A000351 -id:A000351 - cp's OEIS frontend

A304081 Number of ways to write n as p + 2^k + (1+(n mod 2))5^m, where p is an odd prime, and k and m are nonnegative integers with 2^k + (1+(n mod 2))5^m squarefree.

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

Offset: 1

Zhi-Wei Sun, May 06 2018

nonn

Conjecture: a(n) > 0 for all n > 7.
This has been verified for n up to 2*10^10.
See also A303821, A303934, A303949, A304031 and A304122 for related information, and A304034 for a similar conjecture.
The author would like to offer 2500 US dollars as the prize to the first proof of the conjecture, and 250 US dollars as the prize to the first explicit counterexample. - Zhi-Wei Sun, May 08 2018

			a(6) = 1 since 6 = 3 + 2^1 + 5^0 with 3 an odd prime and 2^1 + 5^0 = 3 squarefree.
a(15) = 1 since 15 = 5 + 2^3 + 2*5^0 with 5 an odd prime and 2^3 + 2*5^0 = 2*5 squarefree.
a(35) = 1 since 35 = 29 + 2^2 + 2*5^0 with 29 an odd prime and 2^2 + 2*5^0 = 2*3 squarefree.
a(91) = 1 since 91 = 17 + 2^6 + 2*5^1 with 17 an odd prime and 2^6 + 2*5^1 = 2*37 squarefree.
a(9574899) = 1 since 9574899 = 9050609 + 2^19 + 2*5^0 with 9050609 an odd prime and 2^19 + 2*5^0 = 2*5*13*37*109 squarefree.
a(6447154629) = 2 since 6447154629 = 6447121859 + 2^15 + 2*5^0 with 6447121859 prime and 2^15 + 2*5^0 = 2*5*29*113 squarefree, and 6447154629 = 5958840611 + 2^15 + 2*5^12 with 5958840611 prime and 2^15 + 2*5^12 = 2*17*41*433*809 squarefree.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Mixed sums of primes and other terms, in: Additive Number Theory (edited by D. Chudnovsky and G. Chudnovsky), pp. 341-353, Springer, New York, 2010.
Zhi-Wei Sun, Conjectures on representations involving primes, in: M. Nathanson (ed.), Combinatorial and Additive Number Theory II, Springer Proc. in Math. & Stat., Vol. 220, Springer, Cham, 2017, pp. 279-310. (See also arXiv, arXiv:1211.1588 [math.NT], 2012-2017.)

Cf. A000040, A000079, A000351, A005117, A098983, A118955, A156695, A273812, A302982, A302984, A303233, A303234, A303338, A303363, A303389, A303393, A303399, A303428, A303401, A303432, A303434, A303539, A303540, A303541, A303543, A303601, A303637, A303639, A303656, A303660, A303702, A303821, A303932, A303934, A303949, A304031, A304032, A304034, A304122.

PQ[n_]:=n>2&&PrimeQ[n];
tab={};Do[r=0;Do[If[SquareFreeQ[2^k+(1+Mod[n,2])*5^m]&&PQ[n-2^k-(1+Mod[n,2])*5^m],r=r+1],{k,0,Log[2,n]},{m,0,If[2^k==n,-1,Log[5,(n-2^k)/(1+Mod[n,2])]]}];tab=Append[tab,r],{n,1,90}];Print[tab]

A005057 a(n) = 5^n - 2^n.

Original entry on oeis.org

0, 3, 21, 117, 609, 3093, 15561, 77997, 390369, 1952613, 9764601, 48826077, 244136529, 1220694933, 6103499241, 30517545357, 152587825089, 762939322053, 3814697003481, 19073485803837, 95367430592049, 476837156105973, 2384185786821321, 11920928946689517

Offset: 0

N. J. A. Sloane, Jun 14 1998

Binomial transform of A024036. - Wesley Ivan Hurt, Apr 04 2014

P. P. Patwardhan, Discrete Structures, Technical Publications Pune, 2009 (first ed.), Section 4.27.1.2, p. 110 (Example 4.44-i).

Ivan Panchenko, Table of n, a(n) for n = 0..200
Feryal Alayont and Evan Henning, Edge Covers of Caterpillars, Cycles with Pendants, and Spider Graphs, J. Int. Seq. (2023) Vol. 26, Art. 23.9.4.
Index entries for linear recurrences with constant coefficients, signature (7,-10).

Magma
```
[ 5^n-2^n: n in [0..24] ];
```

A005057:=n->5^n - 2^n; seq(A005057(n), n=0..50); # Wesley Ivan Hurt, Apr 04 2014

Table[5^n - 2^n, {n, 0, 60}] (* Vladimir Joseph Stephan Orlovsky, Jun 27 2011 *)

a(n)=5^n-1<Charles R Greathouse IV, Jun 28 2011

[5^n - 2^n for n in range(0,21)] # Zerinvary Lajos, Jun 04 2009

a(n) = A000351(n) - A000079(n). - R. J. Mathar, May 07 2008
G.f.: 1/(1-5*x)-1/(1-2*x);
E.g.f.: e^(5*x)-e^(2*x). - Mohammad K. Azarian, Jan 14 2009
a(n) = 7*a(n-1)-10*a(n-2), a(0)=0, a(1)=3. - Vincenzo Librandi, Dec 30 2010
a(n+1) = 3 * A016127(n). - Vladimir Joseph Stephan Orlovsky, Jun 28 2011

A067483 Powers of 5 with initial digit 5.

Original entry on oeis.org

5, 59604644775390625, 582076609134674072265625, 5684341886080801486968994140625, 55511151231257827021181583404541015625, 542101086242752217003726400434970855712890625

Offset: 1

Amarnath Murthy, Feb 09 2002

Each term also has final digit 5. - Muniru A Asiru, Oct 13 2018

Muniru A Asiru, Table of n, a(n) for n = 1..114

Subsequence of A000351 (powers of 5).

Similar entries with another digit: A067480 (2), A067481 (3), A067482 (4).

k:=5;; Filtered(List([0..100],n->k^n),i->ListOfDigits(i)[1]=k); # Muniru A Asiru, Oct 06 2018

Select[5^Range[70],First[IntegerDigits[#]]==5&]  (* Harvey P. Dale, Apr 01 2011 *)

lista(nn) = {for (n=1, nn, if (digits(p=5^n)[1] == 5, print1(p, ", ")););} \\ Michel Marcus, Oct 14 2018

Edited by Frank Ellermann, Feb 11 2002

One more term from Harvey P. Dale, Apr 01 2011

A070366 a(n) = 5^n mod 9.

Original entry on oeis.org

1, 5, 7, 8, 4, 2, 1, 5, 7, 8, 4, 2, 1, 5, 7, 8, 4, 2, 1, 5, 7, 8, 4, 2, 1, 5, 7, 8, 4, 2, 1, 5, 7, 8, 4, 2, 1, 5, 7, 8, 4, 2, 1, 5, 7, 8, 4, 2, 1, 5, 7, 8, 4, 2, 1, 5, 7, 8, 4, 2, 1, 5, 7, 8, 4, 2, 1, 5, 7, 8, 4, 2, 1, 5, 7, 8, 4, 2, 1, 5, 7, 8, 4, 2, 1, 5, 7, 8, 4, 2, 1, 5, 7, 8, 4, 2, 1, 5, 7, 8, 4, 2, 1, 5, 7

Offset: 0

N. J. A. Sloane, May 12 2002

Period 6: repeat [1, 5, 7, 8, 4, 2].
Also the digital root of 5^n. - Cino Hilliard, Dec 31 2004
Digital root of the powers of any number congruent to 5 mod 9. - Alonso del Arte, Jan 26 2014

Cecil Balmond, Number 9: The Search for the Sigma Code. Munich, New York: Prestel (1998): 203.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,0,-1,1).

Cf. Digital roots of powers of c mod 9: c = 2, A153130; c = 4, A100402; c = 7, A070403; c = 8, A010689.

Cf. A000351, A007953, A010878, A010888.

[Modexp(5, n, 9): n in [0..100]]; // Wesley Ivan Hurt, Jun 28 2016

A070366:=n->power(5,n) mod 9: seq(A070366(n), n=0..100); # Wesley Ivan Hurt, Jun 28 2016

PowerMod[5, Range[0, 120], 9] (* Harvey P. Dale, Mar 27 2011 *)
Table[Mod[5^n, 9], {n, 0, 100}] (* G. C. Greubel, Mar 05 2016 *)

a(n)=lift(Mod(5,9)^n); \\ Charles R Greathouse IV, Sep 24 2015

From R. J. Mathar, Apr 20 2010: (Start)
a(n) = a(n-1) - a(n-3) + a(n-4) for n>3.
G.f.: ( 1+4*x+2*x^2+2*x^3 ) / ( (1-x)*(1+x)*(x^2-x+1) ). (End)
a(n) = 1/2^n (mod 9), n >= 0. - Wolfdieter Lang, Feb 18 2014
a(n) = A010878(A000351(n)). - Michel Marcus, Feb 20 2014
From G. C. Greubel, Mar 05 2016: (Start)
a(n) = a(n-6) for n>5.
E.g.f.: (1/2)*(9*exp(x) - exp(-x) + 2*sqrt(3)*exp(x/2)*sin(sqrt(3)*x/2) - 6*exp(x/2)*cos(sqrt(3)*x/2)). (End)
a(n) = (9 - cos(n*Pi) - 6*cos(n*Pi/3) + 2*sqrt(3)*sin(n*Pi/3))/2. - Wesley Ivan Hurt, Jun 28 2016
a(n) = 2^((-n) mod 6) mod 9. - Joe Slater, Mar 23 2017
a(n) = A007953(5*a(n-1)) = A010888(5*a(n-1)). - Stefano Spezia, Mar 20 2025

A265928 T(n,k)=Number of nXk 0..4 arrays with the absolute differences of each element with its with horizontal and antidiagonal neighbors unique.

Original entry on oeis.org

5, 25, 25, 92, 340, 125, 340, 1740, 4616, 625, 1252, 9016, 17936, 62696, 3125, 4616, 44916, 72772, 174000, 851496, 15625, 17012, 223788, 273616, 542940, 1671744, 11564952, 78125, 62696, 1119424, 1042020, 1546496, 4044156, 15962560, 157071768

Offset: 1

R. H. Hardin, Dec 18 2015

Table starts
.......5...........25...........92.........340........1252........4616
......25..........340.........1740........9016.......44916......223788
.....125.........4616........17936.......72772......273616.....1042020
.....625........62696.......174000......542940.....1546496.....4697060
....3125.......851496......1671744.....4044156.....8821464....22093736
...15625.....11564952.....15962560....30029860....51986544...112139348
...78125....157071768....152267520...225444912...309447168...579039920
..390625...2133318088...1451371264..1691502456..1904101280..3147755448
.1953125..28974227016..13834836992.12779302796.11662822720.16695518060
.9765625.393521606584.131883277312.96726712256.73936333840.93517706688

			Some solutions for n=3 k=4
..2..0..0..1....1..1..3..0....1..0..3..2....0..1..1..4....3..4..1..0
..4..4..1..0....2..4..4..1....4..3..0..1....3..0..0..2....0..1..3..3
..3..3..0..1....0..3..1..0....1..0..3..4....4..1..1..4....4..4..1..0

R. H. Hardin, Table of n, a(n) for n = 1..219

Column 1 is A000351.

Empirical for column k:
k=1: a(n) = 5*a(n-1)
k=2: [order 8]
k=3: [order 10] for n>13
k=4: [order 28] for n>32
k=5: [order 31] for n>39
k=6: [order 42] for n>50
k=7: [order 48] for n>55
Empirical for row n:
n=1: [linear recurrence of order 8] for n>9
n=2: [order 55] for n>59

A066771 a(n) = 5^ncos(2n*arctan(1/2)) or denominator of tan(2narctan(1/2)).

Original entry on oeis.org

1, 3, -7, -117, -527, -237, 11753, 76443, 164833, -922077, -9653287, -34867797, 32125393, 1064447283, 5583548873, 6890111163, -98248054847, -761741108157, -2114245277767, 6358056037323, 91004468168113, 387075408075603, 47340744250793, -9392840736385317

Offset: 0

Barbara Haas Margolius, (b.margolius(AT)csuohio.edu), Jan 17 2002

Let A =
  [ -3/5 -(2/5)i,      -(2/5)i,      -(2/5)i,      -(2/5)i ]
  [      -(2/5)i, -3/5 +(2/5)i,      -(2/5)i,       (2/5)i ]
  [      -(2/5)i,      -(2/5)i, -3/5 +(2/5)i,       (2/5)i ]
  [      -(2/5)i,       (2/5)i,       (2/5)i, -3/5 -(2/5)i ]
be the Cayley transform of the matrix iH, where H =
  [1,  1,  1,  1]
  [1, -1,  1, -1]
  [1,  1, -1, -1]
  [1, -1, -1,  1]
is a Hadamard matrix of order 4 and i is the imaginary unit. Any diagonal entry of the matrix A^n is one of the four complex numbers (+ or -)(X/5^n)(+ or -)(Y/(5^n)i). Then a(n) is the X in [A^n](j,j), j=1,2,3,4. - _Simone Severini, Apr 28 2004
Related to the (3,4,5) Pythagorean triple. Each unsigned term represents a leg in a Pythagorean triple in which the hypotenuse = 5^n. E.g., (3 + 4i)^3 = (-117 + 44i), considered as two legs of a triangle, hypotenuse = 125 = 5^3. - Gary W. Adamson, Aug 06 2006
a(n) = 5^n*cos(nC-nA), where C is the angle opposite side AB and A is the angle opposite side BC in a triangle ABC having sidelengths |BC|=3, |CA|=4, |AB|=5 - Clark Kimberling, Oct 02 2024
If a prime p divides a term, then the indices n such that p divides a(n) comprise an arithmetic sequence; e.g., 7 divides a(4n+2) for n >= 0; 13 divides a(6n+3) for n>= 0. See the Renault paper in References. - Clark Kimberling, Oct 03 2024

Steven R. Finch, Mathematical Constants, Cambridge, 2003, pp. 430-433.

Seiichi Manyama, Table of n, a(n) for n = 0..1000
J. M. Borwein and R. Girgensohn, Addition theorems and binary expansions, Canadian J. Math. 47 (1995) 262-273.
E. Eckert, The group of primitive Pythagorean triangles, Mathematics Magazine 57 (1984) 22-27.
Steven R. Finch, Plouffe's Constant [Broken link]
Steven R. Finch, Plouffe's Constant [From the Wayback machine]
Simon Plouffe, The Computation of Certain Numbers Using a Ruler and Compass, J. Integer Seqs. Vol. 1 (1998), #98.1.3.
Marc Renault, The Period, Rank, and Order of the (a,b)-Fibonacci Sequence mod m, Math. Mag. 86 (2013) pp. 372-380.
Index entries for linear recurrences with constant coefficients, signature (6,-25).

Cf. A066770 5^n sin(2n arctan(1/2)), A000351 powers of 5 and also hypotenuse of right triangle with legs given by A066770 and A066771.
Note that A066770, A066771 and A000351 are primitive Pythagorean triples with hypotenuse 5^n. The offset of A000351 is 0, but the offset is 1 for A066770, A066771.
Cf. A093378.
Cf. A193410, A121622.
Cf. A139030.

a[1] := 4/3; for n from 1 to 40 do a[n+1] := (4/3+a[n])/(1-4/3*a[n]):od: seq(abs(denom(a[n])), n=1..40);# a[n]=tan(2n arctan(1/2))

CoefficientList[Series[(1-3x)/(1-6x+25x^2),{x,0,30}],x] (* or *) LinearRecurrence[{6,-25},{1,3},30] (* Harvey P. Dale, Jul 16 2011 *)

PARI
```
a(n)=real((2+I)^(2*n))
```

G.f.: ( 1-3*x ) / ( 1-6*x+25*x^2 ).
A recursive formula for T(n) = tan(2*n*arctan(1/2)) is T(n+1) = (4/3 + T(n))/(1 - (4/3)*T(n)). Unsigned A(n) is the absolute value of the denominator of T(n).
a(n) is the real part of (2+i)^(2n) = Sum_{k=0..n} 4^(n-k)*(-1)^k*C(2n, 2k). - Benoit Cloitre, Aug 03 2002
a(n) = real part of (3 + 4i)^n. - Gary W. Adamson, Aug 06 2006
a(n) = 6*a(n-1) - 25*a(n-2). - Gary Detlefs, Jun 10 2010
a(n) = 5^n*cos(n*arccos(3/5)). - Gary Detlefs, Dec 11 2010
a(n) = (-1)^n * hypergeom([1,-n,1/2-n],[1/2,1],-4). - Gerry Martens, Jul 28 2023

A153648 Triangle T(n, k) = T(n-1, k) + T(n-1, k-1) + jprime(j)T(n-2, k-1) with j=3, read by rows.

Original entry on oeis.org

2, 5, 5, 2, 46, 2, 2, 123, 123, 2, 2, 155, 936, 155, 2, 2, 187, 2936, 2936, 187, 2, 2, 219, 5448, 19912, 5448, 219, 2, 2, 251, 8472, 69400, 69400, 8472, 251, 2, 2, 283, 12008, 159592, 437480, 159592, 12008, 283, 2, 2, 315, 16056, 298680, 1638072, 1638072, 298680, 16056, 315, 2

Offset: 1

Roger L. Bagula, Dec 30 2008

			Triangle begins as:
  2;
  5,   5;
  2,  46,     2;
  2, 123,   123,      2;
  2, 155,   936,    155,       2;
  2, 187,  2936,   2936,     187,       2;
  2, 219,  5448,  19912,    5448,     219,      2;
  2, 251,  8472,  69400,   69400,    8472,    251,     2;
  2, 283, 12008, 159592,  437480,  159592,  12008,   283,   2;
  2, 315, 16056, 298680, 1638072, 1638072, 298680, 16056, 315, 2;

G. C. Greubel, Rows n = 1..50 of the triangle, flattened

Sequences with variable (p,q,j): A153516 (0,1,2), A153518 (0,1,3), A153520 (0,1,4), A153521 (0,1,5), this sequence (1,0,3), A153649 (1,1,4), A153650 (1,4,5), A153651 (1,5,6), A153652 (2,1,7), A153653 (2,1,8), A153654 (2,1,9), A153655 (2,1,10), A153656 (2,3,9), A153657 (2,7,10).

Cf. A000351 (powers of 5).

f:= func< n,j | Round(((3-(-1)^n)/2)*NthPrime(j)^(n-1) - 2^((3-(-1)^n)/2)) >;
function T(n,k,p,q,j)
  if n eq 2 then return NthPrime(j);
  elif (n eq 3 and k eq 2 or n eq 4 and k eq 2 or n eq 4 and k eq 3) then return f(n,j);
  elif (k eq 1 or k eq n) then return 2;
  else return T(n-1,k,p,q,j) + T(n-1,k-1,p,q,j) + (p*j+q)*NthPrime(j)*T(n-2,k-1,p,q,j);
  end if; return T;
end function;
[T(n,k,1,0,3): k in [1..n], n in [1..12]]; // G. C. Greubel, Mar 04 2021

T[n_, k_, p_, q_, j_]:= T[n,k,p,q,j]= If[n==2, Prime[j], If[n==3 && k==2 || n==4 && 2<=k<=3, ((3-(-1)^n)/2)*Prime[j]^(n-1) -2^((3-(-1)^n)/2), If[k==1 || k==n, 2, T[n-1,k,p,q,j] + T[n-1,k-1,p,q,j] + (p*j+q)*Prime[j]*T[n-2,k-1,p,q,j] ]]];
Table[T[n,k,1,0,3], {n,12}, {k,n}]//Flatten (* modified by G. C. Greubel, Mar 04 2021 *)

@CachedFunction
def f(n,j): return ((3-(-1)^n)/2)*nth_prime(j)^(n-1) - 2^((3-(-1)^n)/2)
def T(n,k,p,q,j):
    if (n==2): return nth_prime(j)
    elif (n==3 and k==2 or n==4 and 2<=k<=3): return f(n,j)
    elif (k==1 or k==n): return 2
    else: return T(n-1,k,p,q,j) + T(n-1,k-1,p,q,j) + (p*j+q)*nth_prime(j)*T(n-2,k-1,p,q,j)
flatten([[T(n,k,1,0,3) for k in (1..n)] for n in (1..12)]) # G. C. Greubel, Mar 04 2021

T(n, k) = T(n-1, k) + T(n-1, k-1) + j*prime(j)*T(n-2, k-1) with j=3.
From G. C. Greubel, Mar 04 2021: (Start)
T(n,k,p,q,j) = T(n-1,k,p,q,j) + T(n-1,k-1,p,q,j) + (p*j+q)*prime(j)*T(n-2,k-1,p,q,j) with T(2,k,p,q,j) = prime(j), T(3,2,p,q,j) = 2*prime(j)^2 -4, T(4,2,p,q,j) = T(4,3,p,q,j) = prime(j)^2 -2, T(n,1,p,q,j) = T(n,n,p,q,j) = 2 and (p,q,j) = (1,0,3).
Sum_{k=0..n} T(n,k,p,q,j) = 2*prime(j)^(n-1), for (p,q,j) = (1,0,3), = 2*A000351(n-1). (End)

Edited by G. C. Greubel, Mar 04 2021

A303821 Number of ways to write 2*n as p + 2^x + 5^y, where p is a prime, and x and y are nonnegative integers.

Original entry on oeis.org

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

Offset: 1

Zhi-Wei Sun, May 01 2018

nonn

Conjecture: a(n) > 0 for all n > 1. Moreover, for any integer n > 4, we can write 2*n as p + 2^x + 5^y, where p is an odd prime, and x and y are positive integers.
This has been verified for n up to 10^10.
See also A303934 and A304081 for further refinements, and A303932 and A304034 for similar conjectures.

			a(2) = 1 since 2*2 = 2 + 2^0 + 5^0 with 2 prime.
a(3) = 1 since 2*3 = 3 + 2^1 + 5^0 with 3 prime.
a(5616) = 2 since 2*5616 = 9059 + 2^11 + 5^3 = 10979 + 2^7 + 5^3 with 9059 and 10979 both prime.

Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Mixed sums of primes and other terms, in: Additive Number Theory (edited by D. Chudnovsky and G. Chudnovsky), pp. 341-353, Springer, New York, 2010.
Zhi-Wei Sun, Conjectures on representations involving primes, in: M. Nathanson (ed.), Combinatorial and Additive Number Theory II, Springer Proc. in Math. & Stat., Vol. 220, Springer, Cham, 2017, pp. 279-310. (See also arXiv:1211.1588 [math.NT], 2012-2017.)

Cf. A000040, A000079, A000351, A118955, A156695, A273812, A302982, A302984, A303233, A303234, A303338, A303363, A303389, A303393, A303399, A303428, A303401, A303432, A303434, A303539, A303540, A303541, A303543, A303601, A303637, A303639, A303656, A303660, A303702, A303932, A303934, A304034, A304081.

tab={};Do[r=0;Do[If[PrimeQ[2n-2^k-5^m],r=r+1],{k,0,Log[2,2n-1]},{m,0,Log[5,2n-2^k]}];tab=Append[tab,r],{n,1,80}];Print[tab]

A025625 Numbers of form 5^i*10^j, with i, j >= 0.

Original entry on oeis.org

1, 5, 10, 25, 50, 100, 125, 250, 500, 625, 1000, 1250, 2500, 3125, 5000, 6250, 10000, 12500, 15625, 25000, 31250, 50000, 62500, 78125, 100000, 125000, 156250, 250000, 312500, 390625, 500000, 625000, 781250, 1000000, 1250000, 1562500, 1953125

Offset: 1

David W. Wilson

Robert Israel, Table of n, a(n) for n = 1..10000

Subsequence of A000351 and A011557.

N:= 10^8: # for all terms <= N
sort([seq(seq(5^i*10^j, j=0..floor(log[10](N/5^i))),i=0..floor(log[5](N)))]); # Robert Israel, Nov 19 2019

Block[{a = 5, b = 10, n = 10^7}, Sort@ Flatten@ Table[a^p * b^q, {p, 0, Log[a, n]}, {q, 0, Log[b, n/(a^p)]}]] (* Michael De Vlieger, Nov 19 2019 *)

list(lim)=my(v=List(), N); for(n=0, logint(lim\=1, 10), N=10^n; while(N<=lim, listput(v, N); N*=5)); Set(v) \\ Charles R Greathouse IV, Jan 10 2018

Sum_{n>=1} 1/a(n) = (5*10)/((5-1)*(10-1)) = 25/18. - Amiram Eldar, Sep 25 2020
a(n) ~ exp(sqrt(2*log(5)*log(10)*n)) / sqrt(50). - Vaclav Kotesovec, Sep 25 2020
a(n) = 5^A025655(n) * 10^A025687(n). - R. J. Mathar, Jul 06 2025

A052961 Expansion of (1 - 3x) / (1 - 5x + 3*x^2).

Original entry on oeis.org

1, 2, 7, 29, 124, 533, 2293, 9866, 42451, 182657, 785932, 3381689, 14550649, 62608178, 269388943, 1159120181, 4987434076, 21459809837, 92336746957, 397304305274, 1709511285499, 7355643511673, 31649683701868, 136181487974321, 585958388766001, 2521247479907042

Offset: 0

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

a(n) is the number of tilings of a 2 X n rectangle using integer dimension tiles at least one of whose dimensions is 1, so allowable dimensions are 1 X 1, 1 X 2, 1 X 3, 1 X 4, ..., and 2 X 1. - David Callan, Aug 27 2014
a(n+1) counts closed walks on K_2 containing one loop on the index vertex and four loops on the other vertex. Equivalently the (1,1)entry of A^(n+1) where the adjacency matrix of digraph is A=(1,1;1,4). - _David Neil McGrath, Nov 05 2014
A production matrix for the sequence is M =
  1, 1, 0, 0, 0, 0, 0, ...
  1, 0, 4, 0, 0, 0, 0, ...
  1, 0, 0, 4, 0, 0, 0, ...
  1, 0, 0, 0, 4, 0, 0, ...
  1, 0, 0, 0, 0, 4, 0, ...
  1, 0, 0, 0, 0, 0, 4, ...
  ...
Take powers of M and extract the upper left term, getting the sequence starting (1, 1, 2, 7, 29, 124, ...). - Gary W. Adamson, Jul 22 2016
From Gary W. Adamson, Jul 29 2016: (Start)
The sequence is N=1 in an infinite set obtained from matrix powers of [(1,N); (1,4)], extracting the upper left terms.
The infinite set begins:
N=1  (A052961):  1,  2,  7,  29   124,  533,   2293, ...
N=2  (A052984):  1,  3, 13,  59,  269, 1227,   5597, ...
N=3  (A004253):  1,  4, 19,  91,  436, 2089,  10009, ...
N=4  (A000351):  1,  5, 25, 125,  625, 3125,  15625, ...
N=5  (A015449):  1,  6, 31, 161,  836, 4341,  22541, ...
N=6  (A124610):  1,  7, 37, 199, 1069, 5743,  30853, ...
N=7  (A111363):  1,  8, 43, 239, 1324, 7337,  40653, ...
N=8  (A123270):  1,  9, 49, 281, 1601, 9129,  52049, ...
N=9  (A188168):  1, 10, 55, 325, 1900, 11125, 65125, ...
N=10 (A092164):  1, 11, 61, 371, 2221, 13331, 79981, ...
... (End)

Karl V. Keller, Jr., Table of n, a(n) for n = 0..1000
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 1032
Kai Liang, Solving tiling enumeration problems by tensor network contractions, arXiv:2503.17698, March 2025.
Index entries for linear recurrences with constant coefficients, signature (5,-3).

Column k=2 of A254414.

Cf. A000351, A004253, A015449, A052984, A092164, A111363, A123270, A124610, A188168.

a:=[1,2];; for n in [3..30] do a[n]:=5*a[n-1]-3*a[n-2]; od; a; # G. C. Greubel, Oct 23 2019

I:=[1,2]; [n le 2 select I[n] else 5*Self(n-1)-3*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 17 2014

spec:= [S,{S = Sequence(Union(Prod(Sequence(Union(Z,Z,Z)),Z),Z))}, unlabeled ]: seq(combstruct[count ](spec,size = n), n = 0..20);
seq(coeff(series((1-3*x)/(1-5*x+3*x^2), x, n+1), x, n), n = 0..30); # G. C. Greubel, Oct 23 2019

CoefficientList[Series[(1-3x)/(1-5x+3x^2),{x,0,30}],x] (* or *) LinearRecurrence[{5,-3},{1,2},30] (* Harvey P. Dale, Nov 23 2013 *)

my(x='x+O('x^30)); Vec((1-3*x)/(1-5*x+3*x^2)) \\ G. C. Greubel, Oct 23 2019

def A052961_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P((1-3*x)/(1-5*x+3*x^2)).list()
A052961_list(30) # G. C. Greubel, Oct 23 2019

G.f.: (1-3*x)/(1-5*x+3*x^2).
a(n) = 5*a(n-1) - 3*a(n-2), with a(0) = 1, a(1) = 2.
a(n) = Sum_{alpha=RootOf(1-5*z+3*z^2)} (-1 + 9*alpha)*alpha^(-1-n)/13.
E.g.f.: (1 + sqrt(13) + (sqrt(13)-1) * exp(sqrt(13)*x)) / (2*sqrt(13) * exp(((sqrt(13)-5)*x)/2)). - Vaclav Kotesovec, Feb 16 2015
a(n) = A116415(n) - 3*A116415(n-1). - R. J. Mathar, Feb 27 2019

cp's OEIS Frontend

A304081 Number of ways to write n as p + 2^k + (1+(n mod 2))*5^m, where p is an odd prime, and k and m are nonnegative integers with 2^k + (1+(n mod 2))*5^m squarefree.

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Mathematica

A005057 a(n) = 5^n - 2^n.

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Magma

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Sage

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A067483 Powers of 5 with initial digit 5.

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GAP

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A070366 a(n) = 5^n mod 9.

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A265928 T(n,k)=Number of nXk 0..4 arrays with the absolute differences of each element with its with horizontal and antidiagonal neighbors unique.

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A066771 a(n) = 5^n*cos(2*n*arctan(1/2)) or denominator of tan(2*n*arctan(1/2)).

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A153648 Triangle T(n, k) = T(n-1, k) + T(n-1, k-1) + j*prime(j)*T(n-2, k-1) with j=3, read by rows.

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A304081 Number of ways to write n as p + 2^k + (1+(n mod 2))5^m, where p is an odd prime, and k and m are nonnegative integers with 2^k + (1+(n mod 2))5^m squarefree.

A066771 a(n) = 5^ncos(2n*arctan(1/2)) or denominator of tan(2narctan(1/2)).

A153648 Triangle T(n, k) = T(n-1, k) + T(n-1, k-1) + jprime(j)T(n-2, k-1) with j=3, read by rows.

A052961 Expansion of (1 - 3x) / (1 - 5x + 3*x^2).