A075153 -id:A075153 - 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

A075421 Trajectory of n under the Reverse and Add! operation carried out in base 4 (presumably) does not reach a palindrome and (presumably) does not join the trajectory of any term m < n.

Original entry on oeis.org

290, 318, 719, 795, 799, 1210, 3903, 4199, 4207, 4219, 4236, 4278, 4279, 4294, 4326, 4333, 4334, 4338, 4402, 4598, 4662, 4726, 5046, 5357, 6157, 6174, 7246, 7247, 7295, 7407, 7549, 8063, 8191, 9211, 12319, 12431, 12463, 12539, 15487, 16519, 16587

Offset: 1

Klaus Brockhaus, Sep 18 2002, revised Jan 28 2004

For 318 (cf. A075153), 266718 (cf. A075466) and 270798 (cf. A075467) one can prove that the base 4 trajectory does not contain a palindrome. A proof for 290 (cf. A075299) has not been found up to now. 4398859679359 is another known candidate (obtained from a remark of David J. Seal, cf. Links) for a term whose trajectory is provably palindrome-free, but is not secured that it does not join the trajectory of some term m < n. - 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 28 for k < 20000). On the other hand, the trajectories of the terms listed above do not join the trajectory of any smaller term within at least 1000 steps.
Base-4 analog of A063048 (base 10) and A075252 (base 2); subsequence of A075420.
From A.H.M. Smeets, Mar 18 2019: (Start)
David J. Seal (see LINKS) observed a cyclic pattern (length 6) in the trajectories that can be represented by an extended right regular grammar with production rules:
S -> S_a | S_b | S_c | S_d | S_e | S_f,
S_a -> 1033202000232 T_a, T_a -> 222 T_a | 2302333113230
S_b -> 2022321332331 T_b, T_b -> 111 T_b | 1223001203131
S_c -> 10002003002212 T_c, T_c -> 222 T_c | 3221333101333
S_d -> 103312202321111 T_d, T_d -> 111 T_d | 1102023122000
S_e -> 110200123122222 T_e, T_e -> 222 T_e | 2231232001301
S_f -> 213301021321111 T_f, T_f -> 111 T_f | 1113213003312
Within the first 471 terms of this sequence we observed three trajectories with a cyclic pattern (length 6) that can be represented by a context-free grammar with production rules:
S -> S_a | S_b | S_c | S_d | S_e | S_f,
S_a -> 10 T_a 00, T_a -> 3 T_a 0 | T_a0,
S_b -> 11 T_b 01, T_b -> 0 T_b 3 | T_b0,
S_c -> 22 T_c 12, T_c -> 0 T_c 3 | T_c0,
S_d -> 10 T_d 000, T_d -> 3 T_d 0 | T_d0,
S_e -> 11 T_e 301, T_e -> 0 T_e 3 | T_e0,
S_f -> 22 T_f 312, T_f -> 0 T_f 3 | T_f0.
The terminating strings in these context-free grammars are given by:
n     2      359                371
a(n)  318    266718             270798
T_a0  33230  33230000001033230  3323001033230
T_b0  03123  03123010001103123  0312302103123
T_c0  01313  01313120002201313  0131320201313
T_d0  33323  33323000001033323  3332300103323
T_e0  03222  03222301001103222  0322201113222
T_f0  02111  02111312002202111  0211112222111
From the fact that both, right regular grammars and context-free grammars occur, we wonder if other trajectories can be represented by context-sensitive grammars as well, by which other trajectories can be proven never to end up in a palindromic string? (End)

			719 is a term since the trajectory of 719 (presumably) does not lead to an integer which occurs in the trajectory of 290 or of 318.

A.H.M. Smeets, Table of n, a(n) for n = 1..471
Klaus Brockhaus, Illustration: Distribution of terms below 2000000
David J. Seal, Results
Index entries for sequences related to Reverse and Add!

Cf. A063048, A075252, A075420, A075299, A075153, A075466, A075467, A091675.

limit = 10^3; (* Assumes that there is no palindrome if none is found before "limit" iterations *)
utraj = {};
Select[Range[0, 17000], (x = NestWhileList[# + IntegerReverse[#, 4] &, #, # !=IntegerReverse[#, 4] & , 1, limit];
   If[Length[x] >= limit  && Intersection[x, utraj] == {},
    utraj = Union[utraj, x]; True,
utraj = Union[utraj, x]]) &] (* Robert Price, Oct 16 2019 *)

A166912 a(n) = 20a(n-1) - 64a(n-2) - 225 for n > 2; a(0) = 106, a(1) = 8075, a(2) = 114235.

Original entry on oeis.org

106, 8075, 114235, 1767675, 28042235, 447713275, 7159562235, 114537594875, 1832539914235, 29320392212475, 469125289738235, 7506000693166075, 120095995320074235, 1921535862038855675, 30744573540292362235

Offset: 0

Klaus Brockhaus, Oct 27 2009

Related to Reverse and Add trajectory of 318 in base 4: A075153(6*n) = 3*a(n).

lim_{n -> infinity} a(n)/a(n-1) = 16.

G. C. Greubel, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (21, -84, 64).

Cf. A075153, A166913, A166914, A166915, A166916, A166917, A167120, A167121, A167122.

Join[{106},LinearRecurrence[{21,-84,64},{8075,114235,1767675},20]] (* Harvey P. Dale, Jun 07 2012 *)

m=15; v=concat([106, 8075, 114235], vector(m-3)); for(n=4, m, v[n]=20*v[n-1]-64*v[n-2]-225); v

a(n) = (1280*16^n + 940*4^n - 15)/3 for n > 0.
G.f.: (106 + 5849*x - 46436*x^2 + 40256*x^3)/((1-x)*(1-4*x)*(1-16*x)).
From G. C. Greubel, May 28 2016: (Start)
a(n) = 21*a(n-1) - 84*a(n-2) + 64*a(n-3).
E.g.f.: (1/3)*(-15*exp(x) + 940*exp(4*x) + 1280*exp(16*x)) - 629. (End)

A166913 a(n) = 20a(n-1) - 64a(n-2) - 150 for n > 2; a(0) = 357, a(1) = 14450, a(2) = 221650.

Original entry on oeis.org

357, 14450, 221650, 3508050, 55975250, 894989650, 14317376850, 229068199250, 3665051866450, 58640672576850, 938250132084050, 15011999596762450, 240191983481869650, 3843071695444596050, 61489146966052263250

Offset: 0

Klaus Brockhaus, Oct 27 2009

Related to Reverse and Add trajectory of 318 in base 4: A075153(6*n+1) = 3*a(n).

lim_{n -> infinity} a(n)/a(n-1) = 16.

G. C. Greubel, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (21, -84, 64).

Cf. A075153, A166912, A166914, A166915, A166916, A166917, A167120, A167121, A167122.

Join[{357},LinearRecurrence[{21,-84,64},{14450,221650,3508050},20]] (* Harvey P. Dale, Jun 18 2014 *)

m=15; v=concat([357, 14450, 221650], vector(m-3)); for(n=4, m, v[n]=20*v[n-1]-64*v[n-2]-150); v

a(n) = (2560*16^n + 600*4^n - 10)/3 for n > 0.
G.f.: (357 + 6953*x - 51812*x^2 + 44352*x^3)/((1-x)*(1-4*x)*(1-16*x)).
a(0)=357, a(1)=14450, a(2)=221650, a(3)=3508050, a(n)=21*a(n-1)- 84*a(n-2)+ 64*a(n-3). - Harvey P. Dale, Jun 18 2014
E.g.f.: (1/3)*(-10*exp(x) + 600*exp(4*x) + 2560*exp(16*x)) - 693. - G. C. Greubel, May 28 2016

A166915 a(n) = 20a(n-1) - 64a(n-2) - 45 for n>1; a(0) = 399, a(1) = 5695.

Original entry on oeis.org

399, 5695, 88319, 1401855, 22384639, 357974015, 5726863359, 91626930175, 1466019348479, 23456263438335, 375300030463999, 6004799749226495, 96076793034833919, 1537228676746182655, 24595658780694282239

Offset: 0

Klaus Brockhaus, Oct 27 2009

Related to Reverse and Add trajectory of 318 in base 4: A075153(6*n+3) = 15*a(n).

lim_{n -> infinity} a(n)/a(n-1) = 16.

G. C. Greubel, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (21, -84, 64).

Cf. A075153, A166912, A166913, A166914, A166916, A166917, A006105, A167031, A167032, A167033.

LinearRecurrence[{21, -84, 64}, {399, 5695, 88319}, 50] (* G. C. Greubel, May 28 2016 *)

m=15; v=concat([399, 5695], vector(m-2)); for(n=3, m, v[n]=20*v[n-1]-64*v[n-2]-45); v

a(n) = (1024*16^n + 176*4^n - 3)/3.
G.f.: (399 - 2684*x + 2240*x^2)/((1-x)*(1-4*x)*(1-16*x)).
From G. C. Greubel, May 28 2016: (Start)
a(n) = 21*a(n-1) - 84*a(n-2) + 64*a(n-3).
E.g.f.: (1/3)*(1024*exp(16*x) + 176*exp(4*x) - 3*exp(x)). (End)

A166916 a(n) = 20a(n-1) - 64a(n-2) - 15 for n > 1; a(0) = 357, a(1) = 5525.

Original entry on oeis.org

357, 5525, 87637, 1399125, 22373717, 357930325, 5726688597, 91626231125, 1466016552277, 23456252253525, 375299985724757, 6004799570269525, 96076792319006037, 1537228673882871125, 24595658769241036117

Offset: 0

Klaus Brockhaus, Oct 27 2009

Related to Reverse and Add trajectory of 318 in base 4: A075153(6*n+4) = 30*a(n).

lim_{n -> infinity} a(n)/a(n-1) = 16.

G. C. Greubel, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (21, -84, 64).

Cf. A075153, A166912, A166913, A166914, A166915, A166917, A006105, A167031, A167032, A167033.

LinearRecurrence[{21,-84,64},{357,5525,87637},20] (* Harvey P. Dale, Sep 24 2012 *)

m=15; v=concat([357, 5525], vector(m-2)); for(n=3, m, v[n]=20*v[n-1]-64*v[n-2]-15); v

a(n) = (1024*16^n + 48*4^n - 1)/3.
G.f.: (357 - 1972*x + 1600*x^2)/((1-x)*(1-4*x)*(1-16*x)).
a(0)=357, a(1)=5525, a(2)=87637, a(n)=21*a(n-1)-84*a(n-2)+64*a(n-3). - Harvey P. Dale, Sep 24 2012
E.g.f.: (1/3)*(1024*exp(16*x) + 48*exp(4*x) - exp(x)). - G. C. Greubel, May 28 2016

A075420 Trajectory of n under the Reverse and Add! operation carried out in base 4 (presumably) does not reach a palindrome.

Original entry on oeis.org

290, 318, 378, 381, 438, 444, 462, 498, 501, 504, 510, 545, 567, 573, 627, 633, 636, 639, 693, 696, 699, 717, 719, 732, 751, 753, 756, 759, 765, 775, 795, 799, 800, 822, 823, 828, 835, 847, 859, 882, 883, 888, 894, 895, 915, 919, 927, 948, 954, 967, 972

Offset: 1

Klaus Brockhaus, Sep 18 2002

Base-4 analog of A023108 (base 10) and A066059 (base 2).

A.H.M. Smeets, Table of n, a(n) for n = 1..20000
Index entries for sequences related to Reverse and Add!

Cf. A023108, A066059, A075421, A075299, A075153.

limit = 10^3; (* Assumes that there is no palindrome if none is found before "limit" iterations *)
Select[Range[1000],
Length@NestWhileList[# + IntegerReverse[#, 4] &, #, # !=
IntegerReverse[#, 4]  &, 1, limit] == limit + 1 &] (* Robert Price, Oct 16 2019 *)

{stop=1000; for(n=1,980,k=n; c=0; while(c0,d=divrem(a,4); a=d[1]; rev=4*rev+d[2]); if(rev==k,c=stop+1,k=k+rev; c++)); if(c==stop,print1(n,",")))}

Offset changed to 1 by A.H.M. Smeets, Feb 10 2019

A166914 a(n) = 20a(n-1) - 64a(n-2) for n > 1; a(0) = 21, a(1) = 340.

Original entry on oeis.org

21, 340, 5456, 87360, 1398016, 22369280, 357912576, 5726617600, 91625947136, 1466015416320, 23456247709696, 375299967549440, 6004799497568256, 96076792028200960, 1537228672719650816, 24595658764588154880

Offset: 0

Klaus Brockhaus, Oct 27 2009

nonn

Related to Reverse and Add trajectory of 318 in base 4: A075153(6*n+2) = 240*a(n).

G. C. Greubel, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (20,-64).

Cf. A075153, A166912, A166913, A166915, A166916, A166917, A166927, A166965, A166984.

[Binomial(4^(n+3), 2)/96: n in [0..30]]; // G. C. Greubel, Oct 02 2024

CoefficientList[Series[(21-80x)/((1-4x)(1-16x)),{x,0,20}],x]  (* or *) LinearRecurrence[{20,-64},{21,340},20] (* Harvey P. Dale, Feb 23 2011 & Mar 30 2012 *)

{m=15; v=concat([21, 340], vector(m-2)); for(n=3, m, v[n]=20*v[n-1]-64*v[n-2]); v}

A166914=BinaryRecurrenceSequence(20,-64,21,340)
[A166914(n) for n in range(31)] # G. C. Greubel, Oct 02 2024

a(n) = (64*16^n - 4^n)/3.
G.f.: (21 - 80*x)/((1-4*x)*(1-16*x)).
Limit_{n -> infinity} a(n)/a(n-1) = 16.
From G. C. Greubel, May 28 2016: (Start)
a(n) = 20*a(n-1) - 64*a(n-2).
E.g.f.: (1/3)*(-exp(4*x) + 64*exp(16*x)). (End)

A166917 a(n) = 20a(n-1) - 64a(n-2) for n > 1; a(0) = 85, a(1) = 1364.

Original entry on oeis.org

85, 1364, 21840, 349504, 5592320, 89478144, 1431654400, 22906486784, 366503854080, 5864061927424, 93824991887360, 1501199874392064, 24019198007050240, 384307168179912704, 6148914691147038720, 98382635059426361344, 1574122160955116748800, 25185954575299047849984

Offset: 0

Klaus Brockhaus, Oct 27 2009

Related to Reverse and Add trajectory of 318 in base 4: A075153(6*n+5) = 240*a(n).

G. C. Greubel, Table of n, a(n) for n = 0..500
Index entries for linear recurrences with constant coefficients, signature (20, -64).

Cf. A075153, A166912, A166913, A166914, A166915, A166916, A166927, A166965, A166984.

[Binomial(4^(n+4), 2)/384: n in [0..30]]; // G. C. Greubel, Oct 02 2024

LinearRecurrence[{20,-64}, {85, 1364}, 50] (* G. C. Greubel, May 28 2016 *)

{m=15; v=concat([85, 1364], vector(m-2)); for(n=3, m, v[n]=20*v[n-1]-64*v[n-2]); v}

A166917=BinaryRecurrenceSequence(20,-64,85,1364)
[A166917(n) for n in range(31)] # G. C. Greubel, Oct 02 2024

a(n) = (256*16^n - 4^n)/3.
G.f.: (85 - 336*x)/((1-4*x)*(1-16*x)).
Limit_{n -> infinity} a(n)/a(n-1) = 16.
E.g.f.: (1/3)*(256*exp(16*x) - exp(4*x)). - G. C. Greubel, May 28 2016

A075466 Trajectory of 266718 under the Reverse and Add! operation carried out in base 4, written in base 10.

Original entry on oeis.org

266718, 1017375, 2019150, 4934715, 20413980, 34239885, 64220175, 127195950, 321080475, 1286586060, 2154739965, 4288508415, 8571775230, 21401016315, 85781907180, 149736661725, 278082371775, 1369020907200, 1433193762225

Offset: 0

Klaus Brockhaus, Sep 18 2002

266718 = A075421(358) is the smallest term > 318 of A075421 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. - The generating function given describes the sequence from a(26) onward; the g.f. for the complete sequence is known but more than twice as big.

			266718 (decimal) = 1001013132 -> 1001013132 + 2313101001 = 3320120133 = 1017375 (decimal).

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

Cf. A075153, A075421, A075467.

NestWhileList[# + IntegerReverse[#, 4] &, 266718,  # !=
IntegerReverse[#, 4] &, 1, 23] (* Robert Price, Oct 18 2019 *)

{m=266718; stop=19; c=0; while(c0,d=divrem(k,4); k=d[1]; rev=4*rev+d[2]); c++; m=m+rev)}

a(0), ..., a(18) as above; a(19) = 2780823717750; a(20) = 5492189757120; a(21) = 5749636151985; a(22) = 11156010444150; a(23) = 21968759028480; a(24) = 22226205423345; a(25) = 44109148986870; for n > 25 and n = 2 (mod 6): a(n) = 5*4^(2*k+14)-83865605*4^k where k = (n-2)/6; n = 3 (mod 6): a(n) = 5*4^(2*k+14)+3941683435*4^k-15 where k = (n-3)/6; n = 4 (mod 6): a(n) = 10*4^(2*k+14)+2515968150*4^k-10 where k = (n-4)/6; n = 5 (mod 6): a(n) = 20*4^(2*k+14)-335462420*4^k where k = (n-5)/6; n = 0 (mod 6): a(n) = 20*4^(2*k+14)+3690086620*4^k-15 where k = (n-6)/6; n = 1 (mod 6): a(n) = 40*4^(2*k+14)+2012774520*4^k-10 where k = (n-7)/6. G.f.: -15*(47049901525664*x^11+23708157972464*x^10+23433347158016*x^9-46912496118440*x^8-23502049861628*x^7-23433347158016*x^6-11908468626600*x^5-6137441522940*x^4-5862630708480*x^3+11771063219370*x^2+5931333412095*x+5862630708480)/((x-1)*(x^2+x+1)*(2*x^3-1)*(2*x^3+1)*(4*x^3-1))

cp's OEIS Frontend

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

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A075421 Trajectory of n under the Reverse and Add! operation carried out in base 4 (presumably) does not reach a palindrome and (presumably) does not join the trajectory of any term m < n.

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A166912 a(n) = 20*a(n-1) - 64*a(n-2) - 225 for n > 2; a(0) = 106, a(1) = 8075, a(2) = 114235.

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A166913 a(n) = 20*a(n-1) - 64*a(n-2) - 150 for n > 2; a(0) = 357, a(1) = 14450, a(2) = 221650.

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A166915 a(n) = 20*a(n-1) - 64*a(n-2) - 45 for n>1; a(0) = 399, a(1) = 5695.

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A166916 a(n) = 20*a(n-1) - 64*a(n-2) - 15 for n > 1; a(0) = 357, a(1) = 5525.

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A075420 Trajectory of n under the Reverse and Add! operation carried out in base 4 (presumably) does not reach a palindrome.

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A166912 a(n) = 20a(n-1) - 64a(n-2) - 225 for n > 2; a(0) = 106, a(1) = 8075, a(2) = 114235.

A166913 a(n) = 20a(n-1) - 64a(n-2) - 150 for n > 2; a(0) = 357, a(1) = 14450, a(2) = 221650.

A166915 a(n) = 20a(n-1) - 64a(n-2) - 45 for n>1; a(0) = 399, a(1) = 5695.

A166916 a(n) = 20a(n-1) - 64a(n-2) - 15 for n > 1; a(0) = 357, a(1) = 5525.

A166914 a(n) = 20a(n-1) - 64a(n-2) for n > 1; a(0) = 21, a(1) = 340.

A166917 a(n) = 20a(n-1) - 64a(n-2) for n > 1; a(0) = 85, a(1) = 1364.