A003593 -id:A003593 - cp's OEIS frontend

A107364 Numbers of the form (3^i)*(13^j).

1, 3, 9, 13, 27, 39, 81, 117, 169, 243, 351, 507, 729, 1053, 1521, 2187, 2197, 3159, 4563, 6561, 6591, 9477, 13689, 19683, 19773, 28431, 28561, 41067, 59049, 59319, 85293, 85683, 123201, 177147, 177957, 255879, 257049, 369603, 371293, 531441

Offset: 1

Douglas Winston (douglas.winston(AT)srupc.com), May 23 2005

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

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

[n: n in [1..10^7] | PrimeDivisors(n) subset [3, 13]]; // Vincenzo Librandi, Jun 27 2016

mx = 540000; Sort@ Flatten@ Table[3^i*13^j, {i, 0, Log[3, mx]}, {j, 0, Log[13, mx/3^i]}] (* Robert G. Wilson v, Aug 17 2012 *)
fQ[n_]:=PowerMod[39, n, n] == 0; Select[Range[2 10^7], fQ] (* Vincenzo Librandi, Jun 27 2016 *)

list(lim)=my(v=List(),N);for(n=0,log(lim)\log(13),N=13^n;while(N<=lim,listput(v,N);N*=3));vecsort(Vec(v)) \\ Charles R Greathouse IV, Jun 28 2011

Sum_{n>=1} 1/a(n) = (3*13)/((3-1)*(13-1)) = 13/8. - Amiram Eldar, Sep 23 2020

a(n) ~ exp(sqrt(2*log(3)*log(13)*n)) / sqrt(39). - Vaclav Kotesovec, Sep 23 2020

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

A022337 Exponent of 5 (value of j) in n-th number of form 3^i*5^j.

Original entry on oeis.org

0, 0, 1, 0, 1, 2, 0, 1, 2, 0, 3, 1, 2, 0, 3, 1, 4, 2, 0, 3, 1, 4, 2, 0, 5, 3, 1, 4, 2, 0, 5, 3, 1, 6, 4, 2, 0, 5, 3, 1, 6, 4, 2, 0, 7, 5, 3, 1, 6, 4, 2, 0, 7, 5, 3, 1, 8, 6, 4, 2, 0, 7, 5, 3, 1, 8, 6, 4, 2, 0, 9, 7, 5, 3, 1, 8, 6, 4, 2, 0, 9, 7, 5, 3, 1, 10, 8, 6, 4, 2, 0, 9, 7, 5, 3, 1, 10, 8, 6, 4, 2, 0, 11, 9, 7, 5

Offset: 1

Clark Kimberling

nonn

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

Cf. A003593, A022336, A112754, A112762, A112765.

s = {}; m = 12; Do[n = 5^k; While[n <= 5^m, AppendTo[s, n]; n *= 3], {k, 0, m}]; IntegerExponent[#, 5] & /@ Union[s] (* Amiram Eldar, Feb 06 2020 *)

a(n) = A112765(A003593(n)) = A112754(n) - A022336(n). - Reinhard Zumkeller, Sep 18 2005

A079002 Numbers n such that the Fibonacci residues F(k) mod n form the complete set (0,1,2,...,n-1).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 9, 10, 14, 15, 20, 25, 27, 30, 35, 45, 50, 70, 75, 81, 100, 125, 135, 150, 175, 225, 243, 250, 350, 375, 405, 500, 625, 675, 729, 750, 875, 1125, 1215, 1250, 1750, 1875, 2025, 2187, 2500, 3125, 3375, 3645, 3750, 4375, 5625, 6075, 6250, 6561

Offset: 1

Benoit Cloitre, Feb 01 2003

nonn

			Fibonacci numbers (A000045) are 0,1,1,2,3,5,8,... and their residues mod 5 are 0,1,1,2,3,0,3,3,4,...; i.e., all possible remainders mod 5 occur in the Fibonacci sequence mod 5, so 5 is in the sequence. This is not true for n=8, so 8 is not in the sequence.

R. L. Graham, D. E. Knuth and O. Patashnick, "Concrete Mathematics", second edition, Addison Wesley, 1994, ex. 6.85, p. 318, p. 562.

T. D. Noe, Table of n, a(n) for n = 1..1000
B. Avila and Y. Chen, On moduli for which the Lucas numbers contain a complete residue system, Fibonacci Quarterly, 51 (2013), 151-152.
S. A. Burr, On moduli for which the Fibonacci numbers contain a complete system of residues, Fibonacci Quarterly, 9 (1971), 497-504.
Cheng Lien Lang and Mong Lung Lang, Fibonacci system and residue completeness, arXiv:1304.2892 [math.NT], 2013.

Cf. A066853, A001175, A003593, A224482, A249104.

Select[Range[10^4], MatchQ[FactorInteger[#], {{1, 1}}|{{2, 1}}|{{2, 2}}| {{3, }}|{{2, 1}, {3, 1}}|{{7, 1}}|{{2, 1}, {7, 1}}|{{5, }}|{{2, 1}, {5, }}|{{2, 2}, {5, }}|{{3, }, {5, }}|{{2, 1}, {3, 1}, {5, }}|{{5, }, {7, 1}}|{{2, 1}, {5, }, {7, 1}}]&] (* _Jean-François Alcover, Sep 01 2018 *)

is(n)=n/=5^valuation(n,5); n==3^valuation(n,3) || setsearch([2,4,6,7,14],n) \\ Charles R Greathouse IV, Apr 23 2013

Consists of the integers of the forms 5^k, 2*5^k, 4*5^k, 3^j*5^k, 6*5^k, 7*5^k and 14*5^k [see Concrete Mathematics].

Corrected by Ron Knott, Jan 05 2005

Entry revised by N. J. A. Sloane, Nov 28 2006, following a suggestion from Martin Fuller

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

A357567 a(n) = 5A005259(n) - 14A005258(n).

Original entry on oeis.org

-9, -17, 99, 5167, 147491, 3937483, 105834699, 2907476527, 81702447651, 2342097382483, 68273597307599, 2018243113678027, 60365426282638091, 1823553517258576723, 55557712038989195099, 1705170989220937925167, 52672595030914982754851, 1636296525812843554700323

Offset: 0

Peter Bala, Oct 19 2022

Conjectures:
1) a(p) == a(1) (mod p^5) for all primes p >= 5.
2) For r >= 2, a(p^r) == a(p^(r-1)) ( mod p^(3*r+3) ) for all primes p >= 5.
These are stronger supercongruences than those satisfied separately by the two types of Apéry numbers A005258 and A005259.
From Peter Bala, Oct 25 2022: (Start)
Additional conjectures:
3) the sequence {u(n): n>= 1} defined by u(n) = (3^42)*A005259(n)^25 - (5^25)* A005258(n)^42 also satisfies the congruences in 1) and 2) above.
4) u(n) == 0 (mod n^5) for integer n of the form 3^i*5^j (see A003593). (End)

			a(11) - a(1) = 2018243113678027 + 17 = (2^2)*(3^2)*(11^5)*17*20476637 == 0 (mod 11^5).

Cf. A005258, A005259, A212334, A352655, A357506, A357507, A357508, A357509, A357568, A357569, A357956, A357957, A357958, A357959, A357960.

seq(add(5*binomial(n,k)^2*binomial(n+k,k)^2 - 14*binomial(n,k)^2*binomial(n+k,k), k = 0..n), n = 0..20);

a(n) = 5*Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k,k)^2 - 14*Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k,k).

For positive integers n and r, a(n*p^r) == a(n*p^(r-1)) ( mod p^(3*r) ) for all primes p >= 5.

A022336 Exponent of 3 (value of i) in n-th number of form 3^i*5^j.

Original entry on oeis.org

0, 1, 0, 2, 1, 0, 3, 2, 1, 4, 0, 3, 2, 5, 1, 4, 0, 3, 6, 2, 5, 1, 4, 7, 0, 3, 6, 2, 5, 8, 1, 4, 7, 0, 3, 6, 9, 2, 5, 8, 1, 4, 7, 10, 0, 3, 6, 9, 2, 5, 8, 11, 1, 4, 7, 10, 0, 3, 6, 9, 12, 2, 5, 8, 11, 1, 4, 7, 10, 13, 0, 3, 6, 9, 12, 2, 5, 8, 11, 14, 1, 4, 7, 10, 13, 0, 3, 6, 9, 12, 15, 2, 5, 8, 11, 14, 1, 4, 7

Offset: 1

Clark Kimberling

nonn

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

Cf. A003593, A007949, A022329, A022337, A112754, A112761.

s = {}; m = 12; Do[n = 5^k; While[n <= 5^m, AppendTo[s, n]; n *= 3], {k, 0, m}]; IntegerExponent[#, 3] & /@ Union[s] (* Amiram Eldar, Feb 06 2020 *)

a(n) = A007949(A003593(n)) = A112754(n) - A022337(n). - Reinhard Zumkeller, Sep 18 2005

A112754 Total number of prime factors of n-th number of the form 3^i*5^j.

Original entry on oeis.org

0, 1, 1, 2, 2, 2, 3, 3, 3, 4, 3, 4, 4, 5, 4, 5, 4, 5, 6, 5, 6, 5, 6, 7, 5, 6, 7, 6, 7, 8, 6, 7, 8, 6, 7, 8, 9, 7, 8, 9, 7, 8, 9, 10, 7, 8, 9, 10, 8, 9, 10, 11, 8, 9, 10, 11, 8, 9, 10, 11, 12, 9, 10, 11, 12, 9, 10, 11, 12, 13, 9, 10, 11, 12, 13, 10, 11, 12, 13, 14, 10, 11, 12, 13, 14, 10, 11, 12

Offset: 1

Reinhard Zumkeller, Sep 18 2005

nonn

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

Cf. A001222, A003593, A022336, A022337, A069352, A112753, A112759.

s = {}; m = 12; Do[n = 5^k; While[n <= 5^m, AppendTo[s, n]; n *= 3], {k, 0, m}]; PrimeOmega[Union[s]] (* Amiram Eldar, Feb 06 2020 *)

a(n) = A001222(A003593(n)) = A022336(n) + A022337(n).

A217584 Numbers k such that d(k^2)/d(k) is an integer, where d(k) is the number of divisors of k.

Original entry on oeis.org

1, 144, 324, 400, 784, 1936, 2025, 2500, 2704, 3600, 3969, 4624, 5625, 5776, 7056, 8100, 8464, 9604, 9801, 13456, 13689, 15376, 15876, 17424, 19600, 21609, 21904, 22500, 23409, 24336, 26896, 29241, 29584, 30625, 35344, 39204, 41616, 42849, 44944, 48400, 51984

Offset: 1

Michel Marcus, Oct 07 2012

nonn

The ratio d(k^2)/d(k) is: 1 for the number 1, 3 for numbers of the form p^4*q^2, 5 for numbers of the form p^4*q^2*r^2 (p, q, r being different primes).
Primes can't be in the sequence. A prime p has two divisors, while p^2 has three divisors: 1, p, p^2. - Alonso del Arte, Oct 07 2012
All the terms are squares since d(m) is odd if and only if m is a square, so d(k^2) is odd and since d(k)|d(k^2), d(k) is also odd, so k is a square. The ratio d(k^2)/d(k) can take values other than 1, 3, and 5: 1587600 is the least term with a ratio 9, and 192099600 is the least term with a ratio 15. - Amiram Eldar, May 23 2020
From Bernard Schott, May 29 2020 and Nov 22 2020: (Start)
This sequence comes from the 3rd problem, proposed by Belarus, during the 39th International Mathematical Olympiad in 1998 at Taipei (Taiwan) [see the link IMO].
If the prime signature of k is (u_1, u_2, ... , u_q) then d(k^2)/d(k) = Product_{i=1..q} (2*u_i+1)/(u_i+1). Two results:
1) If k is a term such that d(k^2)/d(k) = m, then all numbers that have the same prime signature of k are also terms and give the same ratio (see examples below).
2) The set of the integer values of the ratio d(k^2)/d(k) is exactly the set of all positive odd integers (see Marcin E. Kuczma reference).
Some examples:
For numbers with prime signature = (4, 2) (A189988), the ratio is 3 and the smallest such integer is 144 = 2^4 * 3^2.
For numbers with prime signature = (4, 2, 2) (A179746), the ratio is 5 and the smallest such integer is 3600 = 2^4 * 3^2 * 5^2.
For numbers with prime signature = (4, 4, 2, 2) the ratio is 9 and the smallest such integer is 1587600 = 2^4 * 3^4 * 5^2 * 7^2.
For numbers with prime signature = (8, 4, 4, 2, 2) the ratio is 17 and the smallest such integer is 76839840000 = 2^8 * 3^4 * 5^4 * 7^2 * 11^2 (found by David A. Corneth with other prime signatures). (End)

			d(1^2)/d(1) = d(1)/d(1) = 1 an integer, so 1 belongs to the sequence.
144^2 has 45 divisors: 1, 2, 3, 4, 6, 8, 9, 12, ..., 20736, while 144 has 15 divisors: 1, 2, 3, 4, 6, 8, 9, 12, ..., 144; 45/15 = 3 and so 144 is in the sequence.

Marcin E. Kuczma, International Mathematical Olympiads, 1986-1999, The Mathematical Association of America, 2003, pages 134-135.

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..100 from Paolo P. Lava)
The IMO Compendium, Problem 3, 39th IMO 1998.
Kin Y. Li, Problem 3, Mathematical Excalibur, Vol. 4, No. 3, Jan.-Mar. 1999.
Index to sequences related to Olympiads.

Cf. A000005, A003593, A025487, A048691.
Subsequences: A189988 (d(k^2)/d(k) = 3), A179746 (d(k^2)/d(k) = 5).
Cf. A339055 (values taken by d(a(n)^2)/d(a(n))), A339056 (smallest k such that d(k^2)/d(k) = n-th odd).

Select[Range[1000], IntegerQ[DivisorSigma[0, #^2]/DivisorSigma[0, #]] &] (* Alonso del Arte, Oct 07 2012 *)
Select[Range[228]^2, Divisible[DivisorSigma[0, #^2], DivisorSigma[0, #]] &] (* Amiram Eldar, May 23 2020 *)

dn2dn(n)= {for (i=1, n, if (denominator(numdiv(i^2)/numdiv(i))==1, print1(i,", ");););}

A257997 Numbers of the form (2^i)(3^j) or (2^i)(5^j) or (3^i)*(5^j).

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20, 24, 25, 27, 32, 36, 40, 45, 48, 50, 54, 64, 72, 75, 80, 81, 96, 100, 108, 125, 128, 135, 144, 160, 162, 192, 200, 216, 225, 243, 250, 256, 288, 320, 324, 375, 384, 400, 405, 432, 486, 500, 512, 576, 625, 640

Offset: 1

Reinhard Zumkeller, May 16 2015

Union of A003586, A003592 and A003593.

Subsequence of 5-smooth numbers (cf. A051037), having no more than two distinct prime factors: A006530(a(n)) <= 5; A001221(a(n)) <= 2.

			. ----+------+---------     ----+------+-----------
.   1 |   1  |  1            16 |  25  |  5^2
.   2 |   2  |  2            17 |  27  |  3^3
.   3 |   3  |  3            18 |  32  |  2^5
.   4 |   4  |  2^2          19 |  36  |  2^2 * 3^2
.   5 |   5  |  5            20 |  40  |  2^3 * 5
.   6 |   6  |  2 * 3        21 |  45  |  3^2 * 5
.   7 |   8  |  2^3          22 |  48  |  2^4 * 3
.   8 |   9  |  3^2          23 |  50  |  2 * 5^2
.   9 |  10  |  2 * 5        24 |  54  |  2 * 3^3
.  10 |  12  |  2^2 * 3      25 |  64  |  2^6
.  11 |  15  |  3 * 5        26 |  72  |  2^3 * 3^2
.  12 |  16  |  2^4          27 |  75  |  3 * 5^2
.  13 |  18  |  2 * 3^2      28 |  80  |  2^4 * 5
.  14 |  20  |  2^2 * 5      29 |  81  |  3^4
.  15 |  24  |  2^3 * 3      30 |  96  |  2^5 * 3

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Vaclav Kotesovec, Graph - the asymptotic ratio (80000000 terms)

Cf. A003586, A003592, A003593, A051037, A006530, A001221, A258023 (subsequence).

Cf. A337800, A337801.

import Data.List.Ordered (unionAll)
a257997 n = a257997_list !! (n-1)
a257997_list = unionAll [a003586_list, a003592_list, a003593_list]

n = 1000; Join[Table[2^i*3^j, {i, 0, Log[2, n]}, {j, 0, Log[3, n/2^i]}], Table[3^i*5^j, {i, 0, Log[3, n]}, {j, 0, Log[5, n/3^i]}], Table[2^i*5^j, {i, 0, Log[2, n]}, {j, 0, Log[5, n/2^i]}]] // Flatten // Union (* Amiram Eldar, Sep 23 2020 *)

a(n) ~ exp(sqrt(2*log(2)*log(3)*log(5)*n/log(30))). - Vaclav Kotesovec, Sep 22 2020

Sum_{n>=1} 1/a(n) = 29/8. - Amiram Eldar, Sep 23 2020

cp's OEIS Frontend

A107364 Numbers of the form (3^i)*(13^j).

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

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A022337 Exponent of 5 (value of j) in n-th number of form 3^i*5^j.

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A079002 Numbers n such that the Fibonacci residues F(k) mod n form the complete set (0,1,2,...,n-1).

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

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A357567 a(n) = 5*A005259(n) - 14*A005258(n).

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A022336 Exponent of 3 (value of i) in n-th number of form 3^i*5^j.

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A112754 Total number of prime factors of n-th number of the form 3^i*5^j.

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A357567 a(n) = 5A005259(n) - 14A005258(n).

A257997 Numbers of the form (2^i)(3^j) or (2^i)(5^j) or (3^i)*(5^j).