A005054 -id:A005054 - cp's OEIS frontend

A000351 Powers of 5: a(n) = 5^n.

1, 5, 25, 125, 625, 3125, 15625, 78125, 390625, 1953125, 9765625, 48828125, 244140625, 1220703125, 6103515625, 30517578125, 152587890625, 762939453125, 3814697265625, 19073486328125, 95367431640625, 476837158203125, 2384185791015625, 11920928955078125

Offset: 0

N. J. A. Sloane

Same as Pisot sequences E(1, 5), L(1, 5), P(1, 5), T(1, 5). Essentially same as Pisot sequences E(5, 25), L(5, 25), P(5, 25), T(5, 25). See A008776 for definitions of Pisot sequences.
a(n) has leading digit 1 if and only if n = A067497 - 1. - Lekraj Beedassy, Jul 09 2002
With interpolated zeros 0, 1, 0, 5, 0, 25, ... (g.f.: x/(1 - 5*x^2)) second inverse binomial transform of Fibonacci(3n)/Fibonacci(3) (A001076). Binomial transform is A085449. - Paul Barry, Mar 14 2004
Sums of rows of the triangles in A013620 and A038220. - Reinhard Zumkeller, May 14 2006
Sum of coefficients of expansion of (1 + x + x^2 + x^3 + x^4)^n. a(n) is number of compositions of natural numbers into n parts less than 5. a(2) = 25 there are 25 compositions of natural numbers into 2 parts less than 5. - Adi Dani, Jun 22 2011
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 5-colored compositions of n such that no adjacent parts have the same color. - Milan Janjic, Nov 17 2011
Numbers n such that sigma(5n) = 5n + sigma(n). In fact we have this theorem: p is a prime if and only if all solutions of the equation sigma(p*x) = p*x + sigma(x) are powers of p. - Jahangeer Kholdi, Nov 23 2013
From Doug Bell, Jun 22 2015: (Start)
Empirical observation: Where n is an odd multiple of 3, let x = (a(n) + 1)/9 and let y be the decimal expansion of x/a(n); then y*(x+1)/x + 1 = y rotated to the left.
Example:
  a(3) = 125;
  x = (125 + 1)/9 = 14;
  y = 112, which is the decimal expansion of 14/125 = 0.112;
  112*(14 + 1)/14 + 1 = 121 = 112 rotated to the left.
(End)
a(n) is the number of n-digit integers that contain only odd digits (A014261). - Bernard Schott, Nov 12 2022
Number of pyramids in the Sierpinski fractal square-based pyramid at the n-th step, while A279511 gives the corresponding number of vertices (see IREM link with drawings). - Bernard Schott, Nov 29 2022

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
O. M. Cain, The Exceptional Selfcondensability of Powers of Five, arXiv:1910.13829 [math.HO], 2019.
Peter J. Cameron, Sequences realized by oligomorphic permutation groups, J. Integ. Seqs. Vol. 3 (2000), #00.1.5.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 270
IREM Paris-Nord, La pyramide de Sierpinski (in French).
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
Yash Puri and Thomas Ward, Arithmetic and growth of periodic orbits, J. Integer Seqs., Vol. 4 (2001), #01.2.1.
Eric Weisstein's World of Mathematics, Box Fractal
Index entries for linear recurrences with constant coefficients, signature (5).

Cf. A009969 (even bisection), A013710 (odd bisection), A005054 (first differences), A003463 (partial sums).
Cf. A006495, A006496, A159991, A001021.
Cf. A001076, A008776, A013620, A038220, A067497, A085449, A014261.
Sierpinski fractal square-based pyramid: A020858 (Hausdorff dimension), A279511 (number of vertices), this sequence (number of pyramids).

a000351 = (5 ^)
a000351_list = iterate (* 5) 1  -- Reinhard Zumkeller, Oct 31 2012

[5^n : n in [0..30]]; // Wesley Ivan Hurt, Sep 27 2016

[ seq(5^n,n=0..30) ];
A000351:=-1/(-1+5*z); # Simon Plouffe in his 1992 dissertation

Table[5^n, {n, 0, 30}] (* Stefan Steinerberger, Apr 06 2006 *)
5^Range[0, 30] (* Harvey P. Dale, Aug 22 2011 *)

makelist(5^n,n,0,20); /* Martin Ettl, Dec 27 2012 */

a(n)=5^n \\ Charles R Greathouse IV, Jun 10 2011

def a(n): return 5**n
print([a(n) for n in range(24)]) # Michael S. Branicky, Nov 12 2022

(List.fill(50)(5: BigInt)).scanLeft(1: BigInt)( * ) // Alonso del Arte, May 31 2019

a(n) = 5^n.
a(0) = 1; a(n) = 5*a(n-1) for n > 0.
G.f.: 1/(1 - 5*x).
E.g.f.: exp(5*x).
a(n) = A006495(n)^2 + A006496(n)^2.
a(n) = A159991(n) / A001021(n). - Reinhard Zumkeller, May 02 2009
From Bernard Schott, Nov 12 2022: (Start)
Sum_{n>=0} 1/a(n) = 5/4.
Sum_{n>=0} (-1)^n/a(n) = 5/6. (End)
a(n) = Sum_{k=0..n} C(2*n+1,n-k)*A000045(2*k+1). - Vladimir Kruchinin, Jan 14 2025

A001653 Numbers k such that 2*k^2 - 1 is a square.

Original entry on oeis.org

1, 5, 29, 169, 985, 5741, 33461, 195025, 1136689, 6625109, 38613965, 225058681, 1311738121, 7645370045, 44560482149, 259717522849, 1513744654945, 8822750406821, 51422757785981, 299713796309065, 1746860020068409, 10181446324101389, 59341817924539925

Offset: 1

N. J. A. Sloane

Consider all Pythagorean triples (X,X+1,Z) ordered by increasing Z; sequence gives Z values.
The defining equation is X^2 + (X+1)^2 = Z^2, which when doubled gives 2Z^2 = (2X+1)^2 + 1. So the sequence gives Z's such that 2Z^2 = odd square + 1 (A069894).
(x,y) = (a(n), a(n+1)) are the solutions with x < y of x/(yz) + y/(xz) + z/(xy)=3 with z=2. - Floor van Lamoen, Nov 29 2001
Consequently the sum n^2*(2n^2 - 1) of the first n odd cubes (A002593) is also a square. - Lekraj Beedassy, Jun 05 2002
Numbers n such that 2*n^2 = ceiling(sqrt(2)*n*floor(sqrt(2)*n)). - Benoit Cloitre, May 10 2003
Also, number of domino tilings in S_5 X P_2n. - Ralf Stephan, Mar 30 2004. Here S_5 is the star graph on 5 vertices with the edges {1,2}, {1,3}, {1,4}, {1,5}.
If x is in the sequence then so is x*(8*x^2-3). - James R. Buddenhagen, Jan 13 2005
In general, Sum_{k=0..n} binomial(2n-k,k)j^(n-k) = (-1)^n*U(2n,i*sqrt(j)/2), i=sqrt(-1). - Paul Barry, Mar 13 2005
a(n) = L(n,6), where L is defined as in A108299; see also A002315 for L(n,-6). - Reinhard Zumkeller, Jun 01 2005
Define a T-circle to be a first-quadrant circle with integral radius that is tangent to the x- and y-axes. Such a circle has coordinates equal to its radius. Let C(0) be the T-circle with radius 1. Then for n  >0, define C(n) to be the largest T-circle that intersects C(n-1). C(n) has radius a(n) and the coordinates of its points of intersection with C(n-1) are A001108(n) and A055997(n). Cf. A001109. - Charlie Marion, Sep 14 2005
Number of 01-avoiding words of length n on alphabet {0,1,2,3,4,5} which do not end in 0. - Tanya Khovanova, Jan 10 2007
The lower principal convergents to 2^(1/2), beginning with 1/1, 7/5, 41/29, 239/169, comprise a strictly increasing sequence; numerators = A002315 and denominators = {a(n)}. - Clark Kimberling, Aug 26 2008
Apparently Ljunggren shows that 169 is the last square term.
If (p,q) is a solution of the Diophantine equation: X^2 + (X+1)^2 = Y^2 then (p+q) or (p+q+1) are perfect squares. If (p,q) is a solution of the Diophantine equation: X^2 + (X+1)^2 = Y^2 then (p+q) or (p+q)/8 are perfect squares. If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation: X^2 + (X+1)^2 = Y^2 with p < r then s-r = p+q+1. - Mohamed Bouhamida, Aug 29 2009
If (p,q) and (r,s) are two consecutive solutions of the Diophantine equation: X^2 + (X + 1)^2 = Y^2 with p < r then r = 3p+2q+1 and s = 4p+3q+2. - Mohamed Bouhamida, Sep 02 2009
Equals INVERT transform of A005054: (1, 4, 20, 100, 500, 2500, ...) and INVERTi transform of A122074: (1, 6, 40, 268, 1796, ...). - Gary W. Adamson, Jul 22 2010
a(n) is the number of compositions of n when there are 5 types of 1 and 4 types of other natural numbers. - Milan Janjic, Aug 13 2010
The remainder after division of a(n) by a(k) appears to belong to a periodic sequence: 1, 5, ..., a(k-1), 0, a(k)-a(k-1), ..., a(k)-1, a(k)-1, ..., a(k)-a(k-1), 0, a(k-1), ..., 5, 1. See Bouhamida's Sep 01 2009 comment. - Charlie Marion, May 02 2011
Apart from initial 1: subsequence of A198389, see also A198385. - Reinhard Zumkeller, Oct 25 2011
(a(n+1), 2*b(n+1)) and (a(n+2), 2*b(n+1)), n >= 0, with b(n):= A001109(n), give the (u(2*n), v(2*n)) and (u(2*n+1), v(2*n+1)) sequences, respectively, for Pythagorean triples (x,y,z), where x=|u^2-v^2|, y=2*u*v and z=u^2+v^2, with u odd and v even, which are generated from (u(0)=1, v(0)=2) by the substitution rule (u,v) -> (2*v+u,v) if u < v and (u,v) -> (u,2*u+v) if u > v. This leads to primitive triples because gcd(u,v) = 1 is respected. This corresponds to (primitive) Pythagorean triangles with |x-y|=1 (the catheti differ by one length unit). This (u,v) sequence starts with (1,2), (5,2), (5,12), (29,12), (29,70) ... - Wolfdieter Lang, Mar 06 2012
Area of the Fibonacci snowflake of order n. - José Luis Ramírez Ramírez, Dec 13 2012
Area of the 3-generalized Fibonacci snowflake of order n, n >= 3. - José Luis Ramírez Ramírez, Dec 13 2012
For the o.g.f. given by Johannes W. Meijer, Aug 01 2010, in the formula section see a comment under A077445. - Wolfdieter Lang, Jan 18 2013
Positive values of x (or y) satisfying x^2 - 6xy + y^2 + 4 = 0. - Colin Barker, Feb 04 2014
Length of period of the continued fraction expansion of a(n)*sqrt(2) is 1, the corresponding repeating value is A077444(n). - Ralf Stephan, Feb 20 2014
Positive values of x (or y) satisfying x^2 - 34xy + y^2 + 144 = 0. - Colin Barker, Mar 04 2014
The value of the hypotenuse in each triple of the Tree of primitive Pythagorean triples (cf. Wikipedia link) starting with root (3,4,5) and recursively selecting the central branch at each triple node of the tree. - Stuart E Anderson, Feb 05 2015
Positive integers z such that z^2 is a centered square number (A001844). - Colin Barker, Feb 12 2015
The aerated sequence (b(n)) n >= 1 = [1, 0, 5, 0, 29, 0, 169, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -8, Q = 1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047 for the connection with Chebyshev polynomials. - Peter Bala, Mar 25 2015
A002315(n-1)/a(n) is the closest rational approximation of sqrt(2) with a denominator not larger than a(n). These rational approximations together with those obtained from the sequences A001541 and A001542 give a complete set of closest rational approximations of sqrt(2) with restricted numerator or denominator. A002315(n-1)/a(n) < sqrt(2). - A.H.M. Smeets, May 28 2017
Equivalently, numbers x such that (x-1)*x/2 + x*(x+1)/2 = y^2 + (y+1)^2. y-values are listed in A001652. Example: for x=29 and y=20, 28*29/2 + 29*30/2 = 20^2 + 21^2. - Bruno Berselli, Mar 19 2018
From Wolfdieter Lang, Jun 13 2018: (Start)
(a(n), a(n+1)), with a(0):= 1, give all proper positive solutions m1 = m1(n) and m2 = m2(n), with m1 < m2 and n >= 0, of the Markoff triple (m, m1, m2) (see A002559) for m = 2, i.e., m1^2 - 6*m1*m2 + m2^2 = -4. Hence the unique Markoff triple with largest value m = 2 is (1, 1, 2) (for general m from A002559 this is the famous uniqueness conjecture).
For X = m2 - m1 and Y = m2 this becomes the reduced indefinite quadratic form representation X^2 + 4*X*Y - 4*Y^2 = -4, with discriminant 32, and the only proper fundamental solution (X(0), Y(0)) = (0, 1). For all nonnegative proper (X(n), Y(n)) solutions see (A005319(n) = a(n+1) - a(n), a(n+1)), for n >= 0. (End)
Each Pell(2*k+1) = a(k+1) number with k >= 3 appears as largest number of an ordered Markoff (Markov) triple [x, y, m] with smallest value x = 2 as [2, Pell(2*k-1), Pell(2*k+1)]. This known result follows also from all positive proper solutions of the Pell equation q^2 - 2*m^2 = -1 which are q = q(k) = A002315(k)  and m = m(k) = Pell(2*k+1), for k >= 0. y = y(k) = m(k) - 2*q(k) = Pell(2*k-1), with Pell(-1) = 1. The k = 0 and 1 cases do not satisfy x=2 <= y(k) <= m(k). The numbers 1 and 5 appear also as largest Markoff triple members because they are also Fibonacci numbers, and for these triples x=1. - Wolfdieter Lang, Jul 11 2018
All of the positive integer solutions of a*b+1=x^2, a*c+1=y^2, b*c+1=z^2, x+z=2*y, 0 < a < b < c are given by a=A001542(n), b=A005319(n), c=A001542(n+1), x=A001541(n), y=a(n+1), z=A002315(n) with 0 < n. - Michael Somos, Jun 26 2022

			From _Muniru A Asiru_, Mar 19 2018: (Start)
For k=1, 2*1^2 - 1 = 2 - 1 = 1 = 1^2.
For k=5, 2*5^2 - 1 = 50 - 1 = 49 = 7^2.
For k=29, 2*29^2 - 1 = 1682 - 1 = 1681 = 41^2.
... (End)
G.f. = x + 5*x^2 + 29*x^3 + 169*x^4 + 985*x^5 + 5741*x^6 + ... - _Michael Somos_, Jun 26 2022

A. H. Beiler, Recreations in the Theory of Numbers. New York: Dover, pp. 122-125, 1964.
John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 188.
W. Ljunggren, "Zur Theorie der Gleichung x^2+1=Dy^4", Avh. Norske Vid. Akad. Oslo I. 5, 27pp.
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).
P.-F. Teilhet, Query 2376, L'Intermédiaire des Mathématiciens, 11 (1904), 138-139. - N. J. A. Sloane, Mar 08 2022
David Wells, The Penguin Dictionary of Curious and Interesting Numbers (Rev. ed. 1997), p. 91.

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S. Barbero, U. Cerruti, and N. Murru, A Generalization of the Binomial Interpolated Operator and its Action on Linear Recurrent Sequences, J. Int. Seq. 13 (2010) # 10.9.7, proposition 16.
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Index entries for two-way infinite sequences
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (6,-1).

Other two sides are A001652, A046090.
Cf. A001519, A001109, A005054, A122074, A056220, A056869 (subset of primes).
Cf. A000217, A000290, A002315, A002559, A005319, A069894.
Row 6 of array A094954.
Row 1 of array A188647.
Cf. similar sequences listed in A238379.

a:=[1,5];; for n in [3..25] do a[n]:=6*a[n-1]-a[n-2]; od; a; # Muniru A Asiru, Mar 19 2018

a001653 n = a001653_list !! n
a001653_list = 1 : 5 : zipWith (-) (map (* 6) $ tail a001653_list) a001653_list
-- Reinhard Zumkeller, May 07 2013

I:=[1,5]; [n le 2 select I[n] else 6*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Feb 22 2014

a[0]:=1: a[1]:=5: for n from 2 to 26 do a[n]:=6*a[n-1]-a[n-2] od: seq(a[n], n=0..20); # Zerinvary Lajos, Jul 26 2006
A001653:=-(-1+5*z)/(z**2-6*z+1); # Conjectured (correctly) by Simon Plouffe in his 1992 dissertation; gives sequence except for one of the leading 1's

LinearRecurrence[{6,-1}, {1,5}, 40] (* Harvey P. Dale, Jul 12 2011 *)
a[ n_] := -(-1)^n ChebyshevU[2 n - 2, I]; (* Michael Somos, Jul 22 2018 *)
Numerator[{1} ~Join~
Table[FromContinuedFraction[Flatten[Table[{1, 4}, n]]], {n, 1, 40}]]; (* Greg Dresden, Sep 10 2019 *)

{a(n) = subst(poltchebi(n-1) + poltchebi(n), x, 3)/4}; /* Michael Somos, Nov 02 2002 */

a(n)=([5,2;2,1]^(n-1))[1,1] \\ Lambert Klasen (lambert.klasen(AT)gmx.de), corrected by Eric Chen, Jun 14 2018

{a(n) = -(-1)^n * polchebyshev(2*n-2, 2, I)}; /* Michael Somos, Jun 26 2022 */

G.f.: x*(1-x)/(1-6*x+x^2).
a(n) = 6*a(n-1) - a(n-2) with a(1)=1, a(2)=5.
4*a(n) = A077445(n).
Can be extended backwards by a(-n+1) = a(n).
a(n) = sqrt((A002315(n)^2 + 1)/2). [Inserted by N. J. A. Sloane, May 08 2000]
a(n+1) = S(n, 6)-S(n-1, 6), n>=0, with S(n, 6) = A001109(n+1), S(-2, 6) := -1. S(n, x)=U(n, x/2) are Chebyshev's polynomials of the second kind. Cf. triangle A049310. a(n+1) = T(2*n+1, sqrt(2))/sqrt(2), n>=0, with T(n, x) Chebyshev's polynomials of the first kind. [Offset corrected by Wolfdieter Lang, Mar 06 2012]
a(n) = A000129(2n+1). - Ira M. Gessel, Sep 27 2002
a(n) ~ (1/4)*sqrt(2)*(sqrt(2) + 1)^(2*n+1). - Joe Keane (jgk(AT)jgk.org), May 15 2002
a(n) = (((3 + 2*sqrt(2))^(n+1) - (3 - 2*sqrt(2))^(n+1)) - ((3 + 2*sqrt(2))^n - (3 - 2*sqrt(2))^n)) / (4*sqrt(2)). Limit_{n->infinity} a(n)/a(n-1) = 3 + 2*sqrt(2). - Gregory V. Richardson, Oct 12 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i); then q(n, 4) = a(n). - Benoit Cloitre, Nov 10 2002
For n and j >= 1, Sum_{k=0..j} a(k)*a(n) - Sum_{k=0..j-1} a(k)*a(n-1) = A001109(j+1)*a(n) - A001109(j)*a(n-1) = a(n+j); e.g., (1+5+29)*5 - (1+5)*1=169. - Charlie Marion, Jul 07 2003
From Charlie Marion, Jul 16 2003: (Start)
For n >= k >= 0, a(n)^2 = a(n+k)*a(n-k) - A084703(k)^2; e.g., 169^2 = 5741*5 - 144.
For n > 0, a(n) ^2 - a(n-1)^2 = 4*Sum_{k=0..2*n-1} a(k) = 4*A001109(2n); e.g., 985^2 - 169^2 = 4*(1 + 5 + 29 + ... + 195025) = 4*235416.
Sum_{k=0..n} ((-1)^(n-k)*a(k)) = A079291(n+1); e.g., -1 + 5 - 29 + 169 = 144.
A001652(n) + A046090(n) - a(n) = A001542(n); e.g., 119 + 120 - 169 = 70.
(End)
Sum_{k=0...n} ((2k+1)*a(n-k)) = A001333(n+1)^2 - (1 + (-1)^(n+1))/2; e.g., 1*169 + 3*29 + 5*5 + 7*1 = 288 = 17^2 - 1; 1*29 + 3*5 + 5*1 = 49 = 7^2. - Charlie Marion, Jul 18 2003
Sum_{k=0...n} a(k)*a(n) = Sum_{k=0..n} a(2k) and Sum_{k=0..n} a(k)*a(n+1) = Sum_{k=0..n} a(2k+1); e.g., (1+5+29)*29 = 1+29+985 and (1+5+29)*169 = 5+169+5741. - Charlie Marion, Sep 22 2003
For n >= 3, a_{n} = 7(a_{n-1} - a_{n-2}) + a_{n-3}, with a_1 = 1, a_2 = 5 and a_3 = 29. a(n) = ((-1+2^(1/2))/2^(3/2))*(3 - 2^(3/2))^n + ((1+2^(1/2))/2^(3/2))*(3 + 2^(3/2))^n. - Antonio Alberto Olivares, Oct 13 2003
Let a(n) = A001652(n), b(n) = A046090(n) and c(n) = this sequence. Then for k > j, c(i)*(c(k) - c(j)) = a(k+i) + ... + a(i+j+1) + a(k-i-1) + ... + a(j-i) + k - j. For n < 0, a(n) = -b(-n-1). Also a(n)*a(n+2k+1) + b(n)*b(n+2k+1) + c(n)*c(n+2k+1) = (a(n+k+1) - a(n+k))^2; a(n)*a(n+2k) + b(n)*b(n+2k) + c(n)*c(n+2k) = 2*c(n+k)^2. - Charlie Marion, Jul 01 2003
Let a(n) = A001652(n), b(n) = A046090(n) and c(n) = this sequence. Then for n > 0, a(n)*b(n)*c(n)/(a(n)+b(n)+c(n)) = Sum_{k=0..n} c(2*k+1); e.g., 20*21*29/(20+21+29) = 5+169 = 174; a(n)*b(n)*c(n)/(a(n-1)+b(n-1)+c(n-1)) = Sum_{k=0..n} c(2*k); e.g., 119*120*169/(20+21+29) = 1+29+985+33461 = 34476. - Charlie Marion, Dec 01 2003
Also solutions x > 0 of the equation floor(x*r*floor(x/r))==floor(x/r*floor(x*r)) where r=1+sqrt(2). - Benoit Cloitre, Feb 15 2004
a(n)*a(n+3) = 24 + a(n+1)*a(n+2). - Ralf Stephan, May 29 2004
For n >= k, a(n)*a(n+2*k+1) - a(n+k)*a(n+k+1) = a(k)^2-1; e.g., 29*195025-985*5741 = 840 = 29^2-1; 1*169-5*29 = 24 = 5^2-1; a(n)*a(n+2*k)-a(n+k)^2 = A001542(k)^2; e.g., 169*195025-5741^2 = 144 = 12^2; 1*29-5^2 = 4 = 2^2. - Charlie Marion Jun 02 2004
For all k, a(n) is a factor of a((2n+1)*k+n). a((2*n+1)*k+n) = a(n)*(Sum_{j=0..k-1} (-1)^j*(a((2*n+1)*(k-j)) + a((2*n+1)*(k-j)-1))+(-1)^k); e.g., 195025 = 5*(33461+5741-169-29+1); 7645370045 = 169*(6625109+1136689-1).- Charlie Marion, Jun 04 2004
a(n) = Sum_{k=0..n} binomial(n+k, 2*k)4^k. - Paul Barry, Aug 30 2004 [offset 0]
a(n) = Sum_{k=0..n} binomial(2*n+1, 2*k+1)*2^k. - Paul Barry, Sep 30 2004 [offset 0]
For n < k, a(n)*A001541(k) = A011900(n+k)+A053141(k-n-1); e.g., 5*99 = 495 = 493+2. For n >= k, a(n)*A001541(k) = A011900(n+k)+A053141(n-k); e.g., 29*3 = 87 = 85+2. - Charlie Marion, Oct 18 2004
a(n) = (-1)^n*U(2*n, i*sqrt(4)/2) = (-1)^n*U(2*n, i), U(n, x) Chebyshev polynomial of second kind, i=sqrt(-1). - Paul Barry, Mar 13 2005 [offset 0]
a(n) = Pell(2*n+1) = Pell(n)^2 + Pell(n+1)^2. - Paul Barry, Jul 18 2005 [offset 0]
a(n)*a(n+k) = A000129(k)^2 + A000129(2n+k+1)^2; e.g., 29*5741 = 12^2+169^2. - Charlie Marion, Aug 02 2005
Let a(n)*a(n+k) = x. Then 2*x^2-A001541(k)*x+A001109(k)^2 = A001109(2*n+k+1)^2; e.g., let x=29*985; then 2x^2-17x+6^2 = 40391^2; cf. A076218. - Charlie Marion, Aug 02 2005
With a=3+2*sqrt(2), b=3-2*sqrt(2): a(n) = (a^((2n+1)/2)+b^((2n+1)/2))/(2*sqrt(2)). a(n) = A001109(n+1)-A001109(n). - Mario Catalani (mario.catalani(AT)unito.it), Mar 31 2003
If k is in the sequence, then the next term is floor(k*(3+2*sqrt(2))). - Lekraj Beedassy, Jul 19 2005
a(n) = Jacobi_P(n,-1/2,1/2,3)/Jacobi_P(n,-1/2,1/2,1). - Paul Barry, Feb 03 2006 [offset 0]
a(n) = Sum_{k=0..n} Sum_{j=0..n-k} C(n,j)*C(n-j,k)*Pell(n-j+1), where Pell = A000129. - Paul Barry, May 19 2006 [offset 0]
a(n) = round(sqrt(A002315(n)^2/2)). - Lekraj Beedassy, Jul 15 2006
a(n) = A079291(n) + A079291(n+1). - Lekraj Beedassy, Aug 14 2006
a(n+1) = 3*a(n) + sqrt(8*a(n)^2-4), a(1)=1. - Richard Choulet, Sep 18 2007
6*a(n)*a(n+1) = a(n)^2+a(n+1)^2+4; e.g., 6*5*29 = 29^2+5^2+4; 6*169*985 = 169^2+985^2+4. - Charlie Marion, Oct 07 2007
2*A001541(k)*a(n)*a(n+k) = a(n)^2+a(n+k)^2+A001542(k)^2; e.g., 2*3*5*29 = 5^2+29^2+2^2; 2*99*29*5741 = 2*99*29*5741=29^2+5741^2+70^2. - Charlie Marion, Oct 12 2007
[a(n), A001109(n)] = [1,4; 1,5]^n * [1,0]. - Gary W. Adamson, Mar 21 2008
From Charlie Marion, Apr 10 2009: (Start)
In general, for n >= k, a(n+k) = 2*A001541(k)*a(n)-a(n-k);
e.g., a(n+0) = 2*1*a(n)-a(n); a(n+1) = 6*a(n)-a(n-1); a(6+0) = 33461 = 2*33461-33461; a(5+1) = 33461 = 6*5741-985; a(4+2) = 33461 = 34*985-29; a(3+3) = 33461 = 198*169-1.
(End)
G.f.: sqrt(x)*tan(4*arctan(sqrt(x)))/4. - Johannes W. Meijer, Aug 01 2010
Given k = (sqrt(2)+1)^2 = 3+2*sqrt(2) and a(0)=1, then a(n) = a(n-1)*k-((k-1)/(k^n)). - Charles L. Hohn, Mar 06 2011
Given k = (sqrt(2)+1)^2 = 3+2*sqrt(2) and a(0)=1, then a(n) = (k^n)+(k^(-n))-a(n-1) = A003499(n) - a(n-1). - Charles L. Hohn, Apr 04 2011
Let T(n) be the n-th triangular number; then, for n > 0, T(a(n)) + A001109(n-1) = A046090(n)^2.  See also A046090. - Charlie Marion, Apr 25 2011
For k > 0, a(n+2*k-1) - a(n) = 4*A001109(n+k-1)*A002315(k-1); a(n+2*k) - a(n) = 4*A001109(k)*A002315(n+k-1). - Charlie Marion, Jan 06 2012
a(k+j+1) = (A001541(k)*A001541(j) + A002315(k)*A002315(j))/2. - Charlie Marion, Jun 25 2012
a(n)^2 = 2*A182435(n)*(A182435(n)-1)+1. - Bruno Berselli, Oct 23 2012
a(n) = A143608(n-1)*A143608(n) + 1 = A182190(n-1)+1. - Charlie Marion, Dec 11 2012
G.f.: G(0)*(1-x)/(2-6*x), where G(k) = 1 + 1/(1 - x*(8*k-9)/( x*(8*k-1) - 3/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 12 2013
a(n+1) = 4*A001652(n) + 3*a(n) + 2 [Mohamed Bouhamida's 2009 (p,q)(r,s) comment above rewritten]. - Hermann Stamm-Wilbrandt, Jul 27 2014
a(n)^2 = A001652(n-1)^2 + (A001652(n-1)+1)^2. - Hermann Stamm-Wilbrandt, Aug 31 2014
Sum_{n >= 2} 1/( a(n) - 1/a(n) ) = 1/4. - Peter Bala, Mar 25 2015
a(n) = Sum_{k=0..n} binomial(n,k) * 3^(n-k) * 2^k * 2^floor(k/2). - David Pasino, Jul 09 2016
E.g.f.: (sqrt(2)*sinh(2*sqrt(2)*x) + 2*cosh(2*sqrt(2)*x))*exp(3*x)/2. - Ilya Gutkovskiy, Jul 09 2016
a(n+2) = (a(n+1)^2 + 4)/a(n). - Vladimir M. Zarubin, Sep 06 2016
a(n) = 2*A053141(n)+1. - R. J. Mathar, Aug 16 2019
For n>1, a(n) is the numerator of the continued fraction [1,4,1,4,...,1,4] with (n-1) repetitions of 1,4. For the denominators see A005319. - Greg Dresden, Sep 10 2019
a(n) = round(((2+sqrt(2))*(3+2*sqrt(2))^(n-1))/4). - Paul Weisenhorn, May 23 2020
a(n+1) = Sum_{k >= n} binomial(2*k,2*n)*(1/2)^(k+1). Cf. A102591. - Peter Bala, Nov 29 2021
a(n+1) = 3*a(n) + A077444(n). - César Aguilera, Jul 13 2023

Additional comments from Wolfdieter Lang, Feb 10 2000
Better description from Harvey P. Dale, Jan 15 2002
Edited by N. J. A. Sloane, Nov 02 2002

A033887 a(n) = Fibonacci(3*n + 1).

Original entry on oeis.org

1, 3, 13, 55, 233, 987, 4181, 17711, 75025, 317811, 1346269, 5702887, 24157817, 102334155, 433494437, 1836311903, 7778742049, 32951280099, 139583862445, 591286729879, 2504730781961, 10610209857723, 44945570212853, 190392490709135, 806515533049393, 3416454622906707

Offset: 0

N. J. A. Sloane

Binomial transform of A063727, and second binomial transform of (1,1,5,5,25,25,...), which is A074872 with offset 0. - Paul Barry, Jul 16 2003
Equals INVERT transform of A104934: (1, 2, 8, 28, 100, 356, ...) and INVERTi transform of A005054: (1, 4, 20, 100, 500, ...). - Gary W. Adamson, Jul 22 2010
a(n) is the number of compositions of n when there are 3 types of 1 and 4 types of other natural numbers. - Milan Janjic, Aug 13 2010
F(3*n+1) = 3^n*a(n;2/3), where a(n;d), n = 0, 1, ..., d, denote the delta-Fibonacci numbers defined in comments to A000045 (see also the papers by Witula et al.). - Roman Witula, Jul 12 2012
We note that the remark above by Paul Barry can be easily obtained from the following scaling identity for delta-Fibonacci numbers y^n a(n;x/y) = Sum_{k=0..n} binomial(n,k) (y-1)^(n-k) a(k;x) and the fact that a(n;2)=5^floor(n/2). Indeed, for x=y=2 we get 2^n a(n;1) = Sum_{k=0..n} binomial(n,k) a(k;2) and, by A000045: Sum_{k=0..n} binomial(n,k) 2^k a(k;1) = Sum_{k=0..n} binomial(n,k) F(k+1) 2^k = 3^n a(n;2/3) = F(3n+1). - Roman Witula, Jul 12 2012
Except for the first term, this sequence can be generated by Corollary 1 (iv) of Azarian's paper in the references for this sequence. - Mohammad K. Azarian, Jul 02 2015
Number of 1’s in the substitution system {0 -> 110, 1 -> 11100} at step n from initial string "1" (1 -> 11100 -> 111001110011100110110 -> ...). - Ilya Gutkovskiy, Apr 10 2017
The o.g.f. of {F(m*n + 1)}A000045%20and%20L%20=%20A000032.%20-%20_Wolfdieter%20Lang">{n>=0}, for m = 1, 2, ..., is G(m,x) = (1 - F(m-1)*x) / (1 - L(m)*x + (-1)^m*x^2), with F = A000045 and L = A000032. - _Wolfdieter Lang, Feb 06 2023

			a(5) = Fibonacci(3*5 + 1) = Fibonacci(16) = 987. - _Indranil Ghosh_, Feb 04 2017

Indranil Ghosh, Table of n, a(n) for n = 0..1592
Mohammad K. Azarian, Fibonacci Identities as Binomial Sums, International Journal of Contemporary Mathematical Sciences, 7(38) (2012), 1871-1876.
Paul Barry and A. Hennessy, The Euler-Seidel Matrix, Hankel Matrices and Moment Sequences, J. Int. Seq. 13 (2010), #10.8.2, Example 13.
Gary Detlefs and Wolfdieter Lang, Improved Formula for the Multi-Section of the Linear Three-Term Recurrence Sequence, arXiv:2304.12937 [math.CO], 2023.
I. M. Gessel and Ji Li, Compositions and Fibonacci identities, J. Int. Seq. 16 (2013), #13.4.5.
Edyta Hetmaniok, Bożena Piątek, and Roman Wituła, Binomials Transformation Formulae of Scaled Fibonacci Numbers, Open Mathematics, 15(1) (2017), 477-485.
Tanya Khovanova, Recursive Sequences.
Roman Witula, Binomials transformation formulae of scaled Lucas numbers, Demonstratio Mathematica, 46(1) (2013), 15-27.
Roman Witula and Damian Slota, delta-Fibonacci numbers, Appl. Anal. Discr. Math 3 (2009), 310-329, MR2555042.
Index entries for linear recurrences with constant coefficients, signature (4,1).

Cf. A000032, A000045, A104934, A005054, A063727 (inverse binomial transform), A082761 (binomial transform), A001076, A001077.

Cf. A016777, A049310, A049651, A074872, A122070, A167808, A185384.

[Fibonacci(3*n+1): n in [0..100]]; // Vincenzo Librandi, Apr 17 2011

Fibonacci[Range[1,5!,3]] (* Vladimir Joseph Stephan Orlovsky, May 18 2010 *)

a(n)=fibonacci(3*n+1) \\ Charles R Greathouse IV, Feb 03 2014

Vec((1-x)/(1-4*x-x^2) + O(x^100)) \\ Altug Alkan, Dec 10 2015

a(n) = A001076(n) + A001077(n) = A001076(n+1) - A001076(n).
a(n) = 2*A049651(n) + 1.
a(n) = 4*a(n-1) + a(n-2) for n>1, a(0)=1, a(1)=3;
G.f.: (1 - x)/(1 - 4*x - x^2).
a(n) = ((1 + sqrt(5))*(2 + sqrt(5))^n - (1 - sqrt(5))*(2 - sqrt(5))^n)/(2*sqrt(5)).
a(n) = Sum_{k=0..n} Sum_{j=0..n-k} C(n,j)*C(n-j,k)*F(n-j+1). - Paul Barry, May 19 2006
First differences of A001076. - Al Hakanson (hawkuu(AT)gmail.com), May 02 2009
a(n) = A167808(3*n+1). - Reinhard Zumkeller, Nov 12 2009
a(n) = Sum_{k=0..n} C(n,k)*F(n+k+1). - Paul Barry, Apr 19 2010
Let p[1]=3, p[i]=4, (i>1), and A be a Hessenberg matrix of order n defined by: A[i,j] = p[j-i+1] (i <= j), A[i,j]=-1 (i = j+1), and A[i,j] = 0 otherwise. Then, for n >= 1, a(n) = det A. - Milan Janjic, Apr 29 2010
a(n) = Sum_{i=0..n} C(n,n-i)*A063727(i). - Seiichi Kirikami, Mar 06 2012
a(n) = Sum_{k=0..n} A122070(n,k) = Sum_{k=0..n} A185384(n,k). - Philippe Deléham, Mar 13 2012
a(n) = A000045(A016777(n)). - Michel Marcus, Dec 10 2015
a(n) = F(2*n)*L(n+1) + F(n-1)*(-1)^n for n > 0. - J. M. Bergot, Feb 09 2016
a(n) = Sum_{k=0..n} binomial(n,k)*5^floor(k/2)*2^(n-k). - Tony Foster III, Sep 03 2017
2*a(n) = Fibonacci(3*n) + Lucas(3*n). - Bruno Berselli, Oct 13 2017
a(n)^2 is the denominator of continued fraction [4,...,4, 2, 4,...,4], which has n 4's before, and n 4's after, the middle 2. - Greg Dresden and Hexuan Wang, Aug 30 2021
a(n) = i^n*(S(n, -4*i) + i*S(n-1, -4*i)), with i = sqrt(-1), and the Chebyshev S-polynomials (see A049310) with S(n, -1) = 0. From the simplified trisection formula. See the first entry above with A001076. - Gary Detlefs and Wolfdieter Lang, Mar 06 2023
E.g.f.: exp(2*x)*(5*cosh(sqrt(5)*x) + sqrt(5)*sinh(sqrt(5)*x))/5. - Stefano Spezia, May 24 2024

A167374 Triangle, read by rows, given by [ -1,1,0,0,0,0,0,0,0,...] DELTA [1,0,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, -1, 1, 0, -1, 1, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -1, 1

Offset: 0

Philippe Deléham, Nov 02 2009

Riordan array (1-x,1) read by rows; Riordan inverse is (1/(1-x),1). Columns have g.f. (1-x)x^k. Diagonal sums are A033999. Unsigned version in A097806.
Table T(n,k) read by antidiagonals. T(n,1) = 1, T(n,2) = -1, T(n,k) = 0, k > 2. - Boris Putievskiy, Jan 17 2013
Finite difference operator (pair difference): left multiplication by T of a sequence arranged as a column vector gives a running forward difference, a(k+1)-a(k), or first finite difference (modulo sign), of the elements of the sequence. T^n gives the n-th finite difference (mod sign). T is the inverse of the summation matrix A000012 (regarded as lower triangular matrices). - Tom Copeland, Mar 26 2014

			Triangle begins:
   1;
  -1,  1;
   0, -1,  1;
   0,  0, -1,  1;
   0,  0,  0, -1,  1;
   0,  0,  0,  0, -1,  1; ...
Row number r (r>4) contains (r-2) times '0', then '-1' and '1'.
From _Boris Putievskiy_, Jan 17 2013: (Start)
The start of the sequence as a table:
  1  -1  0  0  0  0  0 ...
  1  -1  0  0  0  0  0 ...
  1  -1  0  0  0  0  0 ...
  1  -1  0  0  0  0  0 ...
  1  -1  0  0  0  0  0 ...
  1  -1  0  0  0  0  0 ...
  1  -1  0  0  0  0  0 ...
  ...
(End)

Boris Putievskiy, Transformations [of] Integer Sequences And Pairing Functions arXiv:1212.2732 [math.CO], 2012.

Cf. A000012, A007318, A029653, A097805, A118800, A130595, A156644.

A167374 := proc(n,k)
    if k> n or k < n-1 then
        0;
    elif k = n then
        1;
    else
        -1 ;
    end if;
end proc: # R. J. Mathar, Sep 07 2016

Table[PadLeft[{-1, 1}, n], {n, 13}] // Flatten (* or *)
MapIndexed[Take[#1, First@ #2] &, CoefficientList[Series[(1 - x)/(1 - x y), {x, 0, 12}], {x, y}]] // Flatten (* Michael De Vlieger, Nov 16 2016 *)
T[n_, k_] := If[ k<0 || k>n, 0, Boole[n==k] - Boole[n==k+1]]; (* Michael Somos, Oct 01 2022 *)

{T(n, k) = if( k<0 || k>n, 0, (n==k) - (n==k+1))}; /* Michael Somos, Oct 01 2022 */

Sum_{k, 0<=k<=n} T(n,k)*x^k = A000007(n), A011782(n), A025192(n), A002001(n), A005054(n), A052934(n), A055272(n), A055274(n), A055275(n), A055268(n), A055276(n) for x = 1,2,3,4,5,6,7,8,9,10,11 respectively .
From Boris Putievskiy, Jan 17 2013: (Start)
a(n) = floor((A002260(n)+2)/(A003056(n)+2))*(-1)^(A002260(n)+A003056(n)+1),  n>0.
a(n) = floor((i+2)/(t+2))*(-1)^(i+t+1), n > 0, where
i = n - t*(t+1)/2,
t = floor((-1 + sqrt(8*n-7))/2). (End)
T*A000012 = Identity matrix. T*A007318 = A097805. T*(A007318)^(-1)= signed A029653. - Tom Copeland, Mar 26 2014
G.f.: (1-x)/(1-x*y). - R. J. Mathar, Aug 11 2015
T = A130595*A156644 = M*T^(-1)*M = M*A000012*M, where M(n,k) = (-1)^n A130595(n,k). Note that M = M^(-1). Cf. A118800 and A097805. - Tom Copeland, Nov 15 2016

A200139 Triangle T(n,k), read by rows, given by (1,1,0,0,0,0,0,0,0,...) DELTA (1,0,0,0,0,0,0,0,0,...) where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 4, 8, 5, 1, 8, 20, 18, 7, 1, 16, 48, 56, 32, 9, 1, 32, 112, 160, 120, 50, 11, 1, 64, 256, 432, 400, 220, 72, 13, 1, 128, 576, 1120, 1232, 840, 364, 98, 15, 1, 256, 1280, 2816, 3584, 2912, 1568, 560, 128, 17, 1, 512, 2816, 6912, 9984, 9408, 6048, 2688, 816, 162, 19, 1

Offset: 0

Philippe Deléham, Nov 13 2011

Riordan array ((1-x)/(1-2x),x/(1-2x)).
Product A097805*A007318 as infinite lower triangular arrays.
Product A193723*A130595 as infinite lower triangular arrays.
T(n,k) is the number of ways to place n unlabeled objects into any number of labeled bins (with at least one object in each bin) and then designate k of the bins. - Geoffrey Critzer, Nov 18 2012
Apparently, rows of this array are unsigned diagonals of A028297. - Tom Copeland, Oct 11 2014
Unsigned A118800, so my conjecture above is true. - Tom Copeland, Nov 14 2016

			Triangle begins:
   1
   1,   1
   2,   3,   1
   4,   8,   5,   1
   8,  20,  18,   7,   1
  16,  48,  56,  32,   9,   1
  32, 112, 160, 120,  50,  11,   1

Cf. A118800 (signed version), A081277, A039991, A001333 (antidiagonal sums), A025192 (row sums); diagonals: A000012, A005408, A001105, A002492, A072819l; columns: A011782, A001792, A001793, A001794, A006974, A006975, A006976.

Cf. A007318, A028297, A059260, A097805, A118801, A130595.

nn=15;f[list_]:=Select[list,#>0&];Map[f,CoefficientList[Series[(1-x)/(1-2x-y x) ,{x,0,nn}],{x,y}]]//Grid  (* Geoffrey Critzer, Nov 18 2012 *)

T(n,k) = 2*T(n-1,k)+T(n-1,k-1) with T(0,0)=T(1,0)=T(1,1)=1 and T(n,k)=0 for k<0 or for n

T(n,k) = A011782(n-k)*A135226(n,k) = 2^(n-k)*(binomial(n,k)+binomial(n-1,k-1))/2.

Sum_{k, 0<=k<=n} T(n,k)*x^k = A000007(n), A011782(n), A025192(n), A002001(n), A005054(n), A052934(n), A055272(n), A055274(n), A055275(n), A052268(n), A055276(n), A196731(n) for n=-1,0,1,2,3,4,5,6,7,8,9,10 respectively.

G.f.: (1-x)/(1-(2+y)*x).

T(n,k) = Sum_j>=0 T(n-1-j,k-1)*2^j.

T = A007318*A059260, so the row polynomials of this entry are given umbrally by p_n(x) = (1 + q.(x))^n, where q_n(x) are the row polynomials of A059260 and (q.(x))^k = q_k(x). Consequently, the e.g.f. is exp[tp.(x)] = exp[t(1+q.(x))] = e^t exp(tq.(x)) = [1 + (x+1)e^((x+2)t)]/(x+2), and p_n(x) = (x+1)(x+2)^(n-1) for n > 0. - Tom Copeland, Nov 15 2016

T^(-1) = A130595*(padded A130595), differently signed A118801. Cf. A097805. - Tom Copeland, Nov 17 2016

The n-th row polynomial in descending powers of x is the n-th Taylor polynomial of the rational function (1 + x)/(1 + 2*x) * (1 + 2*x)^n about 0. For example, for n = 4, (1 + x)/(1 + 2*x) * (1 + 2*x)^4 = (8*x^4 + 20*x*3 + 18*x^2 + 7*x + 1) + O(x^5). - Peter Bala, Feb 24 2018

A055842 Expansion of (1-x)^2/(1-5*x).

Original entry on oeis.org

1, 3, 16, 80, 400, 2000, 10000, 50000, 250000, 1250000, 6250000, 31250000, 156250000, 781250000, 3906250000, 19531250000, 97656250000, 488281250000, 2441406250000, 12207031250000, 61035156250000, 305175781250000, 1525878906250000, 7629394531250000

Offset: 0

Barry E. Williams, May 30 2000

First differences of A005054.
For n>=2, a(n) is equal to the number of functions f:{1,2,...,n}->{1,2,3,4,5} such that for fixed, different x_1, x_2 in {1,2,...,n} and fixed y_1, y_2 in {1,2,3,4,5} we have f(x_1)<>y_1 and f(x_2)<> y_2. - Milan Janjic, Apr 19 2007
a(n) is the number of generalized compositions of n when there are 4 *i-1 different types of i, (i=1,2,...). - Milan Janjic, Aug 26 2010

A. H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, pp. 194-196.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Milan Janjic, Enumerative Formulas for Some Functions on Finite Sets
Index entries for linear recurrences with constant coefficients, signature (5).

Cf. A000351, A005054.

Concatenation([1,3], List([2..30], n-> 16*5^(n-2) )); # G. C. Greubel, Jan 21 2020

[1,3] cat [16*5^(n-2): n in [2..30]]; // G. C. Greubel, Jan 21 2020

R:=PowerSeriesRing(Rationals(), 25); Coefficients(R!( (1-x)^2/(1-5*x))); // Marius A. Burtea, Jan 21 2020

seq( `if`(n<2, 2*n+1, 16*5^(n-2)), n=0..30); # G. C. Greubel, Jan 21 2020

Join[{1,3},16 5^(Range[2,30]-2)] (* Harvey P. Dale, Apr 03 2013 *)

Vec((1-x)^2/(1-5*x) + O(x^30)) \\ Altug Alkan, Mar 13 2016

[1,3]+[16*5^(n-2) for n in (2..30)] # G. C. Greubel, Jan 21 2020

a(n) = 16*5^(n-2), a(0)=1, a(1)=3.
a(n) = 5*a(n-1) + (-1)^n*binomial(2,2-n).
G.f.: (1-x)^2/(1-5*x).
a(n) = Sum_{k=0..n} A201780(n,k)*3^k. - Philippe Deléham, Dec 05 2011
E.g.f.: (9 - 5*x + 16*exp(x))/25. - G. C. Greubel, Jan 21 2020

A060458 Maximum value seen in the final n decimal digits of 2^j for all values of j.

Original entry on oeis.org

8, 96, 992, 9984, 99968, 999936, 9999872, 99999744, 999999488, 9999998976, 99999997952, 999999995904, 9999999991808, 99999999983616, 999999999967232, 9999999999934464, 99999999999868928, 999999999999737856, 9999999999999475712, 99999999999998951424, 999999999999997902848

Offset: 1

Labos Elemer, Apr 09 2001

Consider the final n decimal digits of 2^j for all values of j. They are periodic. Sequence gives maximal value seen in these n digits.

With f(n) = a(n+1) - a(n), the difference f(n) - a(n) is always 8*10^n meaning that a(n) becomes its own "first differences" sequence when each term is prefixed a digit '8'. For higher order differences, the prefix 8 becomes: 8*10^n*Sum_{k=0..m-1} 9^k where m is the order. - R. J. Cano, May 11 2014

			Maximum of the last 4 digits of powers of 2 is 9984=10000-16. It occurs at 2^254. 2^254 = 289480223.....01978282409984 (with 77 digits, last 4 ones are ...9984). The period length of the last-4-digit segment is A005054(4)=500. For n=4 period: amplitude=9984, phase=254.

Index entries for sequences related to final digits of numbers.
Index entries for linear recurrences with constant coefficients, signature (12,-20).

Cf. A000079, A003463, A005054, A060460, A016134.

[10^n-2^n : n in [1..20]]; // Wesley Ivan Hurt, Sep 25 2014

A060458:=n->10^n-2^n: seq(A060458(n), n=1..20); # Wesley Ivan Hurt, Sep 25 2014

RecurrenceTable[{a[n] == 12 a[n - 1] - 20 a[n - 2], a[0] == 0, a[1] == 8}, a[n], {n, 1, 20}]  (* Geoffrey Critzer, Dec 15 2011*)

a(n)=sum(j=0,n-1,2^(3*n-2*j)*binomial(n,j)) \\ R. J. Cano, May 15 2014

A060458(n)=(5^n-1)<M. F. Hasler, Oct 31 2014

[10^n - 2^n for n in range(1,19)] # Zerinvary Lajos, Jun 05 2009

a(n) = 10^n - 2^n = 2^n*(5^n - 1).
From Geoffrey Critzer, Dec 15 2011: (Start)
a(n) = 12*a(n-1) - 20*a(n-2).
O.g.f.: 1/(1-10*x) - 1/(1-2*x). (End)
a(n) = f(n,0) where f(x,y) = Sum_{j=0..x+y-1} (2^(3*x-2*j)*binomial(x,j)). - R. J. Cano, May 15 2014
a(n) = 2^(n+2)*A003463(n). - R. J. Cano, Sep 25 2014
a(n) = 8*A016134(n-1). - R. J. Mathar, Mar 10 2022
E.g.f.: exp(2*x)*(exp(8*x) - 1). - Elmo R. Oliveira, Mar 26 2025

Edited by M. F. Hasler, Oct 31 2014

More terms from Elmo R. Oliveira, Mar 26 2025

A193723 Mirror of the fusion triangle A193722.

Original entry on oeis.org

1, 2, 1, 6, 5, 1, 18, 21, 8, 1, 54, 81, 45, 11, 1, 162, 297, 216, 78, 14, 1, 486, 1053, 945, 450, 120, 17, 1, 1458, 3645, 3888, 2295, 810, 171, 20, 1, 4374, 12393, 15309, 10773, 4725, 1323, 231, 23, 1, 13122, 41553, 58320, 47628, 24948, 8694, 2016, 300, 26, 1

Offset: 0

Clark Kimberling, Aug 04 2011

A193723 is obtained by reversing the rows of the triangle A193722.
Triangle T(n,k), read by rows, given by [2,1,0,0,0,0,0,0,0,...] DELTA [1,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 04 2011
From Philippe Deléham, Nov 14 2011: (Start)
Riordan array ((1-x)/(1-3x), x/(1-3x)).
Product A200139*A007318 as infinite lower triangular arrays. (End)

			First six rows:
    1;
    2,   1;
    6,   5,   1;
   18,  21,   8,   1;
   54,  81,  45,  11,   1;
  162, 297, 216,  78,  14,   1;

Cf. A084938, A193722, A052924 (antidiagonal sums), Diagonals: A000012, A016789, A081266, Columns: A025192, A081038.

z = 9; a = 1; b = 1; c = 1; d = 2;
p[n_, x_] := (a*x + b)^n ; q[n_, x_] := (c*x + d)^n
t[n_, k_] := Coefficient[p[n, x], x^k]; t[n_, 0] := p[n, x] /. x -> 0;
w[n_, x_] := Sum[t[n, k]*q[n + 1 - k, x], {k, 0, n}]; w[-1, x_] := 1
g[n_] := CoefficientList[w[n, x], {x}]
TableForm[Table[Reverse[g[n]], {n, -1, z}]]
Flatten[Table[Reverse[g[n]], {n, -1, z}]] (* A193722 *)
TableForm[Table[g[n], {n, -1, z}]]
Flatten[Table[g[n], {n, -1, z}]] (* A193723 *)

Write w(n,k) for the triangle at A193722.  The triangle at A193723 is then given by w(n,n-k).
T(n,k) = T(n-1,k-1) + 3*T(n-1,k) with T(0,0)=T(1,1)=1 and T(1,0)=2. - Philippe Deléham, Oct 05 2011
From Philippe Deléham, Nov 14 2011: (Start)
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A011782(n), A025192(n), A002001(n), A005054(n), A052934(n), A055272(n), A055274(n), A055275(n), A052268(n), A055276(n), A196731(n) for x=-2,-1,0,1,2,3,4,5,6,7,8,9 respectively.
T(n,k) = Sum_{j>=0} T(n-1-j,k-1)*3^j.
G.f.: (1-x)/(1-(3+y)*x). (End)

Showing 1-10 of 25 results. Next

cp's OEIS Frontend

A110595 a(1)=5. For n > 1, a(n) = 4*5^(n-1) = A005054(n).

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A352378 Irregular triangle read by rows: T(n,k) is the (n-th)-to-last digit of 2^p such that p == k + A123384(n-1) (mod A005054(n)); k >= 0.

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A000351 Powers of 5: a(n) = 5^n.

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A001653 Numbers k such that 2*k^2 - 1 is a square.

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A033887 a(n) = Fibonacci(3*n + 1).

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A167374 Triangle, read by rows, given by [ -1,1,0,0,0,0,0,0,0,...] DELTA [1,0,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938.

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