A153130 -id:A153130 - cp's OEIS frontend

A153237 a(n) = A000079(n) - A153130(n).

0, 0, 0, 0, 9, 27, 63, 126, 252, 504, 1017, 2043, 4095, 8190, 16380, 32760, 65529, 131067, 262143, 524286, 1048572, 2097144, 4194297, 8388603, 16777215, 33554430, 67108860, 134217720, 268435449, 536870907, 1073741823, 2147483646

Offset: 0

Paul Curtz, Dec 21 2008

Harvey P. Dale, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-2,-1,3,-2).

LinearRecurrence[{3,-2,-1,3,-2},{0,0,0,0,9},40] (* Harvey P. Dale, Dec 26 2021 *)

a(n) = 9 *A153234(n). G.f. 9*x^4 / ( (x-1)*(2*x-1)*(1+x)*(x^2-x+1) ). - R. J. Mathar, Dec 17 2012

Definition corrected by Omar E. Pol, Dec 24 2008

Edited by N. J. A. Sloane, Dec 31 2008

A158916 Inverse binomial transform of A153130.

Original entry on oeis.org

1, 1, 1, 1, -8, 19, -35, 64, -125, 253, -512, 1027, -2051, 4096, -8189, 16381, -32768, 65539, -131075, 262144, -524285, 1048573, -2097152, 4194307, -8388611, 16777216, -33554429, 67108861, -134217728, 268435459, -536870915, 1073741824, -2147483645

Offset: 0

Paul Curtz, Mar 30 2009

sign

a(n)= A154589(n)+ A099838(n+4).

Index entries for linear recurrences with constant coefficients, signature (-3, -3, -2).

LinearRecurrence[{-3,-3,-2},{1,1,1,1},40] (* Harvey P. Dale, Feb 04 2019 *)

a(n)= -3a(n-1)-3a(n-2)-2a(n-3), n > 3.

G.f.: (4*x+7*x^2+9*x^3+1)/((2*x+1)*(1+x+x^2)). [R. J. Mathar, May 17 2009]

Edited and extended by R. J. Mathar, May 17 2009

A004000 RATS: Reverse Add Then Sort the digits applied to previous term, starting with 1.

Original entry on oeis.org

1, 2, 4, 8, 16, 77, 145, 668, 1345, 6677, 13444, 55778, 133345, 666677, 1333444, 5567777, 12333445, 66666677, 133333444, 556667777, 1233334444, 5566667777, 12333334444, 55666667777, 123333334444, 556666667777, 1233333334444, 5566666667777, 12333333334444

Offset: 1

N. J. A. Sloane

It is conjectured that no matter what the starting term is, repeatedly applying RATS leads either to this sequence or into a cycle of finite length, such as those in A066710 and A066711.

			668 -> 668 + 866 = 1534 -> 1345.

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

Alois P. Heinz, Table of n, a(n) for n = 1..2002 (first 200 terms from T. D. Noe)
R. K. Guy, Conway's RATS and other reversals, Amer. Math. Monthly, 96 (1989), 425-428.
Eric Weisstein's World of Mathematics, RATS Sequence.

Cf. A036839, A066710, A066711, A066713, A164338, A161593, A114611, A114612, A209878, A209879, A209880.

a004000_list = iterate a036839 1  -- Reinhard Zumkeller, Mar 14 2012

[ n eq 1 select 1 else Seqint(Reverse(Sort(Intseq(p + Seqint(Reverse(Intseq(p))) where p is Self(n-1))))) : n in [1..10]]; // Sergei Haller (sergei(AT)sergei-haller.de), Dec 21 20061

read transforms; RATS := n -> digsort(n + digrev(n)); b := [1]; t := [1]; for n from 1 to 50 do t := RATS(t); b := [op(b),t]; od: b;

NestList[FromDigits[Sort[IntegerDigits[#+FromDigits[Reverse[ IntegerDigits[#]]]]]]&,1,30] (* Harvey P. Dale, Nov 29 2011 *)

step(n)=fromdigits(vecsort(digits(n+fromdigits(Vecrev(digits(n)))))) \\ Charles R Greathouse IV, Jun 23 2017

l = [0, 1]
for n in range(2, 51):
    x = str(l[n - 1])
    l.append(int(''.join(sorted(str(int(x) + int(x[::-1]))))))
print(l[1:]) # Indranil Ghosh, Jul 05 2017

Let a(n) = k, form m by Reversing the digits of k, Add m to k Then Sort the digits of the sum into increasing order to get a(n+1).
a(n+1) = A036839(a(n)). - Reinhard Zumkeller, Mar 14 2012
A010888(a(n)) = A153130(n-1). - Ivan N. Ianakiev, Nov 27 2014
a(2n-1) = (37 * 10^(n-3) + 3332)/3, n >= 11; a(2n) = (167 * 10^(n-3) + 3331)/3, n >= 10. - Jianing Song, May 06 2021

Entry revised by N. J. A. Sloane, Jan 19 2002

A010689 Periodic sequence: Repeat 1, 8.

Original entry on oeis.org

1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1, 8, 1

Offset: 0

N. J. A. Sloane

Also the digital root of 8^n. Also the decimal expansion of 2/11 = 0.181818181818... - Cino Hilliard, Dec 31 2004
Interleaving of A000012 and A010731. - Klaus Brockhaus, Apr 02 2010
Continued fraction expansion of (2 + sqrt(6))/4. - Klaus Brockhaus, Apr 02 2010
Digital root of the powers of any number congruent to 8 mod 9. - Alonso del Arte, Jan 26 2014

			0.18181818181818181818181818181818181818181...

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

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

Cf. A000012 (all 1's sequence), A010731 (all 8's sequence), A174925 (decimal expansion of (2 + sqrt(6))/4). [Klaus Brockhaus, Apr 02 2010]
Cf. Digital roots of powers of c mod 9: c = 2, A153130; c = 4, A100402; c = 5, A070366; c = 7, A070403.
Cf. sequences listed in Comments section of A283393.
Cf. A010888.

&cat[ [1, 8]: n in [0..52] ]; // Klaus Brockhaus, Apr 02 2010

&cat [[1,8]^^60]; // Bruno Berselli, Mar 10 2017

Table[Mod[8^n, 9], {n, 0, 99}] (* Alonso del Arte, Jan 26 2014 *)
PadRight[{},120,{1,8}] (* Harvey P. Dale, Jun 03 2015 *)

A010689(n):=if evenp(n) then 1 else 8$
makelist(A010689(n),n,0,30); /* Martin Ettl, Nov 09 2012 */

a(n)=1; if(n%2==1, 8, 1) \\ Felix Fröhlich, Aug 11 2014

[power_mod(8,n,9)for n in range(0,105)] # Zerinvary Lajos, Nov 27 2009

From Paul Barry, Sep 16 2004: (Start)
G.f.: (1 + 8*x)/((1 - x)*(1 + x)).
a(n) = (9 - 7*(-1)^n)/2.
a(n) = 8^(ceiling(n/2) - floor(n/2)).
a(n) = gcd((n-1)^3, (n+1)^3). (End)
E.g.f.: cosh(x) + 8*sinh(x). - Stefano Spezia, Feb 09 2025
a(n) = A010888(8*a(n-1)). - Stefano Spezia, Mar 20 2025

Definition edited and keywords cons, cofr added by Klaus Brockhaus, Apr 02 2010

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

A002114 Glaisher's H' numbers.

Original entry on oeis.org

1, 11, 301, 15371, 1261501, 151846331, 25201039501, 5515342166891, 1538993024478301, 533289474412481051, 224671379367784281901, 113091403397683832932811, 67032545884354589043714301, 46211522130188693681603906171

Offset: 1

N. J. A. Sloane

a(n) mod 9 = 1,2,4,8,7,5 repeated period 6 (A153130, see also A001370). a(n) mod 10 = 1. - Paul Curtz, Sep 10 2009

A. Fletcher, J. C. P. Miller, L. Rosenhead and L. J. Comrie, An Index of Mathematical Tables. Vols. 1 and 2, 2nd ed., Blackwell, Oxford and Addison-Wesley, Reading, MA, 1962, Vol. 1, p. 76.
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).

Vincenzo Librandi, Table of n, a(n) for n = 1..100
J. W. L. Glaisher, On a set of coefficients analogous to the Eulerian numbers, Proc. London Math. Soc., 31 (1899), 216-235.
Vladimir Kruchinin, Composition of ordinary generating functions, arXiv:1009.2565 [math.CO], 2010. [Author's named corrected by _N. J. A. Sloane_, Dec 12 2021]
Index entries for sequences related to Glaisher's numbers

a := n -> (-1)^n*6^(2*n)*(Zeta(0,-n*2,1/3)-Zeta(0,-n*2, 5/6)):
seq(a(n), n=1..14);

Select[Rest[With[{nn=28},CoefficientList[Series[1/(2 (2Cos[x]-1)), {x,0,nn}], x]Range[0,nn]!]],#!=0&] (* Harvey P. Dale, Jul 27 2011 *)
FullSimplify[Table[(-1)^(s+1) * BernoulliB[2*s] * (Zeta[2*s + 1, 1/6] - Zeta[2*s + 1, 5/6]) / (4*Pi*Sqrt[3]*Zeta[2*s]), {s, 1, 20}]]  (* Vaclav Kotesovec, May 05 2020 *)

a(n) := sum(sum(binomial(k,j)*(-1)^(k-j+1)*1/2^(j-1)*sum((-1)^(n)*binomial(j,i)*(2*i-j)^(2*n),i,0,floor((j-1)/2)),j,0,k)*(-2)^(k-1),k,1,2*n); /* Vladimir Kruchinin, Aug 05 2010 */

H'(n) = H(n)/3, where H(n)=2^(2n+1)*I(n) (see A002112) and e.g.f. for (-1)^n*I(n) is (3/2)/(1+exp(x)+exp(-x)) (see A047788, A047789).
H'(n) = A000436(n)/2^(2n+1). - Philippe Deléham, Jan 17 2004
For n > 0, H'(n) = Sum{k = 0..n, T(n, k)*9^(n-k)*2^(k-1) }; where DELTA is the operator defined in A084938, T(n, k) is the triangle, read by rows, given by :[0, 1, 0, 4, 0, 9, 0, 16, 0, 25, ...] DELTA [1, 0, 10, 0, 28, 0, 55, 0, 90, ..]= {1}; {0, 1}; {0, 1, 1}; {0, 1, 12, 1}; {0, 1, 63, 123, 1}; {0, 1, 274, 2366, 1234, 1}; ... For 1, 10, 28, 55, 90, 136, ... see A060544 or A060544. - Philippe Deléham, Jan 17 2004
E.g.f. 1/2*1/(2*cos(x)-1). a(n)=sum(sum(binomial(k,j)*(-1)^(k-j+1)*1/2^(j-1)*sum((-1)^(n)*binomial(j,i)*(2*i-j)^(2*n),i,0,floor((j-1)/2)),j,0,k)*(-2)^(k-1),k,1,2*n), n>0. - Vladimir Kruchinin, Aug 05 2010
E.g.f.: E(x)= x^2/(G(0)-x^2) ; G(k)= 2*(2*k+1)*(k+1) - x^2 + 2*x^2*(2*k+1)*(k+1)/G(k+1); (continued fraction Euler's kind, 1-step ). - Sergei N. Gladkovskii, Jan 03 2012
If E(x)=Sum(k=0,1,..., a(k+1)*x^(2k+2)), then A002114(k) = a(k+1)*(2*k+2)!. - Sergei N. Gladkovskii, Jan 09 2012
a(n) ~ (2*n)! * 3^(2*n+1/2) / Pi^(2*n+1). - Vaclav Kotesovec, Feb 26 2014
a(n) = (-1)^n*6^(2*n)*(zeta(-n*2,1/3)-zeta(-n*2,5/6)), where zeta(a, z) is the generalized Riemann zeta function.
From Vaclav Kotesovec, May 05 2020: (Start)
a(n) = (2*n)! * (zeta(2*n+1, 1/6) - zeta(2*n+1, 5/6)) / (sqrt(3)*(2*Pi)^(2*n+1)).
a(n) = (-1)^(n+1) * Bernoulli(2*n) * (zeta(2*n+1, 1/6) - zeta(2*n+1, 5/6)) / (4*Pi*sqrt(3)*zeta(2*n)). (End)
Conjectural e.g.f.: Sum_{n >= 1} (-1)^n*Product_{k = 1..n} (1 - exp(A007310(k)*z) ) =  z + 11*z^2/2! + 301*z^3/3! + .... - Peter Bala, Dec 09 2021

A005015 a(n) = 11*2^n.

Original entry on oeis.org

11, 22, 44, 88, 176, 352, 704, 1408, 2816, 5632, 11264, 22528, 45056, 90112, 180224, 360448, 720896, 1441792, 2883584, 5767168, 11534336, 23068672, 46137344, 92274688, 184549376, 369098752, 738197504, 1476395008, 2952790016, 5905580032, 11811160064

Offset: 0

N. J. A. Sloane

The first differences are the sequence itself. - Alexandre Wajnberg & Eric Angelini, Sep 07 2005
11 times powers of 2. - Omar E. Pol, Dec 16 2008
A144472 = -1,2,9,13,31,57,.... a(n) = A144472(n+1)+A144472(n+2). Also a(n) = A144472(n+3)-A144472(n+1). A144472(n+1) is a Jacobsthal sequence from 2 and 9: A144472(n+3) = A144472(n+2)+2*A144472(n+1). Note a(n) mod 9 = period 6: repeat 2,4,8,7,5,1 = A153130(n+1). - Paul Curtz, Jan 06 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (2).

Row sums of (10, 1)-Pascal triangle A093645.

Cf. A000079, A144472, A153130.

[11*2^n: n in [0..40]]; // Vincenzo Librandi, Aug 14 2011

11*2^Range[0, 60] (* Vladimir Joseph Stephan Orlovsky, Jun 09 2011 *)
NestList[2#&,11,30] (* Harvey P. Dale, Jun 11 2021 *)

a(n)=11<Charles R Greathouse IV, Oct 07 2015

G.f.: 11/(1-2*x).
a(n) = 2*a(n-1), n>0; a(0)=11. - Philippe Deléham, Nov 23 2008
a(n) = A000079(n)*11. - Omar E. Pol, Dec 16 2008
E.g.f.: 11*exp(2*x). - Elmo R. Oliveira, Aug 16 2024

A021823 Decimal expansion of 1/819.

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane

Partial sums of A010892. - Paul Barry, Jun 06 2003
Expansion in any base b >= 3 of 1/((b-1)*(b^2-b+1)) = 1/(b^3-2b^2+2b-1). E.g., 1/14 in base 3, 1/39 in base 4, 1/84 in base 5, etc. - Franklin T. Adams-Watters, Nov 07 2006
a(n) is the second least significant digit in the ternary representation of 2^n (cf. A004642). - Alexandre Herrera, Oct 09 2023

			0.0012210012210012210...

Index entries for linear recurrences with constant coefficients, signature (2,-2,1).

Cf. A077859, A027444.

Cf. A004642, A153130 (2^n mod 9).

Join[{0,0},RealDigits[1/819,10,120][[1]]] (* or *) PadRight[{},120,{0,0,1,2,2,1}] (* or *) LinearRecurrence[{2,-2,1},{0,0,1},120] (* Harvey P. Dale, Aug 19 2012 *)

a(n)=1/819. \\ Charles R Greathouse IV, Sep 24 2015

a(n) = a(n-1)-a(n-2)+1 = 2-a(n-3) = a(n-6). - Henry Bottomley, Apr 12 2000
a(n) = Sum_{k=1..floor(n/2)} (-1)^(k+1)*binomial(n-k, k) = 1-((-1)^floor(n/3)+(-1)^(floor((n+1)/3)))/2. - Vladeta Jovovic, Feb 10 2003
G.f.: x^2/(1-2x+2x^2-x^3)=x^2/((1-x)(x^2-x+1)). - Paul Barry, Jun 06 2003
a(n+2) = sum{k=0..n, binomial(n-2k, n-k)}. - Paul Barry, Jan 15 2005
a(0)=0, a(1)=0, a(2)=1, a(n)=2*a(n-1)-2*a(n-2)+a(n-3). - Harvey P. Dale, Aug 19 2012

A167762 a(n) = 2a(n-1)+3a(n-2)-6*a(n-3) starting a(0)=a(1)=0, a(2)=1.

Original entry on oeis.org

0, 0, 1, 2, 7, 14, 37, 74, 175, 350, 781, 1562, 3367, 6734, 14197, 28394, 58975, 117950, 242461, 484922, 989527, 1979054, 4017157, 8034314, 16245775, 32491550, 65514541, 131029082, 263652487, 527304974, 1059392917, 2118785834, 4251920575, 8503841150

Offset: 0

Paul Curtz, Nov 11 2009

Inverse binomial transform yields two zeros followed by A077917 (a signed variant of A127864).
a(n) mod 10 is zero followed by a sequence with period length 8: 0, 1, 2, 7, 4, 7, 4, 5 (repeat).
a(n) is the number of length n+1 binary words with some prefix w such that w contains three more 1's than 0's and no prefix of w contains three more 0's than 1's. - Geoffrey Critzer, Dec 13 2013
From Gus Wiseman, Oct 06 2023: (Start)
Also the number of subsets of {1..n} with two distinct elements summing to n + 1. For example, the a(2) = 1 through a(5) = 14 subsets are:
  {1,2}  {1,3}    {1,4}      {1,5}
         {1,2,3}  {2,3}      {2,4}
                  {1,2,3}    {1,2,4}
                  {1,2,4}    {1,2,5}
                  {1,3,4}    {1,3,5}
                  {2,3,4}    {1,4,5}
                  {1,2,3,4}  {2,3,4}
                             {2,4,5}
                             {1,2,3,4}
                             {1,2,3,5}
                             {1,2,4,5}
                             {1,3,4,5}
                             {2,3,4,5}
                             {1,2,3,4,5}
The complement is counted by A038754.
Allowing twins gives A167936, complement A108411.
For n instead of n + 1 we have A365544, complement A068911.
The version for all subsets (not just pairs) is A366130.
(End)

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2,3,-6).

Cf. A024495, A167710.
First differences are A167936, complement A108411.
Cf. A004526, A004737, A008967, A038754, A046663, A068911, A088809, A093971, A365376, A365544, A366130.

LinearRecurrence[{2,3,-6},{0,0,1},40] (* Harvey P. Dale, Sep 17 2013 *)
CoefficientList[Series[x^2/((2 x - 1) (3 x^2 - 1)), {x, 0, 50}], x] (* Vincenzo Librandi, Sep 17 2013 *)
Table[Length[Select[Subsets[Range[n]],MemberQ[Total/@Subsets[#,{2}],n+1]&]],{n,0,10}] (* Gus Wiseman, Oct 06 2023 *)

a(n) mod 9 = A153130(n), n>3 (essentially the same as A154529, A146501 and A029898).
a(n+1)-2*a(n) = 0 if n even, = A000244((1+n)/2) if n odd.
a(2*n) = A005061(n). a(2*n+1) = 2*A005061(n).
G.f.: x^2/((2*x-1)*(3*x^2-1)). a(n) = 2^n - A038754(n). - R. J. Mathar, Nov 12 2009
G.f.: x^2/(1-2*x-3*x^2+6*x^3). - Philippe Deléham, Nov 11 2009

Edited and extended by R. J. Mathar, Nov 12 2009

A100402 Digital root of 4^n.

Original entry on oeis.org

1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7, 1, 4, 7

Offset: 0

Cino Hilliard, Dec 31 2004

Equals A141725 mod 9. - Paul Curtz, Sep 15 2008
Sequence is the digital root of A016777. - Odimar Fabeny, Sep 13 2010
Digital root of the powers of any number congruent to 4 mod 9. - Alonso del Arte, Jan 26 2014
Period 3: repeat [1, 4, 7]. - Wesley Ivan Hurt, Aug 26 2014
From Timothy L. Tiffin, Dec 02 2023: (Start)
The period 3 digits of this sequence are the same as those of A070403 (digital root of 7^n) but the order is different: [1, 4, 7] vs. [1, 7, 4].
The digits in this sequence appear in the decimal expansions of the following rational numbers: 49/333, 490/333, 4900/333, .... (End)

			4^2 = 16, digitalroot(16) = 7, the third entry.

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

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

Cf. Digital roots of powers of c mod 9: c = 2, A153130; c = 5, A070366; c = 7, A070403; c = 8, A010689.

Cf. A000302, A001022, A009966, A009975, A009984, A010872, A010888, A070403, A087752, A121013, A141725, A016777.

Table[PowerMod[4, n, 9], {n, 0, 100}] (* Timothy L. Tiffin, Dec 03 2023 *)
StringRepeat["1, 4, 7, ", 100] (* Timothy L. Tiffin, Dec 03 2023 *)

a(n)=[1,4,7][1+n%3]; \\ Joerg Arndt, Aug 26 2014

[power_mod(4, n, 9) for n in range(0, 105)] # Zerinvary Lajos, Nov 25 2009

a(n) = 4^n mod 9. - Zerinvary Lajos, Nov 25 2009
From R. J. Mathar, Apr 13 2010: (Start)
a(n) = a(n-3) for n>2.
G.f.: (1+4*x+7*x^2)/ ((1-x)*(1+x+x^2)). (End)
a(n) = A010888(A000302(n)). - Michel Marcus, Aug 25 2014
a(n) = 3*A010872(n) + 1. - Robert Israel, Aug 25 2014
a(n) = 4 - 3*cos(2*n*Pi/3) - sqrt(3)*sin(2*n*Pi/3). - Wesley Ivan Hurt, Jun 30 2016
a(n) = A153130(2n). - Timothy L. Tiffin, Dec 01 2023
a(n) = A010888(A001022(n)) = A010888(A009966(n)) = A010888(A009975(n)) = A010888(A009984(n)) = A010888(A087752(n)) = A010888(A121013(n)). - Timothy L. Tiffin, Dec 02 2023
a(n) = A010888(4*a(n-1)). - Stefano Spezia, Mar 20 2025

cp's OEIS Frontend

A153237 a(n) = A000079(n) - A153130(n).

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A158916 Inverse binomial transform of A153130.

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A004000 RATS: Reverse Add Then Sort the digits applied to previous term, starting with 1.

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A010689 Periodic sequence: Repeat 1, 8.

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

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A002114 Glaisher's H' numbers.

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A167762 a(n) = 2a(n-1)+3a(n-2)-6*a(n-3) starting a(0)=a(1)=0, a(2)=1.