236, 243, 244, 245, 259, 261, 268, 275, 279, 284, 314, 316, 325, 332, 338, 341, 343, 356, 363, 365, 369, 381, 387, 388

Offset: 1

Bernard Schott, Jul 17 2017

These numbers could be called 2-Brazilian numbers.
The smallest number of this sequence is 15 which is also the smallest odd composite Brazilian in A257521 with 15 = 11111_2 = 33_4. The number 15 is highly Brazilian in A329383.
Following the Goormaghtigh conjecture, only two primes, 31 and 8191, which are both Mersenne numbers, are Brazilian in two different bases (A119598).

			18 = 2 * 9 = 22_8 = 3 * 6 = 33_5.
26 = 2 * 13 = 2 * 111_3 = 222_3 = 22_12.
31 = 11111_2 = 111_5;
8191 = 1111111111111_2 = 111_90.

Wikipedia, Goormaghtigh conjecture

Cf. A085104, A125134, A220570, A220571, A257521, A284758, A288783, A290016, A290017, A290018, A329383.

bresilienbaseb:=proc(n,b)
local r,q,coupleq:
if n0 then
return [couple[1]+1,r]
else
return [0,0]
end if
end if
end proc;
bresil:=proc(n)
local b,L,k,t:
k:=0:
for b from 2 to (n-2) do
t:=bresilienbase(n,b):
if t[1]>0 then
k:=k+1
L[k]:=[b,t[1],t[2]]:
end if:
end do:
seq(L[i],i=1..k);
end proc;
nbbresil:=n->nops([bresil(n)]);
#Numbers 2 times Brazilian
for n from 1 to 100 do if nbbresil(n)=2 then print(n,bresil(n)) else fi; od:

Mathematica

Flatten@ Position[#, 2] &@ Table[Count[Range[2, n - 2], ?(And[Length@ # != 1, Length@ Union@ # == 1] &@ IntegerDigits[n, #] &)], {n, 400}] (* _Michael De Vlieger, Jul 18 2017 *)

A125598 a(n) = ((n+1)^(n-1) - 1)/n.

Original entry on oeis.org

0, 1, 5, 31, 259, 2801, 37449, 597871, 11111111, 235794769, 5628851293, 149346699503, 4361070182715, 139013933454241, 4803839602528529, 178901440719363487, 7143501829211426575, 304465936543600121441

Offset: 1

Views

Graph

List

Refs

Listen

History

Text

Internal format

Author

Alexander Adamchuk, Nov 26 2006

Keywords

nonn

Comments

Odd prime p divides a(p-2).

a(n) is prime for n = {3,4,6,74, ...}; prime terms are {5, 31, 2801, ...}.

a(n) is the (n-1)-th generalized repunit in base (n+1). For example, a(5) = 259 which is 1111 in base 6. - Mathew Englander, Oct 20 2020

Links

G. C. Greubel, Table of n, a(n) for n = 1..350

Crossrefs

Cf. A000272 (n^(n-2)), A125599.

Cf. other sequences of generalized repunits, such as A125118, A053696, A055129, A060072, A031973, A173468, A023037, A119598, A085104, and A162861.

Programs

Magma

[((n+1)^(n-1) -1)/n: n in [1..25]]; // G. C. Greubel, Aug 15 2022

Mathematica

Table[((n+1)^(n-1)-1)/n, {n,25}]

Sage

[gaussian_binomial(n,1,n+2) for n in range(0,18)] # Zerinvary Lajos, May 31 2009

Formula

a(n) = ((n+1)^(n-1) - 1)/n.

a(n) = (A000272(n+1) - 1)/n.

a(2k-1)/(2k+1) = A125599(k) for k>0.

From Mathew Englander, Dec 17 2020: (Start)

a(n) = (A060072(n+1) - A083069(n-1))/2.

For n > 1, a(n) = Sum_{k=0..n-2} (n+1)^k.

For n > 1, a(n) = Sum_{j=0..n-2} n^j*C(n-1,j+1). (End)

A325658 Brazilian composites of the form 1 + b + b^2 + b^3 + ... + b^k, b > 1, k > 1.

Original entry on oeis.org

15, 21, 40, 57, 63, 85, 91, 111, 121, 133, 156, 183, 255, 259, 273, 341, 343, 364, 381, 400, 507, 511, 553, 585, 651, 703, 781, 813, 820, 871, 931, 993, 1023, 1057, 1111, 1191, 1261, 1333, 1365, 1407, 1464, 1555, 1561, 1641, 1807, 1885, 1893, 1981, 2047, 2071, 2163, 2257, 2353

Offset: 1

Views

Graph

List

Refs

Listen

History

Text

Internal format

Author

Bernard Schott, May 12 2019

Keywords

nonn
base

Comments

Composites that are repunits in base b >= 2 with three or more digits. If the Goormaghtigh conjecture is true, there are no composite numbers which can be represented as a string of three or more 1's in a base >= 2 in more than one way (A119598).

Only three known perfect powers belong to this sequence: 121, 343 and 400 (A208242).

Except for 121, each term of this sequence have also at least one Brazilian representation with only 2 digits.

Examples

121 = (11111)_3, 133 = (111)_11 = (77)_18.

Links

Robert Israel, Table of n, a(n) for n = 1..10000

Yann Bugeaud and T. N. Shorey, On the Diophantine Equation (x^m - 1)/(x-1) = (y^n - 1)/(y-1), Pacific Journal of Mathematics, Vol. 207, No 1, November 2002.

Sean A. Irvine, Java program (github)

Michel Waldschmidt, Lecture on the abc conjecture and some of its consequences, 6th World Conference on 21st Century Mathematics 2013, Lahore, p. 14 (Goormaghtigh conjecture).

Wikipedia, Goormatigh conjecture.

Crossrefs

Complement of A085104 with respect to A053696.

Intersection of A053696 and A220571.

Cf. A119598, A208242, A325659.

Programs

Maple

N:= 3000: Res:= NULL: for m from 2 while 1+m+m^2 <= N do for k from 2 do v:= (m^(k+1)-1)/(m-1); if v > N then break fi; if not isprime(v) then Res:= Res, v fi od od: sort(convert({Res},list)); # Robert Israel, May 13 2019

PARI

lista(nn) = {forcomposite(n=1, nn, for(b=2, sqrtint(n), my(d=digits(n, b), sd=Set(d)); if ((#d >= 3) && (#sd == 1) && (sd[1] == 1), print1(n, ", "); break);););} \\ Michel Marcus, May 18 2019

A341434 a(n) is the number of bases 1 < b < n in which n is divisible by its product of digits.

Original entry on oeis.org

0, 0, 1, 1, 1, 2, 2, 3, 2, 2, 1, 5, 2, 3, 4, 6, 1, 5, 1, 5, 4, 4, 1, 9, 2, 2, 4, 5, 1, 7, 3, 9, 4, 2, 3, 12, 1, 2, 3, 10, 1, 7, 2, 7, 7, 2, 1, 15, 2, 5, 3, 6, 1, 10, 3, 10, 4, 3, 1, 14, 1, 2, 7, 14, 3, 8, 1, 6, 3, 6, 1, 20, 2, 3, 8, 7, 3, 7, 1, 16, 7, 2, 1, 14

Offset: 1

Views

Graph

List

Refs

Listen

History

Text

Internal format

Author

Amiram Eldar, Feb 11 2021

Keywords

nonn
base

Examples

a(3) = 1 since 3 is divisible by its product of digits only in base 2: 3 = 11_2 and 1*1 | 3. a(6) = 2 since 6 is divisible by its product of digits in 2 bases: in base 4, 6 = 12_4 and 1*2 | 6, and in base 5, 6 = 11_5 and 1*1 | 6.

Links

Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A007602, A068953, A080221, A085104, A088323, A119598.

Programs

Mathematica

q[n_, b_] := (p = Times @@ IntegerDigits[n, b]) > 0 && Divisible[n, p]; a[n_] := Count[Range[2, n], _?(q[n, #] &)]; Array[a, 100]

PARI

a(n) = sum(b=2, n-1, my(x=vecprod(digits(n, b))); x && !(n%x)); \\ Michel Marcus, Feb 12 2021

Formula

a(n) > 0 for all numbers n > 2 since n in base b = n-1 is 11.

a(n) > 1 for all even numbers > 4 since n in base b = n-2 is 12. Similarly, a(n) > 1 for all composite numbers > 4 since if n = k*m, then n is divisible by its product of digits in bases n-m and n-k.

a(p) > 1 for primes p in A085104.

a(p) > 2 for primes p in A119598 (i.e., 31, 8191, ...).

a(n) >= A088323(n), with equality if n = 4 or if n is a prime.

Showing 1-10 of 10 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.
Content is available under The OEIS End-User License Agreement.

cp's OEIS Frontend

A065764 Sum of divisors of square numbers.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

GAP

Magma

Maple

Mathematica

MuPAD

PARI

Python

Sage

Formula

A053696 Numbers that can be represented as a string of three or more 1's in a base >= 2.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

PARI

PARI

Formula

A060072 a(n) = (n^(n-1) - 1)/(n-1) for n>1, a(1) = 0.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

SageMath

Formula

Extensions

A167783 Numbers that are repdigits with length > 2 in more than one base.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Extensions

A326381 Numbers m such that beta(m) = tau(m)/2 + 1 where beta(m) is the number of Brazilian representations of m and tau(m) is the number of divisors of m.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

PARI

A088323 Number of numbers b>1 such that n is a repunit in base b representation.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs