A003598 -id:A003598 - cp's OEIS frontend

A108687 Numbers of the form (9^i)*(11^j), with i, j >= 0.

1, 9, 11, 81, 99, 121, 729, 891, 1089, 1331, 6561, 8019, 9801, 11979, 14641, 59049, 72171, 88209, 107811, 131769, 161051, 531441, 649539, 793881, 970299, 1185921, 1449459, 1771561, 4782969, 5845851, 7144929, 8732691, 10673289, 13045131

Offset: 1

Douglas Winston (douglas.winston(AT)srupc.com), Jun 17 2005

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Subsequence of A003597.

Cf. A025633, A025634, A107788, A107764, A003596, A107988, A003598, A003599, A108090.

import Data.Set (singleton, deleteFindMin, insert)
a108687 n = a108687_list !! (n-1)
a108687_list = f $ singleton (1,0,0) where
   f s = y : f (insert (9 * y, i + 1, j) $ insert (11 * y, i, j + 1) s')
         where ((y, i, j), s') = deleteFindMin s
-- Reinhard Zumkeller, May 15 2015

f[upto_]:=With[{max9=Floor[Log[9,upto]],max11=Floor[Log[11,upto]]}, Select[Union[Times@@{9^First[#],11^Last[#]}&/@Tuples[{Range[0, max9], Range[0, max11]}]], #<=upto&]]; f[14000000]  (* Harvey P. Dale, Mar 11 2011 *)

from sympy import integer_log
def A108687(n):
    def bisection(f,kmin=0,kmax=1):
        while f(kmax) > kmax: kmax <<= 1
        kmin = kmax >> 1
        while kmax-kmin > 1:
            kmid = kmax+kmin>>1
            if f(kmid) <= kmid:
                kmax = kmid
            else:
                kmin = kmid
        return kmax
    def f(x): return n+x-sum(integer_log(x//11**i,9)[0]+1 for i in range(integer_log(x,11)[0]+1))
    return bisection(f,n,n) # Chai Wah Wu, Mar 25 2025

Sum_{n>=1} 1/a(n) = (9*11)/((9-1)*(11-1)) = 99/80. - Amiram Eldar, Sep 24 2020

a(n) ~ exp(sqrt(2*log(9)*log(11)*n)) / sqrt(99). - Vaclav Kotesovec, Sep 24 2020

A025635 Numbers of form 9^i*10^j, with i, j >= 0.

Original entry on oeis.org

1, 9, 10, 81, 90, 100, 729, 810, 900, 1000, 6561, 7290, 8100, 9000, 10000, 59049, 65610, 72900, 81000, 90000, 100000, 531441, 590490, 656100, 729000, 810000, 900000, 1000000, 4782969, 5314410, 5904900, 6561000, 7290000, 8100000, 9000000

Offset: 1

David W. Wilson

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A025612, A025616, A025621, A025625, A025629, A025632, A025634, A108761, A003596, A003597, A107988, A003598, A108698, A003599, A107788, A108687, A108779, A108090.

import Data.Set (singleton, deleteFindMin, insert)
a025635 n = a025635_list !! (n-1)
a025635_list = f $ singleton (1,0,0) where
   f s = y : f (insert (9 * y, i + 1, j) $ insert (10 * y, i, j + 1) s')
         where ((y, i, j), s') = deleteFindMin s
-- Reinhard Zumkeller, May 15 2015

With[{max = 10^7}, Flatten[Table[9^i*10^j, {i, 0, Log[9, max]}, {j, 0, Log[10, max/9^i]}]] // Sort] (* Amiram Eldar, Mar 29 2025 *)

list(lim)=my(v=List(), N); for(n=0, logint(lim\=1, 10), N=10^n; while(N<=lim, listput(v, N); N*=9)); Set(v) \\ Charles R Greathouse IV, Jan 10 2018

from sympy import integer_log
def A025635(n):
    def bisection(f,kmin=0,kmax=1):
        while f(kmax) > kmax: kmax <<= 1
        kmin = kmax >> 1
        while kmax-kmin > 1:
            kmid = kmax+kmin>>1
            if f(kmid) <= kmid:
                kmax = kmid
            else:
                kmin = kmid
        return kmax
    def f(x): return n+x-sum(integer_log(x//10**i,9)[0]+1 for i in range(integer_log(x,10)[0]+1))
    return bisection(f,n,n) # Chai Wah Wu, Mar 25 2025

Sum_{n>=1} 1/a(n) = 5/4. - Amiram Eldar, Mar 29 2025

a(n) = 9^A025683(n) * 10^A025691(n). - R. J. Mathar, Jul 06 2025

A107466 Numbers of the form (5^i)*(13^j).

Original entry on oeis.org

1, 5, 13, 25, 65, 125, 169, 325, 625, 845, 1625, 2197, 3125, 4225, 8125, 10985, 15625, 21125, 28561, 40625, 54925, 78125, 105625, 142805, 203125, 274625, 371293, 390625, 528125, 714025, 1015625, 1373125, 1856465, 1953125, 2640625

Offset: 1

Douglas Winston (douglas.winston(AT)srupc.com), May 27 2005

nonn

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

Cf. A003586, A003592, A003593, A003591, A003594, A003595, A003596, A003597, A003598, A003599, A107326, A107364.

mx = 2700000; Sort@ Flatten@ Table[5^i*13^j, {i, 0, Log[5, mx]}, {j, 0, Log[13, mx/5^i]}] (* Robert G. Wilson v, Aug 17 2012 *)

list(lim)=my(v=List(),N);for(n=0,log(lim)\log(13),N=13^n;while(N<=lim,listput(v,N);N*=5));vecsort(Vec(v)) \\ Charles R Greathouse IV, Jun 28 2011

from sympy import integer_log
def A107466(n):
    def bisection(f,kmin=0,kmax=1):
        while f(kmax) > kmax: kmax <<= 1
        kmin = kmax >> 1
        while kmax-kmin > 1:
            kmid = kmax+kmin>>1
            if f(kmid) <= kmid:
                kmax = kmid
            else:
                kmin = kmid
        return kmax
    def f(x): return n+x-sum(integer_log(x//13**i,5)[0]+1 for i in range(integer_log(x,13)[0]+1))
    return bisection(f,n,n) # Chai Wah Wu, Mar 25 2025

Sum_{n>=1} 1/a(n) = (5*13)/((5-1)*(13-1)) = 65/48. - Amiram Eldar, Sep 23 2020

a(n) ~ exp(sqrt(2*log(5)*log(13)*n)) / sqrt(65). - Vaclav Kotesovec, Sep 23 2020

A108698 Numbers of the form (6^i)*(11^j), with i, j >= 0.

Original entry on oeis.org

1, 6, 11, 36, 66, 121, 216, 396, 726, 1296, 1331, 2376, 4356, 7776, 7986, 14256, 14641, 26136, 46656, 47916, 85536, 87846, 156816, 161051, 279936, 287496, 513216, 527076, 940896, 966306, 1679616, 1724976, 1771561, 3079296, 3162456

Offset: 1

Douglas Winston (douglas.winston(AT)srupc.com), Jun 19 2005

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A025626, A025627, A025628, A025629, A064476, A107710, A003596, A003597, A107988, A003598, A108687, A003599, A107788, A108090.

import Data.Set (singleton, deleteFindMin, insert)
a108698 n = a108698_list !! (n-1)
a108698_list = f $ singleton (1,0,0) where
   f s = y : f (insert (6 * y, i + 1, j) $ insert (11 * y, i, j + 1) s')
         where ((y, i, j), s') = deleteFindMin s
-- Reinhard Zumkeller, May 15 2015

n = 10^6; Flatten[Table[6^i*11^j, {i, 0, Log[6, n]}, {j, 0, Log[11, n/6^i]}]] // Sort (* Amiram Eldar, Oct 07 2020 *)

Sum_{n>=1} 1/a(n) = (6*11)/((6-1)*(11-1)) = 33/25. - Amiram Eldar, Oct 07 2020

a(n) ~ exp(sqrt(2*log(6)*log(11)*n)) / sqrt(66). - Vaclav Kotesovec, Oct 07 2020

A108056 Numbers of the form (7^i)*(13^j).

Original entry on oeis.org

1, 7, 13, 49, 91, 169, 343, 637, 1183, 2197, 2401, 4459, 8281, 15379, 16807, 28561, 31213, 57967, 107653, 117649, 199927, 218491, 371293, 405769, 753571, 823543, 1399489, 1529437, 2599051, 2840383, 4826809, 5274997, 5764801, 9796423, 10706059, 18193357, 19882681

Offset: 1

Douglas Winston (douglas.winston(AT)srupc.com), Jun 02 2005

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

Cf. A003586, A003592, A003593, A003591, A003594, A003595, A003596, A003597, A003598, A003599, A107326, A107364, A107466.

n = 10^7; Flatten[Table[7^i*13^j, {i, 0, Log[7, n]}, {j, 0, Log[13, n/7^i]}]] // Sort (* Amiram Eldar, Sep 23 2020 *)

list(lim)=my(v=List(),N);for(n=0,log(lim)\log(13),N=13^n;while(N<=lim,listput(v,N);N*=7));vecsort(Vec(v)) \\ Charles R Greathouse IV, Jun 28 2011

from sympy import integer_log
def A108056(n):
    def bisection(f,kmin=0,kmax=1):
        while f(kmax) > kmax: kmax <<= 1
        while kmax-kmin > 1:
            kmid = kmax+kmin>>1
            if f(kmid) <= kmid:
                kmax = kmid
            else:
                kmin = kmid
        return kmax
    def f(x): return n+x-sum(integer_log(x//13**i,7)[0]+1 for i in range(integer_log(x,13)[0]+1))
    return bisection(f,n,n) # Chai Wah Wu, Oct 22 2024

Sum_{n>=1} 1/a(n) = (7*13)/((7-1)*(13-1)) = 91/72. - Amiram Eldar, Sep 23 2020

a(n) ~ exp(sqrt(2*log(7)*log(13)*n)) / sqrt(91). - Vaclav Kotesovec, Sep 23 2020

More terms from Amiram Eldar, Sep 23 2020

A108779 Numbers of the form (10^i)*(11^j), with i, j >= 0.

Original entry on oeis.org

1, 10, 11, 100, 110, 121, 1000, 1100, 1210, 1331, 10000, 11000, 12100, 13310, 14641, 100000, 110000, 121000, 133100, 146410, 161051, 1000000, 1100000, 1210000, 1331000, 1464100, 1610510, 1771561, 10000000, 11000000, 12100000, 13310000

Offset: 1

Douglas Winston (douglas.winston(AT)srupc.com), Jun 26 2005

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Cf. A025612, A025616, A025621, A025625, A025629, A025632, A025634, A025635, A108761, A003596, A003597, A107988, A003598, A108698, A003599, A107788, A108687, A108090.

import Data.Set (singleton, deleteFindMin, insert)
a108779 n = a108779_list !! (n-1)
a108779_list = f $ singleton (1,0,0) where
   f s = y : f (insert (10 * y, i + 1, j) $ insert (11 * y, i, j + 1) s')
         where ((y, i, j), s') = deleteFindMin s
-- Reinhard Zumkeller, May 15 2015

n = 10^7; Flatten[Table[10^i*11^j, {i, 0, Log10[n]}, {j, 0, Log[11, n/10^i]}]] // Sort (* Amiram Eldar, Sep 25 2020 *)

Sum_{n>=1} 1/a(n) = (10*11)/((10-1)*(11-1)) = 11/9. - Amiram Eldar, Sep 25 2020

a(n) ~ exp(sqrt(2*log(10)*log(11)*n)) / sqrt(110). - Vaclav Kotesovec, Sep 25 2020

A045576 Numbers k that divide 3^k + 2^k.

Original entry on oeis.org

1, 5, 25, 55, 125, 275, 605, 625, 1375, 3025, 3125, 6655, 6875, 15125, 15625, 30025, 31375, 33275, 34375, 73205, 75625, 78125, 150125, 156875, 166375, 171875, 330275, 345125, 366025, 378125, 390625, 439835, 750625, 784375, 805255, 831875

Offset: 1

David W. Wilson

nonn

For any j>=0, 5*A003598(j) is a term of the sequence. - Benoit Cloitre, Mar 08 2002
From Robert Israel, Jun 29 2017: (Start)
This is a semigroup: if m and n are in the sequence, then so is m*n.
If n is in the sequence and is divisible by prime p, then so is p*n.
The only prime powers in the sequence are the powers of 5.
Conjecture: Every member of the sequence except 1 is of the form p*m where p is prime and m is in the sequence. (End)
There are infinitely many primes p that divide some term in the sequence. Proof: Define the set A as all primes p such that a k where p divides 2^(5^k) + 3^(5^k) exists is finite. Since 2^(5^(k+1)) + 3^(5^(k+1)) is  2^(5^k) + 3^(5^k) multiplied by some positive integer a. It can be verified that gcd(a, 2^(5^k) + 3^(5^k)) is 5, so 2^(5^(k+1)) + 3^(5^(k+1)) has a larger number of different prime factors than 2^(5^k) + 3^(5^k). Therefore, A is infinite. For each prime q in A, suppose that q divides 2^(5^x) + 3^(5^x) for some x, then 5^x also divides it, so 5^x*q divides it as well, hence 5^x*q is a term of the sequence. The original theorem is proved. - Yifan Xie, Nov 14 2024

Robert Israel and Giovanni Resta, Table of n, a(n) for n = 1..1000 (first 205 terms from Robert Israel)

Cf. A003598, A007689.

select(t -> 3 &^ t + 2 &^ t mod t = 0, [seq(i,i=1..10^6,2)]); # Robert Israel, Jun 29 2017

isok(n) = ((3^n+2^n) % n) == 0; \\ Michel Marcus, Jun 29 2017

isok(n)=(Mod(2,n)^n+Mod(3,n)^n)==0; \\ significantly more efficient
for(n=1,10^6,if(isok(n),print1(n,", "))); \\ Joerg Arndt, Aug 13 2017

A056739 Numbers k such that k | 10^k + 9^k + 8^k + 7^k + 6^k + 5^k + 4^k + 3^k + 2^k + 1^k.

Original entry on oeis.org

1, 5, 11, 25, 55, 121, 125, 275, 365, 605, 625, 925, 1331, 1375, 2365, 3025, 3125, 6655, 6875, 14641, 15125, 15625, 22625, 27565, 32125, 33275, 34375, 73205, 75625, 78125, 123365, 161051, 166375, 171875, 366025, 378125, 390625, 541717, 660605

Offset: 1

Robert G. Wilson v, Aug 25 2000

nonn

Contains A003598. In general n=p^i * q^j => n | Sum_{k=1..2*p} k^n, where p and q=2*p+1 are prime (see Meyer ref).

All terms == 1 or 5 (mod 6). The only prime terms are 5 and 11. - Robert Israel, Jun 25 2025

Robert Israel, Table of n, a(n) for n = 1..140
Christian Meyer, On conjecture no. 22 arising from the OEIS

Cf. A001557, A003598, A057291, A057292.

filter:= n ->   10 &^n + 9 &^ n + 8 &^ n + 7 &^ n + 6&^ n + 5&^n + 4&^n + 3&^n + 2&^n + 1 mod n = 0:
select(filter, [seq(seq(6*i + j, j=[1,5]),i=0..10^6)]); # Robert Israel, Jun 25 2025

Do[ If[ Mod[ PowerMod[ 10, n, n ] + PowerMod[ 9, n, n ] + PowerMod[ 8, n, n ] + PowerMod[ 7, n, n ] + PowerMod[ 6, n, n ] + PowerMod[ 5, n, n ] + PowerMod[ 4, n, n ] + PowerMod[ 3, n, n ] + PowerMod[ 2, n, n ] + 1, n ] == 0, Print[ n ] ], {n, 1, 10^6} ]
Select[Range[700000],Divisible[Total[Range[10]^#],#]&] (* Harvey P. Dale, Nov 24 2014 *)
Select[Range[700000],Mod[Total[PowerMod[Range[10],#,#]],#]==0&] (* Harvey P. Dale, Feb 23 2023 *)

A108201 Numbers of the form (5^i)*(12^j), with i, j >= 0.

Original entry on oeis.org

1, 5, 12, 25, 60, 125, 144, 300, 625, 720, 1500, 1728, 3125, 3600, 7500, 8640, 15625, 18000, 20736, 37500, 43200, 78125, 90000, 103680, 187500, 216000, 248832, 390625, 450000, 518400, 937500, 1080000, 1244160, 1953125, 2250000, 2592000

Offset: 1

Douglas Winston (douglas.winston(AT)srupc.com), Jun 15 2005

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

Cf. A025622, A003595, A025623, A025624, A025625, A003598, A107466, A064476.

Take[Union[5^First[#] 12^Last[#]&/@Tuples[Range[0,20],2]],50] (* Harvey P. Dale, Mar 23 2012 *)

Sum_{n>=1} 1/a(n) = 15/11. - Amiram Eldar, Mar 29 2025

A036305 Composite numbers whose prime factors contain no digits other than 1 and 5.

Original entry on oeis.org

25, 55, 121, 125, 275, 605, 625, 755, 1331, 1375, 1661, 3025, 3125, 3775, 5755, 6655, 6875, 7555, 8305, 12661, 14641, 15125, 15625, 16621, 18271, 18875, 22801, 28775, 33275, 34375, 37775, 41525, 57755, 63305, 73205, 75625, 77555, 77755, 78125

Offset: 1

Patrick De Geest, Dec 15 1998

All terms are a product of at least two terms of A020453. - David A. Corneth, Oct 09 2020

David A. Corneth, Table of n, a(n) for n = 1..10000 (first 1000 terms from Alois P. Heinz)
Index entries for sequences related to prime factors.

Cf. A003598, A020453, A036302-A036325.

Sum_{n>=1} 1/a(n) = Product_{p in A020453} (p/(p - 1)) - Sum_{p in A020453} 1/p - 1 = 0.0873463128... . - Amiram Eldar, May 18 2022

cp's OEIS Frontend

A108687 Numbers of the form (9^i)*(11^j), with i, j >= 0.

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A025635 Numbers of form 9^i*10^j, with i, j >= 0.

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A107466 Numbers of the form (5^i)*(13^j).

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A108698 Numbers of the form (6^i)*(11^j), with i, j >= 0.

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A108056 Numbers of the form (7^i)*(13^j).

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A108779 Numbers of the form (10^i)*(11^j), with i, j >= 0.

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Haskell

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A045576 Numbers k that divide 3^k + 2^k.

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A056739 Numbers k such that k | 10^k + 9^k + 8^k + 7^k + 6^k + 5^k + 4^k + 3^k + 2^k + 1^k.

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