cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-5 of 5 results.

A175502 a(1) = 1. a(n) = the smallest integer not yet occurring such that each unordered pair {d(a(k)),d(a(k-1))} occurs only once at most, for 2<= k <= n, where d(m) = the number of divisors of m.

Original entry on oeis.org

1, 2, 3, 4, 6, 5, 12, 8, 10, 16, 7, 24, 9, 18, 20, 30, 14, 36, 11, 48, 15, 60, 13, 64, 21, 120, 17, 144, 22, 180, 19, 192, 25, 49, 72, 28, 80, 40, 42, 81, 32, 100, 54, 84, 90, 112, 121, 168, 44, 240, 23, 360, 26, 320, 45, 252, 56, 210, 96, 196
Offset: 1

Views

Author

Leroy Quet, May 31 2010

Keywords

Comments

This sequence seems likely to be a permutation of the positive integers. Is it?

Crossrefs

Extensions

More terms from Sean A. Irvine, Jun 14 2011

A221364 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4*n+3))/(1 - x^(4*n+1)) when x = 1/2*(3 - sqrt(5)).

Original entry on oeis.org

1, 1, 1, 5, 1, 16, 1, 45, 1, 121, 1, 320, 1, 841, 1, 2205, 1, 5776, 1, 15125, 1, 39601, 1, 103680, 1, 271441, 1, 710645, 1, 1860496, 1, 4870845, 1, 12752041, 1, 33385280, 1, 87403801, 1, 228826125, 1, 599074576, 1, 1568397605, 1, 4106118241, 1, 10749957120, 1, 28143753121
Offset: 0

Views

Author

Peter Bala, Jan 15 2013

Keywords

Comments

The function F(x) := Product_{n >= 0} (1 - x^(4*n+3))/(1 - x^(4*n+1)) is analytic for |x| < 1. When x is a quadratic irrational of the form x = 1/2*(N - sqrt(N^2 - 4)), N an integer greater than 2, the real number F(x) has a predictable simple continued fraction expansion. The first examples of these expansions, for N = 2, 4, 6 and 8, are due to Hanna. See A174500 through A175503. The present sequence is the case N = 3. See also A221365 (N = 5), A221366 (N = 7), A221369 (N = 9).
If we denote the present sequence by [1, c(1), 1, c(2), 1, c(3), ...] then for k = 1, 2, ..., the simple continued fraction expansion of F((1/2*(3 - sqrt(5)))^k) is given by the sequence [1; c(k), 1, c(2*k), 1, c(3*k), 1, ...]. Examples are given below.

Examples

			F(1/2*(3 - sqrt(5))) = 1.53879 34992 88095 08323 ... = 1 + 1/(1 + 1/(1 + 1/(5 + 1/(1 + 1/(16 + 1/(1 + 1/(45 + ...))))))).
F((1/2*(3 - sqrt(5)))^2) = 1.16725 98258 10214 95210 ... = 1 + 1/(5 + 1/(1 + 1/(45 + 1/(1 + 1/(320 + 1/(1 + 1/(2205 + ...))))))).
F((1/2*(3 - sqrt(5)))^3) = 1.05883 42773 67371 19975 ... = 1 + 1/(16 + 1/(1 + 1/(320 + 1/(1 + 1/(5776 + 1/(1 + 1/(103680 + ...))))))).
		

Crossrefs

Cf. A001906, A002878, A004146, A049684, A081070, A081071, A174500 (N = 4), A221365 (N = 5), A221366 (N = 7), A221369 (N = 9).

Formula

a(2*n-1) = (1/2*(3 + sqrt(5)))^n + (1/2*(3 - sqrt(5)))^n - 2 = A004146(n); a(2*n) = 1.
a(4*n+1) = A081071(n) = A002878(n)^2;
a(4*n-1) = A081070(n) = 5*A049684(n) = 5*(A001906(n))^2.
a(n) = 4*a(n-2)-4*a(n-4)+a(n-6). G.f.: -(x^4+x^3-3*x^2+x+1) / ((x-1)*(x+1)*(x^2-x-1)*(x^2+x-1)). - Colin Barker, Jan 20 2013

Extensions

More terms from Michel Marcus, Feb 21 2025

A221365 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4*n+3))/(1 - x^(4*n+1)) when x = 1/2*(5 - sqrt(21)).

Original entry on oeis.org

1, 3, 1, 21, 1, 108, 1, 525, 1, 2523, 1, 12096, 1, 57963, 1, 277725, 1, 1330668, 1, 6375621, 1, 30547443, 1, 146361600, 1, 701260563, 1, 3359941221, 1, 16098445548, 1, 77132286525, 1, 369562987083, 1, 1770682648896, 1
Offset: 0

Views

Author

Peter Bala, Jan 15 2013

Keywords

Comments

The function F(x) := Product_{n >= 0} (1 - x^(4*n+3))/(1 - x^(4*n+1)) is analytic for |x| < 1. When x is a quadratic irrational of the form x = 1/2*(N - sqrt(N^2 - 4)), N an integer greater than 2, the real number F(x) has a predictable simple continued fraction expansion. The first examples of these expansions, for N = 2, 4, 6 and 8, are due to Hanna. See A174500 through A175503. The present sequence is the case N = 5. See also A221364 (N = 3), A221366 (N = 7) and A221367 (N = 9).
If we denote the present sequence by [1, c(1), 1, c(2), 1, c(3), ...] then for k = 1, 2, ..., the simple continued fraction expansion of F((1/2*(5 - sqrt(21)))^k) is given by the sequence [1; c(k), 1, c(2*k), 1, c(3*k), 1, ...].

Examples

			F(1/2*(5 - sqrt(21))) = 1.25274 83510 08359 27965 ... = 1 + 1/(3 + 1/(1 + 1/(21 + 1/(1 + 1/(108 + 1/(1 + 1/(525 + ...))))))).
F((1/2*(5 - sqrt(21)))^2) = 1.04545 84663 16495 30047 ... = 1 + 1/(21 + 1/(1 + 1/(525 + 1/(1 + 1/(12096 + 1/(1 + 1/(277725 + ...))))))).
F((1/2*(5 - sqrt(21)))^3) = 1.00917 43188 83793 73068 ... = 1 + 1/(108 + 1/(1 + 1/(12096 + 1/(1 + 1/(1330668 + 1/(1 + 1/(146361600 + ...))))))).
		

Crossrefs

Cf. A004254, A030221, A054493, A174500 (N = 4), A221364 (N = 3), A221366 (N = 7), A221369 (N = 9).

Programs

  • Mathematica
    LinearRecurrence[{0,6,0,-6,0,1},{1,3,1,21,1,108},40] (* Harvey P. Dale, Jun 06 2023 *)

Formula

a(2*n-1) = (1/2*(5 + sqrt(21)))^n + (1/2*(5 - sqrt(21)))^n - 2 = 3*A054493(n); a(2*n) = 1.
a(4*n+1) = 3*(A030221(n))^2; a(4*n-1) = 21*(A004254(n))^2.
a(n) = 6*a(n-2)-6*a(n-4)+a(n-6). G.f.: -(x^4+3*x^3-5*x^2+3*x+1) / ((x-1)*(x+1)*(x^4-5*x^2+1)). - Colin Barker, Jan 20 2013

A221366 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4*n+3))/(1 - x^(4*n+1)) when x = (1/2)*(7 - 3*sqrt(5)).

Original entry on oeis.org

1, 5, 1, 45, 1, 320, 1, 2205, 1, 15125, 1, 103680, 1, 710645, 1, 4870845, 1, 33385280, 1, 228826125, 1, 1568397605, 1, 10749957120, 1, 73681302245, 1, 505019158605, 1, 3461452808000, 1, 23725150497405, 1
Offset: 0

Views

Author

Peter Bala, Jan 15 2013

Keywords

Comments

The function F(x) := Product_{n >= 0} (1 - x^(4*n+3))/(1 - x^(4*n+1)) is analytic for |x| < 1. When x is a quadratic irrational of the form x = 1/2*(N - sqrt(N^2 - 4)), N an integer greater than 2, the real number F(x) has a predictable simple continued fraction expansion. The first examples of these expansions, for N = 2, 4, 6 and 8, are due to Hanna. See A174500 through A175503. The present sequence is the case N = 7. See also A221364 (N = 3), A221365 (N = 5) and A221367 (N = 9).
If we denote the present sequence by [1, c(1), 1, c(2), 1, c(3), ...] then for k = 1, 2, ..., the simple continued fraction expansion of F((1/2*(7 - sqrt(45)))^k) is given by the sequence [1; c(k), 1, c(2*k), 1, c(3*k), 1, ...]. Examples are given below.

Examples

			F(1/2*(7 - sqrt(45))) = 1.16725 98258 10214 95210 ... = 1 + 1/(5 + 1/(1 + 1/(45 + 1/(1 + 1/(320 + 1/(1 + 1/(2205 + ...))))))).
F((1/2*(7 - sqrt(45)))^2) = 1.02173 93445 69104 86504 ... = 1 + 1/(45 + 1/(1 + 1/(2205 + 1/(1 + 1/(103680 + 1/(1 + 1/(4870845 + ...))))))).
F((1/2*(7 - sqrt(45)))^3) = 1.00311 52648 91110 10148 ... = 1 + 1/(320 + 1/(1 + 1/(103680 + 1/(1 + 1/(33385280 + 1/(1 + 1/(10749957120 + ...))))))).
		

Crossrefs

Cf. A174500 (N = 4), A221364 (N = 3), A221365 (N = 5), A221369 (N = 9).

Programs

  • Mathematica
    LinearRecurrence[{0,8,0,-8,0,1},{1,5,1,45,1,320},40] (* or *) Riffle[ LinearRecurrence[{8,-8,1},{5,45,320},20],1,{1,-1,2}] (* Harvey P. Dale, Jan 04 2018 *)

Formula

a(2*n-1) = (1/2*(7 + sqrt(45)))^n + (1/2*(7 - sqrt(45)))^n - 2 = A081070(n); a(2*n) = 1.
a(4*n-1) = 45*A049682(n) = 45*(A004187(n))^2;
a(4*n+1) = 5*(A033890(n))^2.
a(n) = 8*a(n-2)-8*a(n-4)+a(n-6). G.f.: -(x^4+5*x^3-7*x^2+5*x+1) / ((x-1)*(x+1)*(x^2-3*x+1)*(x^2+3*x+1)). - Colin Barker, Jan 20 2013

A221367 The simple continued fraction expansion of F(x) := Product_{n >= 0} (1 - x^(4*n+3))/(1 - x^(4*n+1)) when x = (1/2)*(9 - sqrt(77)).

Original entry on oeis.org

1, 7, 1, 77, 1, 700, 1, 6237, 1, 55447, 1, 492800, 1, 4379767, 1, 38925117, 1, 345946300, 1, 3074591597, 1, 27325378087, 1, 242853811200, 1, 2158358922727, 1, 19182376493357, 1, 170483029517500, 1, 1515164889164157, 1, 13466000972959927, 1
Offset: 0

Views

Author

Peter Bala, Jan 15 2013

Keywords

Comments

The function F(x) := Product_{n >= 0} (1 - x^(4*n+3))/(1 - x^(4*n+1)) is analytic for |x| < 1. When x is a quadratic irrational of the form x = 1/2*(N - sqrt(N^2 - 4)), N an integer greater than 2, the real number F(x) has a predictable simple continued fraction expansion. The first examples of these expansions, for N = 2, 4, 6 and 8, are due to Hanna. See A174500 through A175503. The present sequence is the case N = 9. See also A221364 (N = 3), A221365 (N = 5) and A221366 (N = 7).
If we denote the present sequence by [1, c(1), 1, c(2), 1, c(3), ...] then for k = 1, 2, ..., the simple continued fraction expansion of F((1/2*(9 - sqrt(77)))^k) is given by the sequence [1; c(k), 1, c(2*k), 1, c(3*k), 1, ...]. Examples are given below.

Examples

			F(1/2*(9 - sqrt(77))) = 1.12519 81018 34502 81936 ... = 1 + 1/(7 + 1/(1 + 1/(77 + 1/(1 + 1/(700 + 1/(1 + 1/(6237 + ...))))))).
F((1/2*(9 - sqrt(77)))^2) = 1.01282 05391 65421 74656 ... = 1 + 1/(77 + 1/(1 + 1/(6237 + 1/(1 + 1/(492800 + 1/(1 + 1/(38925117 + ...))))))).
F((1/2*(9 - sqrt(77)))^3) = 1.00142 65335 27667 24640 ... = 1 + 1/(700 + 1/(1 + 1/(492800 + 1/(1 + 1/(345946300 + 1/(1 + 1/(242853811200 + ...))))))).
		

Crossrefs

Cf. A018193, A057081, A174500 (N = 4), A221364 (N = 3), A221365 (N = 5), A221366 (N = 7).

Formula

a(2*n-1) = (1/2*(9 + sqrt(77)))^n + (1/2*(9 - sqrt(77)))^n - 2; a(2*n) = 1.
a(4*n-1) = 77*(A018913(n))^2; a(4*n+1) = 7*(A057081(n))^2.
a(n) = 10*a(n-2)-10*a(n-4)+a(n-6). G.f.: -(x^4+7*x^3-9*x^2+7*x+1) / ((x-1)*(x+1)*(x^4-9*x^2+1)). - Colin Barker, Jan 20 2013
Showing 1-5 of 5 results.