A234959 -id:A234959 - cp's OEIS frontend

A243758 a(n) = Product_{i=1..n} A234959(i).

1, 1, 1, 1, 1, 1, 6, 6, 6, 6, 6, 6, 36, 36, 36, 36, 36, 36, 216, 216, 216, 216, 216, 216, 1296, 1296, 1296, 1296, 1296, 1296, 7776, 7776, 7776, 7776, 7776, 7776, 279936, 279936, 279936, 279936, 279936, 279936, 1679616, 1679616, 1679616, 1679616, 1679616, 1679616, 10077696

Offset: 0

Tom Edgar, Jun 10 2014

This is the generalized factorial for A234959.

a(0) = 1 as it represents the empty product.

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Tyler Ball, Tom Edgar, and Daniel Juda, Dominance Orders, Generalized Binomial Coefficients, and Kummer's Theorem, Mathematics Magazine, Vol. 87, No. 2, April 2014, pp. 135-143.

Cf. A060828, A060818, A234959, A242954, A243757.

a243758 n = a243758_list !! n
a243758_list = scanl (*) 1 a234959_list
-- Reinhard Zumkeller, Feb 09 2015

Table[Product[6^IntegerExponent[k, 6], {k, 1, n}], {n, 0, 20}] (* G. C. Greubel, Dec 24 2016 *)

valp(n,p)=my(s); while(n\=p, s+=n); s
a(n)=6^valp(n,6) \\ Charles R Greathouse IV, Oct 03 2016

S=[0]+[6^valuation(i,6) for i in [1..100]]
[prod(S[1:i+1]) for i in [0..99]]

a(n) = Product_{i=1..n} A234959(i).

a(n) = 6^(A054895(n)).

A122841 Greatest k such that 6^k divides n.

Original entry on oeis.org

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

Offset: 1

Reinhard Zumkeller, Sep 13 2006

nonn

See A054895 for the partial sums. - Hieronymus Fischer, Jun 08 2012

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

Cf. A122840, A007814, A007949, A054895, A234959.

a122841 = f 0 where
   f y x = if r > 0 then y else f (y + 1) x'
           where (x', r) = divMod x 6
-- Reinhard Zumkeller, Nov 10 2013

Table[IntegerExponent[n, 6], {n, 1, 100}] (* Amiram Eldar, Sep 14 2020 *)

a(n) = valuation(n, 6); \\ Michel Marcus, Jan 17 2022

From Hieronymus Fischer, Jun 03 2012: (Start)
With m = floor(log_6(n)), frac(x) = x-floor(x):
a(n) = Sum_{j=1..m} (1 - ceiling(frac(n/6^j))).
a(n) = m + Sum_{j=1..m} (floor(-frac(n/6^j))).
a(n) = A054895(n) - A054895(n-1).
G.f.: Sum_{j>0} x^6^j/(1-x^6^j). (End)
a(A047253(n)) = 0; a(A008588(n)) > 0; a(A044102(n)) > 1. - Reinhard Zumkeller, Nov 10 2013
6^a(n) = A234959(n), n >= 1. - Wolfdieter Lang, Jun 30 2014
Asymptotic mean: lim_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1/5. - Amiram Eldar, Jan 17 2022
a(n) = min(A007814(n), A007949(n)). - Jianing Song, Jul 23 2022

A268354 Highest power of 7 dividing n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 49, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 49, 1, 1, 1, 1, 1, 1, 7

Offset: 1

Tom Edgar, Feb 02 2016

The generalized binomial coefficients produced by this sequence provide an analog to Kummer's Theorem using arithmetic in base 7.

			Since 14 = 7 * 2, a(14) = 7. Likewise, since 7 does not divide 13, a(13) = 1.

Antti Karttunen, Table of n, a(n) for n = 1..20790
Tyler Ball, Tom Edgar, and Daniel Juda, Dominance Orders, Generalized Binomial Coefficients, and Kummer's Theorem, Mathematics Magazine, Vol. 87, No. 2, April 2014, pp. 135-143.
Tom Edgar and Michael Z. Spivey, Multiplicative functions, generalized binomial coefficients, and generalized Catalan numbers, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.6.

Cf. A000420, A001620, A006519, A038500, A234957, A214411, A234959, A060904, A242603, A268357.

[7^Valuation(n,7): n in [1..150]]; // Vincenzo Librandi, Feb 03 2016

7^Table[IntegerExponent[n, 7], {n, 150}] (* Vincenzo Librandi, Feb 03 2016 *)

a(n) = 7^valuation(n, 7) \\ Michel Marcus, Feb 05 2016

Sage
```
[7^valuation(i, 7) for i in [1..100]]
```

a(n) = 7^valuation(n,7).
a(n) = 7^A214411(n).
Completely multiplicative with a(7) = 7, a(p) = 1 for prime p and p <> 7. - Andrew Howroyd, Jul 20 2018
From Peter Bala, Feb 21 2019: (Start)
a(n) = gcd(n,7^n).
a(n) = n/A242603(n).
O.g.f.: x/(1 - x) + 6*Sum_{n >= 1} 7^(n-1)*x^(7^n)/ (1 - x^(7^n)). (End)
Sum_{k=1..n} a(k) ~ (6/(7*log(7)))*n*log(n) + (4/7 + 6*(gamma-1)/(7*log(7)))*n, where gamma is Euler's constant (A001620). - Amiram Eldar, Nov 15 2022
Dirichlet g.f.: zeta(s)*(7^s-1)/(7^s-7). - Amiram Eldar, Jan 03 2023

More terms from Antti Karttunen, Dec 22 2017

A268357 Highest power of 11 dividing n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 121, 1

Offset: 1

Tom Edgar, Feb 02 2016

The generalized binomial coefficients produced by this sequence provide an analog to Kummer's Theorem using arithmetic in base 11.

This first index where this differs from A109014 is 121; a(121) = 121 and A109014(121) = 11.

			Since 22 = 11 * 2, a(22) = 11. Likewise, since 11 does not divide 21, a(21) = 1.

Amiram Eldar, Table of n, a(n) for n = 1..10000
Tyler Ball, Tom Edgar, and Daniel Juda, Dominance Orders, Generalized Binomial Coefficients, and Kummer's Theorem, Mathematics Magazine, Vol. 87, No. 2, April 2014, pp. 135-143.
Tom Edgar and Michael Z. Spivey, Multiplicative functions, generalized binomial coefficients, and generalized Catalan numbers, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.6.

Cf. A001620, A006519, A038500, A234957, A234959, A109014, A268354, A060904.

[11^Valuation(n,11): n in [1..130]]; // Vincenzo Librandi, Feb 03 2016

Table[11^IntegerExponent[n, 11], {n, 130}] (* Bruno Berselli, Feb 03 2016 *)

[11^valuation(i, 11) for i in [1..130]]

a(n) = 11^valuation(n,11).
Completely multiplicative with a(11) = 11, a(p) = 1 for prime p and p<>11. - Andrew Howroyd, Jul 20 2018
From Peter Bala, Feb 21 2019: (Start)
a(n) = gcd(n,11^n).
O.g.f.: x/(1 - x) + 10*Sum_{n >= 1} 11^(n-1)*x^(11^n)/ (1 - x^(11^n)). (End)
Sum_{k=1..n} a(k) ~ (10/(11*log(11)))*n*log(n) + (6/11 + 10*(gamma-1)/(11*log(11)))*n, where gamma is Euler's constant (A001620). - Amiram Eldar, Nov 15 2022
Dirichlet g.f.: zeta(s)*(11^s-1)/(11^s-11). - Amiram Eldar, Jan 03 2023

A340346 The largest divisor of n that is a term of A055932 (numbers divisible by all primes smaller than their largest prime factor).

Original entry on oeis.org

1, 2, 1, 4, 1, 6, 1, 8, 1, 2, 1, 12, 1, 2, 1, 16, 1, 18, 1, 4, 1, 2, 1, 24, 1, 2, 1, 4, 1, 30, 1, 32, 1, 2, 1, 36, 1, 2, 1, 8, 1, 6, 1, 4, 1, 2, 1, 48, 1, 2, 1, 4, 1, 54, 1, 8, 1, 2, 1, 60, 1, 2, 1, 64, 1, 6, 1, 4, 1, 2, 1, 72, 1, 2, 1, 4, 1, 6, 1, 16, 1, 2, 1, 12

Offset: 1

Peter Munn, Jan 04 2021

			For n=2: the largest divisor of 2 is 2, and 2 qualifies as divisible by all primes smaller than its largest prime factor, 2 (since there are no smaller primes). So a(2) = 2.
For n=42: of 42's divisors, no multiples of 7 qualify as being divisible by all primes smaller than their largest prime factor (since that factor is 7 and no divisor of 42 is divisible by 5, a smaller prime). The largest of 42's other divisors is 6, which qualifies (since it is divisible by 2, the only prime smaller than 6's largest prime factor, 3). So a(42) = 6.

Antti Karttunen, Table of n, a(n) for n = 1..8191
Antti Karttunen, Data supplement: n, a(n) computed for n = 1..65537
Index entries for sequences computed from indices in prime factorization

A003961, A006519, A055932, A064989, A341629 are used in a definition of this sequence.
Sequences with related definitions: A327832, A328479.
Cf. A234959.

a[?OddQ] = 1; a[n] := Module[{f = FactorInteger[n]}, ind = Position[PrimePi /@ First /@ f - Range @ Length[f], ?(# > 0 &)]; If[ind == {}, n, Times @@ Power @@@ f[[1 ;; ind[[1, 1]] - 1]]]]; Array[a, 100] (* _Amiram Eldar, Jan 14 2021 *)

is(n) = my(f=factor(n)[, 1]~); f==primes(#f); \\ A055932
a(n) = vecmax(select(is, divisors(n))); \\ Michel Marcus, Jan 19 2021

A341629(n) = if(1==n,1,my(f=factor(n)[, 1]~); (primepi(f[#f])==#f));
A340346(n) = fordiv(n,d,if(A341629(n/d),return(n/d))); \\ Antti Karttunen, Feb 25 2021

For n >= 1, a(2n-1) = 1, a(2n) = A006519(2n) * A003961(a(A064989(2n))).

For n >= 1, lcm(A006519(n), A234959(n)) | a(n).

A339748 a(n) = (6^(valuation(n, 6) + 1) - 1) / 5.

Original entry on oeis.org

1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 43, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 43, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1

Offset: 1

Ilya Gutkovskiy, Dec 15 2020

nonn

Sum of powers of 6 dividing n.

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

Cf. A038712, A080278, A088842, A122841, A234959, A323921, A339747.

Table[(6^(IntegerExponent[n, 6] + 1) - 1)/5, {n, 1, 100}]
nmax = 100; CoefficientList[Series[Sum[6^k x^(6^k)/(1 - x^(6^k)), {k, 0, Floor[Log[6, nmax]] + 1}], {x, 0, nmax}], x] // Rest

G.f.: Sum_{k>=0} 6^k * x^(6^k) / (1 - x^(6^k)).
L.g.f.: -log(Product_{k>=0} (1 - x^(6^k))).
Dirichlet g.f.: zeta(s) / (1 - 6^(1 - s)).

A264981 Highest power of 9 dividing n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 1, 81, 1, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 1, 9

Offset: 1

Antti Karttunen, Dec 07 2015

The generalized binomial coefficients produced by this sequence provide an analog to Kummer's Theorem using arithmetic in base 9. - Tom Edgar, Feb 02 2016

			Since 18 = 9 * 2, a(18) = 9. Likewise, since 9 does not divide 17, a(17) = 1. - _Tom Edgar_, Feb 02 2016

Antti Karttunen, Table of n, a(n) for n = 1..6561
Tyler Ball, Tom Edgar, and Daniel Juda, Dominance Orders, Generalized Binomial Coefficients, and Kummer's Theorem, Mathematics Magazine, Vol. 87, No. 2, April 2014, pp. 135-143.
Tom Edgar and Michael Z. Spivey, Multiplicative functions, generalized binomial coefficients, and generalized Catalan numbers, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.6.

Cf. A001620, A264979.

Similar sequences for other bases: A006519 (2), A038500 (3), A234957 (4), A060904 (5), A234959 (6).

Table[9^Length@ TakeWhile[Reverse@ IntegerDigits[n, 9], # == 0 &], {n, 99}] (* Michael De Vlieger, Dec 09 2015 *)
9^Table[IntegerExponent[n, 9], {n, 150}] (* Vincenzo Librandi, Feb 03 2016 *)

a(n) = 9^valuation(n, 9); \\ Michel Marcus, Dec 08 2015

[9^valuation(i, 9) for i in [1..100]] # Tom Edgar, Feb 02 2016

(define (A264981 n) (let loop ((k 9)) (if (not (zero? (modulo n k))) (/ k 9) (loop (* 9 k)))))

a(n) = 9^valuation(n,9). - Tom Edgar, Feb 02 2016
G.f.: x/(1 - x) + 8 * Sum_{k>=1} 9^(k-1)*x^(9^k)/(1 - x^(9^k)). - Ilya Gutkovskiy, Jul 10 2019
From Amiram Eldar, Dec 31 2022: (Start)
Multiplicative with a(3^e) = 3^(2*floor(e/2)), and a(p^e) = 1 if p != 3.
Dirichlet g.f.: zeta(s)*(9^s-1)/(9^s-9).
Sum_{k=1..n} a(k) ~ (4/(9*log(3)))*n*log(n) + (5/9 + 4*(gamma-1)/(9*log(3)))*n, where gamma is Euler's constant (A001620). (End)

Keyword:mult added by Andrew Howroyd, Jul 20 2018

A254730 Triangle read by rows: T(n,k) = A243758(n)/(A243758(k)*A243758(n-k)).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 6, 6, 6, 6, 6, 1, 1, 1, 6, 6, 6, 6, 1, 1, 1, 1, 1, 6, 6, 6, 1, 1, 1, 1, 1, 1, 1, 6, 6, 1, 1, 1, 1, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 6, 6, 6, 6, 6, 1, 6, 6, 6

Offset: 0

Tom Edgar, Feb 06 2015

These are the generalized binomial coefficients associated with A234959.
The exponent of T(n,k) is the number of 'carries' that occur when adding k and n-k in base 6 using the traditional addition algorithm.
If T(n,k) != 0 mod 6, then n dominates k in base 6.

			The first six terms in A234959 are 1, 1, 1, 1, 1 and 6 and so T(4,2) = 1*1*1*1/((1*1)*(1*1))=1 and T(6,3) = 6*1*1*1*1*1/((1*1*1)*(1*1*1))=6.
The triangle begins:
1
1, 1
1, 1, 1
1, 1, 1, 1
1, 1, 1, 1, 1
1, 1, 1, 1, 1, 1
1, 6, 6, 6, 6, 6, 1
1, 1, 6, 6, 6, 6, 1, 1
1, 1, 1, 6, 6, 6, 1, 1, 1
1, 1, 1, 1, 6, 6, 1, 1, 1, 1
1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
1, 6, 6, 6, 6, 6, 1, 6, 6, 6, 6, 6, 1
1, 1, 6, 6, 6, 6, 1, 1, 6, 6, 6, 6, 1, 1
1, 1, 1, 6, 6, 6, 1, 1, 1, 6, 6, 6, 1, 1, 1
1, 1, 1, 1, 6, 6, 1, 1, 1, 1, 6, 6, 1, 1, 1, 1
1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
Tyler Ball, Tom Edgar, and Daniel Juda, Dominance Orders, Generalized Binomial Coefficients, and Kummer's Theorem, Mathematics Magazine, Vol. 87, No. 2, April 2014, pp. 135-143.
Tyler Ball and Daniel Juda, Dominance over N, Rose-Hulman Undergraduate Mathematics Journal, Vol. 13, No. 2, Fall 2013.
Tom Edgar and Michael Z. Spivey, Multiplicative functions, generalized binomial coefficients, and generalized Catalan numbers, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.6.

Cf. A234959, A243758, A242849, A082907, A254609.

import Data.List (inits)
a254730 n k = a254730_tabl !! n !! k
a254730_row n = a254730_tabl !! n
a254730_tabl = zipWith (map . div)
   a243758_list $ zipWith (zipWith (*)) xss $ map reverse xss
   where xss = tail $ inits a243758_list
-- Reinhard Zumkeller, Feb 09 2015

P=[0]+[6^valuation(i,6) for i in [1..100]]
[m for sublist in [[mul(P[1:n+1])/(mul(P[1:k+1])*mul(P[1:(n-k)+1])) for k in [0..n]] for n in [0..len(P)-1]] for m in sublist]

T(n,k) = A243758(n)/(A243758(k)*A243758(n-k)).
T(n,k) = Product_{i=1..n} A234959(i)/(Product_{i=1..k} A234959(i)*Product_{i=1..n-k} A234959(i)).
T(n,k) = A234959(n)/n*(k/A234959(k)*T(n-1,k-1)+(n-k)/A234959(n-k)*T(n-1,k)).

A268355 Highest power of 8 dividing n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 64, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 1, 1

Offset: 1

Tom Edgar, Feb 02 2016

The generalized binomial coefficients produced by this sequence provide an analog to Kummer's Theorem using arithmetic in base 8.

			Since 16 = 8 * 2, a(16) = 8. Likewise, since 8 does not divide 15, a(15) = 1.

Antti Karttunen, Table of n, a(n) for n = 1..65537
Tyler Ball, Tom Edgar, and Daniel Juda, Dominance Orders, Generalized Binomial Coefficients, and Kummer's Theorem, Mathematics Magazine, Vol. 87, No. 2, April 2014, pp. 135-143.
Tom Edgar and Michael Z. Spivey, Multiplicative functions, generalized binomial coefficients, and generalized Catalan numbers, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.6.

Cf. A001620, A006519, A038500, A234957, A244413, A234959.

seq(8^floor(padic:-ordp(n,2)/3), n=1..100); # Robert Israel, Feb 03 2016

8^Table[IntegerExponent[n, 8], {n, 150}] (* Vincenzo Librandi, Feb 03 2016 *)

a(n) = 8^valuation(n, 8); \\ Michel Marcus, Feb 05 2016

def A268355(n): return (m:=(~n&n-1))+1>>(m.bit_length()%3) # Chai Wah Wu, Jul 09 2022

Sage
```
[8^valuation(i, 8) for i in [1..100]]
```

a(n) = 8^valuation(n,8).
a(n) = 8^A244413(n).
G.f.: Sum_{m>=0} 8^m * Sum_{j=1..7} x^(j*8^m)/(1-x^(8^(m+1))). - Robert Israel, Feb 03 2016
From Amiram Eldar, Dec 31 2022: (Start)
Multiplicative with a(2^e) = 2^(3*floor(e/3)), and a(p^e) = 1 if p >= 3.
Dirichlet g.f.: zeta(s)*(8^s-1)/(8^s-8).
Sum_{k=1..n} a(k) ~ (7/(24*log(2)))*n*log(n) + (9/16 + 7*(gamma-1)/(24*log(2)))*n, where gamma is Euler's constant (A001620). (End)

Keyword:mult added by Andrew Howroyd, Jul 20 2018

cp's OEIS Frontend

A243758 a(n) = Product_{i=1..n} A234959(i).

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A122841 Greatest k such that 6^k divides n.

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A268354 Highest power of 7 dividing n.

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A268357 Highest power of 11 dividing n.

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A340346 The largest divisor of n that is a term of A055932 (numbers divisible by all primes smaller than their largest prime factor).

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A339748 a(n) = (6^(valuation(n, 6) + 1) - 1) / 5.

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A264981 Highest power of 9 dividing n.

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