A000351 -id:A000351 - cp's OEIS frontend

A009964 Powers of 20.

1, 20, 400, 8000, 160000, 3200000, 64000000, 1280000000, 25600000000, 512000000000, 10240000000000, 204800000000000, 4096000000000000, 81920000000000000, 1638400000000000000, 32768000000000000000

Offset: 0

N. J. A. Sloane

Same as Pisot sequences E(1, 20), L(1, 20), P(1, 20), T(1, 20). Essentially same as Pisot sequences E(20, 400), L(20, 400), P(20, 400), T(20, 400). See A008776 for definitions of Pisot sequences.
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 20-colored compositions of n such that no adjacent parts have the same color. - Milan Janjic, Nov 17 2011
a(n) gives the number of small cubes in the n-th iteration of the Menger sponge fractal. - Felix Fröhlich, Jul 09 2016
Equivalently, the number of vertices in the n-Menger sponge graph.

T. D. Noe, Table of n, a(n) for n = 0..100
Allan Bickle, Degrees of Menger and Sierpinski Graphs, Congr. Num. 227 (2016) 197-208.
Allan Bickle, MegaMenger Graphs, The College Mathematics Journal, 49 1 (2018) 20-26.
Tanya Khovanova, Recursive Sequences
Eric Weisstein's World of Mathematics, Menger Sponge
Eric Weisstein's World of Mathematics, Menger Sponge Graph
Eric Weisstein's World of Mathematics, Vertex Count
Wikipedia, Menger sponge
Index entries for linear recurrences with constant coefficients, signature (20).

Cf. A291066 (edge count).
Cf. A000079, A011557; A000302, A000351; A000244, A159991.
Cf. A291066, A083233, and A332705 on the surface area of the n-Menger sponge graph.

List([0..20],n->20^n); # Muniru A Asiru, Nov 21 2018

[20^n: n in [0..100]] // Vincenzo Librandi, Nov 21 2010

[20^n$n=0..20]; # Muniru A Asiru, Nov 21 2018

20^Range[0, 10] (* or *) LinearRecurrence[{20}, {1}, 20] (* Eric W. Weisstein, Aug 17 2017 *)

makelist(20^n,n,0,30); /* Martin Ettl, Nov 05 2012 */

a(n)=20^n \\ Charles R Greathouse IV, Jun 19 2015

powers(20,12) \\ Charles R Greathouse IV, Jun 19 2015

[20**n for n in range(21)] # Stefano Spezia, Nov 21 2018

[20^n for n in range(21)] # Zerinvary Lajos, Apr 29 2009

G.f.: 1/(1-20*x).
E.g.f.: exp(20*x).
a(n) = A159991(n)/A000244(n). - Reinhard Zumkeller, May 02 2009
From Vincenzo Librandi, Nov 21 2010: (Start)
a(n) = 20^n.
a(n) = 20*a(n-1) for n > 0, a(0) = 1. (End)
a(n) = A000079(n)*A011557(n) = A000302(n)*A000351(n). - Felix Fröhlich, Jul 09 2016

A024049 a(n) = 5^n - 1.

Original entry on oeis.org

0, 4, 24, 124, 624, 3124, 15624, 78124, 390624, 1953124, 9765624, 48828124, 244140624, 1220703124, 6103515624, 30517578124, 152587890624, 762939453124, 3814697265624, 19073486328124, 95367431640624

Offset: 0

N. J. A. Sloane

Numbers whose base 5 representation is 44444.......4. - Zerinvary Lajos, Feb 03 2007

For n > 0, a(n) is the sum of divisors of 3*5^(n-1). - Patrick J. McNab, May 27 2017

			For n = 5, a(5) = 4*5 + 16*10 + 64*10 + 256*5 + 1024*1 = 3124. - _Bruno Berselli_, Nov 11 2015

Vincenzo Librandi, Table of n, a(n) for n = 0..400
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. A000351, A003463, A125831, A248722.

[5^n-1: n in [0..30]]; // Vincenzo Librandi, Jun 06 2011

5^Range[0,50]-1 (* Vladimir Joseph Stephan Orlovsky, Feb 20 2011 *)
LinearRecurrence[{6,-5},{0,4},30] (* Harvey P. Dale, Apr 06 2019 *)

a(n)=5^n-1 \\ Charles R Greathouse IV, Apr 17 2012

G.f.: 1/(1-5*x) - 1/(1-x) = 4*x/((1-5*x)*(1-x)). - Mohammad K. Azarian, Jan 14 2009
E.g.f.: exp(5*x) - exp(x). - Mohammad K. Azarian, Jan 14 2009
a(n+1) = 5*a(n) + 4. - Reinhard Zumkeller, Nov 22 2009
a(n) = Sum_{i=1..n} 4^i*binomial(n,n-i) for n>0, a(0)=0. - Bruno Berselli, Nov 11 2015
a(n) = A000351(n) - 1. - Sean A. Irvine, Jun 19 2019
Sum_{n>=1} 1/a(n) = A248722. - Amiram Eldar, Nov 13 2020
a(n) = 2*A125831(n) = 4*A003463(n). - Elmo R. Oliveira, Dec 10 2023

A048898 One of the two successive approximations up to 5^n for the 5-adic integer sqrt(-1).

Original entry on oeis.org

0, 2, 7, 57, 182, 2057, 14557, 45807, 280182, 280182, 6139557, 25670807, 123327057, 123327057, 5006139557, 11109655182, 102662389557, 407838170807, 3459595983307, 3459595983307, 79753541295807, 365855836217682, 2273204469030182, 2273204469030182, 49956920289342682

Offset: 0

Michael Somos, Jul 26 1999

This is the root congruent to 2 mod 5.
Or, residues modulo 5^n of a 5-adic solution of x^2+1=0.
The radix-5 expansion of a(n) is obtained from the n rightmost digits in the expansion of the following pentadic integer:
  ...422331102414131141421404340423140223032431212 = u
The residues modulo 5^n of the other 5-adic solution of x^2+1=0 are given by A048899 which corresponds to the pentadic integer -u:
  ...022113342030313303023040104021304221412013233 = -u
The digits of u and -u are given in A210850 and A210851, respectively. - Wolfdieter Lang, May 02 2012
For approximations for p-adic square roots see also the W. Lang link under A268922. - Wolfdieter Lang, Apr 03 2016
From Jianing Song, Sep 06 2022: (Start)
For n > 0, a(n)-1 is one of the four solutions to x^4 == -4 (mod 5^n), the one that is congruent to 1 modulo 5.
For n > 0, a(n)+1 is one of the four solutions to x^4 == -4 (mod 5^n), the one that is congruent to 3 modulo 5. (End)

			a(0)=0 because 0 satisfies any equation in integers modulo 1.
a(1)=2 because 2 is one solution of x^2+1=0 modulo 5. (The other solution is 3, which gives rise to A048899.)
a(2)=7 because the equation (5y+a(1))^2+1=0 modulo 25 means that y is 1 modulo 5.

J. H. Conway, The Sensual Quadratic Form, p. 118, Mathematical Association of America, 1997, The Carus Mathematical Monographs, Number 26.
K. Mahler, Introduction to p-Adic Numbers and Their Functions, Cambridge, 1973, p. 35.

Seiichi Manyama, Table of n, a(n) for n = 0..1431 (terms 0..200 from Vincenzo Librandi)
Peter Bala, Using Lucas polynomials to find the p -adic square roots of -1, -2 and -3, Dec 2022.
G. P. Michon, On the witnesses of a composite integer, Numericana.
G. P. Michon, Introduction to p-adic integers, Numericana.

The two successive approximations up to p^n for p-adic integer sqrt(-1): this sequence and A048899 (p=5), A286840 and A286841 (p=13), A286877 and A286878 (p=17).
Cf. A000351 (powers of 5), A034939(n) = Min(a(n), A048899(n)).
Different from A034935.

[n le 2 select 2*(n-1) else Self(n-1)^5 mod 5^(n-1): n in [1..30]]; // Vincenzo Librandi, Feb 29 2016

a[0] = 0; a[1] = 2; a[n_] := a[n] = Mod[a[n-1]^5, 5^n]; Table[a[n], {n, 0, 21}] (* Jean-François Alcover, Nov 24 2011, after PARI *)
Join[{0}, RecurrenceTable[{a[1] == 2, a[n] == Mod[a[n-1]^5, 5^n]}, a, {n, 25}]] (* Vincenzo Librandi, Feb 29 2016 *)

PARI
```
{a(n) = if( n<2, 2, a(n-1)^5) % 5^n}
```

a(n) = lift(sqrt(-1 + O(5^n))); \\ Kevin Ryde, Dec 22 2020

If n>0, a(n) = 5^n - A048899(n).
From Wolfdieter Lang, Apr 28 2012: (Start)
Recurrence: a(n) = a(n-1)^5 (mod 5^n), a(1) = 2, n>=2. See the J.-F. Alcover Mathematica program and the PARI program below.
a(n) == 2^(5^(n-1)) (mod 5^n), n>=1.
a(n)*a(n-1) + 1 == 0 (mod 5^(n-1)), n>=1.
(a(n)^2 + 1)/5^n = A210848(n), n>=0.
(End)
Another recurrence: a(n) = modp(a(n-1) + a(n-1)^2 + 1, 5^n), n >= 2, a(1) = 2. Here modp(a, m) is the representative from {0, 1, ..., |m|-1} of the residue class a modulo m. Note that a(n) is in the residue class of a(n-1) modulo 5^(n-1) (see Hensel lifting). - Wolfdieter Lang, Feb 28 2016
a(n) == L(5^n,2) (mod 5^n), where L(n,x) denotes the n-th Lucas polynomial of A114525. - Peter Bala, Nov 20 2022

Additional comments from Gerard P. Michon, Jul 15 2009
Edited by N. J. A. Sloane, Jul 25 2009
Name clarified by Wolfdieter Lang, Feb 19 2016

A009969 Powers of 25.

Original entry on oeis.org

1, 25, 625, 15625, 390625, 9765625, 244140625, 6103515625, 152587890625, 3814697265625, 95367431640625, 2384185791015625, 59604644775390625, 1490116119384765625, 37252902984619140625, 931322574615478515625, 23283064365386962890625, 582076609134674072265625, 14551915228366851806640625, 363797880709171295166015625, 9094947017729282379150390625

Offset: 0

N. J. A. Sloane

Same as Pisot sequences E(1, 25), L(1, 25), P(1, 25), T(1, 25). Essentially same as Pisot sequences E(25, 625), L(25, 625), P(25, 625), T(25, 625). See A008776 for definitions of Pisot sequences.
A000005(a(n)) = A005408(n+1). - Reinhard Zumkeller, Mar 04 2007
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 25-colored compositions of n such that no adjacent parts have the same color. - Milan Janjic, Nov 17 2011

T. D. Noe, Table of n, a(n) for n = 0..100
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (25).

Bisection of A000351 (powers of 5).

Cf. A218728 (partial sums).

[25^n: n in [0..100]] // Vincenzo Librandi, Nov 21 2010

25^Range[0,20] (* or *) NestList[25#&,1,20] (* Harvey P. Dale, Dec 12 2016 *)

a(n)=25^n \\ Charles R Greathouse IV, Sep 24 2015

[lucas_number1(n,25,0) for n in range(1, 17)] # Zerinvary Lajos, Apr 29 2009

G.f.: 1/(1-25*x). - Philippe Deléham, Nov 23 2008
E.g.f.: exp(25*x). - Zerinvary Lajos, Apr 29 2009
a(n) = 25^n; a(n) = 25*a(n-1), n > 0; a(0)=1. - Vincenzo Librandi, Nov 21 2010
a(n) = A000351(2n) = 5^(2n). - M. F. Hasler, Sep 02 2021

A008839 Numbers k such that the decimal expansion of 5^k contains no zeros.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 9, 10, 11, 17, 18, 30, 33, 58

Offset: 1

Olivier Gérard

Probably 58 is last term.

Searched for k up to 10^10. - David Radcliffe, Dec 27 2015

			Here is 5^58, conjecturally the largest power of 5 that does not contain a 0:
34694469519536141888238489627838134765625. - _N. J. A. Sloane_, Feb 10 2023, corrected by _Patrick De Geest_, Jun 09 2024

M. F. Hasler, Zeroless powers, OEIS Wiki, Mar 07 2014.
W. Schneider, NoZeros: Powers n^k without Digit Zero [Cached copy]
Eric Weisstein's World of Mathematics, Zero.

Cf. A000351 (5^n).
For the zeroless numbers (powers x^n), see A238938, A238939, A238940, A195948, A238936, A195908, A195946, A195945, A195942, A195943, A103662.
For the corresponding exponents, see A007377, A008839, A030700, A030701, A008839, A030702, A030703, A030704, A030705, A030706, A195944.
For other related sequences, see A305925, A052382, A027870, A102483, A103663.

[ n: n in [0..500] | not 0 in Intseq(5^n) ]; // Vincenzo Librandi Oct 19 2012

Do[ If[ Union[ RealDigits[ 5^n ][[1]]] [[1]] != 0, Print[ n ]], {n, 0, 25000}]

for(n=0,99,vecmin(digits(5^n))&& print1(n",")) \\ M. F. Hasler, Mar 07 2014

Definition corrected and initial term 0 added by M. F. Hasler, Sep 25 2011
Further edits by M. F. Hasler, Mar 08 2014
Keyword:fini removed by Jianing Song, Jan 28 2023 as finiteness is only conjectured.

A085362 a(0)=1; for n>0, a(n) = 25^(n-1) - (1/2)Sum_{i=1..n-1} a(i)*a(n-i).

Original entry on oeis.org

1, 2, 8, 34, 150, 678, 3116, 14494, 68032, 321590, 1528776, 7301142, 35003238, 168359754, 812041860, 3926147730, 19022666310, 92338836390, 448968093320, 2186194166950, 10659569748370, 52037098259090, 254308709196660

Offset: 0

Mario Catalani (mario.catalani(AT)unito.it), Jun 25 2003

Number of bilateral Schroeder paths (i.e. lattice paths consisting of steps U=(1,1), D=(1,-1) and H=(2,0)) from (0,0) to (2n,0) and with no H-steps at even (zero, positive or negative) levels. Example: a(2)=8 because we have UDUD, UUDD, UHD, UDDU and their reflections in the x-axis. First differences of A026375. - Emeric Deutsch, Jan 28 2004
From G. C. Greubel, May 22 2020: (Start)
This sequence is part of a class of sequences, for m >= 0, with the properties:
a(n) = 2*m*(4*m+1)^(n-1) - (1/2)*Sum_{k=1..n-1} a(k)*a(n-k).
a(n) = Sum_{k=0..n} m^k * binomial(n-1, n-k) * binomial(2*k, k).
a(n) = (2*m) * Hypergeometric2F1(-n+1, 3/2; 2; -4*m), for n > 0.
n*a(n) = 2*((2*m+1)*n - (m+1))*a(n-1) - (4*m+1)*(n-2)*a(n-2).
(4*m + 1)^n = Sum_{k=0..n} Sum_{j=0..k} a(j)*a(k-j).
G.f.: sqrt( (1 - t)/(1 - (4*m+1)*t) ).
This sequence is the case of m=1. (End)

Vincenzo Librandi, Table of n, a(n) for n = 0..200
László Németh, Tetrahedron trinomial coefficient transform, arXiv:1905.13475 [math.CO], 2019.

Bisection of A026392.
Cf. A000351 (5^n), A026375, A085363, A085364, A085455.
Cf. A110170, A162478, A359758, A360132.
Cf. A063886, A360317, A360318, A360319, A360321, A360322.
Essentially the same as A026387.

R:=PowerSeriesRing(Rationals(), 30); Coefficients(R!( Sqrt((1-x)/(1-5*x)) )); // G. C. Greubel, May 23 2020

a := n -> `if`(n=0,1,2*hypergeom([3/2, 1-n], [2], -4)):
seq(simplify(a(n)), n=0..22); # Peter Luschny, Jan 30 2017

CoefficientList[Series[Sqrt[(1-x)/(1-5x)], {x, 0, 25}], x]

my(x='x+O('x^66)); Vec(sqrt((1-x)/(1-5*x))) \\ Joerg Arndt, May 10 2013

def A085362_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P( sqrt((1-x)/(1-5*x)) ).list()
A085362_list(30) # G. C. Greubel, May 23 2020

G.f.: sqrt((1-x)/(1-5*x)).
Sum_{i=0..n} (Sum_{j=0..i} a(j)*a(i-j)) = 5^n.
D-finite with recurrence: a(n) = (2*(3*n-2)*a(n-1)-5*(n-2)*a(n-2))/n; a(0)=1, a(1)=2. - Emeric Deutsch, Jan 28 2004
a(n) ~ 2*5^(n-1/2)/sqrt(Pi*n). - Vaclav Kotesovec, Oct 14 2012
G.f.: G(0), where G(k)= 1 + 4*x*(4*k+1)/( (4*k+2)*(1-x) - 2*x*(1-x)* (2*k+1)*(4*k+3)/(x*(4*k+3) + (1-x)*(k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 22 2013
a(n) = Sum_{k=0..n} binomial(2*k,k)*binomial(n-1,n-k). - Vladimir Kruchinin, May 30 2016
a(n) = 2*hypergeom([3/2, 1-n], [2], -4) for n>0. - Peter Luschny, Jan 30 2017
a(0) = 1; a(n) = (2/n) * Sum_{k=0..n-1} (n+k) * a(k). - Seiichi Manyama, Mar 28 2023
From Seiichi Manyama, Aug 22 2025: (Start)
a(n) = (1/4)^n * Sum_{k=0..n} 5^k * binomial(2*k,k) * binomial(2*(n-k),n-k)/(1-2*(n-k)).
a(n) = Sum_{k=0..n} (-1)^k * 5^(n-k) * binomial(2*k,k)/(1-2*k) * binomial(n-1,n-k). (End)

A303389 Number of ways to write n as a(a+1)/2 + b(b+1)/2 + 5^c + 5^d, where a,b,c,d are nonnegative integers with a <= b and c <= d.

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 1, 3, 2, 2, 2, 4, 3, 2, 2, 3, 3, 3, 2, 2, 2, 4, 3, 2, 1, 5, 4, 3, 2, 5, 5, 5, 5, 3, 3, 5, 5, 4, 4, 4, 5, 5, 2, 5, 3, 5, 4, 7, 2, 4, 6, 6, 5, 4, 4, 5, 8, 4, 4, 4, 7, 6, 4, 3, 4, 8, 4, 7, 3, 3, 6, 8, 2, 5, 6, 5, 4, 6, 4, 3

Offset: 1

Zhi-Wei Sun, Apr 23 2018

nonn

Conjecture: a(n) > 0 for all n > 1. In other words, any integers n > 1 can be written as the sum of two triangular numbers and two powers of 5.
This has been verified for all n = 2..10^10.
See A303393 for the numbers of the form x*(x+1)/2 + 5^y with x and y nonnegative integers.
See also A303401, A303432 and A303540 for similar conjectures.

			a(4) = 1 with 4 = 1*(1+1)/2 + 1*(1+1)/2 + 5^0 + 5^0.
a(5) = 1 with 5 = 0*(0+1)/2 + 2*(2+1)/2 + 5^0 + 5^0.
a(7) = 1 with 7 = 0*(0+1)/2 + 1*(1+1)/2 + 5^0 + 5^1.
a(25) = 1 with 25 = 0*(0+1)/2 + 5*(5+1)/2 + 5^1 + 5^1.

Zhi-Wei Sun, Table of n, a(n) for n = 1..100000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190.
Zhi-Wei Sun, New conjectures on representations of integers (I), Nanjing Univ. J. Math. Biquarterly 34(2017), no. 2, 97-120.
Zhi-Wei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017-2018.

Cf. A000217, A000351, A271518, A273812, A281976, A299924, A299537, A299794, A300219, A300362, A300396, A300441, A301376, A301391, A301471, A301472, A302920, A302981, A302982, A302983, A302984, A302985, A303233, A303234, A303235, A303338, A303363, A303393, A303401, A303432, A303540.

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
f[n_]:=f[n]=FactorInteger[n];
g[n_]:=g[n]=Sum[Boole[Mod[Part[Part[f[n],i],1],4]==3&&Mod[Part[Part[f[n],i],2],2]==1],{i,1,Length[f[n]]}]==0;
QQ[n_]:=QQ[n]=(n==0)||(n>0&&g[n]);
tab={};Do[r=0;Do[If[QQ[4(n-5^j-5^k)+1],Do[If[SQ[8(n-5^j-5^k-x(x+1)/2)+1],r=r+1],{x,0,(Sqrt[4(n-5^j-5^k)+1]-1)/2}]],{j,0,Log[5,n/2]},{k,j,Log[5,n-5^j]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A303393 Numbers of the form x*(x+1)/2 + 5^y with x and y nonnegative integers.

Original entry on oeis.org

1, 2, 4, 5, 6, 7, 8, 11, 15, 16, 20, 22, 25, 26, 28, 29, 31, 33, 35, 37, 40, 41, 46, 50, 53, 56, 60, 61, 67, 70, 71, 79, 80, 83, 91, 92, 96, 103, 106, 110, 116, 121, 125, 126, 128, 130, 131, 135, 137, 140, 141, 145, 146, 153, 154, 158, 161, 170, 172, 176

Offset: 1

Zhi-Wei Sun, Apr 23 2018

nonn

The author's conjecture in A303389 has the following equivalent version: Each integer n > 1 can be expressed as the sum of two terms of the current sequence.

This has been verified for all n = 2..2*10^8.

			a(1) = 1 with 1 = 0*(0+1)/2 + 5^0.
a(2) = 2 with 2 = 1*(1+1)/2 + 5^0.
a(3) = 4 with 4 = 2*(2+1)/2 + 5^0.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190.
Zhi-Wei Sun, New conjectures on representations of integers (I), Nanjing Univ. J. Math. Biquarterly 34(2017), no. 2, 97-120.
Zhi-Wei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017-2018.

Cf. A000217, A000351, A271518, A273812, A281976, A299924, A299537, A299794, A300219, A300362, A300396, A300441, A301376, A301391, A301471, A301472, A302920, A302981, A302982, A302983, A302984, A302985, A303233, A303234, A303235, A303338, A303363, A303389.

TQ[n_]:=TQ[n]=IntegerQ[Sqrt[8n+1]];
tab={};Do[Do[If[TQ[m-5^k],tab=Append[tab,m];Goto[aa]],{k,0,Log[5,m]}];Label[aa],{m,1,176}];Print[tab]

A303399 Number of ordered pairs (a, b) with 0 <= a <= b such that n - 5^a - 5^b can be written as the sum of two triangular numbers.

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 1, 2, 2, 2, 2, 3, 3, 2, 2, 2, 3, 2, 2, 2, 2, 3, 2, 2, 1, 4, 3, 3, 2, 5, 4, 4, 4, 3, 3, 4, 4, 3, 4, 4, 4, 3, 2, 4, 3, 3, 3, 5, 2, 4, 5, 4, 4, 4, 4, 3, 5, 3, 4, 4, 4, 4, 4, 3, 3, 5, 4, 5, 3, 3, 5, 5, 2, 4, 6, 3, 3, 4, 4, 3

Offset: 1

Zhi-Wei Sun, Apr 23 2018

nonn

Conjecture: a(n) > 0 for all n > 1.
This is equivalent to the author's conjecture in A303389. It has been verified that a(n) > 0 for all n = 2..6*10^9.
Note that a nonnegative integer m is the sum of two triangular numbers if and only if 4*m + 1 can be written as the sum of two squares.

			a(6) = 2 with 6 - 5^0 - 5^0 = 1*(1+1)/2 + 2*(2+1)/2 and 6 - 5^0 - 5^1 = 0*(0+1)/2 + 0*(0+1)/2.
a(7) = 1 with 7 - 5^0 - 5^1 = 0*(0+1)/2 + 1*(1+1)/2.
a(25) = 1 with 25 - 5^1 - 5^1 = 0*(0+1)/2 + 5*(5+1)/2.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190.
Zhi-Wei Sun, New conjectures on representations of integers (I), Nanjing Univ. J. Math. Biquarterly 34(2017), no. 2, 97-120.
Zhi-Wei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017-2018.

Cf. A000217, A000351, A271518, A273812, A281976, A299924, A300219, A300441, A301376, A301391, A301471, A301472, A302920, A302981, A302982, A302983, A302984, A302985, A303233, A303234, A303235, A303338, A303363, A303389, A303393.

f[n_]:=f[n]=FactorInteger[n];
g[n_]:=g[n]=Sum[Boole[Mod[Part[Part[f[n],i],1],4]==3&&Mod[Part[Part[f[n],i],2],2]==1],{i,1,Length[f[n]]}]==0;
QQ[n_]:=QQ[n]=(n==0)||(n>0&&g[n]);
tab={};Do[r=0;Do[If[QQ[4(n-5^j-5^k)+1],r=r+1],{j,0,Log[5,n/2]},{k,j,Log[5,n-5^j]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A005054 a(0) = 1; a(n) = 4*5^(n-1) for n >= 1.

Original entry on oeis.org

1, 4, 20, 100, 500, 2500, 12500, 62500, 312500, 1562500, 7812500, 39062500, 195312500, 976562500, 4882812500, 24414062500, 122070312500, 610351562500, 3051757812500, 15258789062500, 76293945312500, 381469726562500, 1907348632812500, 9536743164062500

Offset: 0

N. J. A. Sloane

Consider the sequence formed by the final n decimal digits of {2^k: k >= 0}. For n=1 this is 1, 2, 4, 8, 6, 2, 4, ... (A000689) with period 4. For any n this is periodic with period a(n). Cf. A000855 (n=2), A126605 (n=3, also n=4). - N. J. A. Sloane, Jul 08 2022
First differences of A000351.
Length of repeating cycle of the final n+1 digits in Fermat numbers. - Lekraj Beedassy, Robert G. Wilson v and Eric W. Weisstein, Jul 05 2004
Number of n-digit endings for a power of 2 whose exponent is greater than or equal to n. - J. Lowell
For n>=1, a(n) is equal to the number of functions f:{1,2,...,n}->{1,2,3,4,5} such that for a fixed x in {1,2,...,n} and a fixed y in {1,2,3,4,5} we have f(x) != y. - Aleksandar M. Janjic and Milan Janjic, Mar 27 2007
Equals INVERT transform of A033887: (1, 3, 13, 55, 233, ...) and INVERTi transform of A001653: (1, 5, 29, 169, 985, 5741, ...). - Gary W. Adamson, Jul 22 2010
a(n) = (n+1) terms in the sequence (1, 3, 4, 4, 4, ...) dot (n+1) terms in the sequence (1, 1, 4, 20, 100, ...). Example: a(4) = 500 = (1, 3, 4, 4, 4) dot (1, 1, 4, 20, 100) = (1 + 3 + 16, + 80 + 400), where (1, 3, 16, 80, 400, ...) = A055842, finite differences of A005054 terms. - Gary W. Adamson, Aug 03 2010
a(n) is the number of compositions of n when there are 4 types of each natural number. - Milan Janjic, Aug 13 2010
Apart from the first term, number of monic squarefree polynomials over F_5 of degree n. - Charles R Greathouse IV, Feb 07 2012
For positive integers that can be either of two colors (designated by ' or ''), a(n) is the number of compositions of 2n that are cardinal palindromes; that is, palindromes that only take into account the cardinality of the numbers and not their colors. Example: 3', 2'', 1', 1, 2', 3'' would count as a cardinal palindrome. - Gregory L. Simay, Mar 01 2020
a(n) is the length of the period of the sequence Fibonacci(k) (mod 5^(n-1)) (for n>1) and the length of the period of the sequence Lucas(k) (mod 5^n) (Kramer and Hoggatt, 1972). - Amiram Eldar, Feb 02 2022

T. Koshy, "The Ends Of A Fermat Number", pp. 183-4 Journal Recreational Mathematics, vol. 31(3) 2002-3 Baywood NY.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 458.
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets.
Judy Kramer and V. E. Hoggatt, Jr., Special Cases of Fibonacci Periodicity (Part 1, Part 2), The Fibonacci Quarterly, Vol. 10, No. 5 (1972), pp. 519-522, 530.
Eric Weisstein's World of Mathematics, Fermat Number.
Index entries for linear recurrences with constant coefficients, signature (5).

Cf. A000010, A000032, A000045, A000351, A000215, A055842.
Cf. A001653, A033887. - Gary W. Adamson, Jul 22 2010
See also A000689, A000855, A126605, A216099, A216164. - N. J. A. Sloane, Jul 08 2022

[(4*5^n+0^n)/5: n in [0..30]]; // Vincenzo Librandi, Jun 08 2013

a:= n-> ceil(4*5^(n-1)):
seq(a(n), n=0..30);  # Alois P. Heinz, Jul 08 2022

CoefficientList[Series[(1 - x) / (1 - 5 x), {x, 0, 50}], x] (* Vincenzo Librandi, Jun 08 2013 *)

Vec((1-x)/(1-5*x) + O(x^100)) \\ Altug Alkan, Dec 07 2015

a(n) = (4*5^n + 0^n) / 5. - R. J. Mathar, May 13 2008
G.f.: (1-x)/(1-5*x). - Philippe Deléham, Nov 02 2009
G.f.: 1/(1 - 4*Sum_{k>=1} x^k).
a(n) = 5*a(n-1) for n>=2. - Vincenzo Librandi, Dec 31 2010
a(n) = phi(5^n) = A000010(A000351(n)).
E.g.f.: (4*exp(5*x)+1)/5. - Paul Barry, Apr 20 2003
a(n + 1) = (((1 + sqrt(-19))/2)^n + ((1 - sqrt(-19))/2)^n)^2 - (((1 + sqrt(-19))/2)^n - ((1 - sqrt(-19))/2)^n)^2. - Raphie Frank, Dec 07 2015

Better definition from R. J. Mathar, May 13 2008

Edited by N. J. A. Sloane, Jul 08 2022

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