A047201 -id:A047201 - cp's OEIS frontend

A337833 Minimum m coprime to 5 such that the convergence speed of m^^m := m^(m^^(m-1)) is equal to n >= 0, where A317905(n) represents the convergence speed of m^^m (and m = A047201(n), the n-th non-multiple of 5).

1, 2, 7, 57, 182, 3124, 1068, 32318, 390624, 280182, 3626068, 23157318, 120813568, 1220703124, 1097376068, 11109655182, 49925501068, 762939453124, 355101282318, 19073486328124, 15613890344818, 365855836217682, 2384185791015624

Offset: 0

Marco Ripà, Sep 24 2020

Let "s" denote the last digit of m, and V(m(s)) its convergence speed. For any n, the smallest bases that are not congruent to 5 modulo 10 (as in A337392) cannot be such that s = 6, since V(m(6)) = V(m(4)) + 2.

			For n = 19, a(19) = 19073486328124 is the smallest base (radix-10) of the tetration m^^m which is characterized by a congruence speed of 19.

Marco Ripà, La strana coda della serie n^n^...^n, Trento, UNI Service, Nov 2011. ISBN 978-88-6178-789-6

Marco Ripà, On the constant congruence speed of tetration, Notes on Number Theory and Discrete Mathematics, 2020, 26(3), 245-260.
Marco Ripà, The congruence speed formula, Notes on Number Theory and Discrete Mathematics, 2021, 27(4), 43-61.

Cf. A317824, A317903, A317905, A321130, A337392.

A124476 Smallest multiple of A047201(n) (i.e., numbers not divisible by 5) with only digits 6 and 7.

Original entry on oeis.org

67, 76, 6777, 76, 7776, 6776, 776, 6777, 6677, 7776, 676, 6776, 7776, 6766, 7776, 76, 67767, 6776, 667, 7776, 676, 6777, 6776, 667, 67766, 7776, 66777777, 6766, 7776, 666777, 76, 77766, 7667, 76776, 676777, 6776, 67666, 667776, 7776, 67767, 67677

Offset: 1

Lekraj Beedassy, Dec 17 2006

Nick Hobson, Solution to Puzzle 80:"Sixes and sevens" (generalization of)

f[n_] := Block[{k = n},While[Union @ IntegerDigits[k] != {6, 7}, k += n];k];f /@ Select[Range[52], Mod[ #, 5] > 0 &] (* Ray Chandler, Dec 18 2006 *)

Corrected and extended by Ray Chandler, Dec 18 2006

A141845 a(n) = 5*a(n-1) + A047201(n), a(1) = 1. A047201 = numbers not divisible by 5: (1, 2, 3, 4, 6, 7, 8, 9, 11, ...).

Original entry on oeis.org

1, 7, 38, 194, 976, 4887, 24443, 122224, 611131, 3055667, 15278348, 76391754, 381958786, 1909793947, 9548969753, 47744848784, 238724243941, 1193621219727, 5968106098658, 29840530493314, 149202652466596, 746013262333007

Offset: 1

Gary W. Adamson, Jul 11 2008

nonn

			a(4) = 194 = 5*a(3) + A047201(4) = 5*38 + 4.

Index entries for linear recurrences with constant coefficients, signature (6,-5,0,1,-6,5).

Cf. A047201.

LinearRecurrence[{6,-5,0,1,-6,5},{1,7,38,194,976,4887},30] (* Harvey P. Dale, Sep 10 2024 *)

G.f.: (1+x+x^2+x^3+x^4)/((1-x)^2(1+x)(1+x^2)(1-5x)). - R. J. Mathar, Sep 06 2008

Extended and corrected by R. J. Mathar, Sep 06 2008

A000032 Lucas numbers beginning at 2: L(n) = L(n-1) + L(n-2), L(0) = 2, L(1) = 1.

Original entry on oeis.org

2, 1, 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, 843, 1364, 2207, 3571, 5778, 9349, 15127, 24476, 39603, 64079, 103682, 167761, 271443, 439204, 710647, 1149851, 1860498, 3010349, 4870847, 7881196, 12752043, 20633239, 33385282, 54018521, 87403803

Offset: 0

N. J. A. Sloane, May 24 1994

Cf. A000204 for Lucas numbers beginning with 1.
Also the number of independent vertex sets and vertex covers for the cycle graph C_n for n >= 2. - Eric W. Weisstein, Jan 04 2014
Also the number of matchings in the n-cycle graph C_n for n >= 3. - Eric W. Weisstein, Oct 01 2017
Also the number of maximal independent vertex sets (and maximal vertex covers) for the n-helm graph for n >= 3. - Eric W. Weisstein, May 27 2017
Also the number of maximal independent vertex sets (and maximal vertex covers) for the n-sunlet graph for n >= 3. - Eric W. Weisstein, Aug 07 2017
This is also the Horadam sequence (2, 1, 1, 1). - Ross La Haye, Aug 18 2003
For distinct primes p, q, L(p) is congruent to 1 mod p, L(2p) is congruent to 3 mod p and L(pq) is congruent 1 + q(L(q) - 1) mod p. Also, L(m) divides F(2km) and L((2k + 1)m), k, m >= 0.
a(n) = Sum_{k=0..ceiling((n - 1)/2)} P(3; n - 1 - k, k), n >= 1, with a(0) = 2. These are the sums over the SW-NE diagonals in P(3; n, k), the (3, 1) Pascal triangle A093560. Observation by Paul Barry, Apr 29 2004. Proof via recursion relations and comparison of inputs. Also SW-NE diagonal sums of the (1, 2) Pascal triangle A029635 (with T(0, 0) replaced by 2).
Suppose psi = log(phi) = A002390. We get the representation L(n) = 2*cosh(n*psi) if n is even; L(n) = 2*sinh(n*psi) if n is odd. There is a similar representation for Fibonacci numbers (A000045). Many Lucas formulas now easily follow from appropriate sinh- and cosh-formulas. For example: the identity cosh^2(x) - sinh^2(x) = 1 implies L(n)^2 - 5*F(n)^2 = 4*(-1)^n (setting x = n*psi). - Hieronymus Fischer, Apr 18 2007
From John Blythe Dobson, Oct 02 2007, Oct 11 2007: (Start)
The parity of L(n) follows easily from its definition, which shows that L(n) is even when n is a multiple of 3 and odd otherwise.
The first six multiplication formulas are:
L(2n) = L(n)^2 - 2*(-1)^n;
L(3n) = L(n)^3 - 3*(-1)^n*L(n);
L(4n) = L(n)^4 - 4*(-1)^n*L(n)^2 + 2;
L(5n) = L(n)^5 - 5*(-1)^n*L(n)^3 + 5*L(n);
L(6n) = L(n)^6 - 6*(-1)^n*L(n)^4 + 9*L(n)^2 - 2*(-1)^n.
Generally, L(n) | L(mn) if and only if m is odd.
In the expansion of L(mn), where m represents the multiplier and n the index of a known value of L(n), the absolute values of the coefficients are the terms in the m-th row of the triangle A034807. When m = 1 and n = 1, L(n) = 1 and all the terms are positive and so the row sums of A034807 are simply the Lucas numbers. (End)
From John Blythe Dobson, Nov 15 2007: (Start)
The comments submitted by Miklos Kristof on Mar 19 2007 for the Fibonacci numbers (A000045) contain four important identities that have close analogs in the Lucas numbers:
For a >= b and odd  b, L(a + b) + L(a - b) = 5*F(a)*F(b).
For a >= b and even b, L(a + b) + L(a - b) = L(a)*L(b).
For a >= b and odd  b, L(a + b) - L(a - b) = L(a)*L(b).
For a >= b and even b, L(a + b) - L(a - b) = 5*F(a)*F(b).
A particularly interesting instance of the difference identity for even b is L(a + 30) - L(a - 30) = 5*F(a)*832040, since 5*832040 is divisible by 100, proving that the last two digits of Lucas numbers repeat in a cycle of length 60 (see A106291(100)). (End)
From John Blythe Dobson, Nov 15 2007: (Start)
The Lucas numbers satisfy remarkable difference equations, in some cases best expressed using Fibonacci numbers, of which representative examples are the following:
L(n) - L(n - 3)  = 2*L(n - 2);
L(n) - L(n - 4)  = 5*F(n - 2);
L(n) - L(n - 6)  = 4*L(n - 3);
L(n) - L(n - 12) = 40*F(n - 6);
L(n) - L(n - 60) = 4160200*F(n - 30).
These formulas establish, respectively, that the Lucas numbers form a cyclic residue system of length 3 (mod 2), of length 4 (mod 5), of length 6 (mod 4), of length 12 (mod 40) and of length 60 (mod 4160200). The divisibility of the last modulus by 100 accounts for the fact that the last two digits of the Lucas numbers begin to repeat at L(60).
The divisibility properties of the Lucas numbers are very complex and still not fully understood, but several important criteria are established in Zhi-Hong Sun's 2003 survey of congruences for Fibonacci numbers. (End)
Sum_{n>0} a(n)/(n*2^n) = 2*log(2). - Jaume Oliver Lafont, Oct 11 2009
A010888(a(n)) = A030133(n). - Reinhard Zumkeller, Aug 20 2011
The powers of phi, the golden ratio, approach the values of the Lucas numbers, the odd powers from above and the even powers from below. - Geoffrey Caveney, Apr 18 2014
Inverse binomial transform is (-1)^n * a(n). - Michael Somos, Jun 03 2014
Lucas numbers are invariant to the following transformation for all values of the integers j and n, including negative values, thus: L(n) = (L(j+n) + (-1)^n * L(j-n))/L(j). The same transformation applied to all sequences of the form G(n+1) = m * G(n) + G(n-1) yields Lucas numbers for m = 1, except where G(j) = 0, regardless of initial values which may be nonintegers. The corresponding sequences for other values of m are: for m = 2, 2*A001333; for m = 3, A006497; for m = 4, 2*A001077; for m = 5, A087130; for m = 6, 2*A005667; for m = 7, A086902. The invariant ones all have G(0) = 2, G(1) = m. A related family of sequences is discussed at A059100. - Richard R. Forberg, Nov 23 2014
If x=a(n), y=a(n+1), z=a(n+2), then -x^2 - z*x - 3*y*x - y^2 + y*z + z^2 = 5*(-1)^(n+1). - Alexander Samokrutov, Jul 04 2015
A conjecture on the divisibility of infinite subsequences of Lucas numbers by prime(n)^m, m >= 1, is given in A266587, together with the prime "entry points". - Richard R. Forberg, Dec 31 2015
A trapezoid has three lengths of sides in order L(n-1), L(n+1), L(n-1). For increasing n a very close approximation to the maximum area will have the fourth side equal to 2*L(n). For a trapezoid with sides L(n-1), L(n-3), L(n-1), the fourth side will be L(n). - J. M. Bergot, Mar 17 2016
Satisfies Benford's law [Brown-Duncan, 1970; Berger-Hill, 2017]. - N. J. A. Sloane, Feb 08 2017
Lucas numbers L(n) and Fibonacci numbers F(n), being related by the formulas F(n) = (F(n-1) + L(n-1))/2 and L(n) = 2 F(n+1) - F(n), are a typical pair of "autosequences" (see the link to OEIS Wiki). - Jean-François Alcover, Jun 09 2017
For n >= 3, the Lucas number L(n) is the dimension of a commutative Hecke algebra of affine type A_n with independent parameters. See Theorem 1.4, Corollary 1.5, and the table on page 524 in the link "Hecke algebras with independent parameters". - Jia Huang, Jan 20 2019
From Klaus Purath, Apr 19 2019: (Start)
While all prime numbers appear as factors in the Fibonacci numbers, this is not the case with the Lucas numbers. For example, L(n) is never divisible by the following prime numbers < 150: 5, 13, 17, 37, 53, 61, 73, 89, 97, 109, 113, 137, 149 ... See A053028. Conjecture: Three properties can be determined for these prime numbers:
First observation: The prime factors > 3 occur in the Fibonacci numbers with an odd index.
Second observation: These are the prime numbers p congruent to 2, 3 (modulo 5), which occur both in Fibonacci(p+1) and in Fibonacci((p+1)/2) as prime factors, or the prime numbers p congruent to 1, 4 (modulo 5), which occur both in Fibonacci((p-1)/2) and in Fibonacci((p-1)/(2^k)) with k >= 2.
Third observation: The Pisano period lengths of these prime numbers, given in A001175, are always divisible by 4, but not by 8. In contrast, those of the prime factors of Lucas numbers are divisible either by 2, but not by 4, or by 8. (See also comment in A053028 by N. J. A. Sloane, Feb 21 2004). (End)
L(n) is the sum of 4*k consecutive terms of the Fibonacci sequence (A000045) divided by Fibonacci(2*k): (Sum_{i=0..4*k-1, k>=1} F(n+i))/F(2*k) = L(n+2*k+1). Sequences extended to negative indices, following the rule a(n-1) = a(n+1) - a(n). - Klaus Purath, Sep 15 2019
If one forms a sequence (A) of the Fibonacci type with the initial values A(0) = A022095(n) and A(1) = A000285(n), then A(n+1) = L(n+1)^2 always applies. - Klaus Purath, Sep 29 2019
From Kai Wang, Dec 18 2019: (Start)
L((2*m+1)k)/L(k) = Sum_{i=0..m-1} (-1)^(i*(k+1))*L((2*m-2*i)*k) + (-1)^(m*k).
Example: k=5, m=2, L(5)=11, L(10)=123, L(20)=15127, L(25)=167761. L(25)/L(5) = 15251, L(20) + L(10) + 1 = 15127 + 123 + 1 = 15251. (End)
From Peter Bala, Dec 23 2021: (Start)
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) ( mod p^k ) hold for all prime p and positive integers n and k.
For a positive integer k, the sequence (a(n))n>=1 taken modulo k becomes a purely periodic sequence. For example, taken modulo 11, the sequence becomes [1, 3, 4, 7, 0, 7, 7, 3, 10, 2, 1, 3, 4, 7, 0, 7, 7, 3, 10, 2, ...], a periodic sequence with period 10. (End)
For any sequence with recurrence relation b(n) = b(n-1) + b(n-2), it can be shown that the recurrence relation for every k-th term is given by: b(n) = A000032(k) * b(n-k) + (-1)^(k+1) * b(n-2k), extending to negative indices as necessary. - Nick Hobson, Jan 19 2024
For n >= 3, L(n) is the number of (n-1)-digit numbers where all consecutive pairs of digits have a difference of at least 8. - Edwin Hermann, Apr 19 2025

			G.f. = 2 + x + 3*x^2 + 4*x^3 + 7*x^4 + 11*x^5 + 18*x^6 + 29*x^7 + ...

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Index entries for sequences related to Chebyshev polynomials..
Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2).
Index entries for linear recurrences with constant coefficients, signature (1,1).
Index entries for two-way infinite sequences.
Index entries for "core" sequences.
Index entries for sequences related to Benford's law.

Cf. A000204. A000045(n) = (2*L(n + 1) - L(n))/5.
First row of array A103324.
a(n) = A101220(2, 0, n), for n > 0.
a(k) = A090888(1, k) = A109754(2, k) = A118654(2, k - 1), for k > 0.
Cf. A131774, A001622, A002878 (L(2n+1)), A005248 (L(2n)), A006497, A080039, A049684 (summation of Fibonacci(4n+2)), A106291 (Pisano periods), A057854 (complement), A354265 (generalized Lucas numbers).
Cf. sequences with formula Fibonacci(n+k)+Fibonacci(n-k) listed in A280154.
Subsequence of A047201.

a000032 n = a000032_list !! n
a000032_list = 2 : 1 : zipWith (+) a000032_list (tail a000032_list)
-- Reinhard Zumkeller, Aug 20 2011

Magma
```
[Lucas(n): n in [0..120]];
```

with(combinat): A000032 := n->fibonacci(n+1)+fibonacci(n-1);
seq(simplify(2^n*(cos(Pi/5)^n+cos(3*Pi/5)^n)), n=0..36)

a[0] := 2; a[n] := Nest[{Last[#], First[#] + Last[#]} &, {2, 1}, n] // Last
Array[2 Fibonacci[# + 1] - Fibonacci[#] &, 50, 0] (* Joseph Biberstine (jrbibers(AT)indiana.edu), Dec 26 2006 *)
Table[LucasL[n], {n, 0, 36}] (* Zerinvary Lajos, Jul 09 2009 *)
LinearRecurrence[{1, 1}, {2, 1}, 40] (* Harvey P. Dale, Sep 07 2013 *)
LucasL[Range[0, 20]] (* Eric W. Weisstein, Aug 07 2017 *)
CoefficientList[Series[(-2 + x)/(-1 + x + x^2), {x, 0, 20}], x] (* Eric W. Weisstein, Sep 21 2017 *)

{a(n) = if(n<0, (-1)^n * a(-n), if( n<2, 2-n, a(n-1) + a(n-2)))};

{a(n) = if(n<0, (-1)^n * a(-n), polsym(x^2 - x - 1, n)[n+1])};

{a(n) = real((2 + quadgen(5)) * quadgen(5)^n)};

a(n)=fibonacci(n+1)+fibonacci(n-1) \\ Charles R Greathouse IV, Jun 11 2011

polsym(1+x-x^2, 50) \\ Charles R Greathouse IV, Jun 11 2011

def A000032_gen(): # generator of terms
    a, b = 2, 1
    while True:
        yield a
        a, b = b, a+b
it = A000032_gen()
A000032_list = [next(it) for  in range(50)] # _Cole Dykstra, Aug 02 2022

from sympy import lucas
def A000032(n): return lucas(n) # Chai Wah Wu, Sep 23 2023

[(i:=3)+(j:=-1)] + [(j:=i+j)+(i:=j-i) for  in range(100)] # _Jwalin Bhatt, Apr 02 2025

[lucas_number2(n,1,-1) for n in range(37)] # Zerinvary Lajos, Jun 25 2008

G.f.: (2 - x)/(1 - x - x^2).
L(n) = ((1 + sqrt(5))/2)^n + ((1 - sqrt(5))/2)^n = phi^n + (1-phi)^n.
L(n) = L(n - 1) + L(n - 2) = (-1)^n * L( - n).
L(n) = Fibonacci(2*n)/Fibonacci(n) for n > 0. - Jeff Burch, Dec 11 1999
E.g.f.: 2*exp(x/2)*cosh(sqrt(5)*x/2). - Len Smiley, Nov 30 2001
L(n) = F(n) + 2*F(n - 1) = F(n + 1) + F(n - 1). - Henry Bottomley, Apr 12 2000
a(n) = sqrt(F(n)^2 + 4*F(n + 1)*F(n - 1)). - Benoit Cloitre, Jan 06 2003 [Corrected by Gary Detlefs, Jan 21 2011]
a(n) = 2^(1 - n)*Sum_{k=0..floor(n/2)} C(n, 2k)*5^k. a(n) = 2T(n, i/2)( - i)^n with T(n, x) Chebyshev's polynomials of the first kind (see A053120) and i^2 = - 1. - Paul Barry, Nov 15 2003
L(n) = 2*F(n + 1) - F(n). - Paul Barry, Mar 22 2004
a(n) = (phi)^n + ( - phi)^( - n). - Paul Barry, Mar 12 2005
From Miklos Kristof, Mar 19 2007: (Start)
Let F(n) = A000045 = Fibonacci numbers, L(n) = a(n) = Lucas numbers:
L(n + m) + (-1)^m*L(n - m) = L(n)*L(m).
L(n + m) - (-1)^m*L(n - m) = 8*F(n)*F(m).
L(n + m + k) + (-1)^k*L(n + m - k) + (-1)^m*(L(n - m + k) + (-1)^k*L(n - m - k)) = L(n)*L(m)*L(k).
L(n + m + k) - (-1)^k*L(n + m - k) + (-1)^m*(L(n - m + k) - (-1)^k*L(n - m - k)) = 5*F(n)*L(m)*F(k).
L(n + m + k) + (-1)^k*L(n + m - k) - (-1)^m*(L(n - m + k) + (-1)^k*L(n - m - k)) = 5*F(n)*F(m)*L(k).
L(n + m + k) - (-1)^k*L(n + m - k) - (-1)^m*(L(n - m + k) - (-1)^k*L(n - m - k)) = 5*L(n)*F(m)*F(k). (End)
Inverse: floor(log_phi(a(n)) + 1/2) = n, for n>1. Also for n >= 0, floor((1/2)*log_phi(a(n)*a(n+1))) = n. Extension valid for all integers n: floor((1/2)*sign(a(n)*a(n+1))*log_phi|a(n)*a(n+1)|) = n {where sign(x) = sign of x}. - Hieronymus Fischer, May 02 2007
Let f(n) = phi^n + phi^(-n), then L(2n) = f(2n) and L(2n + 1) = f(2n + 1) - 2*Sum_{k>=0} C(k)/f(2n + 1)^(2k + 1) where C(n) are Catalan numbers (A000108). - Gerald McGarvey, Dec 21 2007, modified by Davide Colazingari, Jul 01 2016
Starting (1, 3, 4, 7, 11, ...) = row sums of triangle A131774. - Gary W. Adamson, Jul 14 2007
a(n) = trace of the 2 X 2 matrix [0,1; 1,1]^n. - Gary W. Adamson, Mar 02 2008
From Hieronymus Fischer, Jan 02 2009: (Start)
For odd n: a(n) = floor(1/(fract(phi^n))); for even n>0: a(n) = ceiling(1/(1 - fract(phi^n))). This follows from the basic property of the golden ratio phi, which is phi - phi^(-1) = 1 (see general formula described in A001622).
a(n) = round(1/min(fract(phi^n), 1 - fract(phi^n))), for n>1, where fract(x) = x - floor(x). (End)
E.g.f.: exp(phi*x) + exp(-x/phi) with phi: = (1 + sqrt(5))/2 (golden section). 1/phi = phi - 1. See another form given in the Smiley e.g.f. comment. - Wolfdieter Lang, May 15 2010
L(n)/L(n - 1) -> A001622. - Vincenzo Librandi, Jul 17 2010
a(n) = 2*a(n-2) + a(n-3), n>2. - Gary Detlefs, Sep 09 2010
L(n) = floor(1/fract(Fibonacci(n)*phi)), for n odd. - Hieronymus Fischer, Oct 20 2010
L(n) = ceiling(1/(1 - fract(Fibonacci(n)*phi))), for n even. - Hieronymus Fischer, Oct 20 2010
L(n) = 2^n * (cos(Pi/5)^n + cos(3*Pi/5)^n). - Gary Detlefs, Nov 29 2010
L(n) = (Fibonacci(2*n - 1)*Fibonacci(2*n + 1) - 1)/(Fibonacci(n)*Fibonacci(2*n)), n != 0. - Gary Detlefs, Dec 13 2010
L(n) = sqrt(A001254(n)) = sqrt(5*Fibonacci(n)^2 - 4*(-1)^(n+1)). - Gary Detlefs, Dec 26 2010
L(n) = floor(phi^n) + ((-1)^n + 1)/2 = A014217(n) +((-1)^n+1)/2, where phi = A001622. - Gary Detlefs, Jan 20 2011
L(n) = Fibonacci(n + 6) mod Fibonacci(n + 2), n>2. - Gary Detlefs, May 19 2011
For n >= 2, a(n) = round(phi^n) where phi is the golden ratio. - Arkadiusz Wesolowski, Jul 20 2012
a(p*k) == a(k) (mod p) for primes p. a(2^s*n) == a(n)^(2^s) (mod 2) for s = 0,1,2.. a(2^k) ==  - 1 (mod 2^k). a(p^2*k) == a(k) (mod p) for primes p and s = 0,1,2,3.. [Hoggatt and Bicknell]. - R. J. Mathar, Jul 24 2012
From Gary Detlefs, Dec 21 2012: (Start)
L(k*n) = (F(k)*phi + F(k - 1))^n + (F(k + 1) - F(k)*phi)^n.
L(k*n) = (F(n)*phi + F(n - 1))^k + (F(n + 1) - F(n)*phi)^k.
where phi = (1 + sqrt(5))/2, F(n) = A000045(n).
(End)
L(n) = n * Sum_{k=0..floor(n/2)} binomial(n - k,k)/(n - k), n>0 [H. W. Gould]. - Gary Detlefs, Jan 20 2013
G.f.: G(0), where G(k) = 1 + 1/(1 - (x*(5*k-1))/((x*(5*k+4)) - 2/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 15 2013
L(n) = F(n) + F(n-1) + F(n-2) + F(n-3). - Bob Selcoe, Jun 17 2013
L(n) = round(sqrt(L(2n-1) + L(2n-2))). - Richard R. Forberg, Jun 24 2014
L(n) = (F(n+1)^2 - F(n-1)^2)/F(n) for n>0. - Richard R. Forberg, Nov 17 2014
L(n+2) = 1 + A001610(n+1) = 1 + Sum_{k=0..n} L(k). - Tom Edgar, Apr 15 2015
L(i+j+1) = L(i)*F(j) + L(i+1)*F(j+1) with F(n)=A000045(n). - J. M. Bergot, Feb 12 2016
a(n) = (L(n+1)^2 + 5*(-1)^n)/L(n+2). - J. M. Bergot, Apr 06 2016
Dirichlet g.f.: PolyLog(s,-1/phi) + PolyLog(s,phi), where phi is the golden ratio. - Ilya Gutkovskiy, Jul 01 2016
L(n) = F(n+2) - F(n-2). - Yuchun Ji, Feb 14 2016
L(n+1) = A087131(n+1)/2^(n+1) = 2^(-n)*Sum_{k=0..n} binomial(n,k)*5^floor((k+1)/2). - Tony Foster III, Oct 14 2017
L(2*n) = (F(k+2*n) + F(k-2*n))/F(k); n >= 1, k >= 2*n. - David James Sycamore, May 04 2018
From Greg Dresden and Shaoxiong Yuan, Jul 16 2019: (Start)
L(3n + 4)/L(3n + 1) has continued fraction: n 4's followed by a single 7.
L(3n + 3)/L(3n) has continued fraction: n 4's followed by a single 2.
L(3n + 2)/L(3n - 1) has continued fraction: n 4's followed by a single -3. (End)
From Klaus Purath, Sep 15 2019: (Start)
All involved sequences extended to negative indices, following the rule a(n-1) = a(n+1) - a(n).
L(n) = (2*L(n+2) - L(n-3))/5.
L(n) = (2*L(n-2) + L(n+3))/5.
L(n) = F(n-3) + 2*F(n).
L(n) = 2*F(n+2) - 3*F(n).
L(n) = (3*F(n-1) + F(n+2))/2.
L(n) = 3*F(n-3) + 4*F(n-2).
L(n) = 4*F(n+1) - F(n+3).
L(n) = (F(n-k) + F(n+k))/F(k) with odd k>0.
L(n) = (F(n+k) - F(n-k))/F(k) with even k>0.
L(n) = A001060(n-1) - F(n+1).
L(n) = (A022121(n-1) - F(n+1))/2.
L(n) = (A022131(n-1) - F(n+1))/3.
L(n) = (A022139(n-1) - F(n+1))/4.
L(n) = (A166025(n-1) - F(n+1))/5.
The following two formulas apply for all sequences of the Fibonacci type.
(a(n-2*k) + a(n+2*k))/a(n) = L(2*k).
(a(n+2*k+1) - a(n-2*k-1))/a(n) = L(2*k+1). (End)
L(n) = F(n-k)*L(k+1) + F(n-k-1)*L(k), for all k >= 0, where F(n) = A000045(n). - Michael Tulskikh, Dec 06 2019
F(n+2*m) = L(m)*F(n+m) + (-1)^(m-1)*F(n) for all n >= 0 and m >= 0. - Alexander Burstein, Mar 31 2022
a(n) = i^(n-1)*cos(n*c)/cos(c) = i^(n-1)*cos(c*n)*sec(c), where c = Pi/2 + i*arccsch(2). - Peter Luschny, May 23 2022
From Yike Li and Greg Dresden, Aug 25 2022: (Start)
L(2*n) = 5*binomial(2*n-1,n) - 2^(2*n-1) + 5*Sum_{j=1..n/5} binomial(2*n,n+5*j) for n>0.
L(2*n+1) = 2^(2n) - 5*Sum_{j=0..n/5} binomial(2*n+1,n+5*j+3). (End)
From Andrea Pinos, Jul 04 2023: (Start)
L(n) ~ Gamma(1/phi^n) + gamma.
L(n) = Re(phi^n + e^(i*Pi*n)/phi^n). (End)
L(n) = ((Sum_{i=0..n-1} L(i)^2) - 2)/L(n-1). - Jules Beauchamp, May 03 2025
From Peter Bala, Jul 09 2025: (Start)
The following series telescope:
For  k >= 1, Sum_{n >= 1} (-1)^((k+1)*(n+1)) * a(2*n*k)/(a((2*n-1)*k)*a((2*n+1)*k)) = 1/a(k)^2.
For positive even k, Sum_{n >= 1} 1/(a(k*n) - (a(k) + 2)/a(k*n)) = 1/(a(k) - 2) and
Sum_{n >= 1} (-1)^(n+1)/(a(k*n) + (a(k) - 2)/a(k*n)) = 1/(a(k) + 2).
For positive odd k, Sum_{n >= 1} 1/(a(k*n) - (-1)^n*(a(2*k) + 2)/a(k*n)) = (a(k) + 2)/(2*(a(2*k) - 2)) and
Sum_{n >= 1} (-1)^(n+1)/(a(k*n) - (-1)^n*(a(2*k) + 2)/a(k*n)) = (a(k) - 2)/(2*(a(2*k) - 2)). (End)

A008587 Multiples of 5: a(n) = 5 * n.

Original entry on oeis.org

0, 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100, 105, 110, 115, 120, 125, 130, 135, 140, 145, 150, 155, 160, 165, 170, 175, 180, 185, 190, 195, 200, 205, 210, 215, 220, 225, 230, 235, 240, 245, 250, 255, 260, 265, 270, 275

Offset: 0

N. J. A. Sloane

1/31 = 0.0322580645... = (1/2)^5 + (1/2)^10 + (1/2)^15 + ... - Gary W. Adamson, Mar 14 2009
Complement of A047201; A079998(a(n))=1; A011558(a(n))=0. - Reinhard Zumkeller, Nov 30 2009
The y-intercept of a line perpendicular to y=mx,where m is the slope a/b and in this case a=2 and b=1, is a^2 + b^2 or 5, the first value of the list given. The remaining value are multiples of the first number of the list. - Larry J Zimmermann, Aug 21 2010

Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, page 85.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 317
Luis Manuel Rivera, Integer sequences and k-commuting permutations, arXiv preprint arXiv:1406.3081 [math.CO], 2014-2015.
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. index to numbers of the form n*(d*n+10-d)/2 in A140090.

Cf. A008706, A011558, A020714, A030308, A047201, A079998.

[5*n: n in [0..60]]; // Vincenzo Librandi, Jun 10 2013

Range[0, 500, 5] (* Vladimir Joseph Stephan Orlovsky, May 26 2011 *)
NestList[#+5&,0,60] (* Harvey P. Dale, Dec 18 2023 *)

makelist(5*n,n,0,20); /* Martin Ettl, Dec 17 2012 */

a(n)=5*n \\ Charles R Greathouse IV, Mar 22 2016

From R. J. Mathar, May 26 2008: (Start)
O.g.f.: 5x/(1-x)^2.
a(n) = A008706(n), n > 0. (End)
a(n) = Sum_{k>=0} A030308(n,k)*A020714(k). - Philippe Deléham, Oct 17 2011
E.g.f.: 5*x*exp(x). - Stefano Spezia, Aug 19 2024

A035959 Number of partitions of n in which no parts are multiples of 5.

Original entry on oeis.org

1, 1, 2, 3, 5, 6, 10, 13, 19, 25, 34, 44, 60, 76, 100, 127, 164, 205, 262, 325, 409, 505, 628, 769, 950, 1156, 1414, 1713, 2081, 2505, 3026, 3625, 4352, 5192, 6200, 7364, 8756, 10357, 12258, 14450, 17034, 20006, 23500, 27510, 32200, 37582, 43846, 51022

Offset: 0

Olivier Gérard

Also number of partitions with at most 4 parts of size 1 and differences between parts at distance 6 are greater than 1.
Also number of partitions of n where no part appears more than four times.
Case k=7, i=5 of Gordon Theorem.

			G.f. = 1 + x + 2*x^2 + 3*x^3 + 5*x^4 + 6*x^5 + 10*x^6 + 13*x^7 + 19*x^8 + ...
G.f. = q + q^7 + 2*q^13 + 3*q^19 + 5*q^25 + 6*q^31 + 10*q^37 + 13*q^43 + ...
a(6) counts these partitions: 6, 42, 411, 33, 321, 3111, 2211, 21111, 111111. - _Clark Kimberling_, Mar 09 2014

G. E. Andrews, The Theory of Partitions, Addison-Wesley, 1976, p. 109.

Seiichi Manyama, Table of n, a(n) for n = 0..10000 (terms 0..1000 from T. D. Noe)
Riccardo Aragona, Roberto Civino, and Norberto Gavioli, A modular idealizer chain and unrefinability of partitions with repeated parts, arXiv:2301.06347 [math.RA], 2023.
Vaclav Kotesovec, A method of finding the asymptotics of q-series based on the convolution of generating functions, arXiv:1509.08708 [math.CO], Sep 30 2015, p. 15.
G. N. Watson, Ramanujans Vermutung ueber Zerfaellungsanzahlen, J. Reine Angew. Math. (Crelle), 179 (1938), 97-128. See the expression Y = C/B in the notation of p. 106. [Added by _N. J. A. Sloane_, Nov 13 2009]
Eric Weisstein's World of Mathematics, Partition Function b_k.
Wikipedia, Glaisher's Theorem

Cf. A061198, A061199, A047201.
Cf. A000009 (m=2), A000726 (m=3), A001935 (m=4), A219601 (m=6), A035985 (m=7), A261775 (m=8), A104502 (m=9), A261776 (m=10).
Cf. A096938, A133563, A320607.
Number of r-regular partitions for r = 2 through 12: A000009, A000726, A001935, A035959, A219601, A035985, A261775, A104502, A261776, A328545, A328546.

a035959 = p a047201_list where
   p _      0 = 1
   p ks'@(k:ks) m = if m < k then 0 else p ks' (m - k) + p ks m
-- Reinhard Zumkeller, Dec 17 2011

max = 47; f[x_] := (x^5-1)/(x-1); g[x_] := Product[f[x^k], {k, 1, max}]; CoefficientList[ Series[g[x], {x, 0, max}], x] (* Jean-François Alcover, Nov 29 2011, after Michael Somos *)
t = Flatten[Table[5 n + r, {n, 0, 60}, {r, 1, 4}]]; p[n_] := IntegerPartitions[n, All, t]; Table[p[n], {n, 0, 8}] (* shows partitions *)
a[n_] := Length@p@n; a /@ Range[0, 50] (* Clark Kimberling, Mar 09 2014 *)
nmax = 50; CoefficientList[Series[Product[(1 - x^(5*k))/(1 - x^k), {k, 1, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Aug 31 2015 *)
QP = QPochhammer; s = QP[q^5]/QP[q] + O[q]^50; CoefficientList[s, q] (* Jean-François Alcover, Nov 25 2015, after Michael Somos *)
Table[Count[IntegerPartitions@n, x_ /; ! MemberQ [Mod[x, 5], 0, 2] ], {n, 0, 47}] (* Robert Price, Jul 28 2020 *)
Table[Count[IntegerPartitions[n],?(NoneTrue[Mod[#,5]==0&])],{n,0,50}] (* _Harvey P. Dale, Dec 25 2021 *)

{a(n) = if( n<0, 0, polcoeff( eta(x^5 + x * O(x^n)) / eta(x + x * O(x^n)), n))}; /* Michael Somos, May 28 2006 */

G.f.: Product_{j>=1} (1 + x^j + x^2j + x^3j + x^4j). - Jon Perry, Mar 30 2004
G.f.: Product_{n>0, n==1, 2, 3, 4 mod 5} 1/(1-q^n).
Given g.f. A(x) then B(x) = x * A(x^3)^2 satisfies 0 = f(B(x), B(x^2)) where f(u, v) = u^3 + v^3 - u*v - 5*u^2*v^2. - Michael Somos, May 28 2006
Given g.f. A(x) then B(x) = x * A(x^3)^2 satisfies 0 = f(B(x), B(x^2), B(x^4)) where f(u, v, w) = v + 5*v^2*(u + w) - (u^2 + u*w + w^2). - Michael Somos, May 28 2006
Euler transform of period 5 sequence [ 1, 1, 1, 1, 0, ...]. - Michael Somos, May 28 2006
G.f.: Product_{k > 0} P5(x^k) where P5 is 5th cyclotomic polynomial.
Convolution inverse is A145466. - Michael Somos, Jun 26 2014
a(n) ~ 2*Pi * BesselI(1, 2*sqrt((6*n + 1)/5) * Pi/3) / (5*sqrt(6*n + 1)) ~ exp(2*Pi*sqrt(2*n/15)) / (3^(1/4) * 10^(3/4) * n^(3/4)) * (1 + (Pi/(3*sqrt(15)) - 3*sqrt(15)/(16*Pi)) / sqrt(2*n) + (Pi^2/540 - 225/(1024*Pi^2) - 5/32) / n). - Vaclav Kotesovec, Aug 31 2015, extended Jan 14 2017
a(n) = (1/n)*Sum_{k=1..n} A116073(k)*a(n-k), a(0) = 1. - Seiichi Manyama, Mar 25 2017
G.f.: exp(Sum_{k>=1} x^k*(1 + x^k + x^(2*k) + x^(3*k))/(k*(1 - x^(5*k)))). - Ilya Gutkovskiy, Aug 15 2018

A191702 Dispersion of A008587 (5,10,15,20,25,30,...), by antidiagonals.

Original entry on oeis.org

1, 5, 2, 25, 10, 3, 125, 50, 15, 4, 625, 250, 75, 20, 6, 3125, 1250, 375, 100, 30, 7, 15625, 6250, 1875, 500, 150, 35, 8, 78125, 31250, 9375, 2500, 750, 175, 40, 9, 390625, 156250, 46875, 12500, 3750, 875, 200, 45, 11, 1953125, 781250, 234375, 62500, 18750

Offset: 1

Clark Kimberling, Jun 12 2011

For a background discussion of dispersions and their fractal sequences, see A191426. For dispersions of congruence sequences mod 3 or mod 4, see A191655, A191663, A191667.
...
Each of the sequences (5n, n>1), (5n+1, n>1), (5n+2, n>=0), (5n+3, n>=0), (5n+4, n>=0), generates a dispersion. Each complement (beginning with its first term >1) also generates a dispersion. The ten sequences and dispersions are listed here:
...
A191702=dispersion of A008587 (5k, k>=1)
A191703=dispersion of A016861 (5k+1, k>=1)
A191704=dispersion of A016873 (5k+2, k>=0)
A191705=dispersion of A016885 (5k+3, k>=0)
A191706=dispersion of A016897 (5k+4, k>=0)
A191707=dispersion of A047201 (1, 2, 3, 4 mod 5 and >1)
A191708=dispersion of A047202 (0, 2, 3, 4 mod 5 and >1)
A191709=dispersion of A047207 (0, 1, 3, 4 mod 5 and >1)
A191710=dispersion of A032763 (0, 1, 2, 4 mod 5 and >1)
A191711=dispersion of A001068 (0, 1, 2, 3 mod 5 and >1)
...
EXCEPT for at most 2 initial terms (so that column 1 always starts with 1):
A191702 has 1st col A047201, all else A008587
A191703 has 1st col A047202, all else A016861
A191704 has 1st col A047207, all else A016873
A191705 has 1st col A032763, all else A016885
A191706 has 1st col A001068, all else A016897
A191707 has 1st col A008587, all else A047201
A191708 has 1st col A042968, all else A047203
A191709 has 1st col A042968, all else A047207
A191710 has 1st col A042968, all else A032763
A191711 has 1st col A042968, all else A001068
...
Regarding the dispersions A191670-A191673, there is a formula for sequences of the type
"(a or b or c or d mod m)", used in Mathematica programs for A191707-A191711):  if f(n)=(n mod 3), then
  (a,b,c,d,a,b,c,d,a,b,c,d,...) is given by
   a*f(n+3)+b*f(n+2)+c*f(n+1)+d*f(n), so that for n>=1,
"(a, b, c, d mod m)" is given by
   a*f(n+3)+b*f(n+2)+c*f(n+1)+d*f(n)+m*floor((n-1)/4)).

			Northwest corner:
  1...5....25....125...625
  2...10...50....250...1250
  3...15...75....375...1875
  4...20...100...500...2500
  6...30...150...750...3750

Ivan Neretin, Table of n, a(n) for n = 1..5050 (first 100 antidiagonals, flattened)

Cf. A047201, A008587, A191707, A191426.

(* Program generates the dispersion array T of the increasing sequence f[n] *)
r = 40; r1 = 12;  c = 40; c1 = 12;
f[n_] := 5n
Table[f[n], {n, 1, 30}]  (* A008587 *)
mex[list_] := NestWhile[#1 + 1 &, 1, Union[list][[#1]] <= #1 &, 1, Length[Union[list]]]
rows = {NestList[f, 1, c]};
Do[rows = Append[rows, NestList[f, mex[Flatten[rows]], r]], {r}];
t[i_, j_] := rows[[i, j]];
TableForm[Table[t[i, j], {i, 1, 10}, {j, 1, 10}]]
(* A191702 *)
Flatten[Table[t[k, n - k + 1], {n, 1, c1}, {k, 1, n}]] (* A191702  *)

T(i,j) = T(i,1)*T(1,j) = (i-1+floor((i+3)/4))*5^(j-1), i>=1, j>=1.

A079998 The characteristic function of the multiples of five.

Original entry on oeis.org

1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1

Offset: 0

Vladimir Baltic, Feb 10 2003

Number of permutations satisfying -k <= p(i) - i <= r and p(i) - i not in I, i = 1..n, with k = 2, r = 3, I = {-1, 0, 1, 2}.
a(n) = 1 if n = 5k, a(n) = 0 otherwise. Also, number of permutations satisfying -k <= p(i) - i <= r and p(i) - i not in I, i = 1..n, with k = 1, r = 4, I = {0, 1, 2, 3}.
a(n) is also the number of partitions of n with each part being five (a(0) = 1 because the empty partition has no parts to test equality with five). Hence a(n) is also the number of 2-regular graphs on n vertices with each component having girth exactly five. - Jason Kimberley, Oct 02 2011
This sequence is the Euler transformation of A185015. - Jason Kimberley, Oct 02 2011

D. H. Lehmer, Permutations with strongly restricted displacements. Combinatorial theory and its applications, II (Proc. Colloq., Balatonfured, 1969), pp. 755-770. North-Holland, Amsterdam, 1970.

Antti Karttunen, Table of n, a(n) for n = 0..16385
Vladimir Baltic, On the number of certain types of strongly restricted permutations, Applicable Analysis and Discrete Mathematics Vol. 4, No 1 (April, 2010), 119-135.
Benoit Cloitre, A study of a family of self-referential sequences, arXiv:2506.18103 [math.GM], 2025. See p. 10.
Index entries for characteristic functions
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,1).

Cf. A011558, A008587, A002524-A002529, A072827, A072850-A072856, A079955-A080014.

Characteristic function of multiples of g: A000007 (g = 0), A000012 (g = 1), A059841 (g = 2), A079978 (g = 3), A121262 (g = 4), this sequence (g = 5), A079979 (g = 6), A082784 (g = 7). - Jason Kimberley, Oct 14 2011

[Binomial(n-1,4) mod 5 : n in [0..100]]; // Wesley Ivan Hurt, Oct 06 2014

A079998:=n->binomial(n-1,4) mod 5: seq(A079998(n), n=0..100); # Wesley Ivan Hurt, Oct 06 2014

Table[Mod[Binomial[n - 1, 4], 5], {n, 0, 100}] (* Wesley Ivan Hurt, Oct 06 2014 *)
Table[Boole[Divisible[n, 5]], {n, 0, 99}] (* Alonso del Arte, Nov 29 2014 *)
PadRight[{},120,{1,0,0,0,0}] (* Harvey P. Dale, Jul 11 2023 *)

a(n)=!(n%5) \\ Charles R Greathouse IV, Mar 07 2012

(define (A079998 n) (if (zero? (modulo n 5)) 1 0)) ;; Antti Karttunen, Dec 21 2017

Recurrence: a(n) = a(n-5). G.f.: -1/(x^5 - 1).
a(n) = 1 - A011558(n); a(A008587(n)) = 1; a(A047201(n)) = 0. - Reinhard Zumkeller, Nov 30 2009
a(n) = floor(1/2*cos(2*n*Pi/5) + 1/2). - Gary Detlefs, May 16 2011
a(n) = floor(n/5) - floor((n-1)/5). - Tani Akinari, Oct 21 2012
a(n) = binomial(n - 1, 4) mod 5. - Wesley Ivan Hurt, Oct 06 2014

More terms from Antti Karttunen, Dec 21 2017

A011558 Expansion of (x + x^3)/(1 + x + ... + x^4) mod 2.

Original entry on oeis.org

0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0, 1, 1, 1, 1, 0

Offset: 0

N. J. A. Sloane

Multiplicative with a(5^e) = 0, a(p^e) = 1 otherwise. - David W. Wilson, Jun 12 2005
Characteristic function of numbers coprime to 5. - Reinhard Zumkeller, Nov 30 2009
From R. J. Mathar, Jul 15 2010: (Start)
The sequence is the principal Dirichlet character mod 5. (The other real character mod 5 is A080891.)
Associated Dirichlet L-functions are for example L(2,chi) = Sum_{n>=1} a(n)/n^2 = 1.5791367... = (psi'(1/5) + psi'(2/5) + psi'(3/5) + psi'(4/5))/25 or L(3,chi) = Sum_{n>=1} a(n)/n^3 = 1.192440... = -(psi''(1/5) + psi''(2/5) + psi''(3/5) + psi''(4/5))/250, where psi' and psi'' are the trigamma and tetragamma functions. (End)
a(n) is for n >= 1 also the characteristic function for rational g-adic integers (+n/5)A047201).%20See%20the%20definition%20in%20the%20Mahler%20reference,%20p.%207%20and%20also%20p.%2010.%20-%20_Wolfdieter%20Lang">g and also (-n/5)_g for all integers g >= 2 without a factor of 5 (A047201). See the definition in the Mahler reference, p. 7 and also p. 10. - _Wolfdieter Lang, Jul 11 2014
Conjecture: a(n+1) is the number of ways of partitioning n into distinct parts of A084215. - R. J. Mathar, Mar 01 2023

			G.f. = x + x^2 + x^3 + x^4 + x^6 + x^7 + x^8 + x^9 + x^11 + x^12 + ...

Arthur Gill, Linear Sequential Circuits, McGraw-Hill, 1966, Eq. (17-10).
K. Mahler, p-adic numbers and their functions, 2nd ed., Cambridge University press, 1981.

Antti Karttunen, Table of n, a(n) for n = 0..65537
Michael Gilleland, Some Self-Similar Integer Sequences
R. Gold, Characteristic linear sequences and their coset functions, J. SIAM Applied. Math., 14 (1966), 980-985.
Index entries for characteristic functions
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,1).

Cf. A000035, A011655, A109720 coprimality with 2, 3, 7, respectively.
Cf. A168185, A145568, A168184, A168182, A168181, A097325, A166486.
Cf. A080891, A100047.

seq(n&^4 mod 5, n=0..50); # Gary Detlefs, Mar 20 2010

Mod[#,2]&/@CoefficientList[Series[(x+x^3)/(1+x+x^2+x^3+x^4) ,{x,0,100}], x] (* or *) Flatten[Table[{0,1,1,1,1},{30}]] (* Harvey P. Dale, May 15 2011 *)
a[ n_] := Sign@Mod[ n, 5]; (* Michael Somos, May 24 2015 *)

a(n)=!!(n%5) \\ Charles R Greathouse IV, Sep 23 2012

{a(n) = n%5>0}; /* Michael Somos, May 24 2015 */

(define (A011558 n) (if (zero? (modulo n 5)) 0 1)) ;; Antti Karttunen, Dec 21 2017

O.g.f.: x*(1+x+x^2+x^3)/(1-x^5). - Wolfdieter Lang, Feb 05 2009
From Reinhard Zumkeller, Nov 30 2009: (Start)
a(n) = 1 - A079998(n).
a(A047201(n))=1, a(A008587(n))=0.
A033437(n) = Sum_{k=0..n} a(k)*(n-k). (End)
a(n) = n^4 mod 5. - Gary Detlefs, Mar 20 2010
Sum_{n>=1} a(n)/n^s = L(s,chi) = (1-1/5^s)*Riemann_zeta(s), s > 1. - R. J. Mathar, Jul 31 2010
For the general case. The characteristic function of numbers that are not multiples of m is a(n) = floor((n-1)/m) - floor(n/m) + 1, m,n > 0. - Boris Putievskiy, May 08 2013
a(n) = sgn(n mod 5). - Wesley Ivan Hurt, Jun 30 2013
Euler transform of length 5 sequence [ 1, 0, 0, -1, 1]. - Michael Somos, May 24 2015
Moebius transform is length 5 sequence [ 1, 0, 0, 0, -1]. - Michael Somos, May 24 2015
G.f.: f(x) - f(x^5) where f(x) := x / (1 - x). - Michael Somos, May 24 2015
|a(n)| = |A080891(n)| = |A100047(n)|. - Michael Somos, May 24 2015

More terms from Antti Karttunen, Dec 21 2017

A123865 a(n) = n^4 - 1.

Original entry on oeis.org

0, 15, 80, 255, 624, 1295, 2400, 4095, 6560, 9999, 14640, 20735, 28560, 38415, 50624, 65535, 83520, 104975, 130320, 159999, 194480, 234255, 279840, 331775, 390624, 456975, 531440, 614655, 707280, 809999, 923520, 1048575, 1185920, 1336335

Offset: 1

Reinhard Zumkeller, Oct 16 2006

a(n) mod 5 = 0 iff n mod 5 > 0: a(A008587(n)) = 4; a(A047201(n)) = 0; a(n) mod 5 = 4*(1-A079998(n)).

A129292(n) = number of divisors of a(n) that are not greater than n. - Reinhard Zumkeller, Apr 09 2007

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Milan Janjic and Boris Petkovic, A Counting Function, arXiv 1301.4550 [math.CO], 2013.
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000583, A005563, A024002, A123866, A123867, A123868, A219086.

List([1..40], n-> n^4 -1); # G. C. Greubel, Aug 08 2019

[n^4 - 1: n in [1..40]]; // Vincenzo Librandi, May 01 2011

seq(n^4 -1, n=1..40); # G. C. Greubel, Aug 08 2019

Table[n^4-1,{n,40}] (* Vladimir Joseph Stephan Orlovsky, Apr 15 2011 *)

vector(40, n, n^4 -1) \\ G. C. Greubel, Aug 08 2019

[n^4 -1 for n in (1..40)] # G. C. Greubel, Aug 08 2019

G.f.: x^2*(15 + 5*x + 5*x^2 - x^3)/(1-x)^5. - Colin Barker, Jan 10 2012
-4*a(n+1) = -4*n*(n+2)*(n^2+2*n+2) = (n+n*i)*(n+2+n*i)*(n+(n+2)*i)*(n+2+(n+2)*i), where i is the imaginary unit. - Jon Perry, Feb 05 2014
From Vaclav Kotesovec, Feb 14 2015: (Start)
Sum_{n>=2} 1/a(n) = 7/8 - Pi*coth(Pi)/4 = A256919.
Sum_{n>=2} (-1)^n / a(n) = 1/8 - Pi/(4*sinh(Pi)). (End)
a(n) = A005563(A005563(n)). - Bruno Berselli, May 28 2015
E.g.f.: 1 + (-1 + x + 7*x^2 + 6*x^3 + x^4)*exp(x). - G. C. Greubel, Aug 08 2019
Product_{n>=2} (1 + 1/a(n)) = 4*Pi*csch(Pi). - Amiram Eldar, Jan 20 2021

cp's OEIS Frontend

A337833 Minimum m coprime to 5 such that the convergence speed of m^^m := m^(m^^(m-1)) is equal to n >= 0, where A317905(n) represents the convergence speed of m^^m (and m = A047201(n), the n-th non-multiple of 5).

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A124476 Smallest multiple of A047201(n) (i.e., numbers not divisible by 5) with only digits 6 and 7.

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A141845 a(n) = 5*a(n-1) + A047201(n), a(1) = 1. A047201 = numbers not divisible by 5: (1, 2, 3, 4, 6, 7, 8, 9, 11, ...).

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A000032 Lucas numbers beginning at 2: L(n) = L(n-1) + L(n-2), L(0) = 2, L(1) = 1.

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A008587 Multiples of 5: a(n) = 5 * n.

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A035959 Number of partitions of n in which no parts are multiples of 5.

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