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User: David Johnson-Davies

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David Johnson-Davies has authored 7 sequences.

A330581 a(0) = 2; thereafter a(n) = a(n - 1)^n + 1.

Original entry on oeis.org

2, 3, 10, 1001, 1004006004002, 1020191144865623440455270145683555422808365843606721760320033

Offset: 0

David Johnson-Davies, Dec 18 2019

nonn

Note that this could be extended backwards to a(-1), and any nonzero value x for a(-1) would work, since x^0 + 1 = 2.

David Johnson-Davies, Recursive functions without a base case

Cf. A000522, A182242.

(defun a (n) (+ (if (zerop n) 1 (expt (a (- n 1)) n)) 1))

a:= proc(n) option remember; `if`(n<0, %,
      1 + a(n-1)^n)
    end:
seq(a(n), n=0..5);  # Alois P. Heinz, Dec 18 2019

a[0] = 2; a[n_] := a[n] = a[n - 1]^n + 1; Array[a, 6, 0] (* Amiram Eldar, Dec 19 2019 *)
nxt[{n_,a_}]:={n+1,a^(n+1)+1}; NestList[nxt,{0,2},5][[;;,2]] (* Harvey P. Dale, Dec 10 2023 *)

a(n) = if(n==0, 2, a(n-1)^n+1) \\ Felix Fröhlich, Dec 18 2019

Edited by N. J. A. Sloane, Dec 27 2019

A048122 a(1) = 1, a(2) = 2, a(n) = a(n-1)^3 - a(n-2)^3.

Original entry on oeis.org

1, 2, 7, 335, 37595032, 53136308104856854277393, 150028625136232090492356612601560283583272160844416809510162344944689

Offset: 1

David Johnson-Davies

Next terms are 205 and 614 digits long.

Cf. A048112.

RecurrenceTable[{a[1]==1,a[2]==2,a[n]==a[n-1]^3-a[n-2]^3},a,{n,8}] (* Harvey P. Dale, Mar 20 2013 *)

More terms from Patrick De Geest, Jun 15 1999

A048572 a(n) = sum of digits of a(n-1)*a(n-2).

Original entry on oeis.org

1, 2, 2, 4, 8, 5, 4, 2, 8, 7, 11, 14, 10, 5, 5, 7, 8, 11, 16, 14, 8, 4, 5, 2, 1, 2, 2, 4, 8, 5, 4, 2, 8, 7, 11, 14, 10, 5, 5, 7, 8, 11, 16, 14, 8, 4, 5, 2, 1, 2, 2, 4, 8, 5, 4, 2, 8, 7, 11, 14, 10, 5, 5, 7, 8, 11, 16, 14, 8, 4, 5, 2

Offset: 1

David Johnson-Davies

This sequence is periodic with period 24. - T. D. Noe, Feb 15 2008

Index entries for linear recurrences with constant coefficients, signature (0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1).

Nest[Append[#,Total[IntegerDigits[Last[#]#[[-2]]]]]&,{1,2},80]  (* Harvey P. Dale, Apr 21 2011 *)
PadRight[{},100,{1,2,2,4,8,5,4,2,8,7,11,14,10,5,5,7,8,11,16,14,8,4,5,2}] (* Harvey P. Dale, Jan 18 2015 *)

a(n) = a(n-24) for n >= 25. - Wesley Ivan Hurt, Mar 09 2023

A048112 a(1) = 1, a(2) = 1, a(3) = 1, a(n) = a(n-3) * (a(n-2) + a(n-1)).

Original entry on oeis.org

1, 1, 1, 2, 3, 5, 16, 63, 395, 7328, 486549, 195081415, 1433122040192, 697379012200764243, 136045964066902820948075525, 194970470582876525091946983913863955456

Offset: 1

David Johnson-Davies

Cf. A048122.

RecurrenceTable[{a[1]==a[2]==a[3]==1,a[n]==a[n-3] (a[n-2]+a[n-1])},a,{n,20}] (* Harvey P. Dale, Jan 14 2012 *)
nxt[{a_,b_,c_}]:={b,c,a(b+c)}; NestList[nxt,{1,1,1},20][[;;,1]] (* Harvey P. Dale, Jul 14 2024 *)

More terms from Patrick De Geest, Jun 15 1999

A048784 a(n) = tau(binomial(2*n,n)), where tau = number of divisors (A000005).

Original entry on oeis.org

1, 2, 4, 6, 8, 18, 24, 32, 48, 48, 48, 128, 96, 192, 384, 480, 384, 768, 1152, 1536, 2304, 2048, 2048, 3840, 3456, 4608, 6144, 3840, 8192, 20480, 10240, 12288, 18432, 36864, 36864, 49152, 24576, 32768, 98304, 92160, 73728, 245760, 262144

Offset: 0

David Johnson-Davies

nonn

Charles R Greathouse IV, Table of n, a(n) for n = 0..10000
G. V. Fedorov, Number of divisors of the central binomial coefficient, Moscow Univ. Math. Bull., Vol. 68 (2013), pp. 194-197.

Cf. A000005, A000984, A002162.

A048784 := proc(n)
    numtheory[tau](binomial(2*n,n)) ;
end proc:
seq(A048784(n),n=0..30) ; # R. J. Mathar, Jul 12 2024

f[n_] := DivisorSigma[0, Binomial[2 n, n]]; Table[f@n, {n, 0, 42}] (* Robert G. Wilson v, Apr 08 2009 *)

fv(n,p)=my(s);while(n\=p,s+=n);s
a(n)=my(s=1);forprime(p=2,2*n,s*=fv(2*n,p)-2*fv(n,p)+1);s \\ Charles R Greathouse IV, Aug 21 2013

a(n) = A000005(A000984(n)). - Michel Marcus, Aug 21 2013

log(a(n)) = log(2) * (pi(2*n)-pi(n)) + log(2) * (n/log(n)) * Sum_{k=0..T} c_k/log(n)^k + O(n/log(n)^(T+2)) for any T >= 0, where c_k = Sum_{m>=1} Integral_{m+1/2..m+1} log(t)^m/t^2 dt. In particular for T = 0, log(a(n)) = 2 * log(2)^2 * (n/log(n)) + O(n/log(n)^2) (Fedorov, 2013). - Amiram Eldar, Dec 10 2024

A048691 a(n) = d(n^2), where d(k) = A000005(k) is the number of divisors of k.

Original entry on oeis.org

1, 3, 3, 5, 3, 9, 3, 7, 5, 9, 3, 15, 3, 9, 9, 9, 3, 15, 3, 15, 9, 9, 3, 21, 5, 9, 7, 15, 3, 27, 3, 11, 9, 9, 9, 25, 3, 9, 9, 21, 3, 27, 3, 15, 15, 9, 3, 27, 5, 15, 9, 15, 3, 21, 9, 21, 9, 9, 3, 45, 3, 9, 15, 13, 9, 27, 3, 15, 9, 27, 3, 35, 3, 9, 15, 15, 9, 27, 3, 27

Offset: 1

David Johnson-Davies

Inverse Moebius transform of A034444: Sum_{d|n} 2^omega(d), where omega(n) = A001221(n) is the number of distinct primes dividing n.
Number of elements in the set {(x,y): x|n, y|n, gcd(x,y)=1}.
Number of elements in the set {(x,y): lcm(x,y)=n}.
Also gives total number of positive integral solutions (x,y), order being taken into account, to the optical or parallel resistor equation 1/x + 1/y = 1/n. Indeed, writing the latter as X*Y=N, with X=x-n, Y=y-n, N=n^2, the one-to-one correspondence between solutions (X, Y) and (x, y) is obvious, so that clearly, the solution pairs (x, y) are tau(N)=tau(n^2) in number. - Lekraj Beedassy, May 31 2002
Number of ordered pairs of positive integers (a,c) such that n^2 - ac = 0. Therefore number of quadratic equations of the form ax^2 + 2nx + c = 0 where a,n,c are positive integers and each equation has two equal (rational) roots, -n/a. (If a and c are positive integers, but, instead, the coefficient of x is odd, it is impossible for the equation to have equal roots.) - Rick L. Shepherd, Jun 19 2005
Problem A1 on the 21st Putnam competition in 1960 (see John Scholes link) asked for the number of pairs of positive integers (x,y) such that xy/(x+y) = n: the answer is a(n); for n = 4, the a(4) = 5 solutions (x,y) are (5,20), (6,12), (8,8), (12,6), (20,5). - Bernard Schott, Feb 12 2023
Numbers k such that a(k)/d(k) is an integer are in A217584 and the corresponding quotients are in A339055. - Bernard Schott, Feb 15 2023

A. M. Gleason et al., The William Lowell Putnam Mathematical Competitions, Problems & Solutions:1938-1960 Soln. to Prob. 1 1960, p. 516, MAA, 1980.
Ross Honsberger, More Mathematical Morsels, Morsel 43, pp. 232-3, DMA No. 10 MAA, 1991.
Loren C. Larson, Problem-Solving Through Problems, Prob. 3.3.7, p. 102, Springer 1983.
Alfred S. Posamentier and Charles T. Salkind, Challenging Problems in Algebra, Prob. 9-9 pp. 143 Dover NY, 1988.
D. O. Shklarsky et al., The USSR Olympiad Problem Book, Soln. to Prob. 123, pp. 28, 217-8, Dover NY.
Wacław Sierpiński, Elementary Theory of Numbers, pp. 71-2, Elsevier, North Holland, 1988.
James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 91.
Charles W. Trigg, Mathematical Quickies, Question 194, pp. 53, 168, Dover, 1985.

Enrique Pérez Herrero, Table of n, a(n) for n = 1..10000
Ovidiu Bagdasar, On Some Functions Involving the lcm and gcd of Integer Tuples.
Umberto Cerruti, Percorsi tra i numeri (in Italian), pages 3-4.
Daniele A. Gewurz and Francesca Merola, Sequences realized as Parker vectors of oligomorphic permutation groups, J. Integer Seq., Vol. 6 (2003), Article 03.1.6.
John Scholes, Problem A1 of 21st Putnam Competition 1960, 2002. [Wayback Machine link]
Wacław Sierpiński, Elementary Theory of Numbers, Warszawa 1964.
Index to sequences related to Olympiads and other Mathematical competitions.

Cf. A000005, A034444, Mobius transf. of A035116, A048785, A061391, A018805, A002088, A015614, A018892, A063647, A182139.
Partial sums give A061503.
For similar LCM sequences, see A070919, A070920, A070921.
Cf. A005408, A124010.
For the earliest occurrence of 2n-1 see A016017.
Cf. A001620, A082633, A217584, A339055.

a048691 = product . map (a005408 . fromIntegral) . a124010_row
-- Reinhard Zumkeller, Jul 12 2012

A048691[n_]:=DivisorSigma[0,n^2] (* Enrique Pérez Herrero, May 30 2010 *)
DivisorSigma[0,Range[80]^2] (* Harvey P. Dale, Apr 08 2015 *)

numlib::tau (n^2)$ n=1..90 // Zerinvary Lajos, May 13 2008

A048691(n)=prod(i=1,#n=factor(n)[,2],n[i]*2+1) /* or, of course, a(n)=numdiv(n^2) */ \\ M. F. Hasler, Dec 30 2007

for(n=1, 100, print1(direuler(p=2, n, (1 - X^2)/(1 - X)^3)[n], ", ")) \\ Vaclav Kotesovec, Aug 21 2021

[sigma(n**2, 0) for n in range(1, 81)] # Stefano Spezia, Jul 14 2025

a(n) = A000005(A000290(n)).
tau(n^2) = Sum_{d|n} mu(n/d)*tau(d)^2, where mu(n) = A008683(n), cf. A061391.
Multiplicative with a(p^e) = 2e+1. - Vladeta Jovovic, Jul 23 2001
Also a(n) = Sum_{d|n} (tau(d)*moebius(n/d)^2), Dirichlet convolution of A000005 and A008966. - Benoit Cloitre, Sep 08 2002
a(n) = A055205(n) + A000005(n). - Reinhard Zumkeller, Dec 08 2009
Dirichlet g.f.: (zeta(s))^3/zeta(2s). - R. J. Mathar, Feb 11 2011
a(n) = Sum_{d|n} 2^omega(d). Inverse Mobius transform of A034444. - Enrique Pérez Herrero, Apr 14 2012
G.f.: Sum_{k>=1} 2^omega(k)*x^k/(1 - x^k). - Ilya Gutkovskiy, Mar 10 2018
Sum_{k=1..n} a(k) ~ n*(6/Pi^2)*(log(n)^2/2 + log(n)*(3*gamma - 1) + 1 - 3*gamma + 3*gamma^2 - 3*gamma_1 + (2 - 6*gamma - 2*log(n))*zeta'(2)/zeta(2) + (2*zeta'(2)/zeta(2))^2 - 2*zeta''(2)/zeta(2)), where gamma is Euler's constant (A001620) and gamma_1 is the first Stieltjes constant (A082633). - Amiram Eldar, Jan 26 2023

Additional comments from Vladeta Jovovic, Apr 29 2001

A048736 Dana Scott's sequence: a(n) = (a(n-2) + a(n-1) * a(n-3)) / a(n-4), a(0) = a(1) = a(2) = a(3) = 1.

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 5, 13, 22, 41, 111, 191, 361, 982, 1693, 3205, 8723, 15042, 28481, 77521, 133681, 253121, 688962, 1188083, 2249605, 6123133, 10559062, 19993321, 54419231, 93843471, 177690281, 483649942, 834032173, 1579219205, 4298430243, 7412446082, 14035282561, 38202222241, 65877982561

Offset: 0

David Johnson-Davies

The recursion has the Laurent property. If a(0), a(1), a(2), a(3) are variables, then a(n) is a Laurent polynomial (a rational function with a monic monomial denominator). - Michael Somos, Feb 05 2012

A generalization is if the recursion is modified to a(n) = (a(n-2) + a(n-1) * b*a(n-3)) / a(n-4) where b is a constant, and with arbitrary nonzero initial values, (a(0), a(1), a(2), a(3)), then a(n) = c*(a(n-3) - a(n-6)) + a(n-9) for all n in Z where c is another constant. - Michael Somos, Oct 28 2021

			G.f. = 1 + x + x^2 + x^3 + 2*x^4 + 3*x^5 + 13*x^6 + 22*x^7 + 41*x^8 + 111*x^9 + ...

Seiichi Manyama, Table of n, a(n) for n = 0..3165 (first 501 terms from T. D. Noe)
Joshua Alman, Cesar Cuenca, and Jiaoyang Huang, Laurent phenomenon sequences, Journal of Algebraic Combinatorics 43(3) (2015), 589-633.
Hal Canary, The Dana Scott Recurrence [From _Jaume Oliver Lafont_, Sep 25 2009]
S. Fomin and A. Zelevinsky, The Laurent phenomenon, arXiv:math/0104241 [math.CO], 2001.
S. Fomin and A. Zelevinsky, The Laurent Phenomenon, Advances in Applied Mathematics, 28 (2002), 119-144.
David Gale, The strange and surprising saga of the Somos sequences, Math. Intelligencer 13(1) (1991), pp. 40-42.
D. Gale, Tracking the Automatic Ant And Other Mathematical Explorations, A Collection of Mathematical Entertainments Columns from The Mathematical Intelligencer, Springer, 1998, p. 4.
Matthew Christopher Russell, Using experimental mathematics to conjecture and prove theorems in the theory of partitions and commutative and non-commutative recurrences, PhD Dissertation, Mathematics Department, Rutgers University, May 2016.
Eric Weisstein's World of Mathematics, Laurent Polynomial
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (0,0,10,0,0,-10,0,0,1)

Cf. A006720, A006721, A006722, A006723, A092420, A072881.
Cf. A192241, A192242 (primes and where they occur).
Cf. A276531.

a048736 n = a048736_list !! n
a048736_list = 1 : 1 : 1 : 1 :
   zipWith div
     (zipWith (+)
       (zipWith (*) (drop 3 a048736_list)
                    (drop 1 a048736_list))
       (drop 2 a048736_list))
     a048736_list
-- Reinhard Zumkeller, Jun 26 2011

I:=[1,1,1,1]; [n le 4 select I[n] else (Self(n-2) + Self(n-1)*Self(n-3)) / Self(n-4): n in [1..30]]; // G. C. Greubel, Feb 20 2018

P:=proc(q) local n,v; v:=[1,1,1,1]; for n from 1 to q do
v:=[op(v),(v[-2]+v[-1]*v[-3])/v[-4]] od: op(v); end: P(35); # Paolo P. Lava, Aug 24 2025

RecurrenceTable[{a[0] == a[1] == a[2] == a[3] == 1, a[n] == (a[n - 2] + a[n - 1]a[n - 3])/a[n - 4]}, a[n], {n, 40}] (* or *) LinearRecurrence[{0, 0, 10, 0, 0, -10, 0, 0, 1}, {1, 1, 1, 1, 2, 3, 5, 13, 22}, 41] (* Harvey P. Dale, Oct 22 2011 *)

Vec((1+x+x^2-9*x^3-8*x^4-7*x^5+5*x^6+3*x^7+2*x^8) / (1-10*x^3+10*x^6-x^9)+O(x^99)) \\ Charles R Greathouse IV, Jul 01 2011

a(n) = 9*a(n-3) - a(n-6) - 3 - ( ceiling(n/3) - floor(n/3) ), with a(0) = a(1) = a(2) = a(3) = 1, a(4) = 2, a(5) = 3. - Michael Somos
From Jaume Oliver Lafont, Sep 17 2009: (Start)
a(n) = 10*a(n-3) - 10*a(n-6) + a(n-9).
G.f.: (1 + x + x^2 - 9*x^3 - 8*x^4 - 7*x^5 + 5*x^6 + 3*x^7 + 2*x^8)/(1 - 10*x^3 + 10*x^6 - x^9). (End)
a(n) = a(3-n) for all n in Z. - Michael Somos, Feb 05 2012

More terms from Michael Somos

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User: David Johnson-Davies

A330581 a(0) = 2; thereafter a(n) = a(n - 1)^n + 1.

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A048122 a(1) = 1, a(2) = 2, a(n) = a(n-1)^3 - a(n-2)^3.

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A048572 a(n) = sum of digits of a(n-1)*a(n-2).

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A048112 a(1) = 1, a(2) = 1, a(3) = 1, a(n) = a(n-3) * (a(n-2) + a(n-1)).

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A048784 a(n) = tau(binomial(2*n,n)), where tau = number of divisors (A000005).

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A048691 a(n) = d(n^2), where d(k) = A000005(k) is the number of divisors of k.

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A048736 Dana Scott's sequence: a(n) = (a(n-2) + a(n-1) * a(n-3)) / a(n-4), a(0) = a(1) = a(2) = a(3) = 1.

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