A000351 -id:A000351 - cp's OEIS frontend

A003464 a(n) = (6^n - 1)/5.

0, 1, 7, 43, 259, 1555, 9331, 55987, 335923, 2015539, 12093235, 72559411, 435356467, 2612138803, 15672832819, 94036996915, 564221981491, 3385331888947, 20311991333683, 121871948002099, 731231688012595, 4387390128075571

Offset: 0

N. J. A. Sloane

a(n) = A125118(n, 5) for n>4. - Reinhard Zumkeller, Nov 21 2006
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=6, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>=1, a(n)=det(A). - Milan Janjic, Feb 21 2010
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=7, (i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n>1, a(n-1)=(-1)^n*charpoly(A,1). - Milan Janjic, Feb 21 2010
Repunits to base 6. A repunit consisting of zero 1's (empty string) gives the empty sum, i.e., 0 (only case where leading zero is shown, for convenience). - Daniel Forgues, Jul 08 2011
3*a(n) is the total number of holes in a certain triangle fractal (start with 6 triangles, 3 holes) after n iterations. See illustration in links. - Kival Ngaokrajang, Feb 21 2015

			a(n) in base 6.................... a(n) in base 10:
0..................................0
1..................................1
11.................................7
111................................43
1111...............................259
11111..............................1555
111111.............................9331
1111111............................55987, etc. - _Philippe Deléham_, Mar 12 2014

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
C. Banderier and D. Merlini, Lattice paths with an infinite set of jumps, FPSAC02, Melbourne, 2002.
Carlos M. da Fonseca and Anthony G. Shannon, A formal operator involving Fermatian numbers, Notes Num. Theor. Disc. Math. (2024) Vol. 30, No. 3, 491-498.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 375
Kival Ngaokrajang, Illustration of initial terms
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
D. C. Santos, E. A. Costa, and P. M. M. C. Catarino, On Gersenne Sequence: A Study of One Family in the Horadam-Type Sequence, Axioms 14, 203, (2025). See p. 4.
Eric Weisstein's World of Mathematics, Repunit.
Index entries for linear recurrences with constant coefficients, signature (7,-6).

[n le 2 select n-1 else 7*Self(n-1) - 6*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 08 2012

a:=n->sum(6^(n-j),j=1..n): seq(a(n), n=1..21); # Zerinvary Lajos, Jan 04 2007
A003464:=1/(6*z-1)/(z-1); # conjectured by Simon Plouffe in his 1992 dissertation
a[0]:=0:a[1]:=1:for n from 2 to 50 do a[n]:=5*a[n-1]+6*a[n-2]+2 od: seq(a[n], n=1..33); # Zerinvary Lajos, Dec 14 2008

(6^Range[20]-1)/5 (* Harvey P. Dale, Dec 14 2010 *)
LinearRecurrence[{7, -6}, {0, 1}, 30] (* Vincenzo Librandi, Nov 08 2012 *)

A003464(n):=floor((6^n-1)/5)$  makelist(A003464(n),n,0,30); /* Martin Ettl, Nov 05 2012 */

PARI
```
for(n=1,10,print1((6^n-1)/5,","));
```

[lucas_number1(n,7,6) for n in range(1, 22)] # Zerinvary Lajos, Apr 23 2009

[gaussian_binomial(n,1,6) for n in range(1,22)] # Zerinvary Lajos, May 28 2009

Binomial transform of A003948. If preceded by 0, then binomial transform of powers of 5, A000351 (preceded by 0). - Paul Barry, Mar 28 2003
a(n) = Sum_{k=1..n} C(n, k)*5^(k-1).
E.g.f.: (exp(6*x) - exp(x))/5. - Paul Barry, Mar 28 2003
G.f.: x/((1-x)*(1-6*x)). - Lambert Klasen (lambert.klasen(AT)gmx.net), Feb 06 2005
a(n) = 6*a(n-1) + 1 with a(1)=1. - Vincenzo Librandi, Nov 17 2010
a(n) = 7*a(n-1) - 6*a(n-2). - Vincenzo Librandi, Nov 08 2012

More terms from Reinhard Zumkeller, Nov 21 2006

G.f. corrected by Philippe Deléham, Mar 11 2014

A027868 Number of trailing zeros in n!; highest power of 5 dividing n!.

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 12, 12, 12, 12, 12, 13, 13, 13, 13, 13, 14, 14, 14, 14, 14, 15, 15, 15, 15, 15, 16, 16, 16, 16, 16, 18, 18, 18, 18, 18, 19

Offset: 0

Warut Roonguthai

Also the highest power of 10 dividing n! (different from A054899). - Hieronymus Fischer, Jun 18 2007
Alternatively, a(n) equals the expansion of the base-5 representation A007091(n) of n (i.e., where successive positions from right to left stand for 5^n or A000351(n)) under a scale of notation whose successive positions from right to left stand for (5^n - 1)/4 or A003463(n); for instance, n = 7392 has base-5 expression 2*5^5 + 1*5^4 + 4*5^3 + 0*5^2 + 3*5^1 + 2*5^0, so that a(7392) = 2*781 + 1*156 + 4*31 + 0*6 + 3*1 + 2*0 = 1845. - Lekraj Beedassy, Nov 03 2010
Partial sums of A112765. - Hieronymus Fischer, Jun 06 2012
Also the number of trailing zeros in A000165(n) = (2*n)!!. - Stefano Spezia, Aug 18 2024

			a(100)  = 24.
a(10^3) = 249.
a(10^4) = 2499.
a(10^5) = 24999.
a(10^6) = 249998.
a(10^7) = 2499999.
a(10^8) = 24999999.
a(10^9) = 249999998.
a(10^n) = 10^n/4 - 3 for 10 <= n <= 15 except for a(10^14) = 10^14/4 - 2. - _M. F. Hasler_, Dec 27 2019

M. Gardner, "Factorial Oddities." Ch. 4 in Mathematical Magic Show: More Puzzles, Games, Diversions, Illusions and Other Mathematical Sleight-of-Mind from Scientific American. New York: Vintage, 1978, pp. 50-65.

Hieronymus Fischer, Table of n, a(n) for n = 0..10000 (first 1000 terms from T. D. Noe)
David S. Hart, James E. Marengo, Darren A. Narayan, and David S. Ross, On the number of trailing zeros in n!, College Math. J., 39(2):139-145, 2008.
Enrique Pérez Herrero, Trailing Zeros in n!, Psychedelic Geometry Blogspot.
Soichi Ikeda and Kaneaki Matsuoka, On transcendental numbers generated by certain integer sequences, Siauliai Math. Semin., 8 (16) 2013, 63-69.
Shyam Sunder Gupta, Fascinating Factorials, Exploring the Beauty of Fascinating Numbers, Springer (2025) Ch. 16, 411-442.
Shu-Chung Liu and Jean C.-C. Yeh, Catalan numbers modulo 2^k, J. Int. Seq. 13 (2010), 10.5.4, eq (5).
Antonio M. Oller-Marcén, A new look at the trailing zeros of n!, arXiv:0906.4868v1 [math.NT], 2009.
Antonio M. Oller-Marcén and J. Maria Grau, On the Base-b Expansion of the Number of Trailing Zeros of b^k!, J. Int. Seq. 14 (2011) 11.6.8
Eric Weisstein's World of Mathematics, Factorial.
Index entries for sequences related to factorial numbers

See A000966 for the missing numbers. See A011371 and A054861 for analogs involving powers of 2 and 3.
Cf. A054899, A007953, A112765, A067080, A098844, A132027, A067080, A098844, A132029, A054999, A112765, A191610, A000351.
Cf. also A000142, A004154.
Cf. A000165, A008904.

a027868 n = sum $ takeWhile (> 0) $ map (n `div`) $ tail a000351_list
-- Reinhard Zumkeller, Oct 31 2012

[Valuation(Factorial(n), 5): n in [0..80]]; // Bruno Berselli, Oct 11 2021

0, seq(add(floor(n/5^i),i=1..floor(log[5](n))), n=1..100); # Robert Israel, Nov 13 2014

Table[t = 0; p = 5; While[s = Floor[n/p]; t = t + s; s > 0, p *= 5]; t, {n, 0, 100} ]
Table[ IntegerExponent[n!], {n, 0, 80}] (* Robert G. Wilson v *)
zOF[n_Integer?Positive]:=Module[{maxpow=0},While[5^maxpow<=n,maxpow++];Plus@@Table[Quotient[n,5^i],{i,maxpow-1}]]; Attributes[zOF]={Listable}; Join[{0},zOF[ Range[100]]] (* Harvey P. Dale, Apr 11 2022 *)

a(n)={my(s);while(n\=5,s+=n);s} \\ Charles R Greathouse IV, Nov 08 2012, edited by M. F. Hasler, Dec 27 2019

a(n)=valuation(n!,5) \\ Charles R Greathouse IV, Nov 08 2012

apply( A027868(n)=(n-sumdigits(n,5))\4, [0..99]) \\ M. F. Hasler, Dec 27 2019

from sympy import multiplicity
A027868, p5 = [0,0,0,0,0], 0
for n in range(5,10**3,5):
    p5 += multiplicity(5,n)
    A027868.extend([p5]*5) # Chai Wah Wu, Sep 05 2014

def A027868(n): return 0 if n<5 else n//5 + A027868(n//5) # David Radcliffe, Jun 26 2016

from sympy.ntheory.factor_ import digits
def A027868(n): return n-sum(digits(n,5)[1:])>>2 # Chai Wah Wu, Oct 18 2024

a(n) = Sum_{i>=1} floor(n/5^i).
a(n) = (n - A053824(n))/4.
From Hieronymus Fischer, Jun 25 2007 and Aug 13 2007, edited by M. F. Hasler, Dec 27 2019: (Start)
G.f.: g(x) = Sum_{k>0} x^(5^k)/(1-x^(5^k))/(1-x).
a(n) = Sum_{k=5..n} Sum_{j|k, j>=5} (floor(log_5(j)) - floor(log_5(j-1))).
G.f.: g(x) = L[b(k)](x)/(1-x) where L[b(k)](x) = Sum_{k>=0} b(k)*x^k/(1-x^k) is a Lambert series with b(k) = 1, if k>1 is a power of 5, else b(k) = 0.
G.f.: g(x) = Sum_{k>0} c(k)*x^k/(1-x), where c(k) = Sum_{j>1, j|k} floor(log_5(j)) - floor(log_5(j - 1)).
Recurrence:
a(n) = floor(n/5) + a(floor(n/5));
a(5*n) = n + a(n);
a(n*5^m) = n*(5^m-1)/4 + a(n).
a(k*5^m) = k*(5^m-1)/4, for 0 <= k < 5, m >= 0.
Asymptotic behavior:
a(n) = n/4 + O(log(n)),
a(n+1) - a(n) = O(log(n)), which follows from the inequalities below.
a(n) <= (n-1)/4; equality holds for powers of 5.
a(n) >= n/4 - 1 - floor(log_5(n)); equality holds for n = 5^m-1, m > 0.
lim inf (n/4 - a(n)) = 1/4, for n -> oo.
lim sup (n/4 - log_5(n) - a(n)) = 0, for n -> oo.
lim sup (a(n+1) - a(n) - log_5(n)) = 0, for n -> oo.
(End)
a(n) <= A027869(n). - Reinhard Zumkeller, Jan 27 2008
10^a(n) = A000142(n) / A004154(n). - Reinhard Zumkeller, Nov 24 2012
a(n) = Sum_{k=1..floor(n/2)} floor(log_5(n/k)). - Ammar Khatab, Feb 01 2025

Examples added by Hieronymus Fischer, Jun 06 2012

A112765 Exponent of highest power of 5 dividing n. Or, 5-adic valuation of n.

Original entry on oeis.org

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

Offset: 1

Reinhard Zumkeller, Sep 18 2005

A027868 gives partial sums.

This is also the 5-adic valuation of Fibonacci(n). See Lengyel link. - Michel Marcus, May 06 2017

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Dario T. de Castro, P-adic Order of Positive Integers via Binomial Coefficients, INTEGERS, Electronic J. of Combinatorial Number Theory, Vol. 22, Paper A61, 2022.
T. Lengyel, The order of the Fibonacci and Lucas numbers, Fibonacci Quart. 33 (1995), no. 3, 234-239. See Lemma 1 p. 235.

Cf. A007814, A007949, A112762, A022337, A122840, A027868, A054899, A122841, A160093, A160094, A196563, A196564.

Cf. A343251, A000351 (positions of records, greedy inverse).

a112765 n = fives n 0 where
   fives n e | r > 0     = e
             | otherwise = fives n' (e + 1) where (n',r) = divMod n 5
-- Reinhard Zumkeller, Apr 08 2011

A112765 := proc(n)
    padic[ordp](n,5) ;
end proc: # R. J. Mathar, Jul 12 2016

a[n_] := IntegerExponent[n, 5]; Array[a, 105] (* Jean-François Alcover, Jan 25 2018 *)

A112765(n)=valuation(n,5); /* Joerg Arndt, Apr 08 2011 */

def a(n):
    k = 0
    while n > 0 and n%5 == 0: n //= 5; k += 1
    return k
print([a(n) for n in range(1, 106)]) # Michael S. Branicky, Aug 06 2021

Totally additive with a(p) = 1 if p = 5, 0 otherwise.
From Hieronymus Fischer, Jun 08 2012: (Start)
With m = floor(log_5(n)), frac(x) = x-floor(x):
a(n) = Sum_{j=1..m} (1 - ceiling(frac(n/5^j))).
a(n) = m + Sum_{j=1..m} (floor(-frac(n/5^j))).
a(n) = A027868(n) - A027868(n-1).
G.f.: Sum_{j>0} x^5^j/(1-x^5^j). (End)
a(5n) = A055457(n). - R. J. Mathar, Jul 17 2012
Asymptotic mean: lim_{m->oo} (1/m) * Sum_{k=1..m} a(k) = 1/4. - Amiram Eldar, Feb 14 2021
a(n) = 5*Sum_{j=1..floor(log(n)/log(5))} frac(binomial(n, 5^j)*5^(j-1)/n). - Dario T. de Castro, Jul 10 2022

A001692 Number of irreducible polynomials of degree n over GF(5); dimensions of free Lie algebras.

Original entry on oeis.org

1, 5, 10, 40, 150, 624, 2580, 11160, 48750, 217000, 976248, 4438920, 20343700, 93900240, 435959820, 2034504992, 9536718750, 44878791360, 211927516500, 1003867701480, 4768371093720, 22706531339280

Offset: 0

N. J. A. Sloane

Exponents in expansion of Hardy-Littlewood constant C_5 = 0.409874885.. = A269843 as a product_{n>=2} zeta(n)^(-a(n)).

Number of aperiodic necklaces with n beads of 5 colors. - Herbert Kociemba, Nov 25 2016

E. R. Berlekamp, Algebraic Coding Theory, McGraw-Hill, NY, 1968, p. 84.
Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003
M. Lothaire, Combinatorics on Words. Addison-Wesley, Reading, MA, 1983, p. 79.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Seiichi Manyama, Table of n, a(n) for n = 0..1435 (terms 0..200 from T. D. Noe)
Alin Bostan, Alexander Marynych, and Kilian Raschel, On the least common multiple of several random~integers, arXiv:1901.03002 [math.PR], 2019.
Jeremie Detrey, P. J. Spaenlehauer, and P. Zimmermann, Computing the rho constant, Preprint 2016.
E. N. Gilbert and J. Riordan, Symmetry types of periodic sequences, Illinois J. Math., 5 (1961), 657-665.
Gareth A. Jones and Alexander K. Zvonkin, Groups of prime degree and the Bateman-Horn conjecture, 2021.
G. Niklasch, Some number theoretical constants: 1000-digit values [Cached copy]
Y. Puri and T. Ward, Arithmetic and growth of periodic orbits, J. Integer Seqs., Vol. 4 (2001), #01.2.1.
Dionisel Y. Regalado and Rodel Azura, An Analytic Approximation to the Density of Twin Primes, Recoletos Multidisciplinary Research Journal (2019) Vol. 6, No. 2.
G. J. Simmons, The number of irreducible polynomials of degree n over GF(p), Amer. Math. Monthly, 77 (1970), 743-745.
G. Viennot, Algèbres de Lie Libres et Monoïdes Libres, Lecture Notes in Mathematics 691, Springer Verlag 1978.
Index entries for sequences related to Lyndon words

Cf. A001037, A054720, A002105.
5th column of A074650. - Alois P. Heinz, Aug 08 2008
Cf. A008683, A000351, A027750.

a001692 n = flip div n $ sum $
            zipWith (*) (map a008683 divs) (map a000351 $ reverse divs)
            where divs = a027750_row n
-- Reinhard Zumkeller, Oct 07 2015

a[0] = 1; a[n_] := Sum[MoebiusMu[d]*5^(n/d)/n, {d, Divisors[n]}]; Table[a[n], {n, 0, 21}] (* Jean-François Alcover, Mar 11 2014 *)
mx=40;f[x_,k_]:=1-Sum[MoebiusMu[i] Log[1-k*x^i]/i,{i,1,mx}];CoefficientList[Series[f[x,5],{x,0,mx}],x] (* Herbert Kociemba, Nov 25 2016 *)

a(n)=if(n,sumdiv(n,d,moebius(d)*5^(n/d))/n,1) \\ Charles R Greathouse IV, Jun 15 2011

a(n) = Sum_{d|n} mu(d)*5^(n/d)/n, for n>0.

G.f.: k=5, 1 - Sum_{i>=1} mu(i)*log(1 - k*x^i)/i. - Herbert Kociemba, Nov 25 2016

A329697 a(n) is the number of iterations needed to reach a power of 2 starting at n and using the map k -> k-(k/p), where p is the largest prime factor of k.

Original entry on oeis.org

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

Offset: 1

Ali Sada and Robert G. Wilson v, Feb 28 2020

From Antti Karttunen, Apr 07 2020: (Start)
Also the least number of iterations of nondeterministic map k -> k-(k/p) needed to reach a power of 2, when any prime factor p of k can be used. The minimal length path to the nearest power of 2 (= 2^A064415(n)) is realized whenever one uses any of the A005087(k) distinct odd prime factors of the current k, at any step of the process. For example, this could be done by iterating with the map k -> k-(k/A078701(k)), i.e., by using the least odd prime factor of k (instead of the largest prime).
Proof: Viewing the prime factorization of changing k as a multiset ("bag") of primes, we see that liquefying any odd prime p with step p -> (p-1) brings at least one more 2 to the bag, while applying p -> (p-1) to any 2 just removes it from the bag, but gives nothing back. Thus the largest (and thus also the nearest) power of 2 is reached by eliminating - step by step - all odd primes from the bag, but none of 2's, and it doesn't matter in which order this is done.
The above implies also that the sequence is totally additive, which also follows because both A064097 and A064415 are. That A064097(n) = A329697(n) + A054725(n) for all n > 1 can be also seen by comparing the initial conditions and the recursion formulas of these three sequences.
For any n, A333787(n) is either the nearest power of 2 reached (= 2^A064415(n)), or occurs on some of the paths from n to there.
(End)
A003401 gives the numbers k where a(k) = A005087(k). See also A336477. - Antti Karttunen, Mar 16 2021

			The trajectory of 15 is {12, 8}, taking 2 iterations to reach 8 = 2^3. So a(15) is 2.
From _Antti Karttunen_, Apr 07 2020: (Start)
Considering all possible paths from 15 to 1 nondeterministic map k -> k-(k/p), where p can be any prime factor of k, we obtain the following graph:
        15
       / \
      /   \
    10     12
    / \   / \
   /   \ /   \
  5     8     6
   \__  |  __/|
      \_|_/   |
        4     3
         \   /
          \ /
           2
           |
           1.
It can be seen that there's also alternative route to 8 via 10 (with 10 = 15-(15/3), where 3 is not the largest prime factor of 15), but it's not any shorter than the route via 12.
(End)

Antti Karttunen, Table of n, a(n) for n = 1..65537
Michael De Vlieger, Annotated fan style binary tree labeling the index n, with a color code where black represents a(n) = 0, red a(n) = 1, and magenta the largest value in a(n) for n = 1..16383.

Cf. A000265, A003401, A005087, A052126, A053575, A054725, A064097, A064415, A078701, A087436, A147545, A171462, A209229, A333123, A333787, A333790, A334107, A334109, A335875, A334204, A335880, A335881, A336396, A336466, A336469 [= a(phi(n))], A336928 [= a(sigma(n))], A336470, A336477, A339970.
Cf. A000079, A334101, A334102, A334103, A334104, A334105, A334106 for positions of 0 .. 6 in this sequence, and also array A334100.
Cf. A334099 (a right inverse, positions of the first occurrence of each n).
Cf. A334091 (first differences), A335429 (partial sums).
Cf. also A331410 (analogous sequence when using the map k -> k + k/p), A334861, A335877 (their sums and differences), see also A335878 and A335884, A335885.

a[n_] := Length@ NestWhileList[# - #/FactorInteger[#][[-1, 1]] &, n, # != 2^IntegerExponent[#, 2] &] -1; Array[a, 100]

A329697(n) = if(!bitand(n,n-1),0,1+A329697(n-(n/vecmax(factor(n)[, 1])))); \\ Antti Karttunen, Apr 07 2020

up_to = 2^24;
A329697list(up_to) = { my(v=vector(up_to)); v[1] = 0; for(n=2, up_to, v[n] = if(!bitand(n,n-1),0,1+vecmin(apply(p -> v[n-n/p], factor(n)[, 1]~)))); (v); };
v329697 = A329697list(up_to);
A329697(n) = v329697[n]; \\ Antti Karttunen, Apr 07 2020

A329697(n) = if(n<=2,0, if(isprime(n), A329697(n-1)+1, my(f=factor(n)); (apply(A329697, f[, 1])~ * f[, 2]))); \\ Antti Karttunen, Apr 19 2020

From Antti Karttunen, Apr 07-19 2020: (Start)
a(1) = a(2) = 0; and for n > 2, a(p) = 1 + a(p-1) if p is an odd prime and a(n*m) = a(n) + a(m) if m,n > 1. [This is otherwise equal to the definition of A064097, except here we have a different initial condition, with a(2) = 0].
a(2n) = a(A000265(n)) = a(n).
a(p) = 1+a(p-1), for all odd primes p.
If A209229(n) == 1 [when n is a power of 2], a(n) = 0,
  otherwise a(n) = 1 + a(n-A052126(n)) = 1 + a(A171462(n)).
Equivalently, for non-powers of 2, a(n) = 1 + a(n-(n/A078701(n))),
or equivalently, for non-powers of 2, a(n) = 1 + Min a(n - n/p), for p prime and dividing n.
a(n) = A064097(n) - A064415(n), or equally, a(n) = A064097(n) - A054725(n), for n > 1.
a(A019434(n)) = 1, a(A334092(n)) = 2, a(A334093(n)) = 3, etc. for all applicable n.
For all n >= 0, a(A334099(n)) = a(A000244(n)) = a(A000351(n)) = a(A001026(n)) = a(257^n) = a(65537^n) = n.
a(A122111(n)) = A334107(n), a(A225546(n)) = A334109(n).
(End)
From Antti Karttunen, Mar 16 2021: (Start)
a(n) = a(A336466(n)) + A087436(n) = A336396(n) + A087436(n).
a(A053575(n)) = A336469(n) = a(n) - A005087(n).
a(A147545(n)) = A000120(A147545(n)) - 1.
(End)

A001021 Powers of 12.

Original entry on oeis.org

1, 12, 144, 1728, 20736, 248832, 2985984, 35831808, 429981696, 5159780352, 61917364224, 743008370688, 8916100448256, 106993205379072, 1283918464548864, 15407021574586368, 184884258895036416, 2218611106740436992

Offset: 0

N. J. A. Sloane

Same as Pisot sequences E(1, 12), L(1, 12), P(1, 12), T(1, 12). Essentially same as Pisot sequences E(12, 144), L(12, 144), P(12, 144), T(12, 144). See A008776 for definitions of Pisot sequences.
Central terms of the triangle in A100851. - Reinhard Zumkeller, Mar 04 2006
The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n>=1, a(n) equals the number of 12-colored compositions of n such that no adjacent parts have the same color. - Milan Janjic, Nov 17 2011
Starting with 12, this sequence appears in the film "Vollmond" (1998, dir. Fredi Murer), when the children write it on the sidewalk at night. - Alonso del Arte, Dec 21 2011

N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe, Table of n, a(n) for n = 0..100
P. J. Cameron, Sequences realized by oligomorphic permutation groups, J. Integ. Seqs. Vol. 3 (2000), #00.1.5.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 276.
Tanya Khovanova, Recursive Sequences.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Y. Puri and T. Ward, Arithmetic and growth of periodic orbits, J. Integer Seqs., Vol. 4 (2001), #01.2.1.
Index entries for linear recurrences with constant coefficients, signature (12).

Cf. A024140 and A178248, 12^n +/- 1; A016125.

Cf. A000005, A000244, A000302, A000351, A000384, A001222, A008585, A008776, A100851, A159991.

a001021 = (12 ^)
a001021_list = iterate (* 12) 1  -- Reinhard Zumkeller, Mar 31 2012

A001021:=-1/(-1+12*z); # Simon Plouffe in his 1992 dissertation

Table[12^n, {n, 0, 40}] (* Vladimir Joseph Stephan Orlovsky, Feb 15 2011 *)

a(n)=12^n \\ Charles R Greathouse IV, Dec 21 2011

G.f.: 1/(1-12*x).
E.g.f.: exp(12x).
a(n) = 12*a(n-1). - Zerinvary Lajos, Apr 27 2009
a(n) = A159991(n)/A000351(n). - Reinhard Zumkeller, May 02 2009
From Reinhard Zumkeller, Mar 31 2012: (Start)
a(n) = A000302(n) * A000244(n). - Reinhard Zumkeller, Mar 31 2012
A001222(a(n)) = A008585(n); A000005(a(n)) = A000384(a(n)). (End)
a(n) = det(|ps(i+2, j)|, 1 <= i, j <= n), where ps(n, k) are Legendre-Stirling numbers of the first kind. - Mircea Merca, Apr 04 2013

A034474 a(n) = 5^n + 1.

Original entry on oeis.org

2, 6, 26, 126, 626, 3126, 15626, 78126, 390626, 1953126, 9765626, 48828126, 244140626, 1220703126, 6103515626, 30517578126, 152587890626, 762939453126, 3814697265626, 19073486328126, 95367431640626, 476837158203126

Offset: 0

N. J. A. Sloane

a(n) is the deficiency of 3*5^n (see A033879). - Patrick J. McNab, May 28 2017

			G.f. = 2 + 6*x + 26*x^2 + 126*x^3 + 626*x^4 + 3126*x^5 + 15626*x^6 + ...

T. D. Noe, Table of n, a(n) for n = 0..200
Amelia Carolina Sparavigna, The groupoids of Mersenne, Fermat, Cullen, Woodall and other Numbers and their representations by means of integer sequences, Politecnico di Torino, Italy (2019), [math.NT].
Amelia Carolina Sparavigna, Some Groupoids and their Representations by Means of Integer Sequences, International Journal of Sciences (2019) Vol. 8, No. 10.
Index entries for linear recurrences with constant coefficients, signature (6,-5).

Cf. A000051, A000351, A007689, A024049, A033879, A034472, A034478, A034491, A034524, A052539, A062394, A062395, A062396, A062397, A063376, A063481, A074600-A074624, A178248, A228081, A279396.

[5^n+1: n in [0..30]]; // Vincenzo Librandi, Jan 11 2017

Table[5^n + 1, {n, 0, 25}]
LinearRecurrence[{6,-5},{2,6},30] (* Harvey P. Dale, Jul 29 2015 *)

a(n)=5^n+1 \\ Charles R Greathouse IV, Sep 24 2015

[lucas_number2(n,6,5) for n in range(25)] # Zerinvary Lajos, Jul 08 2008

[sigma(5,n) for n in range(25)] # Zerinvary Lajos, Jun 04 2009

[5^n+1 for n in range(30)] # Bruno Berselli, Jan 11 2017

a(n) = 5*a(n-1) - 4 with a(0) = 2.
a(n) = 6*a(n-1) - 5*a(n-2) for n > 1.
From Mohammad K. Azarian, Jan 02 2009: (Start)
G.f.: 1/(1-x) + 1/(1-5*x) = (2-6*x)/((1-x)*(1-5*x)).
E.g.f.: exp(x) + exp(5*x). (End)
a(n) = A279396(n+5,5). - Wolfdieter Lang, Jan 10 2017
From Elmo R. Oliveira, Dec 06 2023: (Start)
a(n) = A000351(n) + 1.
a(n) = 2*A034478(n). (End)

A126473 Number of strings over a 5 symbol alphabet with adjacent symbols differing by three or less.

Original entry on oeis.org

1, 5, 23, 107, 497, 2309, 10727, 49835, 231521, 1075589, 4996919, 23214443, 107848529, 501037445, 2327695367, 10813893803, 50238661313, 233396326661, 1084301290583, 5037394142315, 23402480441009, 108722104190981, 505095858086951, 2346549744920747

Offset: 0

R. H. Hardin, Dec 27 2006

[Empirical] a(base,n) = a(base-1,n) + 7^(n-1) for base >= 3n-2; a(base,n) = a(base-1,n) + 7^(n-1)-2 when base = 3n-3.
From Johannes W. Meijer, Aug 01 2010: (Start)
The a(n) represent the number of n-move routes of a fairy chess piece starting in a given side square (m = 2, 4, 6 or 8) on a 3 X 3 chessboard. This fairy chess piece behaves like a king on the eight side and corner squares but on the central square the king goes crazy and turns into a red king, see A179596.
For the side squares the 512 red kings lead to 47 different red king sequences, see the cross-references for some examples.
The sequence above corresponds to four A[5] vectors with the decimal [binary] values 367 [1,0,1,1,0,1,1,1,1], 463 [1,1,1,0,0,1,1,1,1], 487 [1,1,1,1,0,0,1,1,1] and 493 [1,1,1,1,0,1,1,0,1]. These vectors lead for the corner squares to A179596 and for the central square to A179597.
This sequence belongs to a family of sequences with g.f. (1+x)/(1-4*x-k*x^2). Red king sequences that are members of this family are A003947 (k=0), A015448 (k=1), A123347 (k=2), A126473 (k=3; this sequence) and A086347 (k=4). Other members of this family are A000351 (k=5), A001834 (k=-1), A111567 (k=-2), A048473 (k=-3) and A053220 (k=-4)
Inverse binomial transform of A154244. (End)
Equals the INVERT transform of A055099: (1, 4, 14, 50, 178, ...). - Gary W. Adamson, Aug 14 2010
Number of one-sided n-step walks taking steps from {E, W, N, NE, NW}. - Shanzhen Gao, May 10 2011
For n>=1, a(n) equals the numbers of words of length n-1 on alphabet {0,1,2,3,4} containing no subwords 00 and 11. - Milan Janjic, Jan 31 2015

Shanzhen Gao and Keh-Hsun Chen, Tackling Sequences From Prudent Self-Avoiding Walks, FCS'14, The 2014 International Conference on Foundations of Computer Science.
S. Gao and H. Niederhausen, Sequences Arising From Prudent Self-Avoiding Walks, 2010.
Index entries for linear recurrences with constant coefficients, signature (4,3).

Cf. 5 symbol differing by two or less A126392, one or less A057960.
Cf. Red king sequences side squares [numerical value A[5]]: A086347 [495], A179598 [239], A126473 [367], A123347 [335], A179602 [95], A154964 [31], A015448 [327], A152187 [27], A003947 [325], A108981 [11], A007483 [2]. - Johannes W. Meijer, Aug 01 2010
Cf. A055099.

with(LinearAlgebra): nmax:=19; m:=2; A[5]:= [1,0,1,1,0,1,1,1,1]: A:=Matrix([[0,1,0,1,1,0,0,0,0],[1,0,1,1,1,1,0,0,0],[0,1,0,0,1,1,0,0,0],[1,1,0,0,1,0,1,1,0],A[5],[0,1,1,0,1,0,0,1,1],[0,0,0,1,1,0,0,1,0],[0,0,0,1,1,1,1,0,1],[0,0,0,0,1,1,0,1,0]]): for n from 0 to nmax do B(n):=A^n: a(n):= add(B(n)[m,k],k=1..9): od: seq(a(n), n=0..nmax); # Johannes W. Meijer, Aug 01 2010
# second Maple program:
a:= n-> (M-> M[1,2]+M[2,2])(<<0|1>, <3|4>>^n):
seq(a(n), n=0..24);  # Alois P. Heinz, Jun 28 2021

LinearRecurrence[{4, 3}, {1, 5}, 24] (* Jean-François Alcover, Dec 10 2024 *)

a(n)=([0,1; 3,4]^n*[1;5])[1,1] \\ Charles R Greathouse IV, May 10 2016

From Johannes W. Meijer, Aug 01 2010: (Start)
G.f.: (1+x)/(1-4*x-3*x^2).
a(n) = 4*a(n-1) + 3*a(n-2) with a(0) = 1 and a(1) = 5.
a(n) = ((1+3/sqrt(7))/2)*(A)^(-n) + ((1-3/sqrt(7))/2)*(B)^(-n) with A = (-2 + sqrt(7))/3 and B = (-2-sqrt(7))/3.
Lim_{k->oo} a(n+k)/a(k) = (-1)^(n+1)*A000244(n)/(A015530(n)*sqrt(7)-A108851(n))
(End)
a(n) = A015330(n)+A015330(n+1). - R. J. Mathar, May 09 2023

Edited by Johannes W. Meijer, Aug 10 2010

A015521 a(n) = 3a(n-1) + 4a(n-2), a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 3, 13, 51, 205, 819, 3277, 13107, 52429, 209715, 838861, 3355443, 13421773, 53687091, 214748365, 858993459, 3435973837, 13743895347, 54975581389, 219902325555, 879609302221, 3518437208883, 14073748835533

Offset: 0

Olivier Gérard

Inverse binomial transform of powers of 5 (A000351) preceded by 0. - Paul Barry, Apr 02 2003
Number of walks of length n between any two distinct vertices of the complete graph K_5. Example: a(2)=3 because the walks of length 2 between the vertices A and B of the complete graph ABCDE are: ACB, ADB, AEB. - Emeric Deutsch, Apr 01 2004
The terms of the sequence are the number of segments (sides) per iteration of the space-filling Peano-Hilbert curve. - Giorgio Balzarotti, Mar 16 2006
General form: k=4^n-k. Also: A001045, A078008, A097073, A115341, A015518, A054878. - Vladimir Joseph Stephan Orlovsky, Dec 11 2008
A further inverse binomial transform generates A015441. - Paul Curtz, Nov 01 2009
For n >= 2, a(n) equals the permanent of the (n-1) X (n-1) tridiagonal matrix with 3's along the central diagonal, and 2's along the subdiagonal and the superdiagonal. - John M. Campbell, Jul 19 2011
Pisano period lengths: 1, 1, 2, 2, 10, 2, 6, 2, 6, 10, 10, 2, 6, 6, 10, 2, 4, 6, 18, 10, ... - R. J. Mathar, Aug 10 2012
Sum_{i=0..m} (-1)^(m+i)*4^i, for m >= 0, gives the terms after 0. - Bruno Berselli, Aug 28 2013
The ratio a(n+1)/a(n) converges to 4 as n approaches infinity. - Felix P. Muga II, Mar 09 2014
This is the Lucas sequence U(P=3,Q=-4), and hence for n>=0, a(n+2)/a(n+1) equals the continued fraction 3 + 4/(3 + 4/(3 + 4/(3 + ... + 4/3))) with n 4's. - Greg Dresden, Oct 07 2019
For n > 0, gcd(a(n), a(n+1)) = 1. - Kengbo Lu, Jul 27 2020

			G.f. = x + 3*x^2 + 13*x^3 + 51*x^4 + 205*x^5 + 819*x^6 + 3277*x^7 + 13107*x^8 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
A. Abdurrahman, CM Method and Expansion of Numbers, arXiv:1909.10889 [math.NT], 2019.
Jean-Paul Allouche, Jeffrey Shallit, Zhixiong Wen, Wen Wu, and Jiemeng Zhang, Sum-free sets generated by the period-k-folding sequences and some Sturmian sequences, arXiv:1911.01687 [math.CO], 2019.
J. Borowska and L. Lacinska, Recurrence form of determinant of a heptadiagonal symmetric Toeplitz matrix, J. Appl. Math. Comp. Mech. 13 (2014) 19-16, remark 2 for permanent of tridiagonal Toeplitz matrices a=3, b=2.
Ji Young Choi, A Generalization of Collatz Functions and Jacobsthal Numbers, J. Int. Seq., Vol. 21 (2018), Article 18.5.4.
E. M. García-Caballero, S. G. Moreno, and M. P. Prophet, A complete view of Viète-like infinite products with Fibonacci and Lucas numbers, Applied Mathematics and Computation 247 (2014) 703-711.
Dale Gerdemann, Fractal generated from (3,4) recursion A015521, YouTube Video, Dec 4, 2014.
F. P. Muga II, Extending the Golden Ratio and the Binet-de Moivre Formula, March 2014; Preprint on ResearchGate.
Index entries for linear recurrences with constant coefficients, signature (3,4).

Cf. A001045, A015518, A054878, A078008, A097073, A109200, A115341, A201455, A213127, A247281.

[Floor(4^n/5-(-1)^n/5): n in [0..30]]; // Vincenzo Librandi, Jun 24 2011

seq(round(4^n/5),n=0..25) # Mircea Merca, Dec 28 2010

k=0;lst={k};Do[k=4^n-k;AppendTo[lst, k], {n, 0, 5!}];lst (* Vladimir Joseph Stephan Orlovsky, Dec 11 2008 *)
LinearRecurrence[{3,4}, {0,1}, 30] (* Harvey P. Dale, Jun 26 2012 *)
CoefficientList[Series[x/((1 - 4 x) (1 + x)), {x, 0, 50}], x] (* Vincenzo Librandi, Mar 26 2014 *)

a(n) = 4^n/5-(-1)^n/5; \\ Altug Alkan, Jan 08 2016

first(n) = Vec(x/(1 - 3*x - 4*x^2) + O(x^n), -n) \\ Iain Fox, Dec 30 2017

def A015521(n): return ((1<<(n<<1))|1)//5 # Chai Wah Wu, Jun 28 2023

[lucas_number1(n,3,-4) for n in range(0, 24)] # Zerinvary Lajos, Apr 22 2009

From Paul Barry, Apr 02 2003: (Start)
a(n) = (4^n - (-1)^n)/5.
E.g.f.: (exp(4*x) - exp(-x))/5. (End)
a(n) = Sum_{k=1..n} binomial(n, k)*(-1)^(n+k)*5^(k-1). - Paul Barry, May 13 2003
a(2*n) = 4*a(2*n-1) - 1, a(2*n+1) = 4*a(2*n) + 1. In general this is true for all sequences of the type a(n) + a(n+1) = q^(n): i.e., a(2*n) = q*a(2n-1) - 1 and a(2*n+1) = q*a(2*n) + 1. - Amarnath Murthy, Jul 15 2003
From Emeric Deutsch, Apr 01 2004: (Start)
a(n) = 4^(n-1) - a(n-1).
G.f.: x/(1-3*x - 4*x^2). (End)
a(n+1) = Sum_{k=0..floor(n/2)} binomial(n-k, k)*3^(n-2k)*4^k. - Paul Barry, Jul 29 2004
a(n) = 4*a(n-1) - (-1)^n, n > 0, a(0)=0. - Paul Barry, Aug 25 2004
a(n) = Sum_{k=0..n} A155161(n,k)*2^(n-k), n >= 1. - Philippe Deléham, Jan 27 2009
a(n) = round(4^n/5). - Mircea Merca, Dec 28 2010
The logarithmic generating function 1/5*log((1+x)/(1-4*x)) = x + 3*x^2/2 + 13*x^3/3 + 51*x^4/4 + ... has compositional inverse 5/(4+exp(-5*x)) - 1, the e.g.f. for a signed version of A213127. - Peter Bala, Jun 24 2012
a(n) = (-1)^(n-1)*Sum_{k=0..n-1} A135278(n-1,k)*(-5)^k = (4^n - (-1)^n)/5 = (-1)^(n-1)*Sum_{k=0..n-1} (-4)^k. Equals (-1)^(n-1)*Phi(n,-4), where Phi is the cyclotomic polynomial when n is an odd prime. (For n > 0.) - Tom Copeland, Apr 14 2014
a(n+1) = 2^(2*n) - a(n), a(0) = 0. - Ben Paul Thurston, Dec 25 2015
a(n) = A247281(n)/5. - Altug Alkan, Jan 08 2016
From Kengbo Lu, Jul 27 2020: (Start)
a(n) = 3*Sum_{k=0..n-1} a(k) + 1 if n odd; a(n) = 3*Sum_{k=0..n-1} a(k) if n even.
a(n) = A030195(n) + Sum_{k=0..n-2} a(k)*A030195(n-k-1).
a(n) = A085449(n) + Sum_{k=0..n-1} a(k)*A085449(n-k).
a(n) = F(n) + 2*Sum_{k=0..n-1} a(k)*F(n-k) + 3*Sum_{k=0..n-2} a(k)*F(n-k-1), where F(n) denotes the Fibonacci numbers.
a(n) = F(n) + Sum_{k=0..n-1} a(k)*(L(n-k) + F(n-k+1)), where F(n) denotes the Fibonacci numbers and L(n) denotes the Lucas numbers.
a(n) = 3^(n-1) + 4*Sum_{k=0..n-2} 3^(n-k-2)*a(k).
a(m+n) = a(m)*a(n+1) + 4*a(m-1)*a(n).
a(2*n) = Sum_{i>=0, j>=0} binomial(n-j-1,i)*binomial(n-i-1,j)*3^(2n-2i-2j-1)*4^(i+j). (End)

A087207 A binary representation of the primes that divide a number, shown in decimal.

Original entry on oeis.org

0, 1, 2, 1, 4, 3, 8, 1, 2, 5, 16, 3, 32, 9, 6, 1, 64, 3, 128, 5, 10, 17, 256, 3, 4, 33, 2, 9, 512, 7, 1024, 1, 18, 65, 12, 3, 2048, 129, 34, 5, 4096, 11, 8192, 17, 6, 257, 16384, 3, 8, 5, 66, 33, 32768, 3, 20, 9, 130, 513, 65536, 7, 131072, 1025, 10, 1, 36, 19, 262144, 65, 258

Offset: 1

Mitch Cervinka (puritan(AT)planetkc.com), Oct 26 2003

The binary representation of a(n) shows which prime numbers divide n, but not the multiplicities. a(2)=1, a(3)=10, a(4)=1, a(5)=100, a(6)=11, a(10)=101, a(30)=111, etc.
For n > 1, a(n) gives the (one-based) index of the column where n is located in array A285321. A008479 gives the other index. - Antti Karttunen, Apr 17 2017
From Antti Karttunen, Jun 18 & 20 2017: (Start)
A268335 gives all n such that a(n) = A248663(n); the squarefree numbers (A005117) are all the n such that a(n) = A285330(n) = A048675(n).
For all n > 1 for which the value of A285331(n) is well-defined, we have A285331(a(n)) <= floor(A285331(n)/2), because then n is included in the binary tree A285332 and a(n) is one of its ancestors (in that tree), and thus must be at least one step nearer to its root than n itself.
Conjecture: Starting at any n and iterating the map n -> a(n), we will always reach 0 (see A288569). This conjecture is equivalent to the conjecture that at any n that is neither a prime nor a power of two, we will eventually hit a prime number (which then becomes a power of two in the next iteration). If this conjecture is false then sequence A285332 cannot be a permutation of natural numbers. On the other hand, if the conjecture is true, then A285332 must be a permutation of natural numbers, because all primes and powers of 2 occur in definite positions in that tree. This conjecture also implies the conjectures made in A019565 and A285320 that essentially claim that there are neither finite nor infinite cycles in A019565.
If there are any 2-cycles in this sequence, then both terms of the cycle should be present in A286611 and the larger one should be present in A286612.
(End)
Binary rank of the distinct prime indices of n, where the binary rank of an integer partition y is given by Sum_i 2^(y_i-1). For all prime indices (with multiplicity) we have A048675. - Gus Wiseman, May 25 2024

			a(38) = 129 because 38 = 2*19 = prime(1)*prime(8) and 129 = 2^0 + 2^7 (in binary 10000001).
a(140) = 13, binary 1101 because 140 is divisible by the first, third and fourth primes and 2^(1-1) + 2^(3-1) + 2^(4-1) = 13.

N. J. A. Sloane, Table of n, a(n) for n = 1..10000 [First 1000 terms from _T. D. Noe_]
Index entries for sequences related to binary expansion of n
Index entries for sequences computed from indices in prime factorization

For partial sums see A288566.
Cf. A000040, A000120, A001221, A005117, A008479, A019565, A055396, A285320, A285321, A285329, A285330, A285332.
Sequences with related definitions: A007947, A008472, A027748, A048675, A248663, A276379 (same sequence shown in base 2), A288569, A289271, A297404.
Cf. A286608 (numbers n for which a(n) < n), A286609 (n for which a(n) > n), and also A286611, A286612.
A003986, A003961, A059896 are used to express relationship between terms of this sequence.
Related to A267116 via A225546.
Positions of particular values are: A000079\{1} (1), A000244\{1} (2), A033845 (3), A000351\{1} (4), A033846 (5), A033849 (6), A143207 (7), A000420\{1} (8), A033847 (9), A033850 (10), A033851 (12), A147576 (14), A147571 (15), A001020\{1} (16), A033848 (17).
A048675 gives binary rank of prime indices.
A061395 gives greatest prime index, least A055396.
A112798 lists prime indices, length A001222, reverse A296150, sum A056239.
Binary indices (listed A048793):
- length A000120, complement A023416
- min A001511, opposite A000012
- sum A029931, product A096111
- max A029837 or A070939, opposite A070940
- complement A368494, sum A359400
- opposite complement A371571, sum A359359
- opposite A371572, sum A230877
Cf. A000720, A005940, A018819, A023506, A071814, A225620, A277319, A277905, A304818, A372689, A372890.

a087207 = sum . map ((2 ^) . (subtract 1) . a049084) . a027748_row
-- Reinhard Zumkeller, Jul 16 2013

a[n_] := Total[ 2^(PrimePi /@ FactorInteger[n][[All, 1]] - 1)]; a[1] = 0; Table[a[n], {n, 1, 69}] (* Jean-François Alcover, Dec 12 2011 *)

a(n) = {if (n==1, 0, my(f=factor(n), v = []); forprime(p=2, vecmax(f[,1]), v = concat(v, vecsearch(f[,1], p)!=0);); fromdigits(Vecrev(v), 2));} \\ Michel Marcus, Jun 05 2017

A087207(n)=vecsum(apply(p->1<M. F. Hasler, Jun 23 2017

from sympy import factorint, primepi
def a(n):
    return sum(2**primepi(i - 1) for i in factorint(n))
print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Jun 06 2017

(definec (A087207 n) (if (= 1 n) 0 (+ (A000079 (+ -1 (A055396 n))) (A087207 (A028234 n))))) ;; This uses memoization-macro definec
(define (A087207 n) (A048675 (A007947 n))) ;; Needs code from A007947 and A048675. - Antti Karttunen, Jun 19 2017

Additive with a(p^e) = 2^(i-1) where p is the i-th prime. - Vladeta Jovovic, Oct 29 2003
a(n) gives the m such that A019565(m) = A007947(n). - Naohiro Nomoto, Oct 30 2003
A000120(a(n)) = A001221(n); a(n) = Sum(2^(A049084(p)-1): p prime-factor of n). - Reinhard Zumkeller, Nov 30 2003
G.f.: Sum_{k>=1} 2^(k-1)*x^prime(k)/(1-x^prime(k)). - Franklin T. Adams-Watters, Sep 01 2009
From Antti Karttunen, Apr 17 2017, Jun 19 2017 & Dec 06 2018: (Start)
a(n) = A048675(A007947(n)).
a(1) = 0; for n > 1, a(n) = 2^(A055396(n)-1) + a(A028234(n)).
A000035(a(n)) = 1 - A000035(n). [a(n) and n are of opposite parity.]
A248663(n) <= a(n) <= A048675(n). [XOR-, OR- and +-variants.]
a(A293214(n)) = A218403(n).
a(A293442(n)) = A267116(n).
A069010(a(n)) = A287170(n).
A007088(a(n)) = A276379(n).
A038374(a(n)) = A300820(n) for n >= 1.
(End)
From Peter Munn, Jan 08 2020: (Start)
a(A059896(n,k)) = a(n) OR a(k) = A003986(a(n), a(k)).
a(A003961(n)) = 2*a(n).
a(n^2) = a(n).
a(n) = A267116(A225546(n)).
a(A225546(n)) = A267116(n).
(End)

More terms from Don Reble, Ray Chandler and Naohiro Nomoto, Oct 28 2003

Name clarified by Antti Karttunen, Jun 18 2017

cp's OEIS Frontend

A003464 a(n) = (6^n - 1)/5.

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A027868 Number of trailing zeros in n!; highest power of 5 dividing n!.

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A112765 Exponent of highest power of 5 dividing n. Or, 5-adic valuation of n.

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A001692 Number of irreducible polynomials of degree n over GF(5); dimensions of free Lie algebras.

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A329697 a(n) is the number of iterations needed to reach a power of 2 starting at n and using the map k -> k-(k/p), where p is the largest prime factor of k.

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A015521 a(n) = 3a(n-1) + 4a(n-2), a(0) = 0, a(1) = 1.