A075253 -id:A075253 - 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

A075252 Trajectory of n under the Reverse and Add! operation carried out in base 2 does not reach a palindrome and (presumably) does not join the trajectory of any term m < n.

Original entry on oeis.org

22, 77, 442, 537, 775, 1066, 1081, 1082, 1085, 1115, 1562, 1575, 1587, 2173, 3355, 3599, 3871, 4099, 4153, 4185, 4193, 4202, 4262, 4285, 4402, 4633, 4666, 6163, 6166, 6374, 9241, 9466, 16544, 16546, 16586, 16601, 16613, 16616, 16720, 16748, 16994

Offset: 1

Klaus Brockhaus, Sep 10 2002

Base-2 analog of A063048 (base 10) and A075421 (base 4); subsequence of A066059. - For the trajectory of 22 (cf. A061561) and the trajectory of 77 (cf. A075253) it has been proved that they do not contain a palindrome. A similar proof can be given for most terms of this sequence, but there are a few terms (4262, 17498, 33378, 33898, ...) whose trajectory does not show the kind of regularity that can be utilized for the construction of a proof. - If the trajectory of an integer k joins the trajectory of a smaller integer which is a term of the present sequence, then this occurs after very few 'Reverse and Add!' steps (at most 84 for numbers < 20000). On the other hand, the trajectories of the terms of this sequence do not join the trajectory of any smaller term within at least 1000 steps.
From A.H.M. Smeets, Feb 12 2019: (Start)
Most terms in this sequence eventually give rise to a regular binary pattern. These regular patterns can be represented by contextfree grammars:
S_a -> 10 T_a 00, T_a -> 1 T_a 0 | A_a(n);
S_b -> 11 T_b 01, T_b -> 0 T_b 1 | B_a(n);
S_c -> 10 T_c 000, T_c -> 1 T_c 0 | C_a(n) and
S_d -> 11 T_d 101, T_d -> 0 T_d 1 | D_a(n).
A_22 = 1101, B_22 = 1000, C_22 = 1101, D_22 = 0010 (see also A058042);
A_77 = 1100010, B_77 = 0000101, C_77 = 1101011, D_77 = 0100000 (see also A075253)
Decimal representations for 10 A_a(n) 00 are given by A306514(n).
Binary representations for 10 A_a(n) 00 are given by A306515(n).
(End)

			442 is a term since the trajectory of 442 (presumably) does not lead to an integer which occurs in the trajectory of 22 or of 77.

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

Cf. A063048, A075421, A066059, A058042, A061561, A075253, A306514, A306515.

limit = 10^2; (* Assumes that there is no palindrome if none is found before "limit" iterations *)
utraj = {};
Select[Range[0, 17000], (x = NestWhileList[# + IntegerReverse[#, 2] &, #, # != IntegerReverse[#, 2] & , 1, limit];
   If[Length[x] >= limit  && Intersection[x, utraj] == {},
    utraj = Union[utraj, x]; True,
utraj = Union[utraj, x]]) &] (* Robert Price, Oct 16 2019 *)

A075153 Trajectory of 318 under the Reverse and Add! operation carried out in base 4, written in base 10.

Original entry on oeis.org

318, 1071, 5040, 5985, 10710, 20400, 24225, 43350, 81600, 85425, 165750, 327360, 342705, 664950, 1309440, 1324785, 2629110, 5241600, 5303025, 10524150, 20966400, 21027825, 41973750, 83880960, 84126705, 167925750, 335523840

Offset: 0

Klaus Brockhaus, Sep 05 2002

290 is conjectured (cf. A066450) to be the smallest number such that the Reverse and Add! algorithm in base 4 does not lead to a palindrome. 318 (not 255 since 255 is a base 4 palindrome) is up to now the smallest number whose base 4 trajectory provably does not contain a palindrome. 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 3 in {1, 2}.
lim_{n -> infinity} a(n)/a(n-1) = 1 for n mod 3 = 0.

			318 (decimal) = 10332 -> 10332 + 23301 = 100233 = 1071 (decimal).

Klaus Brockhaus, On the 'Reverse and Add!' algorithm in base 2
K. S. Brown, Digit Reversal Sums Leading to Palindromes
Index entries for sequences related to Reverse and Add!

Cf. A058042 (trajectory of binary number 10110 (decimal 22)), A061561 (A058042 written in base 10), A066450 (conjectured minimal k so that the trajectory of k in base n does not lead to a palindrome).
Cf. A075253 (trajectory of 77 in base 2), A075420 (trajectory of n in base 4 (presumably) does not reach a palindrome), A075421 (trajectory of n in base 4 (presumably) does not reach a palindrome and (presumably) does not join the trajectory of any term m < n), A075299 (trajectory of 290 in base 4), A075466 (trajectory of 266718 in base 4), A075467 (trajectory of 270798 in base 4), A076247 (trajectory of 1059774 in base 4), A076248 (trajectory of 1059831 in base 4), A091675 (trajectory of n in base 4 (presumably) does not join the trajectory of any m < n).
Cf. A166912 (a(6*n)/3), A166913 (a(6*n+1)/3), A166914 (a(6*n+2)/240), A166915 (a(6*n+3)/15), A166916 (a(6*n+4)/30), A166917 (a(6*n+5)/240).

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(318, 26, 4);

NestWhileList[# + IntegerReverse[#, 4] &, 318,  # !=
IntegerReverse[#, 4] &, 1, 26] (* Robert Price, Oct 18 2019 *)

{m=318; stop=29; c=0; while(c0,d=divrem(k,4); k=d[1]; rev=4*rev+d[2]); c++; m=m+rev)}

a(0) = 318; a(1) = 1071; for n > 1 and n = 2 (mod 6): a(n) = 5*4^(2*k+5)-5*4^(k+2) where k = (n-2)/6; n = 3 (mod 6): a(n) = 5*4^(2*k+5)+55*4^(k+2)-15 where k = (n-3)/6; n = 4 (mod 6): a(n) = 10*4^(2*k+5)+30*4^(k+2)-10 where k = (n-4)/6; n = 5 (mod 6): a(n) = 20*4^(2*k+5)-5*4^(k+2) where k = (n-5)/6; n = 0 (mod 6): a(n) = 20*4^(2*k+5)+235*4^(k+2)-15 where k = (n-6)/6; n = 1 (mod 6): a(n) = 40*4^(2*k+5)+150*4^(k+2)-10 where k = (n-7)/6.

G.f.: 3*(106 +357*x +1680*x^2 +1465*x^3 +1785*x^4 -1600*x^5 -1900*x^6 -3400*x^7 -6800*x^8 -9780*x^9 -9860*x^10 +6720*x^11 +10064*x^12 +11088*x^13) / ((1-x)*(1+x+x^2)*(1-2*x^3)*(1+2*x^3)*(1-4*x^3)).

Two comments added, g.f. edited, MAGMA program and cross-references added by Klaus Brockhaus, Oct 26 2009

A075268 Trajectory of 442 under the Reverse and Add! operation carried out in base 2.

Original entry on oeis.org

442, 629, 1326, 2259, 5508, 6585, 11628, 15129, 24912, 26259, 52038, 77337, 155394, 221931, 442374, 639009, 1179738, 1917027, 3539130, 5062869, 10666542, 18285939, 45369156, 54513657, 96444396, 125792217, 207562704, 220034931

Offset: 0

Klaus Brockhaus, Sep 11 2002

22, 77 and 442 are the first terms of A075252. The base 2 trajectory of 442 is completely different from the trajectories of 22 (cf. A061561) and 77 (cf. A075253). Using the formula given below one can prove that it does not contain a palindrome.
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 2*A177420, A177421, 6*A177422, 3*A177423.

			442 (decimal) = 110111010 -> 110111010 + 010111011 = 1001110101 = 629 (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), 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. A177420 (a(4*n)/2), A177421 (a(4*n+1)), A177422 (a(4*n+2)/6), A177423 (a(4*n+3)/3).

a075268 n = a075268_list !! n
a075268_list = iterate a055944 442  -- 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(442, 28, 2);

NestWhileList[# + IntegerReverse[#, 2] &, 442,  # !=
IntegerReverse[#, 2] &, 1, 27] (* 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(442,28);

a(0), ..., a(28) as above; a(29) = 703932681; a(30) =1310348526; a(31) = 2309980455; a(32) = 6143702712; a(33) = 7131271077; a(34) = 12699398352; a(35) = 13441412493; for n > 35 and
n = 0 (mod 4): a(n) = 3*2^(2*k+23)-12576771*2^k where k = (n-16)/4;
n = 1 (mod 4): a(n) = 3*2^(2*k+23)+12576771*2^k-3 where k = (n-17)/4;
n = 2 (mod 4): a(n) = 6*2^(2*k+23)-12576771*2^k where k = (n-18)/4;
n = 3 (mod 4): a(n) = 6*2^(2*k+23)+37730313*2^k-3 where k = (n-19)/4.
G.f.: (442+629*x+372*x^3+1530*x^4-192*x^5-2244*x^6-852*x^7-3784*x^8-8090*x^9 +5046*x^10+29034*x^11+47016*x^12+54354*x^13+79152*x^14+70254*x^15+65196*x^16 +358986*x^17+724128*x^18+334026*x^19+2081820*x^20+6043662*x^21+18678462*x^22+8601966*x^23 -23147244*x^24-15039648*x^25 -31927752*x^26-67877562*x^27+43880046*x^28+297766074*x^29 +396480108*x^30+734881086*x^31+3072255774*x^32+1018370430*x^33-3939844260*x^34-4608944376*x^35 -6616834356*x^36-3107825028*x^37+6655931736*x^38+7777900872*x^39+484428384*x^40 -2233413600*x^41-62899200*x^42+188697600*x^43) / ((1-x)*(1+x)*(1-2*x^2)*(1-2*x^4)).
G.f. for the sequence starting at a(36): 3*x^36*(8455782368+8724086815*x -8321630144*x^2-8589934590*x^3-17045716960*x^4-18118934750*x^5+16911564736*x^6 +17984782524*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 and three comments added, g.f. edited, PARI program revised, MAGMA program and crossrefs added by Klaus Brockhaus, May 07 2010

A077076 Trajectory of 537 under the Reverse and Add! operation carried out in base 2, written in base 10.

Original entry on oeis.org

537, 1146, 1899, 3618, 4713, 9522, 14427, 28386, 37533, 84966, 138123, 353004, 466209, 738024, 833301, 1525224, 1718853, 3048912, 3239469, 6196176, 6583437, 12389280, 12770397, 24975264, 25749789, 49944384, 50706621, 100282176

Offset: 0

Klaus Brockhaus, Oct 25 2002

The base 2 trajectory of 537 = A075252(4) 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 3*A177682, 6*A177683, 3*A177684, 6*A177685.

			537 (decimal) = 1000011001 -> 1000011001 + 1001100001 = 10001111010= 1146 (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), A077077 (trajectory of 775 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. A177682 (a(4*n)/3), A177683 (a(4*n+1)/6), A177684 (a(4*n+2)/3), A177685 (a(4*n+3)/6).

a077076 n = a077076_list !! n
a077076_list = iterate a055944 537  -- 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(537, 27, 2);

NestWhileList[# + IntegerReverse[#, 2] &, 537,  # !=
IntegerReverse[#, 2] &, 1, 27] (* 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(537,27);

a(0), ..., a(11) as above; for n > 11 and
n = 0 (mod 4): a(n) = 3*2^(2*k+13)+18249*2^k-3 where k = (n-4)/4;
n = 1 (mod 4): a(n) = 6*2^(2*k+13)-12102*2^k where k = (n-5)/4;
n = 2 (mod 4): a(n) = 6*2^(2*k+13)+11718*2^k-3 where k = (n-6)/4;
n = 3 (mod 4): a(n) = 12*2^(2*k+13)-11910*2^k where k = (n-7)/4.
G.f.: 3*(179+382*x+96*x^2+60*x^3-328*x^4-444*x^5+1170*x^6+2232*x^7 +1166*x^8+5644*x^9+15402*x^10+46922*x^11+39850*x^12-62920*x^13-132612*x^14 -97532*x^15-34148*x^16+83800*x^17+109224*x^18+21856*x^19) / ((1-x)*(1+x)*(1-2*x^2)*(1-2*x^4)).
G.f. for the sequence starting at a(12): 3*x^12*(155403+246008*x-188442*x^2-229616*x^3-260350*x^4-508920*x^5+293388*x^6+492528*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 and three comments added, g.f. edited, PARI program revised, MAGMA program and crossrefs added by Klaus Brockhaus, May 12 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

A176635 a(n) = 6a(n-1)-8a(n-2) for n > 1; a(0) = 57, a(1) = 242.

Original entry on oeis.org

57, 242, 996, 4040, 16272, 65312, 261696, 1047680, 4192512, 16773632, 67101696, 268421120, 1073713152, 4294909952, 17179754496, 68719247360, 274877448192, 1099510710272, 4398044676096, 17592182374400, 70368736837632

Offset: 0

Klaus Brockhaus, Apr 22 2010

nonn

Related to Reverse and Add trajectory of 77 in base 2: a(n) = A075253(4*n+3)/12, i.e., one twelfth of fourth quadrisection of A075253.
Second binomial transform of 57 followed by 128*A000079.
Third binomial transform of A176636.

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

Cf. A075253 (Reverse and Add trajectory of 77 in base 2), A176632, A176633, A176634, A176636 (repeat 57, 71), A171472.

[64*4^n-7*2^n: n in [0..25]]; // Vincenzo Librandi, Sep 24 2013

CoefficientList[Series[(57 - 100 x)/((1 - 2 x) (1 - 4 x)), {x, 0, 40}], x] (* Vincenzo Librandi, Sep 24 2013 *)
LinearRecurrence[{6,-8},{57,242},30] (* Harvey P. Dale, Jun 08 2016 *)

{m=21; v=concat([57, 242], vector(m-2)); for(n=3, m, v[n]=6*v[n-1]-8*v[n-2]); v}

a(n) = 64*4^n-7*2^n.

G.f.: (57-100*x)/((1-2*x)*(1-4*x)).

A075299 Trajectory of 290 under the Reverse and Add! operation carried out in base 4, written in base 10.

Original entry on oeis.org

290, 835, 1610, 4195, 17060, 23845, 46490, 89080, 138125, 255775, 506510, 1238395, 5127260, 8616205, 15984335, 31949470, 79793675, 315404860, 569392925, 1060061935, 2114961710, 5206421995, 20997654620, 35262166285

Offset: 0

Klaus Brockhaus, Sep 12 2002

290 is conjectured (cf. A066450) to be the smallest number such that the Reverse and Add! algorithm in base 4 does not lead to a palindrome. Unlike 318 (cf. A075153) its trajectory does not exhibit any recognizable regularity, so that the method by which the base 4 trajectory of 318 as well as the base 2 trajectories of 22 (cf. A061561), 77 (cf. A075253), 442 (cf. A075268) etc. can be proved to be palindrome-free (cf. Links), is not applicable here.

			290 (decimal) = 10202 -> 10202 + 20201 = 31003 = 835 (decimal).

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

Cf. A066450, A075153, A058042, A061561, A075253, A075268.

NestWhileList[# + IntegerReverse[#, 4] &, 290,  # !=
IntegerReverse[#, 4] &, 1, 23] (* Robert Price, Oct 18 2019 *)

{m=290; stop=26; c=0; while(c0,d=divrem(k,4); k=d[1]; rev=4*rev+d[2]); c++; m=m+rev)}

A176632 a(n) = 6a(n-1)-8a(n-2)-9 for n > 2; a(0) = 77, a(1) = 897, a(2) = 3333.

Original entry on oeis.org

77, 897, 3333, 12813, 50205, 198717, 790653, 3154173, 12599805, 50365437, 201394173, 805441533, 3221495805, 12885442557, 51540688893, 206160592893, 824638046205, 3298543534077, 13194156834813, 52776592736253

Offset: 0

Klaus Brockhaus, Apr 22 2010

Related to Reverse and Add trajectory of 77 in base 2: a(n) = A075253(4*n), i.e., first quadrisection of A075253.

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

Cf. A075253 (Reverse and Add trajectory of 77 in base 2), A176633, A176634, A176635, A171471.

[77] cat [3*(64*4^n+22*2^n-1): n in [1..25]]; // Vincenzo Librandi, Sep 24 2013

CoefficientList[Series[(77 + 358 x - 1868 x^2 + 1424 x^3)/((1 - x) (1 - 2 x) (1 - 4 x)), {x, 0, 40}], x] (* Vincenzo Librandi, Sep 24 2013 *)
Join[{77},RecurrenceTable[{a[1]==897,a[2]==3333,a[n]==6a[n-1]-8a[n-2]- 9},a[n],{n,20}]] (* Harvey P. Dale, May 21 2019 *)

{m=20; v=concat([77, 897, 3333], vector(m-3)); for(n=4, m, v[n]=6*v[n-1]-8*v[n-2]-9); v}

a(n) = 3*(64*4^n+22*2^n-1) for n > 0, a(0) = 77.
G.f.: (77+358*x-1868*x^2+1424*x^3)/((1-x)*(1-2*x)*(1-4*x)).
G.f. for the sequence starting at a(1): 3*x*(299-982*x+680*x^2)/((1-x)* (1-2*x)*(1-4*x)).

A176633 a(n) = 6a(n-1)-8a(n-2) for n > 2; a(0) = 83, a(1) = 708, a(2) = 2952.

Original entry on oeis.org

83, 708, 2952, 12048, 48672, 195648, 784512, 3141888, 12575232, 50316288, 201295872, 805244928, 3221102592, 12884656128, 51539116032, 206157447168, 824631754752, 3298530951168, 13194131668992, 52776542404608

Offset: 0

Klaus Brockhaus, Apr 22 2010

nonn

Related to Reverse and Add trajectory of 77 in base 2: a(n) = A075253(4*n+1)/2, i.e., one half of second quadrisection of A075253.

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

Cf. A075253 (Reverse and Add trajectory of 77 in base 2), A176632, A176634, A176635, A171470.

[83] cat [6*(32*4^n-5*2^n): n in [1..25]]; // Vincenzo Librandi, Sep 24 2013

CoefficientList[Series[(83 + 210 x - 632 x^2)/((1 - 2 x) (1 - 4 x)), {x, 0, 40}], x] (* Vincenzo Librandi, Sep 24 2013 *)
LinearRecurrence[{6,-8},{83,708,2952},30] (* Harvey P. Dale, Apr 08 2019 *)

{m=20; v=concat([83, 708, 2952], vector(m-3)); for(n=4, m, v[n]=6*v[n-1]-8*v[n-2]); v}

a(n) = 6*(32*4^n-5*2^n) for n > 0, a(1) = 83.
G.f.: (83+210*x-632*x^2)/((1-2*x)*(1-4*x)).
G.f. for the sequence starting at a(1): 12*x*(59-108*x)/((1-2*x)*(1-4*x)).

cp's OEIS Frontend

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

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A075252 Trajectory of n under the Reverse and Add! operation carried out in base 2 does not reach a palindrome and (presumably) does not join the trajectory of any term m < n.

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A075153 Trajectory of 318 under the Reverse and Add! operation carried out in base 4, written in base 10.

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A075268 Trajectory of 442 under the Reverse and Add! operation carried out in base 2.

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A077076 Trajectory of 537 under the Reverse and Add! operation carried out in base 2, written in base 10.

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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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A176635 a(n) = 6a(n-1)-8a(n-2) for n > 1; a(0) = 57, a(1) = 242.

A176632 a(n) = 6a(n-1)-8a(n-2)-9 for n > 2; a(0) = 77, a(1) = 897, a(2) = 3333.

A176633 a(n) = 6a(n-1)-8a(n-2) for n > 2; a(0) = 83, a(1) = 708, a(2) = 2952.