A112961 -id:A112961 - cp's OEIS frontend

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

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

A134923 a(n) = (a(n-1)+a(n-2))^3, for n>=2; a(0)=0,a(1)=1.

Original entry on oeis.org

0, 1, 1, 8, 729, 400315553, 64151935432803278387177768, 264015418305763603932856608512044494366944180663171394053409979316505418715161

Offset: 0

Prashant Garg (prashant.garg1988(AT)yahoo.com), Jan 29 2008

nonn

a(n) = A112961(n)^3. - Lyle Blosser, Nov 29 2024

			a(4) = (a(3)+a(2))^3 = (8+1)^3 = 729.

Cf. A112961.

RecurrenceTable[{a[0]==0,a[1]==1,a[n]==(a[n-1]+a[n-2])^3},a,{n,10}] (* Harvey P. Dale, Aug 05 2016 *)

More terms from Harvey P. Dale, Aug 05 2016

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.

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

A114957 a(n) = ceiling(a(n-1)^(4/3) + a(n-2)^(4/3)), with a(0) = a(1) = 1.

Original entry on oeis.org

1, 1, 2, 4, 9, 26, 96, 517, 4589, 80409, 3546873, 544383737, 445042712531, 3398279290987133, 510914600201184438040, 4084427005585662985398294639, 6528922582874884079540382952631569851, 12202683821888699966029264978793346242448495941305

Offset: 0

Jonathan Vos Post, Feb 21 2006

			a(2) = ceiling(a(0)^(4/3) + a(1)^(4/3)) = ceiling(1^(4/3) + 1^(4/3)) = 2.
a(3) = ceiling(a(1)^(4/3) + a(2)^(4/3)) = ceiling(1^(4/3) + 2^(4/3)) = ceiling(3.5198421) = 4.
a(4) = ceiling(2^(4/3) + 4^(4/3)) = ceiling(8.86944631) = 9.
a(5) = ceiling(4^(4/3) + 9^(4/3)) = ceiling(25.0703586) = 26.
a(6) = ceiling(9^(4/3) + 26^(4/3)) = ceiling(95.7456522) = 96.
a(7) = ceiling(26^(4/3) + 96^(4/3)) = ceiling(516.595167) = 517.
a(8) = ceiling(96^(4/3) + 517^(4/3)) = ceiling(4588.99022) = 4589.

Cf. A000283, A112961, A112969, A114793.

Nest[Append[#,Ceiling[Total[Take[#,-2]^(4/3)]]]&,{1,1},17]  (* Harvey P. Dale, Apr 21 2011 *)

Corrected and extended by Harvey P. Dale, Apr 21 2011

Comments edited by Petros Hadjicostas, Nov 03 2019

cp's OEIS Frontend

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

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Mathematica

Extensions

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

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

A114955 A 2/3-power Fibonacci sequence.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A134923 a(n) = (a(n-1)+a(n-2))^3, for n>=2; a(0)=0,a(1)=1.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Extensions

A113592 Array of quadratic pseudofibonacci sequences, read by antidiagonals.

Views

Author

Keywords

Comments

Examples

Crossrefs

Formula

A114953 A cubic quartic recurrence.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Formula

Extensions

A114954 A 3/2-power Fibonacci sequence.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Formula

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

Views

Author

Keywords

Comments

Examples

Crossrefs