A112969 -id:A112969 - cp's OEIS frontend

A076725 a(n) = a(n-1)^2 + a(n-2)^4, a(0) = a(1) = 1.

1, 1, 2, 5, 41, 2306, 8143397, 94592167328105, 13345346031444632841427643906, 258159204435047592104207508169153297050209383336364487461

Offset: 0

Michael Somos, Oct 29 2002

a(n) and a(n+1) are relatively prime for n >= 0.
The number of independent sets on a complete binary tree with 2^(n-1)-1 nodes. - Jonathan S. Braunhut (jonbraunhut(AT)usa.net), May 04 2004. For example, when n=3, the complete binary tree with 2 levels has 2^2-1 nodes and has 5 independent sets so a(3)=5. The recursion for number of independent sets splits in two cases, with or without the root node being in the set.
a(10) has 113 digits and is too large to include.

			a(2) = a(1)^2 + a(0)^4 = 1^2 + 1^4 = 2.
a(3) = a(2)^2 + a(1)^4 = 2^2 + 1^4 = 5.
a(4) = a(3)^2 + a(2)^4 = 5^2 + 2^4 = 41.
a(5) = a(4)^2 + a(3)^4 = 41^2 + 5^4 = 2306.
a(6) = a(5)^2 + a(4)^4 = 2306^2 + 41^4 = 8143397.
a(7) = a(6)^2 + a(5)^4 = 8143397^2 + 2306^4 = 94592167328105.

Robert Israel, Table of n, a(n) for n = 0..13
Karl Petersen, Ibrahim Salama, Tree shift complexity, arXiv:1712.02251 [math.DS], 2017.
Index entries for sequences of form a(n+1)=a(n)^2 + ...

Cf. A000283, A005154, A112969, A114793.

A[0]:= 1: A[1]:= 1:
for n from 2 to 10 do
  A[n]:= A[n-1]^2 + A[n-2]^4;
od:
seq(A[i],i=0..10); # Robert Israel, Aug 21 2017

RecurrenceTable[{a[n] == a[n-1]^2 + a[n-2]^4, a[0] ==1, a[1] == 1}, a, {n, 0, 10}] (* Vaclav Kotesovec, Dec 18 2014 *)
NestList[{#[[2]],#[[1]]^4+#[[2]]^2}&,{1,1},10][[All,1]] (* Harvey P. Dale, Jul 03 2021 *)

{a(n) = if( n<2, 1, a(n-1)^2 + a(n-2)^4)}

{a=[0,0];for(n=1,99,iferr(a=[a[2],log(exp(a*[4,0;0,2])*[1,1]~)],E,return([n,exp(a[2]/2^n)])))} \\ To compute an approximation of the constant c1 = exp(lim_{n->oo} (log a(n))/2^n). \\ M. F. Hasler, May 21 2017

a=vector(20); a[1]=1;a[2]=2; for(n=3, #a, a[n]=a[n-1]^2+a[n-2]^4); concat(1, a) \\ Altug Alkan, Apr 04 2018

If b(n) = 1 + 1/b(n-1)^2, b(1)=1, then b(n) = a(n)/a(n-1)^2.
Lim_{n->inf} a(n)/a(n-1)^2 = A092526 (constant).
a(n) is asymptotic to c1^(2^n) * c2.
c1 = 1.2897512927198122075..., c2 = 1/A092526 = A263719 = (1/6)*(108 + 12*sqrt(93))^(1/3) - 2/(108 + 12*sqrt(93))^(1/3) = 0.682327803828019327369483739711... is the root of the equation c2*(1 + c2^2) = 1. - Vaclav Kotesovec, Dec 18 2014

Edited by N. J. A. Sloane at the suggestion of Andrew S. Plewe, Jun 15 2007

A112980 a(0) = 0, a(1) = 1; for n>=2: a(n) = a(n-1)^5 + a(n-2)^5.

Original entry on oeis.org

0, 1, 1, 2, 33, 39135425, 91801604643057285538237803582627026018

Offset: 0

Jonathan Vos Post, Jan 02 2006

			a(3) = 1^5 + 1^5 = 2.
a(4) = 1^5 + 2^5 = 33.
a(5) = 2^5 + 33^5 = 39135425.
a(6) = 33^5 + 39135425^5 = 91801604643057285538237803582627026018.

Eric Weisstein's World of Mathematics, Quintic Equation

Cf. A000045, A000283, A112961, A112969.

RecurrenceTable[{a[1]==a[2]==1,a[n]==a[n-1]^5+a[n-2]^5},a,{n,7}] (* Harvey P. Dale, May 01 2012 *)

a(0)=0 prepended by Alois P. Heinz, Sep 15 2023

A113848 a(1) = a(2) = 1, a(n+2) = 2*a(n) + a(n+1)^2.

Original entry on oeis.org

1, 1, 3, 11, 127, 16151, 260855055, 68045359719085327, 4630170979299719971778494028407039, 21438483297549327871400796194793048411084076762817293736211302918175

Offset: 1

Jonathan Vos Post, Jan 24 2006

In this sequence the primes begin a(3) = 3, a(4) = 11, a(5) = 127, a(9) = 4630170979299719971778494028407039.

			a(1) = 1 by definition.
a(2) = 1 by definition.
a(3) = 2*1 + 1^2 = 3.
a(4) = 2*1 + 3^2 = 11.
a(5) = 2*3 + 11^2 = 127.
a(6) = 2*11 + 127^2 = 16151.

Index entries for sequences of form a(n+1)=a(n)^2 + ...

Cf. A000278, A000283, A014253, A063827, A072878, A112957, A112958, A112959, A112960, A112961, A112969, A113785.

Join[{a=1,b=1},Table[c=1*b^2+2*a;a=b;b=c,{n,10}]] (* Vladimir Joseph Stephan Orlovsky, Jan 18 2011 *)
RecurrenceTable[{a[1]==1, a[2]==1, a[n] == 2*a[n-2] + a[n-1]^2}, a, {n, 1, 10}] (* Vaclav Kotesovec, Dec 18 2014 *)

a(1) = a(2) = 1, for n>2: a(n) = 2*a(n-2) + a(n-1)^2. a(1) = a(2) = 1, for n>0: a(n+2) = 2*a(n) + a(n+1)^2.

a(n) ~ c^(2^n), where c = 1.163464453662702696843453679269882816346479873363677551158525103156732040997... . - Vaclav Kotesovec, Dec 18 2014

A114955 A 2/3-power Fibonacci sequence.

Original entry on oeis.org

1, 1, 2, 3, 4, 5, 6, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8

Offset: 0

Jonathan Vos Post, Feb 21 2006

a(n) is also the minimum number of distinct palindromes (not counting the empty string) occurring as substrings of an n-bit binary string. For example, the string 11001 contains the five distinct palindromes 0, 00, 1, 11, and 1001. In fact, every 5-bit binary string contains five distinct palindromes, so a(5) = 5. - Austin Shapiro, Feb 15 2023

			a(2) = ceiling(a(0)^(2/3) + a(1)^(2/3)) = ceiling(1^(2/3) + 1^(2/3)) = 2.
a(3) = ceiling(a(1)^(2/3) + a(2)^(2/3)) = ceiling(1^(2/3) + 2^(2/3)) = ceiling(2.58740105) = 3.
a(4) = ceiling(2^(2/3) + 3^(2/3)) = ceiling(3.66748488) = 4.
a(5) = ceiling(3^(2/3) + 4^(2/3)) = ceiling(4.59992592) = 5.
a(6) = ceiling(4^(2/3) + 5^(2/3)) = ceiling(5.44385984) = 6.
a(7) = ceiling(5^(2/3) + 6^(2/3)) = ceiling(6.22594499) = 7.
a(8) = ceiling(6^(2/3) + 7^(2/3)) = ceiling(6.96123296) = 7.

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

Cf. A000283, A112961, A112969, A114793.

nxt[{a_,b_}]:={b,Ceiling[b^(2/3)+a^(2/3)]}; Transpose[NestList[nxt,{1,1},80]][[1]] (* Harvey P. Dale, Jan 03 2013 *)

{a(n)=if(n<1, n==0, if(n>8, 8, n-(n>7)))} /* Michael Somos, Aug 31 2006 */

a(0) = a(1) = 1, for n>1 a(n) = ceiling(a(n-1)^(2/3) + a(n-2)^(2/3)).
a(n) = 8 for all n>8.
Euler transform of length 8 sequence [ 1, 1, 1, 0, 0, -1, 0, -1]. - Michael Somos, Aug 31 2006
G.f.: (1-x^6)(1-x^8)/((1-x)(1-x^2)(1-x^3)). - Michael Somos, Aug 31 2006

A258169 a(n) = a(n-1)^4/a(n-2) with a(0)=1, a(1)=2.

Original entry on oeis.org

1, 2, 16, 32768, 72057594037927936, 822752278660603021077484591278675252491367932816789931674304512

Offset: 0

Vaclav Kotesovec, May 22 2015

The next term has 235 digits.

In general, the recurrence a(n) = a(n-1)^k / a(n-2) with a(0) = 1, a(1) = m, k > 2, has a solution a(n) = m^(((k+sqrt(k^2-4))^n - (k-sqrt(k^2-4))^n) / (sqrt(k^2-4) * 2^n)).

Cf. A001353, A112969, A208208, A258161.

RecurrenceTable[{a[n]==a[n-1]^4/a[n-2], a[0]==1, a[1]==2},a,{n,0,6}]
nxt[{a_,b_}]:={b,b^4/a}; NestList[nxt,{1,2},5][[All,1]] (* Harvey P. Dale, Sep 04 2022 *)

a(n) = 2^(A001353(n)).

a(n) = 2^(((2+sqrt(3))^n-(2-sqrt(3))^n)/(2*sqrt(3))).

A113592 Array of quadratic pseudofibonacci sequences, read by antidiagonals.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 1, 3, 6, 11, 1, 4, 11, 40, 127, 1, 5, 18, 127, 1612, 16151, 1, 6, 27, 332, 16151, 2598264, 260855055, 1, 7, 38, 739, 110260

Offset: 1

Jonathan Vos Post, Jan 26 2006

Row 1 is A113848. Column 1 is A000012 (the simplest sequence of positive numbers: the all 1's sequence). Column 2 is A000027 (the natural numbers) = n. Column 3 is A010000 = A059100(n+1) = n^2 + 2. Column 4 is 2*n + (n^2 + 2)^2 = n^4 + 4*n^2 + 2*n + 4. Column 5 is 2*(n^2 + 2) + (n^4 + 4*n^2 + 2*n + 4)^2 = n^8 + 8*n^6 + 4*n^5 + 24*n^4 + 16*n^3 + 38*n^2 + 16*n + 20.

			Table (upper left corner):
1...1...3...11...127....16151...260855055...
1...2...6...40...1612...2598624.675284696600...
1...3...11..127..16151..260855055...
1...4...18..332..110260.12157268264...
1...5...27..739..546175...
1...6...38..1456.2120012...
1...7...51..2615.6838327...
1...8...66..4372.19114516...
1...9...83..6907.47706815
1..10..102..10424.108659980...

Cf. A000012, A000027, A000278, A000283, A010000, A014253, A059100, A063827, A072878, A112957, A112958, A112959, A112960, A112961, A112969, A113785.

Antidiagonals of table: T(i, j) = j-th iteration of a(i, 0) = 1, a(i, 1) = i and for j>1: a(i, j) = 2*a(i, j-2) + a(i, j-1)^2.

A114950 a(n) = a(n-1)^4 + a(n-2)^2, with a(0) = a(1) = 1.

Original entry on oeis.org

1, 1, 2, 17, 83525, 48670514501156640914, 5611303368570568119463158581109807779153712597124269146443734128560476495542441

Offset: 0

Jonathan Vos Post, Feb 21 2006

a(7) has 315 digits.

			a(2) = a(1)^4 + a(0)^2 = 1^4 + 1^2 = 2.
a(3) = a(2)^4 + a(1)^2 = 2^4 + 1^2 = 17.
a(4) = a(3)^4 + a(2)^2 = 17^4 + 2^2 = 83525.
a(5) = a(4)^4 + a(3)^2 = 83525^4 + 17^2 = 48670514501156640914.

Index entries for sequences of form a(n+1)=a(n)^2 + ...

Cf. A000283, A112969, A114793.

RecurrenceTable[{a[0] ==1, a[1] == 1, a[n] == a[n-1]^4 + a[n-2]^2}, a, {n, 0, 8}] (* Vaclav Kotesovec, Dec 18 2014 *)

a(n) ~ c^(4^n), where c = 1.045263645117629170027922399491730015846213509999461317320720034161754262379... . - Vaclav Kotesovec, Dec 18 2014

Formula corrected by Vaclav Kotesovec, Dec 18 2014

Missing a(3) added from Vaclav Kotesovec, Dec 18 2014

A114953 A cubic quartic recurrence.

Original entry on oeis.org

1, 1, 2, 9, 745, 413500186, 70701255783138724397185481, 353412074392865080823440901423426679423573814794711467360597541360306163522857

Offset: 0

Jonathan Vos Post, Feb 21 2006

a(6) has 233 digits. This sequence is related to: A112961 "a cubic Fibonacci sequence" a(1) = a(2) = 1; for n>2: a(n) = a(n-1)^3 + a(n-2)^3 A112969 "a quartic Fibonacci sequence" a(1) = a(2) = 1; for n>2: a(n) = a(n-1)^4 + a(n-2)^4, which is the quartic (or biquadratic) analog of the Fibonacci sequence similarly to A000283 being the quadratic analog of the Fibonacci sequence. Primes in this sequence include a(n) for n = 2. Semiprimes in this sequence include a(n) for n = 3, 4, 6.

			a(2) = a(1)^3 + a(0)^4 = 1^3 + 1^4 = 2.
a(3) = a(2)^3 + a(1)^4 = 2^3 + 1^4 = 9.
a(4) = a(3)^3 + a(2)^4 = 9^3 + 2^4 = 745.
a(5) = a(4)^3 + a(3)^4 = 745^3 + 9^4 = 413500186.
a(6) = a(5)^2 + a(4)^4 = 413500186^3 + 745^4 = 70701255783138724397185481.

Cf. A000283, A112961, A112969, A114793.

RecurrenceTable[{a[0] == 1, a[1] == 1, a[n] == a[n-1]^3 + a[n-2]^4}, a, {n, 0, 8}] (* Vaclav Kotesovec, Dec 18 2014 *)

a(0) = a(1) = 1, for n>1 a(n) = a(n-1)^3 + a(n-2)^4.

a(n) ~ c^(3^n), where c = 1.085072477219577474852112080874481159102040272323161792230192441384737595241... . - Vaclav Kotesovec, Dec 18 2014

Formula corrected by Vaclav Kotesovec, Dec 18 2014

A114954 A 3/2-power Fibonacci sequence.

Original entry on oeis.org

1, 1, 2, 4, 11, 45, 339, 6544, 535619, 392527477, 7777266564708, 21689055127418446258, 101009204076980364695686091211

Offset: 0

Jonathan Vos Post, Feb 21 2006

This sequence is related to: A112961 "a cubic Fibonacci sequence" a(1) = a(2) = 1; for n>2: a(n) = a(n-1)^3 + a(n-2)^3 A112969 "a quartic Fibonacci sequence" a(1) = a(2) = 1; for n>2: a(n) = a(n-1)^4 + a(n-2)^4, which is the quartic (or biquadratic) analog of the Fibonacci sequence similarly to A000283 being the quadratic analog of the Fibonacci sequence. Primes in this sequence include a(n) for n = 2, 4. Semiprimes in this sequence include a(n) for n = 3, 6.

			a(2) = ceiling(a(0)^(3/2) + a(1)^(3/2)) = ceiling(1^1.5 + 1^1.5) = 2.
a(3) = ceiling(a(1)^(3/2) + a(2)^(3/2)) = ceiling(1^1.5 + 2^1.5) = ceiling(3.82842712) = 4.
a(4) = ceiling(2^(3/2) + 4^(3/2)) = ceiling(10.8284271) = 11.
a(5) = ceiling((4^(3/2)) + (11^(3/2))) = ceiling(44.4828727) = 45.
a(6) = ceiling((11^(3/2)) + (45^(3/2))) = ceiling(338.35205) = 339.
a(7) = ceiling((45^(3/2)) + (339^(3/2))) = ceiling(6543.52112) = 6544.

Cf. A000283, A112961, A112969, A114793.

RecurrenceTable[{a[0]==a[1]==1,a[n]==Ceiling[Surd[ a[n-1]^3,2]+ Surd[ a[n-2]^3, 2]]},a,{n,15}] (* Harvey P. Dale, Apr 07 2016 *)

a(0) = a(1) = 1, for n>1 a(n) = ceiling(a(n-1)^(3/2) + a(n-2)^(3/2)).

A114956 a(n) = ceiling(a(n-1)^(3/4) + a(n-2)^(3/4)), with a(0) = a(1) = 1.

Original entry on oeis.org

1, 1, 2, 3, 4, 6, 7, 9, 10, 11, 12, 13, 14, 15, 15, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16, 16

Offset: 0

Jonathan Vos Post, Feb 21 2006

a(17) = 16 is exactly 16^(3/4) + 16^(3/4) = 16. This is a fixed point, so a(n) = 16 for all n>14.

			a(2) = ceiling(a(0)^(3/4) + a(1)^(3/4)) = ceiling(1^(3/4) + 1^(3/4)) = 2.
a(3) = ceiling(a(1)^(3/4) + a(2)^(3/4)) = ceiling(1^(3/4) + 2^(3/4)) = ceiling(2.68179283) = 3.
a(4) = ceiling(2^(3/4) + 3^(3/4)) = ceiling(3.96129989) = 4.
a(5) = ceiling(3^(3/4) + 4^(3/4)) = ceiling(5.10793418) = 6.
a(6) = ceiling(4^(3/4) + 6^(3/4)) = ceiling(6.66208575) = 7.

Cf. A000283, A112961, A112969, A114793.

RecurrenceTable[{a[0]==1,a[1]==1,a[n]==Ceiling[a[n-1]^(3/4)+ a[n-2]^(3/4)]}, a[n],{n,80}] (* Harvey P. Dale, Jul 22 2011 *)

Edited by N. J. A. Sloane, May 20 2006

cp's OEIS Frontend

A076725 a(n) = a(n-1)^2 + a(n-2)^4, a(0) = a(1) = 1.

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A112980 a(0) = 0, a(1) = 1; for n>=2: a(n) = a(n-1)^5 + a(n-2)^5.

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A113848 a(1) = a(2) = 1, a(n+2) = 2*a(n) + a(n+1)^2.

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A114955 A 2/3-power Fibonacci sequence.

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A258169 a(n) = a(n-1)^4/a(n-2) with a(0)=1, a(1)=2.

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A113592 Array of quadratic pseudofibonacci sequences, read by antidiagonals.

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A114950 a(n) = a(n-1)^4 + a(n-2)^2, with a(0) = a(1) = 1.

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A114953 A cubic quartic recurrence.