A077076 -id:A077076 - cp's OEIS frontend

A061561 Trajectory of 22 under the Reverse and Add! operation carried out in base 2.

22, 35, 84, 105, 180, 225, 360, 405, 744, 837, 1488, 1581, 3024, 3213, 6048, 6237, 12192, 12573, 24384, 24765, 48960, 49725, 97920, 98685, 196224, 197757, 392448, 393981, 785664, 788733, 1571328, 1574397, 3144192, 3150333, 6288384, 6294525

Offset: 0

N. J. A. Sloane, May 18 2001

Sequence A058042 written in base 10. 22 is the smallest number whose base 2 trajectory does not contain a palindrome.
lim_{n -> infinity} a(n)/a(n-1) = 2 for n mod 2 = 0.
lim_{n -> infinity} a(n)/a(n-1) = 1 for n mod 2 = 1. - Klaus Brockhaus, Dec 09 2009

T. D. Noe, Table of n, a(n) for n=0..500
Klaus Brockhaus, On the 'Reverse and Add!' algorithm in base 2
Index entries for sequences related to Reverse and Add!

Cf. A035522 (trajectory of 1 in base 2), A058042 (trajectory of 22 in base 2, written in base 2), A075253 (trajectory of 77 in base 2), A075268 (trajectory of 442 in base 2), A077076 (trajectory of 537 in base 2), A077077 (trajectory of 775 in base 2), A066059 (trajectory of n in base 2 (presumably) does not reach a palindrome), A075252 (trajectory of n in base 2 does not reach a palindrome and (presumably) does not join the trajectory of any term m < n), A075153 (trajectory of 318 in base 4).

Cf. A171470 (a(4*n)/2), A171471 (a(4*n+1)), A171472 (a(4*n+2)/12), A171473 (a(4*n+3)/3).

m := 22; stop := 36; c := 0; while c < stop do write(m,","); k := bit_length(m); rev := 0; for i := 0 to k-1 do if bit_test(m,i) then rev := bit_set(rev,k-1-i); end; end; inc(c); m := m+rev; end;.

a061561 n = a061561_list !! n
a061561_list = iterate a055944 22  -- Reinhard Zumkeller, Apr 21 2013

trajectory:=function(init, steps, base) a:=init; S:=[a]; for n in [1..steps] do a+:=Seqint(Reverse(Intseq(a,base)),base); Append(~S, a); end for; return S; end function; trajectory(22, 35, 2); // Klaus Brockhaus, Dec 09 2009

binRA[n_] := If[Reverse[IntegerDigits[n, 2]] == IntegerDigits[n, 2], n, FromDigits[Reverse[IntegerDigits[n, 2]], 2] + n]; NestList[binRA, 22, 100] (* Adapted from Ben Branman's code for A213012, Alonso del Arte, Jun 02 2012 *)

{m=22; stop=36; c=0; while(c0,d=divrem(k,2); k=d[1]; rev=2*rev+d[2]); c++; m=m+rev)}

a(0) = 22; a(1) = 35; for n > 1 and n = 2 (mod 4): a(n) = 6*2^(2*k)-3*2^k where k = (n+6)/4; n = 3 (mod 4): a(n) = 6*2^(2*k)+3*2^k-3 where k = (n+5)/4; n = 0 (mod 4): a(n) = 12*2^(2*k)-3*2^k where k = (n+4)/4; n = 1 (mod 4): a(n) = 12*2^(2*k)+9*2^k-3 where k = (n+3)/4. [Klaus Brockhaus, Sep 05 2002]
G.f.: (22+35*x+18*x^2-72*x^4-90*x^5-48*x^6-60*x^7+80*x^8+112*x^9) / ((1-x)*(1+x)*(1-2*x^2)*(1-2*x^4)). [Klaus Brockhaus, Sep 05 2002, edited Dec 09 2009]
a(n+1) = A055944(a(n)). - Reinhard Zumkeller, Apr 21 2013

More terms from Klaus Brockhaus, May 27 2001

A075253 Trajectory of 77 under the Reverse and Add! operation carried out in base 2.

Original entry on oeis.org

77, 166, 267, 684, 897, 1416, 1557, 2904, 3333, 5904, 6189, 11952, 12813, 24096, 24669, 48480, 50205, 97344, 98493, 195264, 198717, 391296, 393597, 783744, 790653, 1569024, 1573629, 3140352, 3154173, 6283776, 6292989, 12572160

Offset: 0

Klaus Brockhaus, Sep 10 2002

22 is the smallest number whose base 2 trajectory (A061561) provably does not contain a palindrome. 77 is the next number (cf. A075252) with a completely different trajectory which has this property. A proof along the lines of Klaus Brockhaus, On the 'Reverse and Add!' algorithm in base 2, can be based on the formula given below.
lim_{n -> infinity} a(n)/a(n-1) = 2 for n mod 2 = 1.
lim_{n -> infinity} a(n)/a(n-1) = 1 for n mod 2 = 0.
Interleaving of A176632, 2*A176633, 3*A176634, 12*A176635.
From A.H.M. Smeets, Feb 11 2019: (Start)
Pattern with cycle length 4 in binary representation, represented by contextfree grammars with production rules:
S_a -> 10 T_a 00, T_a -> 1 T_a 0 | 1100010;
S_b -> 11 T_b 01, T_b -> 0 T_b 1 | 0000101;
S_c -> 10 T_c 000, T_c -> 1 T_c 0 | 1101011;
S_d -> 11 T_d 101, T_d -> 0 T_d 1 | 0100000;
the trajectory is similar to that of 22 (see A058042) except for the stopping strings in T_a, T_b, T_c and T_d. (End)

			267 (decimal) = 100001011 -> 100001011 + 110100001 = 1010101100 = 684 (decimal).

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Klaus Brockhaus, On the 'Reverse and Add!' algorithm in base 2
Index entries for sequences related to Reverse and Add!
Index entries for linear recurrences with constant coefficients, signature (0, 3, 0, 0, 0, -6, 0, 4).

Cf. A061561 (trajectory of 22 in base 2), A075268 (trajectory of 442 in base 2), A077076 (trajectory of 537 in base 2), A077077 (trajectory of 775 in base 2), A066059 (trajectory of n in base 2 presumably does not reach a palindrome), A075252 (trajectory of n in base 2 does not reach a palindrome and presumably does not join the trajectory of any term m < n), A092210 (trajectory of n in base 2 presumably does not join the trajectory of any m < n).

Cf. A176632 (a(4*n)), A176633 (a(4*n+1)/2), A176634 (a(4*n+2)/3), A176635 (a(4*n+3)/12).

a075253 n = a075253_list !! n
a075253_list = iterate a055944 77  -- Reinhard Zumkeller, Apr 21 2013

trajectory:=function(init, steps, base) S:=[init]; a:=S[1]; for n in [1..steps] do a+:=Seqint(Reverse(Intseq(a,base)),base); Append(~S, a); end for; return S; end function; trajectory(77, 31, 2);

seq(coeff(series((77+166*x+36*x^2+186*x^3+96*x^4-636*x^5-672*x^6-348*x^7-44*x^8+632*x^9+504*x^10)/((1-x)*(1+x)*(1-2*x^2)*(1-2*x^4)),x,n+1), x, n), n = 0 .. 40); # Muniru A Asiru, Feb 12 2019

CoefficientList[Series[(77+166*x+36*x^2+186*x^3+96*x^4-636*x^5-672*x^6 -348*x^7-44*x^8 +632*x^9+504*x^10)/((1-x)*(1+x)*(1-2*x^2)*(1-2*x^4)), {x,0,40}], x] (* G. C. Greubel, Feb 11 2019 *)
NestWhileList[# + IntegerReverse[#, 2] &, 77,  # !=
IntegerReverse[#, 2] &, 1, 31] (* Robert Price, Oct 18 2019 *)

{m=77; stop=34; c=0; while(c0,d=divrem(k,2); k=d[1]; rev=2*rev+d[2]); c++; m=m+rev)}

((77+166*x+36*x^2+186*x^3+96*x^4-636*x^5-672*x^6 -348*x^7-44*x^8 +632*x^9+504*x^10)/((1-x)*(1+x)*(1-2*x^2)*(1-2*x^4))).series(x, 40).coefficients(x, sparse=False) # G. C. Greubel, Feb 11 2019

a(0) = 77; a(1) = 166; a(2) = 267; for n > 2 and
n = 3 (mod 4): a(n) = 48*2^(2*k)-21*2^k where k = (n+5)/4;
n = 0 (mod 4): a(n) = 48*2^(2*k)+33*2^k-3 where k = (n+4)/4;
n = 1 (mod 4): a(n) = 96*2^(2*k)-30*2^k where k = (n+3)/4;
n = 2 (mod 4): a(n) = 96*2^(2*k)+6*2^k-3 where k = (n+2)/4.
G.f.: (77+166*x+36*x^2+186*x^3+96*x^4-636*x^5-672*x^6-348*x^7-44*x^8 +632*x^9+504*x^10) / ((1-x)*(1+x)*(1-2*x^2)*(1-2*x^4)).
G.f. for the sequence starting at a(3): 3*x^3*(228+299*x-212*x^2 -378*x^3-448*x^4-446*x^5+432*x^6+524*x^7) / ((1-x)*(1+x)*(1-2*x^2)*(1-2*x^4)).
a(n+1) = A055944(a(n)). - Reinhard Zumkeller, Apr 21 2013

Three comments added, g.f. edited, MAGMA program and crossrefs added by Klaus Brockhaus, Apr 25 2010

A077077 Trajectory of 775 under the Reverse and Add! operation carried out in base 2, written in base 10.

Original entry on oeis.org

775, 1674, 2325, 5022, 8919, 23976, 26757, 47376, 49581, 96048, 102669, 193056, 197469, 388704, 401949, 779328, 788157, 1563840, 1590333, 3131520, 3149181, 6273408, 6326397, 12554496, 12589821, 25129728, 25235709, 50274816, 50345469

Offset: 0

Klaus Brockhaus, Oct 25 2002

The base 2 trajectory of 775 = A075252(5) provably does not contain a palindrome. A proof can be based on the formula given below.
lim_{n -> infinity} a(n)/a(n-1) = 1 for n mod 2 = 1.
lim_{n -> infinity} a(n)/a(n-1) = 2 for n mod 2 = 0.
Interleaving of A177843, 6*A177844, 3*A177845, 6*A177846.

			775 (decimal) = 1100000111 -> 1100000111 + 1110000011 = 11010001010 = 1674 (decimal).

Reinhard Zumkeller, Table of n, a(n) for n = 0..1000
Klaus Brockhaus, On the 'Reverse and Add!' algorithm in base 2
Index entries for sequences related to Reverse and Add!

Cf. A058042 (trajectory of 22 in base 2, written in base 2), A061561 (trajectory of 22 in base 2), A075253 (trajectory of 77 in base 2), A075268 (trajectory of 442 in base 2), A077076 (trajectory of 537 in base 2), A075252 (trajectory of n in base 2 does not reach a palindrome and (presumably) does not join the trajectory of any term m < n).

Cf. A177843 (a(4*n)), A177844 (a(4*n+1)/6), A177845 (a(4*n+2)/3), A177846 (a(4*n+3)/6).

a077077 n = a077077_list !! n
a077077_list = iterate a055944 775  -- Reinhard Zumkeller, Apr 21 2013

trajectory:=function(init, steps, base) a:=init; S:=[a]; for n in [1..steps] do a+:= Seqint(Reverse(Intseq(a,base)),base); Append(~S, a); end for; return S; end function; trajectory(775, 28, 2);

NestWhileList[# + IntegerReverse[#, 2] &, 775,  # !=
IntegerReverse[#, 2] &, 1, 28] (* Robert Price, Oct 18 2019 *)

trajectory(n,steps) = {local(v, k=n); for(j=0, steps, print1(k, ", "); v=binary(k); k+=sum(j=1, #v, 2^(j-1)*v[j]))};
trajectory(775,28);

a(0), ..., a(5) as above; for n > 5 and
n = 2 (mod 4): a(n) = 3*2^(2*k+7)+273*2^k-3 where k = (n+6)/4;
n = 3 (mod 4): a(n) = 6*2^(2*k+7)-222*2^k where k = (n+5)/4;
n = 0 (mod 4): a(n) = 6*2^(2*k+7)+54*2^k-3 where k = (n+4)/4;
n = 1 (mod 4): a(n) = 12*2^(2*k+7)-282*2^k where k = (n+3)/4.
a(n) = -a(n-1)+2*a(n-2)+2*a(n-3)+2*a(n-4)+2*a(n-5)-4*a(n-6)-4*a(n-7)-3 for n > 12; a(0), ..., a(12) as above.
G.f.: (775+1674*x+1944*x^4+8910*x^5+4650*x^6-14508*x^7-19840*x^8-22644*x^9- 1860*x^10+28680*x^11+14328*x^12-2112*x^13) / ((1-x)*(1+x)*(1-2*x^2)*(1-2*x^4)).
G.f. for the sequence starting at a(6): 3*(8919+15792*x-10230*x^2- 15360*x^3-15358*x^4-31696*x^5+16668*x^6+31264*x^7) / ((1-x)*(1+x)*(1-2*x^2)*(1-2*x^4)).
a(n+1) = A055944(a(n)). - Reinhard Zumkeller, Apr 21 2013

Comment edited, three comments and formula added, g.f. edited, PARI program revised, MAGMA program and crossrefs added by Klaus Brockhaus, May 14 2010

A177682 a(n) = 6a(n-1)-8a(n-2)-3 for n > 4; a(0)=179, a(1)=1571, a(2)=12511, a(3)=155403, a(4)=572951.

Original entry on oeis.org

179, 1571, 12511, 155403, 572951, 2194479, 8583263, 33943743, 134996351, 538428159, 2150598143, 8596163583, 34372196351, 137463869439, 549805645823, 2199122919423, 8796292349951, 35184770744319, 140738285666303

Offset: 0

Klaus Brockhaus, May 12 2010

Related to Reverse and Add trajectory of 537 in base 2: a(n) = A077076(4*n)/3, i.e. one third of first quadrisection of A077076.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (7,-14,8).

Cf. A077076 (Reverse and Add trajectory of 537 in base 2), A177683, A177684, A177685.

[179, 1571, 12511] cat [2*4^(n+5)+6083*2^(n-1)-1: n in [3..25]]; // Vincenzo Librandi, Sep 24 2013

Join[{179,1571,12511},LinearRecurrence[{7,-14,8},{155403,572951,2194479},20]] (* Harvey P. Dale, Oct 06 2011 *)
CoefficientList[Series[(179 + 318 x + 4020 x^2 + 88388 x^3 - 352284 x^4 + 259376 x^5)/((1 - x) (1 - 2 x) (1 - 4 x)), {x, 0, 40}], x] (* Vincenzo Librandi, Sep 24 2013 *)

{m=19; v=concat([179, 1571, 12511, 155403, 572951], vector(m-5)); for(n=6 ,m, v[n]=6*v[n-1]-8*v[n-2]-3); v}

a(n) = 2*4^(n+5)+6083*2^(n-1)-1 for n > 2.
G.f.: (179+318*x+4020*x^2+88388*x^3-352284*x^4+259376*x^5) / ((1-x)*(1-2*x)*(1-4*x)).
G.f. for the sequence starting at a(3): x^3*(155403-514870*x+359464*x^2) / ((1-x)*(1-2*x)*(1-4*x)).

A177683 a(n) = 6a(n-1)-8a(n-2) for n > 4; a(0)=191, a(1)=1587, a(2)=14161, a(3)=123004, a(4)=508152.

Original entry on oeis.org

191, 1587, 14161, 123004, 508152, 2064880, 8324064, 33425344, 133959552, 536354560, 2146450944, 8587869184, 34355607552, 137430691840, 549739290624, 2198990209024, 8796026929152, 35184239902720, 140737223983104

Offset: 0

Klaus Brockhaus, May 12 2010

Related to Reverse and Add trajectory of 537 in base 2: a(n) = A077076(4*n+1)/6, i.e., one sixth of second quadrisection of A077076.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6, -8).

Cf. A077076 (Reverse and Add trajectory of 537 in base 2), A177682, A177684, A177685.

[191, 1587, 14161] cat [2*4^(n+5)-2017*2^(n-1): n in [3..25]]; // Vincenzo Librandi, Sep 24 2013

Join[{191,1587,14161},Transpose[NestList[{Last[#],6Last[#]-8First[#]}&,{123004,508152},20]][[1]]]  (* Harvey P. Dale, Mar 06 2011 *)
CoefficientList[Series[(191 + 441 x + 6167 x^2 + 50734 x^3 - 116584 x^4)/((1 - 2 x) (1 - 4 x)), {x, 0, 40}], x] (* Vincenzo Librandi, Sep 24 2013 *)
LinearRecurrence[{6,-8},{191,1587,14161,123004,508152},20] (* Harvey P. Dale, Oct 16 2019 *)

{m=19; v=concat([191, 1587, 14161, 123004, 508152], vector(m-5)); for(n=6 ,m, v[n]=6*v[n-1]-8*v[n-2]); v}

a(n) = 2*4^(n+5)-2017*2^(n-1) for n > 2.
G.f.: (191+441*x+6167*x^2+50734*x^3-116584*x^4) / ((1-2*x)*(1-4*x)).
G.f. for the sequence starting at a(3): 4*x^3*(30751-57468*x) / ((1-2*x)*(1-4*x)).

A177684 a(n) = 6a(n-1)-8a(n-2)-3 for n > 4; a(0)=633, a(1)=4809, a(2)=46041, a(3)=277767, a(4)=1079823.

Original entry on oeis.org

633, 4809, 46041, 277767, 1079823, 4256799, 16902207, 67358847, 268935423, 1074741759, 4296967167, 17183868927, 68727476223, 274893905919, 1099543625727, 4398110507007, 17592314036223, 70369000161279

Offset: 0

Klaus Brockhaus, May 12 2010

Related to Reverse and Add trajectory of 537 in base 2: a(n) = A077076(4*n+2)/3, i.e., one third of third quadrisection of A077076.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (7,-14,8).

Cf. A077076 (Reverse and Add trajectory of 537 in base 2), A177682, A177683, A177685.

[633, 4809, 46041] cat [4^(n+6)+1953*2^n-1: n in [3..25]]; // Vincenzo Librandi, Sep 24 2013

Join[{633,4809,46041},RecurrenceTable[{a[3]==277767,a[4]==1079823,a[n] == 6a[n-1]-8a[n-2]-3},a,{n,20}]] (* or *) Join[{633,4809,46041}, LinearRecurrence[ {7,-14,8},{277767,1079823,4256799},18]] (* Harvey P. Dale, May 03 2012 *)
CoefficientList[Series[3 (211 + 126 x + 7080 x^2 + 5914 x^3 - 86148 x^4 + 72816 x^5)/((1 - x) (1 - 2 x) (1 - 4 x)), {x, 0, 40}], x] (* Vincenzo Librandi, Sep 24 2013 *)

{m=18; v=concat([633, 4809, 46041, 277767, 1079823], vector(m-5)); for(n=6 ,m, v[n]=6*v[n-1]-8*v[n-2]-3); v}

a(n) = 4^(n+6)+1953*2^n-1 for n > 2.
G.f.: 3*(211+126*x+7080*x^2+5914*x^3-86148*x^4+72816*x^5) / ((1-x)*(1-2*x)*(1-4*x)).
G.f. for the sequence starting at a(3): 3*x^3*(92589-288182*x+195592*x^2) / ((1-x)*(1-2*x)*(1-4*x)).
a(0)=633, a(1)=4809, a(2)=46041, a(3)=277767, a(4)=1079823, a(5)=4256799, a(n)=7*a(n-1)-14*a(n-2)+8*a(n-3). - Harvey P. Dale, May 03 2012

A177685 a(n) = 6a(n-1)-8a(n-2) for n > 4; a(0)=603, a(1)=4731, a(2)=58834, a(3)=254204, a(4)=1032696.

Original entry on oeis.org

603, 4731, 58834, 254204, 1032696, 4162544, 16713696, 66981824, 268181376, 1073233664, 4293950976, 17177836544, 68715411456, 274869776384, 1099495366656, 4398013988864, 17592120999936, 70368614088704

Offset: 0

Klaus Brockhaus, May 12 2010

nonn

Related to Reverse and Add trajectory of 537 in base 2: a(n) = A077076(4*n+3)/6, i.e., one sixth of fourth quadrisection of A077076.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6, -8).

Cf. A077076 (Reverse and Add trajectory of 537 in base 2), A177682, A177683, A177684.

[603, 4731, 58834] cat [4^(n+6)-1985*2^(n-1): n in [3..25]]; // Vincenzo Librandi, Sep 24 2013

CoefficientList[Series[(603 + 1113 x + 35272 x^2 - 60952 x^3 - 21856 x^4)/((1 - 2 x) (1 - 4 x)), {x, 0, 40}], x] (* Vincenzo Librandi, Sep 24 2013 *)
LinearRecurrence[{6,-8},{603,4731,58834,254204,1032696},20] (* Harvey P. Dale, May 26 2019 *)

{m=18; v=concat([603, 4731, 58834, 254204, 1032696], vector(m-5)); for(n=6 ,m, v[n]=6*v[n-1]-8*v[n-2]); v}

a(n) = 4^(n+6)-1985*2^(n-1) for n > 2.
G.f.: (603+1113*x+35272*x^2-60952*x^3-21856*x^4) / ((1-2*x)*(1-4*x)).
G.f. for the sequence starting at a(3): 4*x^3*(63551-123132*x) / ((1-2*x)*(1-4*x)).

A213012 Trajectory of 26 under the Reverse and Add! operation carried out in base 2.

Original entry on oeis.org

26, 37, 78, 135, 360, 405, 744, 837, 1488, 1581, 3024, 3213, 6048, 6237, 12192, 12573, 24384, 24765, 48960, 49725, 97920, 98685, 196224, 197757, 392448, 393981, 785664, 788733, 1571328, 1574397, 3144192, 3150333

Offset: 0

Ben Branman, Jun 01 2012

26 is the second-smallest number (after 22) whose base 2 trajectory does not contain a palindrome.
lim_{n -> infinity} a(n)/a(n-1) = 2 for n mod 2 = 0.
lim_{n -> infinity} a(n)/a(n-1) = 1 for n mod 2 = 1. - Branman
In 2001, Brockhaus proved that if the binary Reverse and Add trajectory of an integer contains an integer of one of four specific given  forms, then the trajectory never reaches a palindrome. In the case of 26, that would be 3(2^(2k + 1) - 2^k), with k = 3 corresponding to 360. - Alonso del Arte, Jun 02 2012

			In binary, 26 is 11010.
a(1) = 37 because 11010 + 01011 = 100101, or 37.
a(2) = 78 because 100101 + 101001 = 1001110, or 78.

Klaus Brockhaus, On the'Reverse and Add!' algorithm in base 2
Index entries for sequences related to Reverse and Add!

Cf. A035522, A061561, A066059, A077076, A077077.

binRA[n_] := If[Reverse[IntegerDigits[n, 2]] == IntegerDigits[n, 2], n, FromDigits[Reverse[IntegerDigits[n, 2]], 2] + n]; NestList[binRA, 26, 100]

cp's OEIS Frontend

A061561 Trajectory of 22 under the Reverse and Add! operation carried out in base 2.

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A075253 Trajectory of 77 under the Reverse and Add! operation carried out in base 2.

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A077077 Trajectory of 775 under the Reverse and Add! operation carried out in base 2, written in base 10.

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A177682 a(n) = 6*a(n-1)-8*a(n-2)-3 for n > 4; a(0)=179, a(1)=1571, a(2)=12511, a(3)=155403, a(4)=572951.

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A177683 a(n) = 6*a(n-1)-8*a(n-2) for n > 4; a(0)=191, a(1)=1587, a(2)=14161, a(3)=123004, a(4)=508152.

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A177684 a(n) = 6*a(n-1)-8*a(n-2)-3 for n > 4; a(0)=633, a(1)=4809, a(2)=46041, a(3)=277767, a(4)=1079823.

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A177682 a(n) = 6a(n-1)-8a(n-2)-3 for n > 4; a(0)=179, a(1)=1571, a(2)=12511, a(3)=155403, a(4)=572951.

A177683 a(n) = 6a(n-1)-8a(n-2) for n > 4; a(0)=191, a(1)=1587, a(2)=14161, a(3)=123004, a(4)=508152.

A177684 a(n) = 6a(n-1)-8a(n-2)-3 for n > 4; a(0)=633, a(1)=4809, a(2)=46041, a(3)=277767, a(4)=1079823.

A177685 a(n) = 6a(n-1)-8a(n-2) for n > 4; a(0)=603, a(1)=4731, a(2)=58834, a(3)=254204, a(4)=1032696.